diff --git "a/MathematikAlpha/md/en-international/en-ibero.md" "b/MathematikAlpha/md/en-international/en-ibero.md" deleted file mode 100644--- "a/MathematikAlpha/md/en-international/en-ibero.md" +++ /dev/null @@ -1,1481 +0,0 @@ -# Iberoamerican Mathematical Olympiads $\mathrm{C}=\mathrm{I}$ - -## 1st Iberoamerican 1985 - -## Problem A1 - -Find all integer solutions to: $a+b+c=24, a^{2}+b^{2}+c^{2}=210, a b c=440$. - -## Solution - -$\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=\left((\mathrm{a}+\mathrm{b}+\mathrm{c})^{2}-\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)\right) / 2=183$, so $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are roots of the cubic $\mathrm{x}^{3}-24 \mathrm{x}^{2}$ $+183 x-440=0$. But it easily factorises as $(x-5)(x-8)(x-11)=0$, so the only solutions are permutations of $(5,8,11)$. - -## Problem A2 - -$\mathrm{P}$ is a point inside the equilateral triangle $\mathrm{ABC}$ such that $\mathrm{PA}=5, \mathrm{~PB}=7, \mathrm{PC}=8$. Find $\mathrm{AB}$. - -## Solution - -Answer: $\sqrt{ } 129$. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-01.jpg?height=363&width=389&top_left_y=1469&top_left_x=248) - -Let the side length be $\mathrm{x}$. Using the cosine formula, we have $\cos \mathrm{APB}=\left(74-\mathrm{x}^{2}\right) / 70, \cos \mathrm{APC}$ $=\left(89-x^{2}\right) / 80, \cos \mathrm{BPC}=\left(113-\mathrm{x}^{2}\right) / 112$. But $\cos \mathrm{BPC}=\cos \mathrm{APC} \cos \mathrm{BPC}-\sin \mathrm{APC} \sin$ BPC, so $\left(113-x^{2}\right) / 112=\left(74-x^{2}\right) / 79\left(89-x^{2}\right) / 80-\sqrt{ }\left(\left(1-\left(74-x^{2}\right)^{2} / 70^{2}\right)\left(1-\left(89-x^{2}\right)^{2} / 80^{2}\right)\right)$. - -We isolate the square root term, then square. We multiply through by 25.256 .49 and, after some simplification, we get $x^{6}-138 x^{4}+1161 x^{2}=0$. Hence $x=0, \pm 3, \pm \sqrt{129}$. We discard the zero and negative solutions. $\mathrm{x}=3$ corresponds to a point $\mathrm{P}$ outside the triangle. So the unique solution for a point $\mathrm{P}$ inside the triangle is $\mathrm{x}=\sqrt{ } 129$. - -## Alternative solution by Johannes Tang - -Rotate the triangle about $\mathrm{C}$ through $60^{\circ}$. Let $\mathrm{P}$ go to $\mathrm{P}^{\prime}$. We have $\mathrm{AP}^{\prime}=7, \mathrm{CP}^{\prime}=8$ and angle $\mathrm{PCP}^{\prime}=60^{\circ}$, so $\mathrm{PP}^{\prime} \mathrm{C}$ is equilateral. Hence angle $\mathrm{CPP}^{\prime}=60^{\circ}$. Also $\mathrm{PP}^{\prime}=8$. Using the cosine formula on triangle $\mathrm{APP}^{\prime}$ we find angle $\mathrm{APP}^{\prime}=60^{\circ}$. Hence angle $\mathrm{APC}=120^{\circ}$. Now applying cosine formula to triangle APC, we get result. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-02.jpg?height=534&width=554&top_left_y=270&top_left_x=251) - -Find the roots $r_{1}, r_{2}, r_{3}, r_{4}$ of the equation $4 x^{4}-a x^{3}+b x^{2}-$ $\mathrm{cx}+5=0$, given that they are positive reals satisfying $\mathrm{r}_{1} / 2+\mathrm{r}_{2} / 4+\mathrm{r}_{3} / 5+\mathrm{r}_{4} / 8=1$. - -## Solution - -We have $r_{1} r_{2} r_{3} r_{4}=5 / 4$ and hence $\left(r_{1} / 2\right)\left(r_{2} / 4\right)\left(r_{3} / 5\right)\left(r_{4} / 8\right)$ $=1 / 4^{4}$. But AM/GM gives that $\left(\mathrm{r}_{1} / 2\right)\left(\mathrm{r}_{2} / 4\right)\left(\mathrm{r}_{3} / 5\right)\left(\mathrm{r}_{4} / 8\right) \leq$ $\left(\left(r_{1} / 2+r_{2} / 4+r_{3} / 5+r_{4} / 8\right) / 4\right)^{4}=1 / 4^{4}$ with equality iff $r_{1} / 2=r_{2} / 4=r_{3} / 5=r_{4} / 8$. Hence we must have $r_{1}=1 / 2, r_{2}$ $=1, \mathrm{r}_{3}=5 / 4, \mathrm{r}_{4}=2$. - -## Problem B1 - -The reals $x, y, z$ satisfy $x \neq 1, y \neq 1, x \neq y$, and $\left(y z-x^{2}\right) /(1-x)=\left(x z-y^{2}\right) /(1-y)$. Show that $\left(y x-x^{2}\right) /(1-x)=x+y+z$. - -## Solution - -We have $y z-x^{2}-y^{2} z+y x^{2}=x z-y^{2}-x^{2} z+x y^{2}$. Hence $z\left(y-x-y^{2}+x^{2}\right)=-y^{2}+x y^{2}-x^{2} y+$ $x^{2}$. Hence $z=(x+y-x y) /(x+y-1)$. - -So $\mathrm{yz}=\mathrm{x}+\mathrm{y}+\mathrm{z}-\mathrm{xy}-\mathrm{xz}$, so $\mathrm{yz}-\mathrm{x}^{2}=\mathrm{x}+\mathrm{y}+\mathrm{z}-\mathrm{x}^{2}-\mathrm{xy}-\mathrm{xz}=(\mathrm{x}+\mathrm{y}+\mathrm{z})(1-\mathrm{x})$, so $(\mathrm{yz}-$ $\left.x^{2}\right) /(1-x)=(x+y+z)$ - -## Problem B2 - -The function $\mathrm{f}(\mathrm{n})$ is defined on the positive integers and takes non-negative integer values. It satisfies (1) $f(m n)=f(m)+f(n),(2) f(n)=0$ if the last digit of $n$ is 3 , (3) $f(10)=0$. Find $\mathrm{f}(1985)$. - -## Solution - -If $\mathrm{f}(\mathrm{mn})=0$, then $\mathrm{f}(\mathrm{m})+\mathrm{f}(\mathrm{n})=0($ by $(1))$. But $\mathrm{f}(\mathrm{m})$ and $\mathrm{f}(\mathrm{n})$ are non-negative, so $\mathrm{f}(\mathrm{m})=\mathrm{f}(\mathrm{n})=$ 0 . Thus $f(10)=0$ implies $f(5)=0$. Similarly $f(3573)=0$ by (2), so $f(397)=0$. Hence $f(1985)$ $=\mathrm{f}(5)+\mathrm{f}(397)=0$. - -## Problem B3 - -$\mathrm{O}$ is the circumcenter of the triangle $\mathrm{ABC}$. The lines $\mathrm{AO}, \mathrm{BO}, \mathrm{CO}$ meet the opposite sides at D, E, F respectively. Show that $1 / A D+1 / B E+1 / C F=2 / A O$. - -## Solution - -Projecting onto the altitude from $\mathrm{A}$, we have $\mathrm{AD} \cos (\mathrm{C}-\mathrm{B})=\mathrm{AC} \sin \mathrm{C}=2 \mathrm{R} \sin \mathrm{B} \sin \mathrm{C}$, so $2 R / A D=\cos (C-B) /(\sin B \sin C)$. - -Hence $2 R / A D+2 R / B E+2 R / C F=\cos (C-B) /(\sin B \sin C)+\cos (A-C) /(\sin C \sin A)+\cos (B$ $-A) /(\sin A \sin B)$. So $2 R \sin A \sin B \sin C(1 / A D+1 / B E+1 / C F)=\sin A \cos (B-C)+\sin B$ $\cos (\mathrm{C}-\mathrm{A})+\sin \mathrm{C} \cos (\mathrm{A}-\mathrm{B})=3 \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C}+\sin \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}+\sin \mathrm{B} \cos \mathrm{A} \cos \mathrm{C}+$ -$\sin \mathrm{C} \cos \mathrm{A} \cos \mathrm{B}=3 \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C}+\sin (\mathrm{A}+\mathrm{B}) \cos \mathrm{C}+\sin \mathrm{C} \cos \mathrm{A} \cos \mathrm{B}=3 \sin \mathrm{A} \sin$ $B \sin C+\sin C(\cos C+\cos A \cos B)=3 \sin A \sin B \sin C+\sin C(-\cos (A+B)+\cos A \cos$ B) $=4 \sin A \sin B \sin C$. Hence $1 / A D+1 / B E+1 / C F=2 / R$. - -## 2nd Iberoamerican 1987 - -## Problem A1 - -Find $f(x)$ such that $f(x)^{2} f((1-x) /(1+x))=64 x$ for $x$ not $0, \pm 1$. - -## Solution - -Put $x=(1-y) /(1+y)$, then $(1--x) /(1+x)=y$, so $f((1-y) /(1+y))^{2} f(y)=64(1-y) /(1+y)$. Hence $f($ $(1-x) /(1+x))^{2} f(x)=64(1-x) /(1+x)$. But $f(x)^{4} f((1-x) /(1+x))^{2}=64^{2} x^{2}$, so $f(x)^{3}=64 x^{2}(1+$ $x) /(1-x)$. Hence $f(x)=4\left(x^{2}(1+x) /(1-x)\right)^{1 / 3}$. - -## Problem A2 - -In the triangle $\mathrm{ABC}$, the midpoints of $\mathrm{AC}$ and $\mathrm{AB}$ are $\mathrm{M}$ and $\mathrm{N}$ respectively. $\mathrm{BM}$ and $\mathrm{CN}$ meet at $P$. Show that if it is possible to inscribe a circle in the quadrilateral AMPN (touching every side), then $\mathrm{ABC}$ is isosceles. - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-03.jpg?height=486&width=572&top_left_y=1433&top_left_x=248) - -If the quadrilateral has an inscribed circle then $\mathrm{AM}+\mathrm{PN}=\mathrm{AN}+\mathrm{PM}$ (consider the tangents to the circle from $A, M, P, N$ ). But if $\mathrm{AB}>\mathrm{AC}$, then $\mathrm{BM}>\mathrm{CN}$ (see below). We have $\mathrm{AN}=$ $\mathrm{AB} / 2, \mathrm{PM}=\mathrm{BM} / 3, \mathrm{AM}=\mathrm{AC} / 2, \mathrm{PN}=\mathrm{CN} / 3$, so it follows that $\mathrm{AM}+\mathrm{PN}<\mathrm{AN}+\mathrm{PM}$. Similarly, $\mathrm{AB}<\mathrm{AC}$ implies $\mathrm{AM}+\mathrm{PN}>\mathrm{AN}+\mathrm{PM}$, so the triangle must be isosceles. - -To prove the result about the medians, note that $\mathrm{BM}^{2}=\mathrm{BC}^{2}+\mathrm{CM}^{2}-2 \mathrm{BC} . \mathrm{CM} \cos \mathrm{C}=(\mathrm{BC}-$ $\mathrm{CM} \cos \mathrm{C})^{2}+(\mathrm{CM} \sin \mathrm{C})^{2}$. Similarly, $\mathrm{CN}^{2}=(\mathrm{BC}-\mathrm{BN} \cos \mathrm{B})^{2}+(\mathrm{BN} \sin \mathrm{B})^{2}$. But $\mathrm{MN}$ is parallel to $\mathrm{BC}$, so $\mathrm{CM} \sin \mathrm{C}=\mathrm{BN} \sin \mathrm{B}$. But $\mathrm{AB}>\mathrm{AC}$, so $\mathrm{BN}>\mathrm{CM}$ and $\mathrm{B}<\mathrm{C}$, so $\cos \mathrm{B}>$ $\cos \mathrm{C}$, hence $\mathrm{BN} \cos \mathrm{B}>\mathrm{CM} \cos \mathrm{C}$ and $\mathrm{BC}-\mathrm{CM} \cos \mathrm{C}>\mathrm{BC}-\mathrm{BN} \cos \mathrm{B}$. So $\mathrm{BM}>\mathrm{CN}$. - -## Problem A3 - -Show that if $(2+\sqrt{3})^{k}=1+m+n \sqrt{3}$, for positive integers $m, n, k$ with $k$ odd, then $m$ is a perfect square. - -## Solution - -We have $(2+\sqrt{ } 3)^{4}=97+56 \sqrt{ } 3=14(7+4 \sqrt{ } 3)-1=14(2+\sqrt{ } 3)^{2}-1$. Hence $(2+\sqrt{3})^{k+2}=14$ $(2+\sqrt{3})^{k}-(2+\sqrt{3})^{k-2}$. Thus if $(2+\sqrt{3})^{k}=a_{k}+b_{k} \sqrt{3}$, then $a_{k+2}=14 a_{k}-a_{k-2}$. - -Now suppose the sequence $c_{k}$ satisfies $c_{1}=1, c_{2}=5, c_{k+1}=4 c_{k}-c_{k-1}$. We claim that $c_{k}^{2}-c_{k-}$ ${ }_{1} c_{k+1}=6$. Induction on $\mathrm{k}$. We have $c_{3}=19$, so $c_{2}{ }^{2}-c_{1} c_{3}=25-19=6$. Thus the result is true for $\mathrm{k}=2$. Suppose it is true for $\mathrm{k}$. Then $c_{k+1}=4 c_{k}-c_{k-1}$, so $c_{k+1}{ }^{2}=4 c_{k} c_{k+1}-c_{k-1} c_{k+1}=4 c_{k} c_{k+1}$ $-c_{k}^{2}+6=c_{k}\left(4 c_{k+1}-c_{k}\right)+6=c_{k} c_{k+2}+6$, so the result is true for $k+1$. - -Now put $\mathrm{d}_{\mathrm{k}}=\mathrm{c}_{\mathrm{k}}^{2}+1$. We show that $\mathrm{d}_{\mathrm{k}+2}=14 \mathrm{~d}_{\mathrm{k}+1}-\mathrm{d}_{\mathrm{k}}$. Induction on $\mathrm{k}$. We have $\mathrm{d}_{1}=2, \mathrm{~d}_{2}=$ $26, \mathrm{~d}_{3}=362=14 \mathrm{~d}_{2}-\mathrm{d}_{1}$, so the result is true for $\mathrm{k}=1$. Suppose it is true for $\mathrm{k}$. We have $\mathrm{c}_{\mathrm{k}+3}-$ $4 c_{k+2}+c_{k+1}=0$. Hence $12+2 c_{k+3} c_{k+1}-8 c_{k+2} c_{k+1}+2 c_{k+1}{ }^{2}=12$. Hence $2 c_{k+2}{ }^{2}-8 c_{k+2} c_{k+1}+2$ $c_{k+1}^{2}=12$. Hence $16 c_{k+2}{ }^{2}-8 c_{k+2} c_{k+1}+c_{k+1}{ }^{2}+1=14 c_{k+2}{ }^{2}+14-c_{k+1}{ }^{2}-1$, or $\left(4 c_{k+2}-c_{k+1}\right)^{2}+$ $1=14\left(c_{k+2}^{2}+1\right)-\left(c_{k+1}{ }^{2}+1\right)$, or $\mathrm{c}_{\mathrm{k}+3^{2}}+1=14\left(\mathrm{c}_{\mathrm{k}+2}{ }^{2}+1\right)-\left(\mathrm{c}_{\mathrm{k}+1}{ }^{2}+1\right)$, or $\mathrm{d}_{\mathrm{k}+3}=14 \mathrm{~d}_{\mathrm{k}+2}-\mathrm{d}_{\mathrm{k}+1}$. So the result is true for all $\mathrm{k}$. - -But $\mathrm{a}_{1}=2, \mathrm{a}_{3}=26$ and $\mathrm{a}_{2 \mathrm{k}+3}=14 \mathrm{a}_{2 \mathrm{k}+1}-\mathrm{a}_{2 k-1}$, and $\mathrm{d}_{1}=2, \mathrm{~d}_{2}=26$ and $\mathrm{d}_{\mathrm{k}+1}=14 \mathrm{~d}_{\mathrm{k}}-\mathrm{d}_{\mathrm{k}-1}$. Hence $\mathrm{a}_{2 \mathrm{k}-1}=\mathrm{d}_{\mathrm{k}}=\mathrm{c}_{\mathrm{k}}^{2}+1$. - -## Problem B1 - -Define the sequence $\mathrm{p}_{1}, \mathrm{p}_{2}, \mathrm{p}_{3}, \ldots$ as follows. $\mathrm{p}_{1}=2$, and $\mathrm{p}_{\mathrm{n}}$ is the largest prime divisor of $\mathrm{p}_{1} \mathrm{p}_{2}$ $\ldots \mathrm{p}_{\mathrm{n}-1}+1$. Prove that 5 does not occur in the sequence. - -## Problem B2 - -Show that the roots $r, s, t$ of the equation $x(x-2)(3 x-7)=2$ are real and positive. Find $\tan ^{-1} r$ $+\tan ^{-1} s+\tan ^{-1} t$. - -## Solution - -Put $f(x)=x(x-2)(3 x-7)-2=3 x^{3}-13 x^{2}+14 x-2$. Then $f(0)=-2, f(1)=2$, so there is a root between 0 and 1 . $f(2)=-2$, so there is another root between 1 and 2 . $f(3)=4$, so the third root is between 2 and 3. $\mathrm{f}(\mathrm{x})=0$ has three roots, so they are all real and positive. - -We have $\tan (a+b+c)=(\tan a+\tan b+\tan \mathrm{c}-\tan a \tan b \tan c) /(1-(\tan a \tan b+\tan b \tan \mathrm{c}$ $+\tan \mathrm{c} \tan \mathrm{a}))$. So putting $\mathrm{a}=\tan ^{-1} \mathrm{r}, \mathrm{b}=\tan ^{-1} \mathrm{~s}, \mathrm{c}=\tan ^{-1} \mathrm{t}$, we have, $\tan (\mathrm{a}+\mathrm{b}+\mathrm{c})=((\mathrm{r}+\mathrm{s}+\mathrm{t})$ $-\mathrm{rst}) /(1-(\mathrm{rs}+\mathrm{st}+\mathrm{tr}))=(13 / 3-2 / 3) /(1-14 / 3)=-1$. So $\mathrm{a}+\mathrm{b}+\mathrm{c}=-\pi / 4+\mathrm{k} \pi$. But we know that each of $\mathrm{r}, \mathrm{s}, \mathrm{t}$ is real and positive, so $\mathrm{a}+\mathrm{b}+\mathrm{c}$ lies in the range 0 to $3 \pi / 2$. Hence $\mathrm{a}+\mathrm{b}+\mathrm{c}$ $=3 \pi / 4$ - -## Problem B3 - -$\mathrm{ABCD}$ is a convex quadrilateral. $\mathrm{P}, \mathrm{Q}$ are points on the sides $\mathrm{AD}, \mathrm{BC}$ respectively such that $\mathrm{AP} / \mathrm{PD}=\mathrm{BQ} / \mathrm{QC}=\mathrm{AB} / \mathrm{CD}$. Show that the angle between the lines $\mathrm{PQ}$ and $\mathrm{AB}$ equals the angle between the lines $\mathrm{PQ}$ and $\mathrm{CD}$. - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-05.jpg?height=471&width=823&top_left_y=256&top_left_x=251) - -If $\mathrm{AB}$ is parallel to $\mathrm{CD}$, then it is obvious that $\mathrm{PQ}$ is parallel to both. So assume $\mathrm{AB}$ and $\mathrm{CD}$ meet at $O$. Take $O$ as the origin for vectors. Let $\mathbf{e}$ be a unit vector in the direction OA and $\mathbf{f}$ a unit vector in the direction OC. Take the vector OA to be ae, OB to be be, OC to be cf, and OD to be df. Then OP is $((d-c) a e+(a-b) d \mathbf{f}) /(d-c+a-b)$ and OQ is $((d-c) b e+(a-$ b)cf)/(d - c + a - b). Hence PQ is $(c-d)(a-b)(\mathbf{e}+\mathbf{f}) /(d-c+a-b)$. But $\mathbf{e}$ and $\mathbf{f}$ are unit vectors, so $\mathbf{e}+\mathbf{f}$ makes the same angle with each of them and hence $P Q$ makes the same angle with $A B$ and CD. - -## 3rd Iberoamerican 1988 - -## Problem A1 - -The sides of a triangle form an arithmetic progression. The altitudes also form an arithmetic progression. Show that the triangle must be equilateral. - -## Solution - -Let the sides be $a$, $a+d, a+2 d$ with $d>=0$. Then the altitudes are $k / a \geq k /(a+d) \geq k /(a+2 d)$, where $k$ is twice the area. We claim that $k / a+k /(a+2 d)>2 k /(a+d)$ unless $d=0$. This is equivalent to $(a+d)(a+2 d)+a(a+d)>2 a(a+2 d)$ or $2 d^{2}>0$, which is obviously true. So the altitudes can only form an arithmetic progression if $\mathrm{d}=0$ and hence the triangle is equilateral. - -## Problem A2 - -The positive integers $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{p}, \mathrm{q}$ satisfy $\mathrm{ad}-\mathrm{bc}=1 \mathrm{and} \mathrm{a} / \mathrm{b}>\mathrm{p} / \mathrm{q}>\mathrm{c} / \mathrm{d}$. Show that $\mathrm{q}>=\mathrm{b}+$ $d$ and that if $q=b+d$, then $p=a+c$. - -## Solution - -$\mathrm{p} / \mathrm{q}>\mathrm{c} / \mathrm{d}$ implies $\mathrm{pd}>\mathrm{cq}$ and hence $\mathrm{pd}>=\mathrm{cq}+1$, so $\mathrm{p} / \mathrm{q} \geq \mathrm{c} / \mathrm{d}+1 /(\mathrm{qd})$. Similarly, $\mathrm{a} / \mathrm{b}>\mathrm{p} / \mathrm{q}$ implies $a / b \geq p / q+1 /(b q)$. So $a / b-c / d \geq 1 /(q d)+1 /(q b)=(b+d) /(q b d)$. But $a / b-c / d=1 / b d$. Hence $\mathrm{q} \geq \mathrm{b}+\mathrm{d}$. - -Now assume $\mathrm{q}=\mathrm{b}+\mathrm{d}$. We have $\mathrm{ad}-\mathrm{bc}=1 \leq \mathrm{d}$, so $\mathrm{ad}+\mathrm{cd}-\mathrm{d} \leq \mathrm{bc}+\mathrm{cd}$ and hence ( $\mathrm{a}+\mathrm{c}-$ 1)/(b+d) $\leq c /$ d. So $p \geq a+c$. Similarly $a d-b c \leq b$, so $b c+b+a b \geq a d+a b$, so $(a+c+1) /(b+d)$ $\geq a / b$. So $p \leq a+c$. Hence $p=a+c$. - -## Problem A3 - -$\mathrm{P}$ is a fixed point in the plane. Show that amongst triangles $\mathrm{ABC}$ such that $\mathrm{PA}=3, \mathrm{~PB}=5, \mathrm{PC}$ $=7$, those with the largest perimeter have $\mathrm{P}$ as incenter. - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-06.jpg?height=508&width=611&top_left_y=500&top_left_x=243) - -Given points $\mathrm{P}, \mathrm{B}, \mathrm{C}$ and a fixed circle center $\mathrm{P}$, we show that the point $\mathrm{A}$ on the circle which maximises $\mathrm{AB}+\mathrm{AC}$ is such that $\mathrm{PA}$ bisects angle $\mathrm{BAC}$. Consider a point $\mathrm{A}^{\prime}$ close to $\mathrm{A}$. Then the change in $A B+A C$ as we move $A$ to $A^{\prime}$ is $A A^{\prime}(\sin P A C-\sin P A B)+O\left(A A^{\prime 2}\right)$. So for a maximal configuration we must have $\sin \mathrm{PAC}=\sin \mathrm{PAB}$, otherwise we could get a larger sum by taking A' on one side or the other. This applies to each vertex of the triangle, so $\mathrm{P}$ must be the incenter. - -## Problem B1 - -Points $\mathrm{A}_{1}, \mathrm{~A}_{2}, \ldots, \mathrm{A}_{\mathrm{n}}$ are equally spaced on the side $\mathrm{BC}$ of the triangle $\mathrm{ABC}$ (so that $\mathrm{BA}_{1}=$ $\left.A_{1} A_{2}=\ldots=A_{n-1} A_{n}=A_{n} C\right)$. Similarly, points $B_{1}, B_{2}, \ldots, B_{n}$ are equally spaced on the side CA, and points $\mathrm{C}_{1}, \mathrm{C}_{2}, \ldots, \mathrm{C}_{\mathrm{n}}$ are equally spaced on the side $\mathrm{AB}$. Show that $\left(\mathrm{AA}_{1}{ }^{2}+\mathrm{AA}_{2}{ }^{2}+\ldots+\right.$ $\left.\mathrm{AA}_{\mathrm{n}}{ }^{2}+\mathrm{BB}_{1}{ }^{2}+\mathrm{BB}_{2}{ }^{2}+\ldots+\mathrm{BB}_{\mathrm{n}}{ }^{2}+\mathrm{C}_{1}{ }^{2}+\ldots+\mathrm{CC}_{\mathrm{n}}{ }^{2}\right)$ is a rational multiple of $\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\right.$ $\left.\mathrm{CA}^{2}\right)$. - -## Solution - -Using the cosine formula, $\mathrm{AA}_{\mathrm{k}}{ }^{2}=\mathrm{AB}^{2}+\mathrm{k}^{2} \mathrm{BC}^{2} /(\mathrm{n}+1)^{2}-2 \mathrm{k} A \mathrm{AB} \cdot \mathrm{BC} /(\mathrm{n}+1) \cos \mathrm{B}$. So $\sum \mathrm{AA}_{\mathrm{k}}{ }^{2}=$ $n B^{2}+B C^{2} /(n+1)^{2}\left(1^{2}+2^{2}+\ldots+n^{2}\right)-2 A B \cdot B C \cos B(1+2+\ldots+n) /(n+1)$. Similarly for the other two sides. - -Thus the total sum is $n\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}\right)+\mathrm{n}(2 \mathrm{n}+1) /(6(\mathrm{n}+1))\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}\right)-\mathrm{n}$ $(\mathrm{AB} \cdot \mathrm{BC} \cos \mathrm{B}+\mathrm{BC} \cdot \mathrm{CA} \cos \mathrm{C}+\mathrm{CA} \cdot \mathrm{AB} \cos \mathrm{A})$. $\mathrm{But} \mathrm{AB} \cdot \mathrm{BC} \cos \mathrm{B}=\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}-\mathrm{CA}^{2}\right) / 2$, so $A B \cdot B C \cos B+B C \cdot C A \cos C+C A \cdot A B \cos A=\left(A B^{2}+\mathrm{BC}^{2}+C A^{2}\right) / 2$. Thus the sum is rational multiple of $\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}\right)$. - -## Problem B2 - -Let $\mathrm{k}^{3}=2$ and let $\mathrm{x}, \mathrm{y}, \mathrm{z}$ be any rational numbers such that $\mathrm{x}+\mathrm{y} \mathrm{k}+\mathrm{z} \mathrm{k}^{2}$ is non-zero. Show that there are rational numbers $\mathrm{u}, \mathrm{v}, \mathrm{w}$ such that $\left(\mathrm{x}+\mathrm{yk}+\mathrm{z} \mathrm{k}^{2}\right)\left(\mathrm{u}+\mathrm{v} \mathrm{k}+\mathrm{w}^{2}\right)=1$. - -## Solution - -We need $x u+2 z v+2 y w=1$, $y u+x v+2 z w=0, z u+y v+x w=0$. This is just a straightforward set of linear equations. Solving, we get: $u=\left(x^{2}-2 y z\right) / d, v=\left(2 z^{2}-x y\right) / d, w=$ $\left(y^{2}-x z\right) / d$, were $d=x^{3}+2 y^{3}+4 z^{3}-6 x y z$. - -This would fail if $\mathrm{d}=0$. But if $\mathrm{d}=0$, then multiplying through by a suitable integer we have $6 \mathrm{mnr}=\mathrm{m}^{3}+2 \mathrm{n}^{3}+4 \mathrm{r}^{3}$ for some integers $\mathrm{m}, \mathrm{n}, \mathrm{r}$. But we can divide by any common factor of $\mathrm{m}, \mathrm{n}, \mathrm{r}$ to get them without any common factor. But $6 \mathrm{mnr}, 2 \mathrm{n}^{3}, 4 \mathrm{r}^{3}$ are all even, so $\mathrm{m}$ must be even. Put $m=2 M$. Then $12 M n r=8 M^{3}+2 n^{3}+4 r^{3}$, so $6 M n r=4 M^{3}+n^{3}+2 r^{3}$. But $6 M n r, 4 M^{3}$ and $2 r^{3}$ are all even, so $n$ must be even. Put $n=2 N$. Then $12 M N r=4 M^{3}+8 N^{3}+2 r^{3}$, so $6 M N r$ $=2 \mathrm{M}^{3}+4 \mathrm{~N}^{3}+\mathrm{r}^{3}$, so $\mathrm{r}$ must be even. So $\mathrm{m}, \mathrm{n}, \mathrm{r}$ had a common factor 2 . Contradiction. So $\mathrm{d}$ cannot be zero. - -## Problem B3 - -Let $S$ be the collection of all sets of $n$ distinct positive integers, with no three in arithmetic progression. Show that there is a member of $S$ which has the largest sum of the inverses of its elements (you do not have to find it or to show that it is unique). - -## Solution - -Induction on $\mathrm{n}$. For $\mathrm{n}=1,\{1\}$ is obviously maximal. Now suppose $\mathrm{a}_{1}<\mathrm{a}_{2}<\ldots<\mathrm{a}_{\mathrm{n}}$ is a maximal set for $n$. Take $a_{n+1}$ to be the smallest integer $>a_{n}$ such that $\left\{a_{1}, a_{2}, \ldots, a_{n+1}\right\}$ has no three members in AP. Now consider the sequences $\mathrm{b}_{1}<\mathrm{b}_{2}<\ldots<\mathrm{b}_{\mathrm{n}}$ which have no three in AP and $b_{n+1} \leq a_{n+1}$. There are only finitely many such sequences. So we can find one which is maximal. Suppose it is $\mathrm{c}_{1}<\mathrm{c}_{2}<\ldots<\mathrm{c}_{\mathrm{n}+1}$. Now take whichever of $\mathrm{a}_{\mathrm{i}}, \mathrm{c}_{\mathrm{i}}$ has the larger sum of inverses. It is clearly maximal with respect to sequences whose largest member is $\leq a_{n+1}$. Suppose we have a sequence $x_{1}a_{n+1}$. Then we have $1 / \mathrm{x}_{n+1}<1 / \mathrm{a}_{n+1}$ and, by induction, $1 / \mathrm{x}_{1}+\ldots+1 / \mathrm{x}_{n} \leq 1 / \mathrm{a}_{1}+\ldots+1 / \mathrm{a}_{n}$, so $1 / \mathrm{x}_{1}+\ldots+1 / \mathrm{x}_{\mathrm{n}+1}$ $<1 / \mathrm{a}_{1}+\ldots+1 / \mathrm{a}_{\mathrm{n}+1}$, so it is worse than the sequence we have chosen. - -## 4th Iberoamerican 1989 - -## Problem A1 - -Find all real solutions to: $\mathrm{x}+\mathrm{y}-\mathrm{z}=-1 ; \mathrm{x}^{2}-\mathrm{y}^{2}+\mathrm{z}^{2}=1,-\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}=-1$. - -## Solution - -Answer: $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,-1,1)$ or $(-1,-1,-1)$. - -From the first equation $x=z-y-1$. Substituting in the second equation: $2 z^{2}-2 y z+2 y-2 z=$ 0 , so $(\mathrm{z}-1)(\mathrm{z}-\mathrm{y})=0$. Hence $\mathrm{z}=1$ or $\mathrm{y}=\mathrm{z}$. If $\mathrm{z}=1$, then from the first equation $\mathrm{x}+\mathrm{y}=0$, and hence from the last equation, $\mathrm{x}=1, \mathrm{y}=-1$. If $\mathrm{y}=\mathrm{z}$, then $\mathrm{x}=-1$, and hence from the last - -equation $\mathrm{y}=\mathrm{z}=-1$. - -## Problem A2 - -Given positive real numbers $\mathrm{x}, \mathrm{y}, \mathrm{z}$ each less than $\pi / 2$, show that $\pi / 2+2 \sin \mathrm{x} \cos \mathrm{y}+2 \sin \mathrm{y}$ $\cos z>\sin 2 x+\sin 2 y+\sin 2 z$. - -## Solution - -We have $\sin 2 x+\sin 2 y+\sin 2 z-2 \sin x \cos y-2 \sin y \cos z=2 \sin x(\cos x-\cos y)+2 \sin$ $y(\cos y-\cos z)+2 \sin z \cos z$, so we wish to show that $\sin x(\cos x-\cos y)+\sin y(\cos y-\cos$ $\mathrm{z})+\sin \mathrm{z} \cos \mathrm{z}<\pi / 2(*)$. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-08.jpg?height=608&width=642&top_left_y=464&top_left_x=273) - -We have to consider six cases: (1) $\mathrm{x} \leq \mathrm{y} \leq \mathrm{z}$; (2) $\mathrm{x} \leq \mathrm{z} \leq \mathrm{y}$; (3) $\mathrm{y} \leq \mathrm{x} \leq \mathrm{z}$; (4) $\mathrm{y} \leq \mathrm{z} \leq \mathrm{x}$; (5) $\mathrm{z} \leq$ $\mathrm{x} \leq \mathrm{y} ;$ (6) $\mathrm{z} \leq \mathrm{y} \leq \mathrm{x}$. The first case is obvious from the diagram, because the lhs represents the shaded area, and the rhs represents the whole quarter circle. - -In cases (2) and (5) the second term is negative, and - $\sin \mathrm{y}<-\sin \mathrm{x}$, so the sum of the first two terms is less than $\sin x(\cos x-\cos y)+\sin x(\cos y-\cos z)=\sin x(\cos x-\cos z)$. But by the same argument as the first case the two rectangles represented by $\sin x(\cos x-\cos z)$ and $\sin \mathrm{z} \cos \mathrm{z}$ are disjoint and fit inside the quarter circle. So we have proved (2) and (5). - -In cases (3) and (4), the first term is negative. The remaining two terms represent disjoint rectangles lying inside the quarter circle, so again the inequality holds. - -In case (6) the first two terms are negative. The last term is $1 / 2 \sin 2 z \leq 1 / 2<\pi / 2$, so the inequality certainly holds. - -## Problem A3 - -If $\mathrm{a}, \mathrm{b}, \mathrm{c}$, are the sides of a triangle, show that $(\mathrm{a}-\mathrm{b}) /(\mathrm{a}+\mathrm{b})+(\mathrm{b}-\mathrm{c}) /(\mathrm{b}+\mathrm{c})+(\mathrm{c}-\mathrm{a}) /(\mathrm{a}+\mathrm{c})<$ $1 / 16$. - -## Solution - -Put $\mathrm{f}(\mathrm{a}, \mathrm{b}, \mathrm{c})=(\mathrm{a}-\mathrm{b}) /(\mathrm{a}+\mathrm{b})+(\mathrm{b}-\mathrm{c}) /(\mathrm{b}+\mathrm{c})+(\mathrm{c}-\mathrm{a}) /(\mathrm{a}+\mathrm{c})$. Let $\mathrm{A}, \mathrm{B}, \mathrm{C}$ be the permutation of $\mathrm{a}, \mathrm{b}, \mathrm{c}$, with $\mathrm{A}<=\mathrm{B}<=\mathrm{C}$. If $(\mathrm{A}, \mathrm{B}, \mathrm{C})=(\mathrm{b}, \mathrm{a}, \mathrm{c}),(\mathrm{a}, \mathrm{c}, \mathrm{b})$ or $(\mathrm{c}, \mathrm{b}, \mathrm{a})$, then $\mathrm{f}(\mathrm{a}, \mathrm{b}, \mathrm{c})=\mathrm{X}$, where $X=(B-A) /(B+A)+(C-B) /(C+B)-(C-A) /(A+C)$. If $(A, B, C)=(a, b, c),(b, c$, a) or ( $\mathrm{c}, \mathrm{a}, \mathrm{b})$, then $\mathrm{f}(\mathrm{a}, \mathrm{b}, \mathrm{c})=-\mathrm{X}$. - -Put $\mathrm{B}=\mathrm{A}+\mathrm{h}, \mathrm{C}=\mathrm{B}+\mathrm{k}=\mathrm{A}+\mathrm{h}+\mathrm{k}$, where $\mathrm{h}, \mathrm{k} \geq 0$. Since $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are the sides of a triangle, we also have A + B $>$ C or A $>k$. So $X=h /(2 A+h)+k /(2 A+2 h+k)-(h+k) /(2 A+h+k)$ $=\mathrm{hk}(\mathrm{h}+\mathrm{k}) /((2 \mathrm{~A}+\mathrm{h})(2 \mathrm{~A}+\mathrm{h}+\mathrm{k})(2 \mathrm{~A}+2 \mathrm{~h}+\mathrm{k}))$. This is obviously non-negative. We claim also that it is $<1 / 20$. That is equivalent to: $20 \mathrm{~h}^{2} \mathrm{k}+20 \mathrm{hk}^{2}<(2 \mathrm{~A}+\mathrm{h})(2 \mathrm{~A}+\mathrm{h}+\mathrm{k})(2 \mathrm{~A}+2 \mathrm{~h}+$ k). Since $k2$. - -Note that if $\mathrm{X}$ is any vertex except $\mathrm{A}$ or $\mathrm{B}$, then an even number of segments with endpoint $\mathrm{X}$ must be in the path. - -Let $\mathrm{F}$ be the bottom left-hand vertex. Two sides of the triangle EFG are in the path, so at least one of $\mathrm{EF}$ and $\mathrm{FG}$ is. But $\mathrm{EF}$ and $\mathrm{EG}$ are the only segments with endpoint $\mathrm{F}$, so an even number of them must be in the path, so both are in the path. Hence, again considering EFG, EG is not in the path. Hence, considering EHG, EH and HG are in the path. - -E has an even number of segments on the path, so $\mathrm{CE}$ is not on the path. Hence (considering CEH) CH is on the path. Similarly, GJ is not on the path and HJ is on the path. An even number of segments at $\mathrm{H}$ are on the path, so $\mathrm{DH}$ and $\mathrm{HI}$ are either both on the path or neither is on the path. But (considering DHI) at least one must be, so they both are. Hence DI is not, and $\mathrm{CD}$ is not. - -Since $\mathrm{n}>2, \mathrm{C}$ is not the top left vertex. Considering MCD, MC and MD are both on the path. Considering DLI, DL is on the path. There must be an even number of segments at D, so DP is on the path. Hence MP is not. Now M cannot be the top left vertex (with $n=3$ ) because then it should have an odd number of segments, whereas it would have two (MC and MD). So there must be a vertex $\mathrm{N}$ above $\mathrm{M}$. Considering NMP, MN must be in the path. But now $\mathrm{M}$ has an odd number of segments. Contradiction. - -## Problem B3 - -$\mathrm{f}(\mathrm{x})$ is a polynomial of degree 3 with rational coefficients. If its graph touches the $\mathrm{x}$-axis, show that it has three rational roots. - -## Solution - -Without loss of generality, $f(x)=x^{3}-a x^{2}+b x-c$, where $a, b, c$ are rational. Since the graph touches the $\mathrm{x}$-axis, there is a repeated root, so we may take the roots to be $\mathrm{h}, \mathrm{h}, \mathrm{k}$. Hence $2 \mathrm{~h}+$ $k=a, 2 h k+k^{2}=b, h^{2} k=c$. Hence $a^{2}-3 b=(h-k)^{2}$. Put $r= \pm \sqrt{ }\left(a^{2}-3 b\right)$, where the sign is chosen so that $h=a / 3+r / 3, k=a / 3-2 r / 3$. We need to show that $r$ is rational. If $r$ is zero there is nothing to prove, so assume $\mathrm{r}$ is non-zero. - -We have $9 h^{2}=2 a^{2}-3 b+2 a r$. Hence $27 h^{2} k=-2 a^{3}+9 a b+\left(6 b-2 a^{2}\right)$ r. But $27 h^{2} k=27 c$. So $r=$ $\left(27 c+2 a^{3}-9 a b\right) /\left(2\left(3 b-a^{2}\right)\right)$. Note that $3 b-2 a^{2}$ is non-zero because $r$ is non-zero. So $r$ is a rational combination of $\mathrm{a}, \mathrm{b}, \mathrm{c}$ and hence is rational. - -## 6th Iberoamerican 1991 - -## Problem 1 - -The number 1 or the number -1 is assigned to each vertex of a cube. Then each face is given the product of its four vertices. What are the possible totals for the resulting 14 numbers? - -## Solution - -Answer: $14,6,2,-2,-6,-10$. - -If every vertex is 1 , we get 14 and that is clearly the highest possible total. The lowest possible total cannot be lower than -14, but we cannot even achieve that because if all the vertices are -1 , then all the faces are 1 . - -If we change a vertex, then we also change three faces. If the vertex and the three faces are all initially the same, then we make a change of $\pm 8$. If three are of one kind and one the opposite, then we make a change of $\pm 4$. If two are of one kind and two the opposite, then we make no change. Thus any sequence of changes must take us to $14+4 \mathrm{n}$ for some integer $\mathrm{n}$. But we have already shown that the total is greater than -14 and at most 14 , so the only possibilities are $-10,-6,-2,2,6,10$ and 14 . - -We show that 10 is not possible. If more than 2 vertices are -1 , then the vertex total is at most 2 , there are only 6 faces, so the total is less than 10 . If all vertices are 1 , then the total is 14 . If all but one vertex is 1 , then the total is 6 . So the only possibility for 10 is just two vertices -1 . But however we choose any two vertices, there is always a face containing only one of them, so at least one face is -1 , so the face total is at most 4 and the vertex total is 4 , so the total is less than 10 . The other totals are possible, for example: - -``` -14: all vertices 1 -6: one vertex -1, rest 1 -2: three vertices of one face -1, rest 1 --2: all vertices -1 --6: all vertices but one -1 --10: two opposite corners 1, rest -1 -``` - -Two perpendicular lines divide a square into four parts, three of which have area 1. Show that the fourth part also has area 1. - -## Problem A3 - -$\mathrm{f}$ is a function defined on all reals in the interval $[0,1]$ and satisfies $\mathrm{f}(0)=0, \mathrm{f}(\mathrm{x} / 3)=\mathrm{f}(\mathrm{x}) / 2$, $f(1-x)=1-f(x)$. Find $f(18 / 1991)$. - -## Problem B1 - -Find a number $\mathrm{N}$ with five digits, all different and none zero, which equals the sum of all distinct three digit numbers whose digits are all different and are all digits of $\mathrm{N}$. - -## Solution - -Answer: 35964 - -There are $4.3=12$ numbers with a given digit of $n$ in the units place. Similarly, there are 12 with it in the tens place and 12 with it in the hundreds place. So the sum of the 3 digit numbers is $12.111(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e})$, where $\mathrm{n}=\mathrm{abcde}$. So $8668 \mathrm{a}=332 \mathrm{~b}+1232 \mathrm{c}+1322 \mathrm{~d}+$ 1331e. We can easily see that $\mathrm{a}=1$ is too small and $\mathrm{a}=4$ is too big, so $\mathrm{a}=2$ or 3 . Obviously e must be even. 0 is too small, so $\mathrm{e}=2,4,6$ or 8 . Working mod 11 , we see that $0=2 \mathrm{~b}+2 \mathrm{~d}$, so $\mathrm{b}+\mathrm{d}=11$. Working $\bmod 7$, we see that $2 \mathrm{a}=3 \mathrm{~b}+6 \mathrm{~d}+\mathrm{e}$. Using the $\bmod 11$ result, $\mathrm{b}=2, \mathrm{~d}=$ 9 or $b=3, d=8$ or $b=4, d=7$ or $b=5, d=6$ or $b=6, d=5$ or $b=7, d=4$ or $b=8, d=3$ or $\mathrm{b}=9, \mathrm{~d}=2$. Putting each of these into the $\bmod 7$ result gives $2 \mathrm{a}-\mathrm{e}=4,1,5,2,6,3,0,4$ mod 7. So putting $\mathrm{a}=2$ and remembering that $\mathrm{e}$ must be $2,4,6,8$ and that all digits must be different gives a, b, d, e = 2,4, 7, 6 or 2, 7, 4, 8 or 2, 8, 3, 4 as the only possibilities. It is then straightforward but tiresome to check that none of these give a solution for $\mathrm{c}$. Similarly putting $\mathrm{a}=4$, gives $\mathrm{a}, \mathrm{b}, \mathrm{d}, \mathrm{e}=3,4,7,8$ or $3,5,6,4$ as the only possibilities. Checking, we find the solution above and no others. - -## Problem B2 - -Let $p(m, n)$ be the polynomial $2 m^{2}-6 m n+5 n^{2}$. The range of $p$ is the set of all integers $k$ such that $\mathrm{k}=\mathrm{p}(\mathrm{m}, \mathrm{n})$ for some integers $\mathrm{m}$, $\mathrm{n}$. Find which members of $\{1,2, \ldots, 100\}$ are in the range of $p$. Show that if $h$ and $k$ are in the range of $p$, then so is $h k$. - -## Answer - -1,2,4,5,8,9,10,13,16,17,18, 20,25,26,29,32,34,36,37, 40,41,45,49,50,52,53,58, $61,64,65,68,72,73,74,80,81,82,85,89,90,97,98,100$ - -## Solution - -We have $p(m, n)=(m-2 n)^{2}+(m-n)^{2}$, so $p(2 a-b, a-b)=a^{2}+b^{2}$. Hence the range of $p$ is just the sums of two squares. - -$\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=(a c-b d)^{2}+(a d+b c)^{2}$, which establishes that if $h$ and $k$ are in the range, then so is hk. - -## Problem B3 - -Given three non-collinear points $\mathrm{M}, \mathrm{N}, \mathrm{H}$ show how to construct a triangle which has $\mathrm{H}$ as orthocenter and $\mathrm{M}$ and $\mathrm{N}$ as the midpoints of two sides. - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-15.jpg?height=340&width=414&top_left_y=504&top_left_x=250) - -Take $\mathrm{H}^{\prime}$ so that $\mathrm{M}$ is the midpoint of $\mathrm{HH}^{\prime}$. The circle diameter $\mathrm{NH}^{\prime}$ meets the line through $\mathrm{H}$ perpendicular to MN in two points (in general), either of which we may take as A. Then B is the reflection of $\mathrm{A}$ in $\mathrm{M}$, and $\mathrm{C}$ is the reflection of $\mathrm{A}$ in $\mathrm{N}$. - -To see that this works, note that $\mathrm{M}$ is the midpoint of $\mathrm{HH}^{\prime}$ and $\mathrm{AB}$, so $\mathrm{AHBH}$ ' is a parallelogram. Hence $\mathrm{AH}$ ' is parallel to $\mathrm{BH}$ and hence perpendicular to $\mathrm{AC}$. In other words $\angle \mathrm{NAH}^{\prime}=90^{\circ}$, so A lies on the circle diameter $\mathrm{NH}^{\prime}$. $\mathrm{MN}$ is parallel to $\mathrm{BC}$, so $\mathrm{A}$ lies on the perpendicular to $\mathrm{MN}$ through $\mathrm{H}$. - -## 7th Iberoamerican 1992 - -## Problem A1 - -$a_{n}$ is the last digit of $1+2+\ldots+n$. Find $a_{1}+a_{2}+\ldots+a_{1992}$. - -## Solution - -It is easy to compile the following table, from which we see that $\mathrm{a}_{\mathrm{n}}$ is periodic with period 20 , and indeed the sum for each decade (from 0 to 9 ) is 35 . Thus the sum for 1992 is $199.35+5+$ $6+8=6984$. - -| $\mathrm{n}$ | | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | -| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | -| 17 | 18 | 19 | 20 | | | | | | | | | | | | | | | | -| $a_{\mathrm{n}}$ | | 0 | 1 | 3 | 6 | 0 | 5 | 1 | 8 | 6 | 5 | 5 | 6 | 8 | 1 | 5 | 0 | 6 | -| 3 | 1 | 0 | 0 | | | | | | | | | | 6 | | 1 | 5 | | | -| sum | | 0 | 1 | 4 | 10 | 10 | 15 | 16 | 24 | 30 | 35 | 40 | 46 | 54 | 55 | 60 | 60 | 66 | -| 69 | 70 | 70 | 70 | | | | | | | | | | | | | | | | - -## Problem A2 - -Let $f(x)=a_{1} /\left(x+a_{1}\right)+a_{2} /\left(x+a_{2}\right)+\ldots+a_{n} /\left(x+a_{n}\right)$, where $a_{i}$ are unequal positive reals. Find the sum of the lengths of the intervals in which $f(x) \geq 1$. -Answer -$\sum a_{i}$ - -## Solution - -wlog $a_{1}>a_{2}>\ldots>a_{n}$. The graph of each $a_{i} /\left(x+a_{i}\right)$ is a rectangular hyberbola with asymptotes $x=-a_{i}$ and $y=0$. So it is not hard to see that the graph of $f(x)$ is made up of $n+1$ strictly decreasing parts. For $\mathrm{x}<-\mathrm{a}_{1}, \mathrm{f}(\mathrm{x})$ is negative. For $\mathrm{x} \square\left(-\mathrm{a}_{\mathrm{i}},-\mathrm{a}_{\mathrm{i}+1}\right), \mathrm{f}(\mathrm{x})$ decreases from $\infty$ to $-\infty$. Finally, for $\mathrm{x}>-\mathrm{a}_{\mathrm{n}}, \mathrm{f}(\mathrm{x})$ decreases from $\infty$ to 0 . Thus $\mathrm{f}(\mathrm{x})=1$ at $\mathrm{n}$ values $\mathrm{b}_{1}<\mathrm{b}_{2}<\ldots<\mathrm{b}_{\mathrm{n}}$, and $f(x) \geq 1$ on the $n$ intervals $\left(-a_{1}, b_{1}\right),\left(-a_{2}, b_{2}\right), \ldots,\left(-a_{n}, b_{n}\right)$. So the sum of the lengths of these intervals is $\sum\left(a_{i}+b_{i}\right)$. We show that $\sum b_{i}=0$. - -Multiplying $f(x)=1$ by $\prod\left(x+a_{j}\right)$ we get a polynomial of degree $n$ : - -$$ -\Pi\left(x+a_{j}\right)-\sum_{i}\left(a_{i} \prod_{j \neq i}\left(x+a_{j}\right)\right)=0 -$$ - -The coefficient of $x^{n}$ is 1 and the coefficient of $x^{n-1}$ is $\sum a_{j}-\sum a_{i}=0$. Hence the sum of the roots, which is $\sum b_{i}$, is zero. - -## Problem A3 - -$\mathrm{ABC}$ is an equilateral triangle with side 2 . Show that any point $\mathrm{P}$ on the incircle satisfies $\mathrm{PA}^{2}$ $+\mathrm{PB}^{2}+\mathrm{PC}^{2}=5$. Show also that the triangle with side lengths $\mathrm{PA}, \mathrm{PB}, \mathrm{PC}$ has area $(\sqrt{3}) / 4$. - -## Solution - -Take vectors centered at the center $\mathrm{O}$ of the triangle. Write the vector $\mathrm{OA}$ as $\mathbf{A}$ etc. Then $\mathrm{PA}^{2}$ $+\mathrm{PB}^{2}+\mathrm{PC}^{2}=(\mathbf{P}-\mathbf{A})^{2}+(\mathbf{P}-\mathbf{B})^{2}+(\mathbf{P}-\mathbf{C})^{2}=3 \mathrm{P}^{2}+\left(\mathrm{A}^{2}+\mathrm{B}^{2}+\mathrm{C}^{2}\right)-2 \mathbf{P} \cdot(\mathbf{A}+\mathbf{B}+\mathbf{C})=15 \mathrm{P}^{2}$, since $A^{2}=B^{2}=C^{2}=4 P^{2}$ and $\mathbf{A}+\mathbf{B}+\mathbf{C}=0$. Finally the side is 2 , so an altitude is $\sqrt{3}$ and the inradius is $(\sqrt{3}) / 3=1 / \sqrt{ } 3$, so $\mathrm{PA}^{2}+\mathrm{PB}^{2}+\mathrm{PC}^{2}=15 / 3=5$. - -Take $\mathrm{Q}$ outside the triangle so that $\mathrm{BQ}=\mathrm{BP}$ and $\mathrm{CQ}=\mathrm{AP}$. Then $\mathrm{BQC}$ and $\mathrm{BPA}$ are congruent, so $\angle \mathrm{ABP}=\angle \mathrm{CBQ}$ and hence $\angle \mathrm{PBQ}=60^{\circ}$, so $\mathrm{PBQ}$ is equilateral. Hence $\mathrm{PQ}$ is $\mathrm{PB}$ and $\mathrm{PQC}$ has sides equal to $\mathrm{PA}, \mathrm{PB}, \mathrm{PC}$. If we construct two similar points outside the other two sides then we get a figure with total area equal to 2 area $\mathrm{ABC}$ and to 3 area $\mathrm{PQC}+$ area of three equilateral triangles sides $\mathrm{PA}, \mathrm{PB}, \mathrm{PC}$. Hence 3 area $\mathrm{PQC}=2$ area $\mathrm{ABC}$ - area $\mathrm{ABC}\left(\mathrm{PA}^{2}+\mathrm{PB}^{2}+\mathrm{PC}^{2}\right) / \mathrm{PA}^{2}=(3 / 4)$ area $\mathrm{ABC}=(3 \sqrt{3}) / 4$. So area $\mathrm{PQC}=(\sqrt{3}) / 4$. - -## Problem B1 - -Let $a_{n}, b_{n}$ be two sequences of integers such that: (1) $a_{0}=0, b_{0}=8$; (2) $a_{n+2}=2 a_{n+1}-a_{n}+2$, $b_{n+2}=2 b_{n+1}-b_{n}$, (3) $a_{n}{ }^{2}+b_{n}{ }^{2}$ is a square for $n>0$. Find at least two possible values for $\left(a_{1992}\right.$, $\mathrm{b}_{1992}$ ). - -## Answer - -(1992. 1996, 4. 1992+8), (1992. 1988, -4. 1992+8) - -## Solution - -$a_{n}$ satisfies a standard linear recurrence relation with general solution $a_{n}=n^{2}+A n+k$. But $a_{n}$ $=0$, so $k=0$. Hence $a_{n}=n^{2}+$ An. If you are not familiar with the general solution, then you can guess this solution and prove it by induction. - -Similarly, $b_{n}=B n+8$. Hence $a_{n}{ }^{2}+b_{n}{ }^{2}=n^{4}+2 A n^{3}+\left(A^{2}+B^{2}\right) n^{2}+16 B n+64$. If this is a square, then looking at the constant and $n^{3}$ terms, it must be ( $\left.n^{2}+A n+8\right)$. Comparing the other terms, $\mathrm{A}=\mathrm{B}= \pm 4$. - -## Problem B2 - -Construct a cyclic trapezium ABCD with AB parallel to CD, perpendicular distance $h$ between $\mathrm{AB}$ and $\mathrm{CD}$, and $\mathrm{AB}+\mathrm{CD}=\mathrm{m}$. - -## Problem B3 - -Given a triangle $\mathrm{ABC}$, take $\mathrm{A}^{\prime}$ on the ray $\mathrm{BA}$ (on the opposite side of $\mathrm{A}$ to $\mathrm{B}$ ) so that $\mathrm{AA}^{\prime}=$ $\mathrm{BC}$, and take $\mathrm{A}$ " on the ray $\mathrm{CA}$ (on the opposite side of $\mathrm{A}$ to $\mathrm{C}$ ) so that $\mathrm{AA}=\mathrm{BC}$. Similarly take $\mathrm{B}^{\prime}, \mathrm{B}^{\prime \prime}$ on the rays $\mathrm{CB}, \mathrm{AB}$ respectively with $\mathrm{BB}^{\prime}=\mathrm{BB}^{\prime \prime}=\mathrm{CA}$, and $\mathrm{C}^{\prime}, \mathrm{C}^{\prime \prime}$ on the rays $\mathrm{AB}$, CB. Show that the area of the hexagon $\mathrm{A}^{\prime \prime} \mathrm{A}^{\prime} \mathrm{B}^{\prime \prime} \mathrm{B}^{\prime} \mathrm{C} \mathrm{C}^{\prime}$ ' $^{\text {is }}$ at least 13 times the area of the triangle $\mathrm{ABC}$. - -## 8th Iberoamerican 1993 - -## Problem 1 - -A palindrome is a positive integers which is unchanged if you reverse the order of its digits. For example, 23432. If all palindromes are written in increasing order, what possible prime values can the difference between successive palindromes take? - -## Solution - -Answer: 2,11 . - -Let $\mathrm{x}$ be a palindrome and $\mathrm{x}^{\prime}$ the next highest palindrome. If $\mathrm{x}<101$, then it is easy to see by inspection that $\mathrm{x}^{\prime}-\mathrm{x}=1,2$ or 11 , so the only prime differences are 2 and 11 . - -So assume $x>100$. If $x$ and $x^{\prime}$ have the same final digit, then their difference is divisible by 10 and hence not prime. So they must have different digits. Thus either $\mathrm{x}=\mathrm{d} 9 \ldots 9 \mathrm{~d}$ and $\mathrm{x}^{\prime}=$ $\mathrm{d}^{\prime} 0 \ldots 0 \mathrm{~d}^{\prime}$, where $\mathrm{d}<9$ and $\mathrm{d}^{\prime}=\mathrm{d}+1$, or $\mathrm{x}^{\prime}$ has one more digit than $\mathrm{x}$ and $\mathrm{d}=9, \mathrm{~d}^{\prime}=1$. In the first case $\mathrm{x}^{\prime}-\mathrm{x}=11$. In the second case $\mathrm{x}^{\prime}-\mathrm{x}=2$. So again the only prime differences are 2 and 11. - -## Problem 2 - -Show that any convex polygon of area 1 is contained in some parallelogram of area 2. - -## Solution - -Let the vertices $\mathrm{X}, \mathrm{Y}$ of the polygon be the two which are furthest apart. The polygon must lie between the lines through $\mathrm{X}$ and $\mathrm{Y}$ perpendicular to $\mathrm{XY}$ (for if a vertex $\mathrm{Z}$ lay outside the line through $\mathrm{Y}$, then $\mathrm{ZY}>\mathrm{XY}$ ). Take two sides of a rectangle along these lines and the other two sides as close together as possible. There must be a vertices $\mathrm{U}$ and $\mathrm{V}$ on each of the other two sides. But now the area of the rectangle is twice the area of XUYV, which is at most the area of the polygon. [In the case of a triangle one side of the rectangle will be XY.] - -## Problem A3 - -Find all functions $\mathrm{f}$ on the positive integers with positive integer values such that (1) if $\mathrm{x}<\mathrm{y}$, then $f(x)\mathrm{m}^{2}$ for some $\mathrm{m}$. Then by $(1), \mathrm{f}(\mathrm{f}(\mathrm{m}))>\mathrm{f}\left(\mathrm{m}^{2}=\mathrm{f}(\mathrm{m} \cdot \mathrm{m})=\mathrm{f}(\mathrm{m})^{2}\right.$. But $\mathrm{f}($ $\mathrm{f}(\mathrm{m}))=\mathrm{m}^{2} \mathrm{f}(\mathrm{m})$, so $\mathrm{m}^{2}>\mathrm{f}(\mathrm{m})$. Contradiction. - -Similarly, suppose $\mathrm{f}(\mathrm{m})<\mathrm{m}^{2}$. Then $\mathrm{m}^{2} \mathrm{f}(\mathrm{m})=\mathrm{f}(\mathrm{f}(\mathrm{m}))<\mathrm{f}\left(\mathrm{m}^{2}\right)=\mathrm{f}(\mathrm{m})^{2}$, so $\mathrm{m}^{2}<\mathrm{f}(\mathrm{m})$. - -Contradiction. So we must have $\mathrm{f}(\mathrm{m})=\mathrm{m}^{2}$. - -## Problem B1 - -$\mathrm{ABC}$ is an equilateral triangle. $\mathrm{D}$ is on the side $\mathrm{AB}$ and $\mathrm{E}$ is on the side $\mathrm{AC}$ such that $\mathrm{DE}$ touches the incircle. Show that AD/DB $+\mathrm{AE} / \mathrm{EC}=1$. - -## Solution - -Put $\mathrm{BD}=\mathrm{x}, \mathrm{CE}=\mathrm{y}, \mathrm{BC}=\mathrm{a}$. Then since the two tangents from $\mathrm{B}$ to the incircle are of equal length, and similarly the two tangents from $\mathrm{D}$ and $\mathrm{E}$, we have $\mathrm{ED}+\mathrm{BC}=\mathrm{BD}+\mathrm{CE}$, or $\mathrm{ED}=$ $x+y-a$. By the cosine law, $E D^{2}=A E^{2}+A D^{2}-A E$. AD. Substituting and simplifying, we get $a=3 x y /(x+y)$. Hence $A D / D B=(2 y-x) /(x+y)$ and $A E / E C=(2 x-y) /(x+y)$ with sum 1 . - -## Problem B2 - -If $\mathrm{P}$ and $\mathrm{Q}$ are two points in the plane, let $\mathrm{m}(\mathrm{PQ})$ be the perpendicular bisector of $\mathrm{PQ} . \mathrm{S}$ is a finite set of $n>1$ points such that: (1) if $P$ and $Q$ belong to $S$, then some point of $m(P Q)$ belongs to S, (2) if PQ, $\mathrm{P}^{\prime} \mathrm{Q}^{\prime}, \mathrm{P}^{\prime \prime} \mathrm{Q}^{\prime \prime}$ are three distinct segments, whose endpoints are all in S, then if there is a point in all of $\mathrm{m}(\mathrm{PQ}), \mathrm{m}\left(\mathrm{P}^{\prime} \mathrm{Q}^{\prime}\right), \mathrm{m}\left(\mathrm{P}^{\prime \prime} \mathrm{Q}^{\prime \prime}\right)$ it does not belong to $\mathrm{S}$. What are the possible values of $n$ ? - -## Answer - -$\mathrm{n}=3$ (equilateral triangle), 5 (regular pentagon). - -## Solution - -There are $n(n-1) / 2$ pairs of points. Each has a point of $S$ on its bisector. But each point of $S$ is on at most two bisectors, so $2 n \geq n(n-1) / 2$. Hence $n \leq 5$. - -The equilateral triangle and regular pentagon show that $\mathrm{n}=3,5$ are possible. - -Consider $\mathrm{n}=4$. There are 6 pairs of points, so at least one point of $\mathrm{S}$ must be on two bisectors. wlog $\mathrm{A}$ is on the bisectors of $\mathrm{BC}$ and $\mathrm{BD}$. But then it is also on the bisector of CD. - -Contradiction. - -## Problem B3 - -We say that two non-negative integers are related if their sum uses only the digits 0 and 1 . For example 22 and 79 are related. Let A and B be two infinite sets of non-negative integers such that: (1) if a $\square \mathrm{A}$ and $\mathrm{b} \square \mathrm{B}$, then a and $\mathrm{b}$ are related, (2) if $\mathrm{c}$ is related to every member of A, then it belongs to $\mathrm{B},(3)$ if $\mathrm{c}$ is related to every member of $\mathrm{B}$, then it belongs to $\mathrm{A}$. Show that in one of the sets $\mathrm{A}, \mathrm{B}$ we can find an infinite number of pairs of consecutive numbers. - -## Solution - -Suppose there is a member of A with last digit d. Then every member of B must have one of two possible last digits. Suppose there are members of B with both possibilities. Then every member of A must have last digit d. So either every member of A has the same last digit or every member of B has the same last digit (or both). Suppose every member of A has the same last digit d. - -But now if $n$ belongs to $B$ and $n+d$ has last digit 0 , then $n+1+d$ has last digit 1 . Moreover, if $m$ is any member of $A$, then $m+n$ has last digit 0 and other digits all 0 or 1 . Hence $m+n+1$ last last digit 1 and other digits all 0 or 1 , so $n+1$ must also belong to B. Similarly, if $n$ is in $B$ and $\mathrm{n}+\mathrm{d}$ has last digit 1 , then $\mathrm{n}-1$ must also belong to $\mathrm{B}$. So in either case there are infinitely many pairs of consecutive numbers in $B$. - -## 9th Iberoamerican 1994 - -## Problem A1 - -Show that there is a number $1<\mathrm{b}<1993$ such that if 1994 is written in base $\mathrm{b}$ then all its digits are the same. Show that there is no number $12$. But $83^{2}=6889>1993$, so $\mathrm{b}$ must divide 24 . Hence $\mathrm{b}=2,3,4,6$, 8,12 , or 24 . But we can easily check that none of these work: - -``` -\(1+2+2^{2}+\ldots+2^{9}=1023,1+\ldots+2^{10}=2047\). -\(1+3+\ldots+3^{6}=1093,1+\ldots+3^{\wedge} 7=3280\) -\(1+4+\ldots+4^{5}=1365,1+\ldots+4^{6}=5461\) -``` - -``` -\(1+6+\ldots 6^{4}=1555,1+\ldots+6^{5}=9331\) -\(1+8+8^{2}+8^{3}=585,1+\ldots+8^{4}=4681\) -\(1+12+12^{2}+12^{3}=1885,1+\ldots+12^{4}=22621\) -\(1+24+24^{2}=601,1+\ldots+24^{3}=14425\) -``` - - -## Problem A2 - -$\mathrm{ABCD}$ is a cyclic quadrilateral. A circle whose center is on the side $\mathrm{AB}$ touches the other three sides. Show that $\mathrm{AB}=\mathrm{AD}+\mathrm{BC}$. What is the maximum possible area of $\mathrm{ABCD}$ in terms of $|\mathrm{AB}|$ and $|C D|$ ? - -## Answer - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-20.jpg?height=48&width=848&top_left_y=981&top_left_x=244) - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-20.jpg?height=391&width=337&top_left_y=1204&top_left_x=266) - -Let the circle have center $\mathrm{O}$ on $\mathrm{AB}$ and radius $\mathrm{r}$. Let $\angle \mathrm{OAD}=\theta, \angle \mathrm{OBC}=\varphi$. Since $\mathrm{ABCD}$ is cyclic, $\angle \mathrm{ADC}=180^{\circ}-\varphi$, so $\angle \mathrm{ODA}=90^{\circ}-\varphi / 2$. If $\mathrm{AD}$ touches the circle at $\mathrm{X}$, then $\mathrm{AD}=\mathrm{AX}+$ $\mathrm{XD}=\mathrm{r} \cot \theta+\mathrm{r} \tan (\varphi / 2)$. Similarly, $\mathrm{BC}=\mathrm{r} \cot \varphi+\mathrm{r} \tan (\theta / 2)$. Put $\mathrm{t}=\tan (\theta / 2)$. Then $\cot \theta=$ $\left(1-\mathrm{t}^{2}\right) / 2 \mathrm{t}$, so $\cot \theta+\tan (\theta / 2)=\left(1+\mathrm{t}^{2}\right) / 2 \mathrm{t}=1 / \sin \theta$. Similarly for $\varphi$, so $\mathrm{AD}+\mathrm{BC}=\mathrm{r} / \sin \theta+\mathrm{r} / \sin$ $\varphi=\mathrm{AO}+\mathrm{OB}=\mathrm{AB}$. - -Suppose $\mathrm{AD}$ and $\mathrm{BC}$ meet at $\mathrm{H}$ (we deal below with the case where they are parallel). Then $\mathrm{HCD}$ and $\mathrm{HAB}$ are similar, so area $\mathrm{HCD}=\left(\mathrm{CD}^{2} / \mathrm{AB}^{2}\right)$ area $\mathrm{HAB}$ and area $\mathrm{ABCD}=(1-$ $\left.\mathrm{CD}^{2} / \mathrm{AB}^{2}\right)$ area $\mathrm{HAB}$. Also $\mathrm{AB} / \mathrm{CD}=\mathrm{HA} / \mathrm{HC}=\mathrm{HB} / \mathrm{HD}=(\mathrm{HA}+\mathrm{HB}) /(\mathrm{HC}+\mathrm{HD})=$ $(\mathrm{HA}+\mathrm{HB}) /(\mathrm{HB}-\mathrm{BC}+\mathrm{HA}-\mathrm{DA})=(\mathrm{HA}+\mathrm{HB}) /(\mathrm{HA}+\mathrm{HB}-\mathrm{AB})$. Hence $\mathrm{HA}+\mathrm{HB}=\mathrm{AB} \mathrm{B}^{2} /(\mathrm{AB}-\mathrm{CD})$, which is fixed. Now for fixed $\mathrm{HA}+\mathrm{HB}$ we maximise the area of $\mathrm{HAB}$ by taking $\mathrm{HA}=\mathrm{HB}$ and hence $\mathrm{AD}=\mathrm{BC}$. - -Put $\mathrm{h}=\mathrm{CD}, \mathrm{k}=\mathrm{AB}$. So $\mathrm{k} \cos \theta+\mathrm{h}=\mathrm{k}$. Hence $\cos \theta=(1-\mathrm{h} / \mathrm{k})$. Hence $\sin \theta=\sqrt{ }\left(2 \mathrm{~h} / \mathrm{k}-\mathrm{h}^{2} / \mathrm{k}^{2}\right)$. So area $A B C D=1 / 2(h+k) 1 / 2 k \sin \theta=(h / 2+k / 2) \sqrt{ }\left(h k / 2-h^{2} / 4\right)(*)$. - -If $\mathrm{AD}$ and $\mathrm{BC}$ are parallel then $\mathrm{A}$ and $\mathrm{B}$ must lie on the circle, so that $\angle \mathrm{DAB}=\angle \mathrm{ABC}=90^{\circ}$. But $A B C D$ is cyclic, so it must be a rectangle. Hence $A B=C D$ and area $A B C D=k^{2} / 2$. In this case $\left(^{*}\right)$ still gives the correct answer. - -There is a bulb in each cell of an $\mathrm{n} x \mathrm{n}$ board. Initially all the bulbs are off. If a bulb is touched, that bulb and all the bulbs in the same row and column change state (those that are on, turn off, and those that are off, turn on). Show that it is possible by touching $\mathrm{m}$ bulbs to turn all the bulbs on. What is the minimum possible value of $\mathrm{m}$ ? - -Answer
$\mathrm{n}$ odd, $\mathrm{n}$ is minimum
$n$ even, $n^{2}$ is minimum - -# Solution - -If $\mathrm{n}$ is odd, touch each bulb in the first column. Then bulbs in the first column are each switched $\mathrm{n}$ times, which is odd and so end up on. All other bulbs are switched just once, and so end up on. $\mathrm{n}$ is obviously minimal, because if $\mathrm{m}<\mathrm{n}$, then there is a bulb which is not switched at all (there must be a column with no bulb touched and a row with no bulb touched, so the bulb in that column and row is not switched). - -In $n$ is even, touch each bulb. Then each bulb is switched $2 \mathrm{n}-1$ times, so ends up on. We show that it is not possible to do better. - -Note first that there is no benefit in touching a bulb more than once, so each must be touched zero of one times. Thus we can represent the scheme as an array of $0 \mathrm{~s}$ and $1 \mathrm{~s}$, where 0 means that the corresponding bulb is not touched, and 1 means that it is touched. - -Let A, B, C, D be four values at the corners of a rectangle. We claim that A+B has the same parity as $C+D$. Let $L_{A B}$ be the number of $1 \mathrm{~s}$ in the row $A B$ are touched, similarly $L_{B C}$ (the number of $1 \mathrm{~s}$ in the column $\mathrm{BC}), \mathrm{L}_{\mathrm{CD}}, \mathrm{L}_{\mathrm{DA}}$. Since bulb $\mathrm{A}$ is switched we must have $\mathrm{L}_{\mathrm{AB}}+\mathrm{L}_{\mathrm{DA}}$ + A odd (note that $\mathrm{L}_{A B}+\mathrm{L}_{D A}$ double-counts the no. of touches of A). Similarly, $\mathrm{L}_{B C}+\mathrm{L}_{\mathrm{CD}}+$ $\mathrm{C}$ is odd, so $\mathrm{A}+\mathrm{C}+\left(\mathrm{L}_{\mathrm{AB}}+\mathrm{L}_{B C}+\mathrm{L}_{C D}+\mathrm{L}_{D A}\right)$ is even. Similarly, considering $\mathrm{B}$ and $\mathrm{D}$, we find that $\mathrm{B}+\mathrm{D}+\left(\mathrm{L}_{A B}+\mathrm{L}_{B C}+\mathrm{L}_{C D}+\mathrm{L}_{D A}\right)$ is even, so $\mathrm{A}+\mathrm{C}$ and $\mathrm{B}+\mathrm{D}$ have the same parity. Adding B+C to both, we get that $\mathrm{A}+\mathrm{B}$ and $\mathrm{C}+\mathrm{D}$ have the same parity. It follows that either $\mathrm{A}$ $=\mathrm{D}$ and $\mathrm{B}=\mathrm{C}$, or $\mathrm{A} \neq \mathrm{D}$ and $\mathrm{B} \neq \mathrm{D}$. - -Keeping A and B fixed, we can now vary C (and hence D). It follows that either the row through B matches that through A, or it has every cell different (to the corresponding cell in row A). Similarly for the other rows. So we have $\mathrm{k}$ rows of one type and $\mathrm{n}-\mathrm{k}$ rows which are equal to its "complement". Suppose first that $\mathrm{k}=\mathrm{n}$, so that all rows are the same. If we have all $1 \mathrm{~s}$, then we have a solution. If we have all $0 \mathrm{~s}$, we obviously do not have a solution. So suppose there is a 0 and a 1 in each row. Then the total count at a 1 is $n-1$ higher than at a 0 (because of the extra $\mathrm{n}-1 \mathrm{1}$ s in the same column). So they cannot both be odd (because $\mathrm{n}$ is even). Contradiction. - -Finally suppose there is a row and a complement row. So position $\mathrm{A}$ in one is 1 , then position $B$ in the same column in the other has 0 . If a row has $h \mathrm{~s}$, then a complement row has $\mathrm{n}-\mathrm{h} 1 \mathrm{~s}$. The column has $\mathrm{z} 1$ s, so $\mathrm{A}$ has $\mathrm{z}+\mathrm{h}-1$ or $\mathrm{z}+\mathrm{n}-\mathrm{h}-1$ 1s, and $\mathrm{B}$ has $\mathrm{z}+\mathrm{h}$ or $\mathrm{z}+\mathrm{n}-\mathrm{h} 1$ s. But since $\mathrm{n}$ is even, $z+h$ and $z+n-h$ have the same parity, so A and B have opposite parity. Contradiction. So the only solution for $\mathrm{n}$ even is all $1 \mathrm{~s}$. - -## Problem B1 - -$\mathrm{ABC}$ is an acute-angled triangle. $\mathrm{P}$ is a point inside its circumcircle. The rays $\mathrm{AP}, \mathrm{BP}, \mathrm{CP}$ intersect the circle again at D, E, F. Find P so that DEF is equilateral. - -## Solution - -Let the angle bisector of $\mathrm{A}$ meet $\mathrm{BC}$ at $\mathrm{A}^{\prime}$. Let the perpendicular bisector of $\mathrm{AA}^{\prime}$ meet the line BC at X. Take the circle center X through A and A'. Similarly, let the angle bisector of B meet $\mathrm{AC}$ at $\mathrm{B}^{\prime}$ and let the perpendicular bisector of $\mathrm{BB}$ ' meet the line $\mathrm{AC}$ at $\mathrm{Y}$. Take the circle center $\mathrm{Y}$ through $\mathrm{B}$ and $\mathrm{B}^{\prime}$. The two circles meet at a point $\mathrm{P}$ inside the triangle, which is the desired point. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-22.jpg?height=514&width=534&top_left_y=785&top_left_x=247) - -$\mathrm{PAB}$ and $\mathrm{PED}$ are similar, so $\mathrm{DE} / \mathrm{AB}=\mathrm{PD} / \mathrm{PB}$. Similarly, $\mathrm{DF} / \mathrm{AC}=\mathrm{PD} / \mathrm{PC}$, so $\mathrm{DE} / \mathrm{DF}=$ $(\mathrm{AB} / \mathrm{AC})(\mathrm{PC} / \mathrm{PB})$. Thus we need $\mathrm{PB} / \mathrm{PC}=\mathrm{AB} / \mathrm{AC}$. So $\mathrm{P}$ must lie on the circle of Apollonius, which is the circle we constructed with center X. Similarly, it must lie on the circle of Apollonius with center $\mathrm{Y}$ and hence be one of their points of intersection. It also lies on the third circle and hence we choose the point of intersection inside the triangle. - -## Problem B2 - -$\mathrm{n}$ and $\mathrm{r}$ are positive integers. Find the smallest $\mathrm{k}$ for which we can construct $\mathrm{r}$ subsets $\mathrm{A}_{1}, \mathrm{~A}_{2}$, $\ldots, A_{r}$ of $\{0,1,2, \ldots, n-1\}$ each with $k$ elements such that each integer $0 \leq m2$ is an integer. Consider the pairs (a, b) of relatively prime positive integers, such that $\mathrm{a}<$ $\mathrm{b} \leq \mathrm{n}$ and $\mathrm{a}+\mathrm{b}>\mathrm{n}$. Show that the sum of $1 / a b$ taken over all such pairs is $1 / 2$. - -## Solution - -Induction on $\mathrm{n}$. It is obvious for $\mathrm{n}=3$, because the only pairs are $(1,3)$ and $(2,3)$, and $1 / 3+$ $1 / 6=1 / 2$. Now suppose it is true for $n$. As we move to $n+1$, we introduce the new pairs (a, $\mathrm{n}+1$ ) with a relatively prime to $\mathrm{n}+1$ and we lose the pairs ( $\mathrm{a}, \mathrm{n}+1-\mathrm{a}$ ) with a relatively prime to $\mathrm{n}+1-\mathrm{a}$ and hence to $\mathrm{n}+1$. So for each a relatively prime to $\mathrm{n}+1$ and $<(\mathrm{n}+1) / 2$ we gain ( $a, n+1)$ and $(n+1-a, n+1)$ and lose $(a, n+1-a)$. But $1 / a(n+1)+1 /((n+1-a)(n+1))=(n+1-a+a) /(a(n+1-$ $a)(n+1))=1 /(a(n+1-a))$. - -## Problem B2 - -An equilateral triangle of side $n$ is divided into $n^{2}$ equilateral triangles of side 1 by lines parallel to the sides. Initially, all the sides of all the small triangles are painted blue. Three coins A, B, C are placed at vertices of the small triangles. Each coin in turn is moved a distance 1 along a blue side to an adjacent vertex. The side it moves along is painted red, so once a coin has moved along a side, the side cannot be used again. More than one coin is allowed to occupy the same vertex. The coins are moved repeatedly in the order A, B, C, A, B, C, ... Show that it is possible to paint all the sides red in this way. - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-28.jpg?height=228&width=420&top_left_y=737&top_left_x=247) - -We use induction. It is obvious for $\mathrm{n}=1$ and 2 - see diagram above. Note that A, B, C start and end at vertices of the large triangle. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-28.jpg?height=494&width=557&top_left_y=1027&top_left_x=241) - -Now assume that for $\mathrm{n}$ we can find a solution with $\mathrm{A}, \mathrm{B}, \mathrm{C}$ starting and ending at the vertices of the large triangle. Take $\mathrm{n}+1$. We start with the paths shown which bring A, B, C to $\mathrm{A}^{\prime}, \mathrm{B}^{\prime}$, $\mathrm{C}^{\prime}$ at the vertices of a triangle side $\mathrm{n}-1$. Now by induction we can continue the paths so that we bring A, B, C, back to the vertices of that triangle after tracing out all its edges. Finally, note that for each of the points $\mathrm{A}^{\prime}, \mathrm{B}^{\prime}, \mathrm{C}^{\prime}$ there is a path length 2 over untraced segments to a vertex of the large triangle. So we get a solution for $\mathrm{n}+1$ and hence for all $\mathrm{n}$. - -## Problem B3 - -$A_{1}, A_{2}, \ldots, A_{n}$ are points in the plane. A non-zero real number $k_{i}$ is assigned to each point, so that the square of the distance between $A_{i}$ and $A_{j}($ for $i \neq j)$ is $k_{i}+k_{j}$. Show that $n$ is at most 4 and that if $n=4$, then $1 / k_{1}+1 / k_{2}+1 / k_{3}+1 / k_{4}=0$. - -## Solution - -Suppose we have four points A, B, C, D with associated numbers $a, b, c, d$. Then $\mathrm{AB}^{2}=\mathrm{a}+\mathrm{b}$, $A C^{2}=a+c$, so $A B^{2}-A C^{2}=b-c$. Similarly, $D B^{2}-D C^{2}=b-c$, so $A B^{2}-A C^{2}=D B^{2}-D C^{2}$. Let $X$ be the foot of the perpendicular from $\mathrm{A}$ to $\mathrm{BC}$, and $\mathrm{Y}$ the foot of the perpendicular from D to BC. Then $A B^{2}-A C^{2}=\left(A X^{2}+X B^{2}\right)-\left(A X^{2}+X C^{2}\right)=X B^{2}-X C^{2}$. Similarly for D, so $\mathrm{XB}^{2}-\mathrm{XC}^{2}=\mathrm{YB}^{2}-\mathrm{YC}^{2}$. Hence $\mathrm{X}=\mathrm{Y}$, so $\mathrm{AD}$ is perpendicular to $\mathrm{BC}$. Similarly, $\mathrm{BD}$ is perpendicular to $\mathrm{AC}$, and $\mathrm{CD}$ is perpendicular to $\mathrm{AB}$. Hence $\mathrm{D}$ is the (unique) orthocenter of ABC. So $\mathrm{n}<=4$. - -Suppose $\mathrm{n}=4$, so we have four points A, B, C, D with associated numbers a, b, c, d. We have $\mathrm{AB}^{2}+\mathrm{AC}^{2}-\mathrm{BC}^{2}=(\mathrm{a}+\mathrm{b})+(\mathrm{a}+\mathrm{c})-(\mathrm{b}+\mathrm{c})=2 \mathrm{a}$. But by the cosine formula it is also $2 \mathrm{AB}$ $A C \cos B A C$. Hence $a=A B A C \cos B A C$. Similarly for A, B, D etc. Hence $a b / c d=(A B A C$ $\cos \mathrm{BAC})(\mathrm{BA} \mathrm{BD} \cos \mathrm{ABD}) /((\mathrm{CA} \cdot \mathrm{CD} \cos \mathrm{ACD})(\mathrm{DB} \mathrm{DC} \cos \mathrm{BDC}))=\left(\mathrm{AB}^{2} / \mathrm{CD}^{2}\right)(\cos$ $\mathrm{BAC} / \cos \mathrm{BDC})(\cos \mathrm{ABD} / \cos \mathrm{ACD})$. - -Take $A B C$ to be acute with $D$ inside. Then angle $\mathrm{ABD}=$ angle $\mathrm{ACD}\left(=90^{\circ}-\right.$ angle $\left.\mathrm{BAC}\right)$, and angle $\mathrm{BDC}=90^{\circ}+$ angle $\mathrm{ACD}=180^{\circ}$ - angle $\mathrm{BAC}$. So $\cos \mathrm{BAC} / \cos \mathrm{BDC}=-1$. Thus $a b / c d=-A B^{2} / C D^{2}=-(a+b) /(c+d)$. Hence $a b(c+d)+c d(a+b)=0$, so $1 / a+1 / b+1 / c+$ $1 / \mathrm{d}=0$. - -## 12th Iberoamerican 1997 - -## Problem A1 - -$k>=1$ is a real number such that if $m$ is a multiple of $n$, then [mk] is a multiple of [nk]. Show that $\mathrm{k}$ is an integer. - -## Solution - -Suppose $\mathrm{k}$ is not an integer. Take an integer $\mathrm{n}$ such that $\mathrm{nk}>1$, but $\mathrm{nk}$ is not an integer. Now take a positive integer $\mathrm{c}$ such that $1 /(\mathrm{c}+1)<=\mathrm{nk}-[\mathrm{nk}]<1 / \mathrm{c}$. Then $1<=(\mathrm{c}+1) \mathrm{nk}-(\mathrm{c}+1)[\mathrm{nk}]<$ $1+1 / \mathrm{c}$. Hence $[(c+1) n k]=(c+1)[n k]+1$. Put $m=(c+1) n$. Then $m$ is a multiple of $n$. But if [mk] is a multiple of [nk], then [mk] - (c+1) [nk] = 1 is a multiple of [nk], which is impossible since $\mathrm{nk}>1$. So we have a contradiction. So $\mathrm{k}$ must be an integer. - -## Problem 2 - -I is the incenter of the triangle $\mathrm{ABC}$. A circle with center I meets the side $\mathrm{BC}$ at $\mathrm{D}$ and $\mathrm{P}$, with D nearer to B. Similarly, it meets the side CA at E and Q, with E nearer to C, and it meets AB at $\mathrm{F}$ and $\mathrm{R}$, with $\mathrm{F}$ nearer to $\mathrm{A}$. The lines EF and QR meet at $\mathrm{S}$, the lines FD and RP meet at T, and the lines DE and PQ meet at U. Show that the circumcircles of DUP, ESQ and FTR have a single point in common. - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-29.jpg?height=506&width=629&top_left_y=2003&top_left_x=248) - -$D$ and $P$ are the reflections of $Q$ and $E$ respectively in the line CI. Hence $P Q$ and $D E$ meet at a point on CI. So U lies on CI. So $\angle \mathrm{PIU}=1 / 2 \angle \mathrm{PIE}=\angle \mathrm{PDE}$ (I is center of circle through D, P, -$\mathrm{E})=\angle \mathrm{PDU}$ (same angle). Hence PDIU is cyclic. In other words, I lies on the circumcircle of DUP. Similarly, it lies on the circumcircles of ESQ and FTR. - -But the same argument shows that $\angle \mathrm{DPT}=\angle \mathrm{DIT}$, so DPIT is cyclic. So $\mathrm{T}$ lies on the circle through D, P and I and hence on the circumcircle of DUP. Similarly, for the other circles. So the circumcircles of CUP and FTR meet at T and I. Similarly, the circumcircles of FTR and ESQ meet at $\mathrm{S}$ and $\mathrm{I}$, and the circumcircles of ESQ and DUP meet at $\mathrm{U}$ and I. So the three circumcircles have just one point in common, namely I. - -## Problem A3 - -$n>1$ is an integer. $D_{n}$ is the set of lattice points $(x, y)$ with $|x|,|y|<=n$. If the points of $D_{n}$ are colored with three colors (one for each point), show that there are always two points with the same color such that the line containing them does not contain any other points of $\mathrm{D}_{\mathrm{n}}$. Show that it is possible to color the points of $\mathrm{D}_{\mathrm{n}}$ with four colors (one for each point) so that if any line contains just two points of $\mathrm{D}_{\mathrm{n}}$ then those two points have different colors. - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-30.jpg?height=442&width=494&top_left_y=1184&top_left_x=267) - -Consider the 4 points shown in the diagram. In each case the segment joining them is the diagonal of an $\mathrm{m} \times 1$ parallelogram or rectangle, so it cannot contain any other lattice points. The next points along each line are obviously outside set $\mathrm{D}_{\mathrm{n}}$. That proves the first part. - -The second part is the standard parity argument. Color ( $\mathrm{x}, \mathrm{y}$ ) with color 1 if $\mathrm{x}$ and $\mathrm{y}$ are both even, 2 if $\mathrm{x}$ is even and $\mathrm{y}$ is odd, 3 if $\mathrm{x}$ is odd and $\mathrm{y}$ is even, and 4 if $\mathrm{x}$ and $\mathrm{y}$ are both odd. Then if two points are the same color, that means the first coordinates are the same parity and their second coordinates are the same parity. Hence the midpoint of the segment joining them is also a lattice point and they are not the only two points of $D_{n}$ on the line. - -## Problem B1 - -Let $o(n)$ be the number of $2 n$-tuples $\left(a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}\right)$ such that each $a_{i}, b_{j}=0$ or 1 and $a_{1} b_{1}+a_{2} b_{2}+\ldots+a_{n} b_{n}$ is odd. Similarly, let e(n) be the number for which the sum is even. Show that $o(n) / e(n)=\left(2^{n}-1\right) /\left(2^{n}+1\right)$. - -## Solution - -We prove by induction that $o(n)=2^{2 n-1}-2^{n-1}$. For $n=1$, this reads $o(1)=2^{1}-2^{0}=1$, which is obviously true - the only such 2 -tuple is $(1,1)$. Suppose it is true for n. - -If $\left(a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}\right)$ gives an odd sum, then we can take $\left(a_{n+1}, b_{n+1}\right)$ to be any of ( 0 , $0),(0,1),(1,0)$ and still get an odd sum for $\left(a_{1}, a_{2}, \ldots, a_{n+1}, b_{1}, b_{2}, \ldots, b_{n+1}\right)$. On the other hand if ( $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ ) is even, then we must have $a_{n+1}=b_{n+1}=1$ to get an odd sum. Thus $o(n+1)=3 o(n)+e(n)$. But $o(n)=2_{2 n-1}-2_{n-1}$ and $e(n)=(o(n)+e(n))-o(n)=2^{2 n}-2^{2 n-1}$ $+2^{n-1}=2^{2 n-1}+2^{n-1}$. So o $o(n+1)=4.2^{2 n-1}-2.2^{n-1}=2^{2(n+1)-1}-2^{(n+1)-1}$, which establishes the result for $\mathrm{n}+1$ and hence for all $\mathrm{n}$. - -Hence $e(n)=2^{2 n}-o(n)=2^{2 n-1}+2^{n-1}$ and $o(n) / e(n)=\left(2^{n}-1\right) /\left(2^{n}+1\right)$. - -## Problem B2 - -$\mathrm{ABC}$ is an acute-angled triangle with orthocenter $\mathrm{H} . \mathrm{AE}$ and $\mathrm{BF}$ are altitudes. $\mathrm{AE}$ is reflected in the angle bisector of angle $\mathrm{A}$ and $\mathrm{BF}$ is reflected in the angle bisector of angle $\mathrm{B}$. The two reflections intersect at $\mathrm{O}$. The rays $\mathrm{AE}$ and $\mathrm{AO}$ meet the circumcircle of $\mathrm{ABC}$ at $\mathrm{M}$ and $\mathrm{N}$ respectively. $\mathrm{P}$ is the intersection of $\mathrm{BC}$ and $\mathrm{HN}, \mathrm{R}$ is the intersection of $\mathrm{BC}$ and $\mathrm{OM}$, and $\mathrm{S}$ is the intersection of HR and OP. Show that AHSO is a parallelogram. - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-31.jpg?height=355&width=500&top_left_y=1176&top_left_x=227) - -We show first that $\mathrm{O}$ is the circumcenter of $\mathrm{ABC} . \angle \mathrm{ABF}=90^{\circ}-\mathrm{A}$. The line $\mathrm{BC}$ is the reflection in $B D$ of the line $B A$ and the line $\mathrm{BF}^{\prime}$ is the refection of $\mathrm{BF}$, so angle $\mathrm{CBF}^{\prime}=90^{\circ}$ A. But if $\mathrm{O}^{\prime}$ is the circumcenter, then $\angle \mathrm{BO}^{\prime} \mathrm{C}=2 \angle \mathrm{BAC}=2 \mathrm{~A}$, so $\angle \mathrm{O}^{\prime} \mathrm{BC}=90^{\circ}-\mathrm{A}$. Hence $\mathrm{O}^{\prime}$ lies on $\mathrm{BF}^{\prime}$. Similarly, it lies on $\mathrm{AE}^{\prime}$ (the reflection of $\mathrm{AE}$ in the angle bisector of $\mathrm{A}$ ). Hence $\mathrm{O}=\mathrm{O}^{\prime}$. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-31.jpg?height=465&width=465&top_left_y=1909&top_left_x=247) - -$\angle \mathrm{MBC}=\angle \mathrm{MAC}=90^{\circ}-\mathrm{C}($ since $\mathrm{AH}$ is an altitude $)=\angle \mathrm{FBC}($ since $\mathrm{BF}$ is an altitude $)=$ $\angle \mathrm{HBC}$ (same angle). So triangles HBE and MBE are congruent and HE $=$ EM. [Note: this should be a familiar result.]. -$\mathrm{AN}$ is a diameter, so angle $\mathrm{AMN}=90^{\circ}=$ angle $\mathrm{AEC}$, so $\mathrm{BC}$ and $\mathrm{MN}$ are parallel. Hence $\mathrm{P}$ is the midpoint of $\mathrm{HN}$ and of $\mathrm{BC}$. So OP is perpendicular to $\mathrm{BC}$. So $\mathrm{AH}$ and $\mathrm{OS}$ are parallel. - -Since $\mathrm{R}$ lies on $\mathrm{BC}$, triangles HER and MER are congruent, so $\angle \mathrm{EHR}=\angle \mathrm{EMR}=\angle \mathrm{AMO}$ $($ same angle $)=\angle \mathrm{MAO}$. Hence HS and AO are parallel. So AHSO is a parallelogram. - -## Problem B3 - -Given 1997 points inside a circle of radius 1, one of them the center of the circle. For each point take the distance to the closest (distinct) point. Show that the sum of the squares of the resulting distances is at most 9 . - -## Solution - -Let the points be $\mathrm{P}_{\mathrm{i}}$ for $\mathrm{i}=1,2, \ldots$, 1997. Take $\mathrm{P}_{1}$ to be the center of the given unit circle. Let $x_{i}$ be the distance from $P_{i}$ to the closest of the other 1996 points. Let $C_{i}$ be the circle center $P_{i}$ radius $\mathrm{x}_{\mathrm{i}} / 2$. Then $\mathrm{C}_{\mathrm{i}}$ and $\mathrm{C}_{\mathrm{j}}$ cannot overlap by more than one point because $\mathrm{x}_{\mathrm{i}}$ and $\mathrm{x}_{\mathrm{j}} \leq \mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{j}}$. Also $\mathrm{x}_{\mathrm{i}} \leq 1$, since $\mathrm{P}_{1} \mathrm{P}_{\mathrm{i}} \leq 1$. Thus $\mathrm{C}_{\mathrm{i}}$ is entirely contained in the circle center $\mathrm{P}_{1}$ radius $3 / 2$. Since the circles $\mathrm{C}_{\mathrm{i}}$ do not overlap, their total area cannot exceed the area of the circle radius $3 / 2$. Hence $\left(x_{1}{ }^{2}+x_{2}{ }^{2}+\ldots+x_{1997}{ }^{2}\right) / 4 \leq 9 / 4$. - -## 13th Iberoamerican 1998 - -## Problem A1 - -There are 98 points on a circle. Two players play alternately as follows. Each player joins two points which are not already joined. The game ends when every point has been joined to at least one other. The winner is the last player to play. Does the first or second player have a winning strategy? - -## Solution - -Answer: the first player has a winning strategy. - -Assume there are $n$ points. The first to play so that $n-2$ points each have at least one segment loses, because the other player simply joins the last two points and the game ends. But there are $\mathrm{N}=(\mathrm{n}-3)(\mathrm{n}-4) / 2$ possible plays amongst the first $\mathrm{n}-3$ points to get a segment. For $\mathrm{n}=1$ or $2 \bmod 4, \mathrm{~N}$ is odd and for $\mathrm{n}=0$ or 3 it is even. So the first player wins for $\mathrm{n}=1$ or $2 \bmod 4$ (and in particular for $\mathrm{n}=98$ ) and the second player for $\mathrm{n}=0$ or $3 \bmod 4$. - -## Problem A2 - -The incircle of the triangle $\mathrm{ABC}$ touches $\mathrm{BC}, \mathrm{CA}, \mathrm{AB}$ at $\mathrm{D}, \mathrm{E}, \mathrm{F}$ respectively. $\mathrm{AD}$ meets the circle again at $\mathrm{Q}$. Show that the line $\mathrm{EQ}$ passes through the midpoint of $\mathrm{AF}$ iff $\mathrm{AC}=\mathrm{BC}$. - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-33.jpg?height=416&width=577&top_left_y=266&top_left_x=248) - -$\angle \mathrm{AQM}=\angle \mathrm{EQD}($ opposite angle $)=\angle \mathrm{EDC}(\mathrm{CD}$ tangent to circle $\mathrm{EQD})=\left(180^{\circ}-\angle \mathrm{C}\right) / 2=$ $\angle \mathrm{A} / 2+\angle \mathrm{B} / 2(*)$. - -$\mathrm{MF}^{2}=\mathrm{MQ} \cdot \mathrm{ME}$ (MF tangent to circle $\mathrm{FQE}$ ). So $\mathrm{AM}=\mathrm{AF}$ is equivalent to $\mathrm{AM}^{2}=\mathrm{MQ} \cdot \mathrm{ME}$ or $\mathrm{AM} / \mathrm{MQ}=\mathrm{ME} / \mathrm{AM}$. But since triangles $\mathrm{AMQ}$ and EMA have a common angle $\mathrm{M}, \mathrm{AM} / \mathrm{MQ}=$ $\mathrm{ME} / \mathrm{AM}$ iff they are similar, and hence iff $\angle \mathrm{AQM}=\angle \mathrm{A}$. Using $(*) \mathrm{AM}=\mathrm{AF}$ iff $\angle \mathrm{A}=\angle \mathrm{B}$. - -## Problem A3 - -Find the smallest number $\mathrm{n}$ such that given any $\mathrm{n}$ distinct numbers from $\{1,2,3, \ldots, 999\}$, one can choose four different numbers $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ such that $\mathrm{a}+2 \mathrm{~b}+3 \mathrm{c}=\mathrm{d}$. - -## Solution - -Answer: $\mathrm{n}=835$. - -Consider the set $S=\{166,167, \ldots, 999\}$. The smallest possible value for $a+2 b+3 c$, for distinct $\mathrm{a}, \mathrm{b}, \mathrm{c}$ in $\mathrm{S}$ is $168+2.167+3.166=1000$. So we cannot find distinct $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ in $\mathrm{S}$ with $a+2 b+3 c=d$. So the smallest $n>834$. - -Now suppose $\mathrm{S}$ is any subset of 835 elements which satisfies the condition. Take it elements to be $\mathrm{m}=\mathrm{a}_{1}<\mathrm{a}_{2}<\ldots<\mathrm{a}_{835}=\mathrm{M}$. Obviously $\mathrm{M} \geq \mathrm{m}+834 \geq 835$, so $-3 \mathrm{~m} \geq 3.834-3 \mathrm{M}$ and hence $M-3 \mathrm{~m} \geq 2502-2 \mathrm{M} \geq 2502-2.999=504$. Put $\mathrm{k}=\mathrm{M}-3 \mathrm{~m}$. - -There are at least 167 disjoint pairs (a,b) of numbers taken from $\{1,2, \ldots, 999\}$ with $a+2 b=$ $\mathrm{k}$, namely - -``` -(k- - 2, 1) -(k-4, 2) - -``` - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-33.jpg?height=46&width=206&top_left_y=2210&top_left_x=248) - -``` - -``` - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-33.jpg?height=40&width=55&top_left_y=2287&top_left_x=248) - -``` -(k-334, 167) - note that in the extreme case k = 504 this is (170, 167) -``` - -At least one number from each pair must either (1) be $\mathrm{M}$ or $\mathrm{m}$ or (2) not belong to $\mathrm{S}$ - or otherwise we would have $a+2 b+3 m=M$ for distinct elements $a, b, m$ and $M$ in $S$. None of the numbers can be $M$ and at most one of them can be $m$, so we have at least 166 numbers which are not in S. That means S contains at most $999-166=833$ numbers. Contradiction. - -So $\mathrm{S}$ cannot have 835 elements. Nor can it have more than 835 elements (or we just take a subset of 835 elements, which must also satisfy the condition, and get a contradiction). - -## Problem B1 - -Representatives from $n>1$ different countries sit around a table. If two people are from the same country then their respective right hand neighbors are from different countries. Find the maximum number of people who can sit at the table for each $\mathrm{n}$. - -## Solution - -Answer: $n^{2}$. - -Obviously there cannot be more than $\mathrm{n}^{2}$ people. For if there were, then at least one country would have more than $n$ representatives. But there are only $n$ different countries to choose their right-hand neighbours from. Contradiction. - -Represent someone from country i by i. Then for $\mathrm{n}=2$, the arrangement 1122 works. [It wraps round, so that the second 2 is adjacent to the first 1.] Suppose we have an arrangement for $\mathrm{n}$. Then each of $11,22, \ldots$, nn must occur just once in the arrangement. Replace 11 by $1(n+1) 11,22$ by $2(n+1) 22, \ldots$, and $(n-1)(n-1)$ by $(n-1)(n+1)(n-1)(n-1)$. Finally replace $n n$ by $\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+1) \mathrm{nn}$. It is easy to check that we now have an arrangement for $\mathrm{n}+1$. We have added one additional representative for each of the countries 1 to $\mathrm{n}$ and $\mathrm{n}+1$ representatives for country $\mathrm{n}+1$, so we have indeed got $(\mathrm{n}+1)^{2}$ people in all. We have also given a representative of each country 1 to $n$ a neighbour from country $n+1$ on his right and we have given the ( $n+1$ ) representatives from country $n+1$ neighbours (on their right) from each of the other countries. Otherwise we have left the seating unchanged. - -## Problem B2 - -$\mathrm{P}_{1}, \mathrm{P}_{2}, \ldots, \mathrm{P}_{\mathrm{n}}$ are points in the plane and $\mathrm{r}_{1}, \mathrm{r}_{2}, \ldots, \mathrm{r}_{\mathrm{n}}$ are real numbers such that the distance between $P_{i}$ and $P_{j}$ is $r_{i}+r_{j}$ (for $i$ not equal to $j$ ). Find the largest $n$ for which this is possible. - -## Solution - -Answer: $\mathrm{n}=4$. - -Draw a circle radius $r_{i}$ at $P_{i}$. Then each pair of circles must touch. But that is possible iff $n \leq 4$. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-34.jpg?height=477&width=557&top_left_y=2126&top_left_x=247) - -## Problem B3 - -$\mathrm{k}$ is the positive root of the equation $\mathrm{x}^{2}-1998 \mathrm{x}-1=0$. Define the sequence $\mathrm{x}_{0}, \mathrm{x}_{1}, \mathrm{x}_{2}, \ldots$ by $\mathrm{x}_{0}$ $=1, \mathrm{x}_{\mathrm{n}+1}=\left[\mathrm{k} \mathrm{x}_{\mathrm{n}}\right]$. Find the remainder when $\mathrm{x}_{1998}$ is divided by 1998. - -## Solution - -Put $\mathrm{p}(\mathrm{x})=\mathrm{x}^{2}-1998 \mathrm{x}-1$. Then $\mathrm{p}(1998)=-1, \mathrm{p}(1999)=1998$, so $1998<\mathrm{k}<1999$. Also $\mathrm{k}$ is irrational (using the formula for the root of a quadratic). We have $x_{n}=\left[k x_{n-1}\right]$, so $x_{n}\mathrm{kx}_{\mathrm{n}-1}-1$. Hence $\mathrm{x}_{\mathrm{n}} / \mathrm{k}<\mathrm{x}_{\mathrm{n}-1}<\mathrm{x}_{\mathrm{n}} / \mathrm{k}+1 / \mathrm{k}$, so $\left[\mathrm{x}_{\mathrm{n}} / \mathrm{k}\right]=\mathrm{x}_{\mathrm{n}-1}-1$. - -$\mathrm{k}=(1998 \mathrm{k}+1) / \mathrm{k}=1998+1 / \mathrm{k}$. Hence $\mathrm{kx}_{\mathrm{n}}=1998 \mathrm{x}_{\mathrm{n}}+\mathrm{x} / \mathrm{k}$. Hence $\mathrm{x}_{\mathrm{n}+1}=\left[\mathrm{kx}_{\mathrm{n}}\right]=1998 \mathrm{x}_{\mathrm{n}}+$ $\left[\mathrm{x}_{\mathrm{n}} / \mathrm{k}\right]=1998 \mathrm{x}_{\mathrm{n}}+\mathrm{x}_{\mathrm{n}-1}-1$. Hence $\mathrm{x}_{\mathrm{n}+1}=\mathrm{x}_{\mathrm{n}-1}-1 \bmod 1998$. So $\mathrm{x}_{1998}=1-999=1000 \bmod$ 1998. - -## 14th Iberoamerican 1999 - -## Problem A1 - -Find all positive integers $n<1000$ such that the cube of the sum of the digits of $n$ equals $n^{2}$. - -## Solution - -$\mathrm{n}<1000$, so the sum of the digits is at most 27 , so $\mathrm{n}^{2}$ is a cube not exceeding $27^{3}$. So we are looking for $\mathrm{m}^{3}$ which is also a square. That implies $\mathrm{m}$ is a square. So the only possibilities are $\mathrm{m}=1,4,9,16,25$. Giving $\mathrm{n}=1,8,27,64,125$. The corresponding cubes of the digit sums are $1,512,729,1000,512$, whereas the corresponding squares are $1,64,729,4096,15625$. Thus the only solutions are $\mathrm{n}=1,27$. - -## Problem A2 - -Given two circles $\mathrm{C}$ and $\mathrm{C}^{\prime}$ we say that $\mathrm{C}$ bisects $\mathrm{C}^{\prime}$ if their common chord is a diameter of $\mathrm{C}^{\prime}$. Show that for any two circles which are not concentric, there are infinitely many circles which bisect them both. Find the locus of the centers of the bisecting circles. - -## Solution - -Let $\mathrm{C}, \mathrm{C}^{\prime}$ have center $\mathrm{O}, \mathrm{O}^{\prime}$ respectively and radius $\mathrm{r}, \mathrm{r}^{\prime}$ respectively. Let a circle center $\mathrm{P}$ bisect $C$. Suppose it meets $C$ at $A$ and $B$. Then $A B$ is perpendicular to $O P$ and is a diameter of C. Hence $\mathrm{PA}^{2}=\mathrm{OP}^{2}+\mathrm{r}^{2}$. Conversely, the circle center $\mathrm{P}$, radius $\sqrt{ }\left(\mathrm{OP}^{2}+\mathrm{r}^{2}\right)$ bisects $\mathrm{C}$. So $\mathrm{P}$ will bisect $\mathrm{C}$ and $\mathrm{C}^{\prime}$ iff $\mathrm{OP}^{2}+\mathrm{r}^{2}=\mathrm{OP}^{\prime 2}+\mathrm{r}^{\prime 2}$. - -It is well-known that the locus of points $\mathrm{P}^{\prime}$ with equal tangents to $\mathrm{C}$ and $\mathrm{C}^{\prime}$ is the radical axis. Call the radical axis $\mathrm{R}$. For a point $\mathrm{P}^{\prime}$ on the radical axis we have $\mathrm{P}^{\prime} \mathrm{O}^{2}-\mathrm{r}^{2}=\mathrm{P}^{\prime} \mathrm{O}^{\prime 2}-\mathrm{r}^{\prime 2}$. If we reflect $\mathrm{P}^{\prime}$ in the perpendicular bisector of $\mathrm{OO}^{\prime}$ to get $\mathrm{P}$, then $\mathrm{PO}=\mathrm{P}^{\prime} \mathrm{O}^{\prime}$ and $\mathrm{PO}^{\prime}=\mathrm{P}^{\prime} \mathrm{O}$, so $\mathrm{PO}^{\prime 2}$ $-r^{2}=\mathrm{PO}^{2-r^{2}}$ and hence $\mathrm{PO}^{2}+\mathrm{r}^{2}$. Call the reflection of the $\mathrm{R}$ in the perpendicular bisector of OO' the line $\mathrm{R}^{\prime}$. We have established that points on $\mathrm{R}^{\prime}$ form part of the locus. Conversely, if $\mathrm{P}^{\prime}$ is such that there is a circle center $\mathrm{P}^{\prime}$ bisecting both circles, then $\mathrm{OP}^{\prime 2}+\mathrm{r}^{2}=\mathrm{O}^{\prime} \mathrm{P}^{\prime 2}+\mathrm{r}^{\prime 2}$, so if $\mathrm{P}$ is the reflection of $\mathrm{P}^{\prime}$ then $\mathrm{OP}^{2}-\mathrm{r}^{2}=\mathrm{O}^{2}-\mathrm{r}^{\mathrm{r}^{2}}$ and hence $\mathrm{P}$ lies on the radical axis R. Hence $\mathrm{P}^{\prime}$ must lie on $\mathrm{R}^{\prime}$. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-36.jpg?height=460&width=534&top_left_y=267&top_left_x=247) - -We have $\mathrm{PT}^{2}=\mathrm{PO}^{2}-\mathrm{r}^{2}=\mathrm{PX}^{2}+\mathrm{OX}^{2}-\mathrm{r}^{2}$, and similarly $\mathrm{PT}^{\prime 2}=\mathrm{PX}^{2}+\mathrm{O}^{\prime} \mathrm{X}^{2}-\mathrm{r}^{\prime 2}$. So $\mathrm{PT}=\mathrm{PT}^{\prime}$ iff $O X^{2}-r^{2}=O^{\prime} X^{2}-r^{\prime 2}$. There is evidently a unique point $X$ for which that is true, so the locus of such $\mathrm{P}$ is the line through $\mathrm{X}$ perpendicular to $\mathrm{OO}^{\prime}$ - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-36.jpg?height=503&width=505&top_left_y=1002&top_left_x=250) - -If the circles intersect, then the point $\mathrm{X}$ evidently lies on the line joining the two common points, because $O X^{2}-r^{2}=-X Y^{2}=O^{\prime} X^{2}-r^{\prime 2}$. In any case the midpoint of each common tangent evidently lies on the line, so that provides a way of constructing it. - -## Problem A3 - -Given points $\mathrm{P}_{1}, \mathrm{P}_{2}, \ldots, \mathrm{P}_{\mathrm{n}}$ on a line we construct a circle on diameter $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{j}}$ for each pair $\mathrm{i}, \mathrm{j}$ and we color the circle with one of $\mathrm{k}$ colors. For each $\mathrm{k}$, find all $\mathrm{n}$ for which we can always find two circles of the same color with a common external tangent. - -## Solution - -Answer: $\mathrm{n}>\mathrm{k}+1$. - -There are $\mathrm{n}-1$ circles with diameter $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{i}+1}$. Obviously, each pair has a common tangent. If $\mathrm{n}-1$ $>\mathrm{k}$, then two of them must have the same color. - -If $\mathrm{n}-1 \leq \mathrm{k}$, then color all circles with diameter $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{j}}$ and $\mathrm{i}<\mathrm{j}$ with color $\mathrm{i}$. Then if two circles have the same color, then both have a tangent at one of the points. Hence one lies inside the other and they do not have a common external tangent. - -## Problem B1 - -Show that any integer greater than 10 whose digits are all members of $\{1,3,7,9\}$ has a prime factor $\geq 11$. - -## Solution - -Such a number cannot be divisible by 2 (or its last digit would be even) or by 5 (or its last digit would be 0 or 5). So if the result is false then the number must be of the form $3^{\mathrm{m}} 7^{\mathrm{n}}$ for non-negative integers $\mathrm{m}, \mathrm{n}$. But we claim that a number of this form must have even 10s digit. - -It is easy to prove the claim by induction. It is true for 3 and 7 (the digit is 0 in both cases). But if we multiply such a number by 3 or 7 , then the new 10 s digit has the same parity as the carry from the units digit. But multiplying 1,3,7, 9 by 3 gives a carry of $0,0,2,6$ respectively, which is always even, and multiplying by 7 gives a carry of $0,2,4,6$, which is also always even. So the new number also has an even 10 s digit. - -## Problem B2 - -$\mathrm{O}$ is the circumcenter of the acute-angled triangle $\mathrm{ABC}$. The altitudes are $\mathrm{AD}, \mathrm{BE}$ and $\mathrm{CF}$. The line $\mathrm{EF}$ cuts the circumcircle at $\mathrm{P}$ and $\mathrm{Q}$. Show that $\mathrm{OA}$ is perpendicular to $\mathrm{PQ}$. If $\mathrm{M}$ is the midpoint of $\mathrm{BC}$, show that $\mathrm{AP}^{2}=2 \mathrm{AD} \cdot \mathrm{OM}$. - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-37.jpg?height=576&width=597&top_left_y=1371&top_left_x=250) - -Let $\mathrm{OA}$ and $\mathrm{PQ}$ meet at $\mathrm{T}$. $\angle \mathrm{AEH}=\angle \mathrm{AFH}=90^{\circ}$, so $\mathrm{AEHF}$ is cyclic, so $\angle \mathrm{AFT}=\angle \mathrm{AFE}$ $($ same angle $)=\angle \mathrm{AHE}=90^{\circ}-\angle \mathrm{HAE}=90^{\circ}-\angle \mathrm{DAC}($ same angle $)=\angle \mathrm{C}$. But $\angle \mathrm{TAF}=\angle \mathrm{OAF}$ (same angle) $=90^{\circ}-(1 / 2) \angle \mathrm{AOB}=90^{\circ}-\angle \mathrm{C}$. Hence $\angle \mathrm{AFT}=90^{\circ}$, which establishes that OA and $P Q$ are perpendicular. - -Let the circumradius be $\mathrm{R}$ and let $\mathrm{AA}^{\prime}$ be a diameter. We have $\mathrm{AF}=\mathrm{AC} \cos \mathrm{A}=2 \mathrm{R} \sin \mathrm{B} \cos$ A. Hence $\mathrm{AT}=\mathrm{AF} \cos \mathrm{OAB}=\mathrm{AF} \sin \mathrm{C}=2 \mathrm{R} \cos \mathrm{A} \sin \mathrm{B} \sin \mathrm{C} . \mathrm{Now} \mathrm{PT}^{2}=\mathrm{PT} \cdot \mathrm{TQ}=$ $A T \cdot A^{\prime} T=A T(2 R-A T)$. Hence $A P^{2}=2 R \cdot A T=4 R^{2} \cos A \sin B \sin C$. - -We have $\mathrm{AD}=\mathrm{AC} \sin \mathrm{C}=2 \mathrm{R} \sin \mathrm{B} \sin \mathrm{C}$, and $\mathrm{OM}=\mathrm{OC} \cos \mathrm{COM}=\mathrm{R} \cos \mathrm{A}$. Hence 2 $\mathrm{AD} \cdot \mathrm{OM}=\mathrm{AP}^{2}$. - -## Problem B3 - -Given two points $\mathrm{A}$ and $\mathrm{B}$, take $\mathrm{C}$ on the perpendicular bisector of $\mathrm{AB}$. Define the sequence $C_{1}, C_{2}, C_{3}, \ldots$ as follows. $C_{1}=C$. If $C_{n}$ is not on $A B$, then $C_{n+1}$ is the circumcenter of the triangle $\mathrm{ABC}_{\mathrm{n}}$. If $\mathrm{C}_{\mathrm{n}}$ lies on $\mathrm{AB}$, then $\mathrm{C}_{\mathrm{n}+1}$ is not defined and the sequence terminates. Find all points $\mathrm{C}$ such that the sequence is periodic from some point on. - -## Solution - -Answer: any $\mathrm{C}$ such that $\angle \mathrm{ACB}=180^{\circ} \mathrm{r} / \mathrm{s}$, with $\mathrm{r}$ and $\mathrm{s}$ relatively prime integers and $\mathrm{s}$ not a power of 2 . - -Let $\angle \mathrm{AC}_{\mathrm{n}} \mathrm{B}=\mathrm{x}_{\mathrm{n}}$, where the angle is measured clockwise, so that $\mathrm{x}_{\mathrm{n}}$ is positive on one side of $A B$ and negative on the other side. Then $x_{n}$ uniquely identifies $C_{n}$ on the perpendicular bisector. - -We have $\mathrm{x}_{\mathrm{n}+1}=2 \mathrm{x}_{\mathrm{n}}$. To make this work in all cases we have to take it mod $180^{\circ}$ (so that if $A C_{n} B$ is obtuse, then $C_{n+1}$ lies on the other side of $A B$ ). If $x_{n}$ is eventually periodic then $x_{m+1}$ $=\mathrm{x}_{\mathrm{n}+1}$, for some $\mathrm{n}>\mathrm{m}$, so $\left(2^{\mathrm{n}}-2^{\mathrm{m}}\right) \mathrm{x}_{1}=0 \bmod 180$. Hence $\mathrm{x}_{1}=180 \mathrm{r} / \mathrm{s}$ for some relatively prime integers $\mathrm{r}, \mathrm{s}$. Also $\mathrm{s}$ cannot be a power of 2 for then we would have $\mathrm{x}_{\mathrm{k}}=180 \mathrm{r}$ for some $\mathrm{k}$, in which case the sequence would terminate rather than be periodic. - -Conversely, suppose $\mathrm{x}_{1}=180 \mathrm{r} / \mathrm{s}$, with $\mathrm{r}$ and $\mathrm{s}$ relatively prime and $\mathrm{s}$ not a power of 2 . Then $\mathrm{x}_{\mathrm{n}+1}=1802^{\mathrm{n}} \mathrm{r} / \mathrm{s}$ cannot be $0 \bmod 180$, so the sequence does not terminate. Put $\mathrm{s}=2^{\mathrm{b}} \mathrm{c}$, with $\mathrm{c}$ odd. Let $\mathrm{d}=\varphi(\mathrm{c})$, where $\varphi(\mathrm{m})$ is Euler's phi function, so that $2^{\mathrm{d}}=1 \mathrm{mod} \mathrm{c}$. Then $\mathrm{x}_{\mathrm{b}+1}=180 \mathrm{r} / \mathrm{c}$ $\bmod 180$ and $2^{\mathrm{b}+\mathrm{d}}=2^{\mathrm{b}} \bmod \mathrm{c}$, so $\mathrm{x}_{\mathrm{b}+\mathrm{d}+1}=180 \mathrm{r} / \mathrm{c}$ mod 180 . Hence the sequence is periodic. - -## 15th Iberoamerican 2000 - -## Problem A1 - -Label the vertices of a regular $n$-gon from 1 to $n>3$. Draw all the diagonals. Show that if $n$ is odd then we can label each side and diagonal with a number from 1 to $n$ different from the labels of its endpoints so that at each vertex the sides and diagonals all have different labels. - -## Solution - -Labeling the diagonal/side between $\mathrm{i}$ and $\mathrm{j}$ as $\mathrm{i}+\mathrm{j}$ (reduced if necessary mod $\mathrm{n}$ ) almost works. The labels for all the lines at a given vertex will be different. But the line between i and $n$ will have label i, the same as one endpoint. However, we are not using the label 2i for the lines from vertex i. So for the line between $i$ and $n$ we use $2 i$ instead of $i+n$. The only points that need checking are (1) whether a line from $\mathrm{i}$ to $\mathrm{n}$ has a label different from $\mathrm{n}$, and (2) whether all the lines at $\mathrm{n}$ have different labels. Both points are ok because $\mathrm{n}$ is odd. - -## Problem A2 - -Two circles $\mathrm{C}$ and $\mathrm{C}^{\prime}$ have centers $\mathrm{O}$ and $\mathrm{O}^{\prime}$ and meet at $\mathrm{M}$ and $\mathrm{N}$. The common tangent closer to $\mathrm{M}$ touches $\mathrm{C}$ at $\mathrm{A}$ and $\mathrm{C}^{\prime}$ at $\mathrm{B}$. The line through $\mathrm{B}$ perpendicular to $\mathrm{AM}$ meets the line $\mathrm{OO}^{\prime}$ at D. BO'B' is a diameter of C'. Show that M, D and B' are collinear. - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-39.jpg?height=425&width=625&top_left_y=256&top_left_x=247) - -A neat coordinate solution by Massaki Yamamoto (a competitor) is as follows. - -Take $A B$ as the $x$-axis and the perpendicular line through $M$ as the $y$-axis. Choose the unit of length so that $\mathrm{M}$ has coordinates $(0,1)$. Let $\mathrm{A}$ be $(-\mathrm{m}, 0)$ and $\mathrm{B}$ be $(\mathrm{n}, 0)$. Then considering the right-angled triangle $O^{\prime} M K$, where $K$ is $(n, 1)$ we find that $O^{\prime}$ is $\left(n,\left(n^{2}+1\right) / 2\right)$. Similarly, $O$ is $\left.\left(-\mathrm{m},\left(\mathrm{m}^{2}+1\right) / 2\right)\right)$. - -The gradient of the lie AM is $1 / \mathrm{m}$, so the gradient of the line $\mathrm{BD}$ is $-\mathrm{m}$ and hence its equation is $m x+y=m n$. The gradient of the line OO' is $(n-m) / 2$, so its equation is $2 y-x(n-m)=m n+1$. These intersect at $\left((m n-1) /(m+n),\left(m n^{2}+m\right) /(m+n)\right)$. B' is $\left(n, n^{2}+1\right)$. It is now easy to check that the lines MB' and MD both have gradient $\mathrm{n}$, so $\mathrm{M}, \mathrm{D}, \mathrm{B}^{\prime}$ are collinear. - -## Problem A3 - -Find all solutions to $(\mathrm{m}+1)^{\mathrm{a}}=\mathrm{m}^{\mathrm{b}}+1$ in integers greater than 1 . - -## Answer - -$(\mathrm{m}, \mathrm{a}, \mathrm{b})=(2,2,3)$. - -## Solution - -Taking equation mod $\mathrm{m}+1$ we get $(-1)^{\mathrm{b}}=-1$, so $\mathrm{b}$ is odd. Hence we can divide the rhs by $\mathrm{m}+1$ to get $m^{b-1}-m^{b-2}+\ldots-m+1$. This has an odd number of terms. If $m$ is odd, then each term is odd and so the total is odd, but $(\mathrm{m}+1)^{\mathrm{a}-1}$ is even (note that $\left.\mathrm{a}>1\right)$. Contradicton, so $\mathrm{m}$ is even. - -We have $\mathrm{m}^{\mathrm{b}}=(\mathrm{m}+1)^{\mathrm{a}}-1$. Expanding the rhs by the binomial theorem, and using $\mathrm{b}>1$, we see that $\mathrm{m}$ must divide $\mathrm{a}$. So a is even also. Put $\mathrm{a}=2 \mathrm{~A}, \mathrm{~m}=2 \mathrm{M}$. We can factorise $(\mathrm{m}+1)^{\mathrm{a}}-1$ as ( $\left.(m+1)^{A}+1\right)\left((m+1)^{A}-1\right)$. The two factors have difference 2 , so their gcd divides 2 , but both factors are even, so their gcd is exactly 2. - -If $\mathrm{M}=1$ or a power of 2 , then the smaller factor $3^{\mathrm{A}}-1$ must be 2 , so $\mathrm{A}=1$ and we have $3^{\mathrm{A}}+$ $1=4$, so $(2 M)^{b}=8$. Hence $M=1$ and $b=3$ and we have the solution $(m, a, b)=(2,2,3)$. - -If $\mathrm{M}$ is not a power of 2 , then $\mathrm{M}^{\mathrm{b}}>2^{\mathrm{b}}$, so we must have the larger factor $2 \cdot \mathrm{M}^{\mathrm{b}}$ and the smaller factor $2^{b-1}$. But the larger factor is now $>2^{b+1}$, so the difference between the factors is at least 3. $2^{\mathrm{b}-1}>2$. Contradiction. - -## Problem B1 - -Some terms are deleted from an infinite arithmetic progression $1, \mathrm{x}, \mathrm{y}, \ldots$ of real numbers to leave an infinite geometric progression $1, \mathrm{a}, \mathrm{b}, \ldots$. Find all possible values of a. - -## Solution - -Answer: the positive integers. - -If a is negative, then the terms in the GP are alternately positive and negative, whereas either all terms in the AP from a certain point on are positive or all terms from a certain point on are negative. So a cannot be negative. If a is zero, then all terms in the GP except the first are zero, but at most one term of the AP is zero, so a cannot be zero. Thus a must be positive, so the AP must have infinitely many positive terms and hence $\mathrm{x} \geq 1$. - -Let $\mathrm{d}=\mathrm{x}-1$, so all terms of the AP have the form 1+nd for some positive integer $\mathrm{n}$. Suppose $a=1+m d, a^{2}=1+n d$, then $(1+m d)^{2}=1+n d$, so $d=(n-2 m) / m^{2}$, which is rational. Hence $\mathrm{a}$ is rational. Suppose $\mathrm{a}=\mathrm{b} / \mathrm{c}$, where $\mathrm{b}$ and $\mathrm{c}$ are relatively prime positive integers and $\mathrm{c}>1$. Then the denominator of the nth term of the GP is $\mathrm{c}^{\mathrm{n}}$, which becomes arbitrarily large as $\mathrm{n}$ increases. But if $\mathrm{d}=\mathrm{h} / \mathrm{k}$, then all terms of the AP have denominator at most $\mathrm{k}$. So we cannot have $\mathrm{c}>1$. So a must be a positive integer. - -On the other hand, it is easy to see that any positive integer works. Take $\mathrm{x}=2$, then the AP includes all positive integers and hence includes any GP with positive integer terms. - -## Problem B2 - -Given a pile of 2000 stones, two players take turns in taking stones from the pile. Each player must remove $1,2,3,4$, or 5 stones from the pile at each turn, but may not take the same number as his opponent took on his last move. The player who takes the last stone wins. Does the first or second player have a winning strategy? - -## Solution - -The first player has a winning strategy. He takes 4 on his first move leaving 7 mod $13(2000=$ $153.13+7+4)$. Now we claim that the first player can always leave: (1) $0 \bmod 13$, (2) $3 \bmod$ 13 by taking away 3 , (3) $5 \bmod 13$ by taking away 5 , or (4) $7 \bmod 13$, and that the second player can never leave 0 mod 13. - -Let us look at each of these in turn. If the first player leaves $0 \bmod 13$, then the second player can take 3 and leave 10. In that case the first player takes 5 (a type (3) move). If the second player takes $1,2,4$ or 5 , leaving $12,11,9$ or $8 \bmod 13$, then the first player takes $5,4,2,1$ (respectively) and leaves 7 mod 13 (a type (4) move). - -If the first player leaves $3 \bmod 13$ by taking away 3 , then the second player cannot leave 0 $\bmod 13$, because he cannot take 3 stones. If he takes 1,2 leaving $2,1 \bmod 13$ respectively, then the first player takes 2,1 leaving 0 mod 13 (a type (1) move). If the second player takes 4, 5 leaving 12, 11 mod 13, then the first player takes 5,4 leaving $7 \bmod 13$ (a type (4) move). - -If the first player leaves $5 \bmod 13$ by taking 5 , then the second player cannot leave $0 \bmod 13$, because he cannot take 5 stones. If he takes $1,2,3,4$ stones, leaving 4,3,2, 1 mod 13 , then the first player takes $4,3,2,1$ stones leaving 0 mod 13 (a type (1) move). - -Finally, if the first player leaves $7 \bmod 13$, and the second player takes 1 stone, then the first player takes 3 stones leaving 3 mod 13 (a type (2) move). If the second player takes $2,3,4$, or 5 stones leaving $5,4,3,2 \bmod 13$, then the first player takes $5,4,3,2$ stones leaving $0 \bmod 13$ (a type (1) move). - -So the second player can never leave 0 mod 13 and hence, in particular, can never take the last stone. But we have shown that the first player can always make a move of one of the four types, so can always move and hence must win (since after less than 2000 moves there will be no stones left). - -## Problem B3 - -A convex hexagon is called a unit if it has four diagonals of length 1, whose endpoints include all the vertices of the hexagon. Show that there is a unit of area $\mathrm{k}$ for any $0<\mathrm{k} \leq 1$. What is the largest possible area for a unit? - -## Solution - -Answer: We can get arbitrarily close to (but not achieve) ( $3 \sqrt{3}$ )/4 (approx 1.3) by: - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-41.jpg?height=182&width=302&top_left_y=1234&top_left_x=266) - -To prove the first part, consider the diagram below. Take $\mathrm{AB}=\mathrm{AC}=1$ and angle $\mathrm{BAC}=2 \theta$. Take $\mathrm{DE}=\mathrm{DF}=1$ and take the points of intersection $\mathrm{X}$ and $\mathrm{Y}$ such that $\mathrm{AX}=\mathrm{DX}=\mathrm{AY}=$ $\mathrm{DY}=2 / 3$. It is easy to check that the area of the hexagon is $\sin 2 \theta$. So by taking $\theta$ in the interval $(0, \pi / 4]$ we can get any area $0<\mathrm{k} \leq 1$. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-41.jpg?height=351&width=526&top_left_y=1755&top_left_x=248) - -It is easy to check that there are six possible configurations for the unit diagonals, as shown in the diagram below. -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-41.jpg?height=340&width=684&top_left_y=2331&top_left_x=250) - -Consider case 1. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-42.jpg?height=211&width=260&top_left_y=360&top_left_x=264) - -The area of the hexagon is area AEDC + area AFE + area BAC. The part of the segment BF that lies inside AEDC is wasted. The rest goes to provide height for the triangles on bases AE and $\mathrm{AC}$. So area AFE + area BAC can be maximised by taking $\mathrm{F}$ close to $\mathrm{A}$ and $\angle \mathrm{BAC}$ as close to a right angle as possible, so that the height of the triangle BAC (on the base $\mathrm{AC}$ ) is as large as possible. We can then get arbitrarily close to the area of: - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-42.jpg?height=215&width=303&top_left_y=949&top_left_x=251) - -We obviously make AEB a straight line. Now area $\mathrm{ADE}+$ area $\mathrm{ADC}=$ area $\mathrm{ACE}+$ area CDE. So if we regard every point except $\mathrm{D}$ as fixed, then we maximise the area by taking $\angle \mathrm{EAD}=\angle \mathrm{CAD}$, so that $\mathrm{D}$ is the maximum distance from $\mathrm{CE}$. Thus a maximal configuration must have $\angle \mathrm{AED}=\angle \mathrm{CAD}$. Similarly, it must have $\angle \mathrm{CAD}=\angle \mathrm{CAB}$, so all three angles must be equal. That disposes of case 1 . - -In cases 2 and 6 we find by a similar (but more tedious argument) the same maximum, although in one case we have to use the argument at the end for the final optimisation. In the other cases the maximum is smaller. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-42.jpg?height=169&width=651&top_left_y=1723&top_left_x=246) - -4 - -5 - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-42.jpg?height=160&width=303&top_left_y=1942&top_left_x=842) - -6 - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-42.jpg?height=143&width=297&top_left_y=2116&top_left_x=845) - -However, all these details would take an already long solution way over length. Does anyone have a better approach? - -No. 6 (second case) can be made arbitrarily close to the figure below (with $\mathrm{AB}=\mathrm{AC}=\mathrm{BD}=$ 1). To optimise it, suppose $\angle A C B=\theta$. Area $A B D C=$ area $A B C+$ area $B C D$. If we fix $\theta$, then $\mathrm{BC}$ is fixed, so to maximise area BCD we must take $\angle \mathrm{CBD}=90^{\circ}$. But $\theta$ cannot be optimal unless also $\angle \mathrm{CAD}=90^{\circ}$. We have $\mathrm{BA}=\mathrm{BD}$ and hence $\angle \mathrm{BAD}=\angle \mathrm{BDA}=45^{\circ}-\theta / 2$. Hence -$90^{\circ}=\angle \mathrm{CAD}=\angle \mathrm{BAC}-\angle \mathrm{BAD}=\left(180^{\circ}-2 \theta\right)-\left(45^{\circ}-\theta / 2\right)$. Hence $\theta=30^{\circ}$. So $\angle \mathrm{ACD}=$ $\angle \mathrm{BDC}=60^{\circ}$ and $\angle \mathrm{CAB}=\angle \mathrm{ABD}=120^{\circ}$. It is easy to check that this has area $(3 \sqrt{3}) / 4$. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-43.jpg?height=286&width=532&top_left_y=411&top_left_x=248) - -16th Iberoamerican 2001 - -## Problem A1 - -Show that there are arbitrarily large numbers $n$ such that: (1) all its digits are 2 or more; and (2) the product of any four of its digits divides $n$. - -## Solution - -$3232=16 \times 202$ and $10000=16 \times 625$. So any number with 3232 as its last 4 digits is divisible by 16 . So consider $\mathrm{N}=22223232$. Its sum of digits is 18 , so it is divisible by 9 . Hence it is divisible by $9 \cdot 16=144$. But any four digits have at most four $2 \mathrm{~s}$ and at most two 3s, so the product of any four digits divides 144 and hence $\mathrm{N}$. But now we can extend $\mathrm{N}$ by inserting an additional $9 \mathrm{~m} 2 \mathrm{~s}$ at the front. Its digit sum is increased by $18 \mathrm{~m}$, so it remains divisible by 144 and it is still divisible by the product of any four digits. - -## Alternative solution - -The number 111111111 with nine 1 s is divisible by 9 . Hence the number with twenty-seven $1 \mathrm{~s}$ which equals $111111111 \times 1000000001000000001$ is divisible by 27 . So $\mathrm{N}$, the number with twenty-seven $3 \mathrm{~s}$, is divisible by $3^{4}$. Now the number with $27 \mathrm{n} 3 \mathrm{~s}$ is divisible by $\mathrm{N}$ and hence by $3^{4}$. - -## Problem A2 - -$\mathrm{ABC}$ is a triangle. The incircle has center I and touches the sides $\mathrm{BC}, \mathrm{CA}, \mathrm{AB}$ at $\mathrm{D}, \mathrm{E}, \mathrm{F}$ respectively. The rays $\mathrm{BI}$ and $\mathrm{CI}$ meet the line $\mathrm{EF}$ at $\mathrm{P}$ and $\mathrm{Q}$ respectively. Show that if DPQ is isosceles, then $\mathrm{ABC}$ is isosceles. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-43.jpg?height=385&width=663&top_left_y=2183&top_left_x=248) -$\mathrm{AF}=\mathrm{AE}$, so $\angle \mathrm{AFE}=90^{\circ}-\mathrm{A} / 2$. Hence $\angle \mathrm{BFP}=90^{\circ}+\mathrm{A} / 2$. But $\angle \mathrm{FBP}=\mathrm{B} / 2$, so $\angle \mathrm{FPB}=\mathrm{C} / 2$. But $\mathrm{BFP}$ and $\mathrm{BDP}$ are congruent $(\mathrm{BF}=\mathrm{BD}, \mathrm{BP}$ common, $\angle \mathrm{FBP}=\angle \mathrm{FDP})$, so $\angle \mathrm{DPB}=\mathrm{C} / 2$ and $\angle \mathrm{DPQ}=\mathrm{C}$. Similarly, $\angle \mathrm{DQP}=\mathrm{B}$. Hence $\angle \mathrm{PDQ}=\mathrm{A}$. So $\mathrm{DQP}$ and $\mathrm{ABC}$ are similar. So if one is isosceles, so is the other. - -## Problem A3 - -Let $\mathrm{X}$ be a set with $\mathrm{n}$ elements. Given $\mathrm{k}>2$ subsets of $\mathrm{X}$, each with at least $\mathrm{r}$ elements, show that we can always find two of them whose intersection has at least $r-n k /(4 k-4)$ elements. - -## Problem B1 - -Call a set of 3 distinct elements which are in arithmetic progression a trio. What is the largest number of trios that can be subsets of a set of $n$ distinct real numbers? - -## Answer
$(\mathrm{m}-1) \mathrm{m}$ for $\mathrm{n}=2 \mathrm{~m}$
$\mathrm{m}^{2}$ for $\mathrm{n}=2 \mathrm{~m}+1$ - -## Solution - -Let $\mathrm{X}$ be one of the elements. What is the largest number of trios that can have $\mathrm{X}$ as middle element? Obviously, at most $\max (\mathrm{b}, \mathrm{a})$, where $\mathrm{b}$ is the number of elements smaller than $\mathrm{X}$ and a is the number larger. Thus if $n=2 m$, the no. of trios is at most $0+1+2+\ldots+m-1+m-1+$ $\mathrm{m}-2+\ldots+1+0=(\mathrm{m}-1) \mathrm{m}$. If $\mathrm{n}=2 \mathrm{~m}+1$, then the no. is at most $0+1+2+\ldots+\mathrm{m}-1+\mathrm{m}+\mathrm{m}-$ $1+\ldots+1+0=\mathrm{m}^{2}$ - -These maxima can be achieved by taking the numbers $1,2,3, \ldots, \mathrm{n}$. - -## Problem B2 - -Two players play a game on a $2000 \times 2001$ board. Each has one piece and the players move their pieces alternately. A short move is one square in any direction (including diagonally) or no move at all. On his first turn each player makes a short move. On subsequent turns a player must make the same move as on his previous turn followed by a short move. This is treated as a single move. The board is assumed to wrap in both directions so a player on the edge of the board can move to the opposite edge. The first player wins if he can move his piece onto the same square as his opponent's piece. For example, suppose we label the squares from $(0,0)$ to ( 1999,2000 ), and the first player's piece is initially at $(0,0)$ and the second player's at $(1996$, 3). The first player could move to $(1999,2000)$, then the second player to $(1996,2)$. Then the first player could move to $(1998,1998)$, then the second player to $(1995,1)$. Can the first player always win irrespective of the initial positions of the two pieces? - -## Problem B3 - -Show that a square with side 1 cannot be covered by five squares with side less than $1 / 2$. - -## 17th Iberoamerican 2002 - -## Problem A1 - -The numbers $1,2, \ldots, 2002$ are written in order on a blackboard. Then the 1 st, 4 th, 7 th, $\ldots$, $3 \mathrm{k}+1$ th, $\ldots$ numbers in the list are erased. Then the 1 st, $4 \mathrm{th}, 7 \mathrm{th}, \ldots 3 \mathrm{k}+1$ th numbers in the remaining list are erased (leaving $3,5,8,9,12, \ldots$ ). This process is carried out repeatedly until there are no numbers left. What is the last number to be erased? - -## Solution - -Answer: 1598. - -Let $\mathrm{a}_{\mathrm{n}}$ be the first number remaining after $\mathrm{n}$ iterations, so $\mathrm{a}_{0}=1, \mathrm{a}_{1}=2, \mathrm{a}_{3}=3, \mathrm{a}_{4}=5$ etc. We claim that: - -``` -\(a_{n+1}=3 / 2 a_{n} \quad\) if \(a_{n}\) is even, and - \(3 / 2\left(a_{n}+1\right)-1\) if \(a_{n}\) is odd. -``` - -We use induction on $n$. Suppose $a_{n}=2 N$. Consider the number $3 N$. There are initially $\mathrm{N}$ smaller numbers $=1 \mathrm{mod} 3$. So after the first iteration, it will lie in 2Nth place. Hence, it will lie in first place after $n+1$ iterations. Similarly, suppose $\mathrm{a}_{\mathrm{n}}=2 \mathrm{~N}+1$. Consider $3 N+2$. There are initially $\mathrm{N}+1$ smaller numbers $=1 \bmod 3$. So after the first iteration, it will lie in $2 \mathrm{~N}+1$ st place. Hence, it will lie in first place after $\mathrm{n}+1$ iterations. That completes the induction. - -We may now calculate successively the members of the sequence: $1,2,3,5,8,12,18,27,41$, $62,93,140,210,315,473,710,1065,1598,2397$. Hence 1598 is the last surviving number from $1,2, \ldots, 2002$. - -## Problem A2 - -Given a set of 9 points in the plane, no three collinear, show that for each point $\mathrm{P}$ in the set, the number of triangles containing $\mathrm{P}$ formed from the other 8 points in the set must be even. - -## Solution - -Join each pair of points, thus dividing the plane into polygonal regions. If a point $\mathrm{P}$ moves around within one of the regions then the number of triangles it belongs to does not change. But if it crosses one of the lines then it leaves some triangles and enters others. Suppose the line is part of the segment joining the points $\mathrm{Q}$ and $\mathrm{R}$ of the set. Then it can only enter or leave a triangle QRX for some $\mathrm{X}$ in the set. Suppose $\mathrm{x}$ points in the set lie on the same side of the line $\mathrm{QR}$ as $\mathrm{P}$. Then there are 6 - $\mathrm{x}$ points on the other side of the line $\mathrm{QR}$. So $\mathrm{P}$ leaves $\mathrm{x}$ triangles and enters $6-\mathrm{x}$. Thus the net change is even. Thus if we move $\mathrm{P}$ until it is in the outer infinite region (outside the convex hull of the other 8 points), then we change the number of triangles by an even number. But in the outside region it belongs to no triangles. - -## Problem A3 - -$\mathrm{ABC}$ is an equilateral triangle. $\mathrm{P}$ is a variable interior point such that $\square \mathrm{APC}=120^{\circ}$. The ray $\mathrm{CP}$ meets $\mathrm{AB}$ at $\mathrm{M}$, and the ray $\mathrm{AP}$ meets $\mathrm{BC}$ at $\mathrm{N}$. What is the locus of the circumcenter of the triangle MBN as $\mathrm{P}$ varies? - -## Solution - -Answer: the segment of the perpendicular bisector of $\mathrm{BG}$ (where $\mathrm{G}$ is the center of the triangle) which forms a rectangle with $\mathrm{AC}$. - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-46.jpg?height=580&width=646&top_left_y=498&top_left_x=251) - -$\angle \mathrm{MPN}=\angle \mathrm{APC}=120^{\circ}$ and $\angle \mathrm{MBN}=60^{\circ}$, so $\mathrm{MBNP}$ is cyclic, in other words, $\mathrm{P}$ lies on the circumcircle of BMN. - -$\mathrm{P}$ also lies on the circle $\mathrm{AGC}$, so $\angle \mathrm{CPG}=\angle \mathrm{CAG}$ (if $\mathrm{P}$ is on the same side of $\mathrm{AG}$ as $\mathrm{A})=30^{\circ}$ $=\angle$ MBG. So PMBG is cyclic. In other words, $\mathrm{G}$ also lies on the circumcircle of BMN. If P lies on the other side, the same conclusion follows from considering $\angle$ APG. - -Since B and G lie on the circumcircle, the center O must lie on the perpendicular bisector of BG. But it is clear that the extreme positions of $\mathrm{O}$ occur when $\mathrm{P}$ is at $\mathrm{A}$ and $\mathrm{B}$ and that these are the feet of the perpendiculars from A and B to the perpendicular bisector. - -## Problem B1 - -$\mathrm{ABC}$ is a triangle. $\mathrm{BD}$ is the an angle bisector. $\mathrm{E}, \mathrm{F}$ are the feet of the perpendiculars from $\mathrm{A}$, $C$ respectively to the line $B D . M$ is the foot of the perpendicular from $D$ to the line $B C$. Show that $\angle \mathrm{DME}=\angle \mathrm{DMF}$. - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-46.jpg?height=435&width=659&top_left_y=2104&top_left_x=253) - -Let $\mathrm{H}$ be the foot of the perpendicular from $\mathrm{D}$ to $\mathrm{AB} . \angle \mathrm{AHD}=\angle \mathrm{AED}=90^{\circ}$, so $\mathrm{AHED}$ is cyclic. Hence $\angle \mathrm{DAE}=\angle \mathrm{DHE}$. But $\mathrm{M}$ is the reflection of $\mathrm{H}$ is the line $\mathrm{BD}$, so $\angle \mathrm{DME}=$ $\angle \mathrm{DAE}$. - -$\mathrm{AE}$ is parallel to $\mathrm{CD}$, so $\angle \mathrm{DAE}=\angle \mathrm{DCF}$. $\angle \mathrm{DFC}=\angle \mathrm{DMC}$, so $\mathrm{DMCF}$ is cyclic. Hence $\angle \mathrm{DCF}$ $=\angle \mathrm{DMF}$. Hence $\angle \mathrm{DME}=\angle \mathrm{DMF}$. - -## Problem B2 - -The sequence $a_{n}$ is defined as follows: $a_{1}=56, a_{n+1}=a_{n}-1 / a_{n}$. Show that $a_{n}<0$ for some $n$ such that $0<\mathrm{n}<2002$. - -## Solution - -Note that whilst $a_{n}$ remains positive we have $a_{1}>a_{2}>a_{3}>\ldots>a_{n}$. Hence if $a_{m}$ and $a_{m+n}$ are in this part of the sequence, then $a_{m+1}=a_{m}-1 / a_{m}, a_{m+2}=a_{m+1}-1 / a_{m+1}0$ for all $\mathrm{n}<=2002$. Then we get successively: - -$a_{337}<56-336 / 56=50$ - -$a_{837}<50-500 / 50=40$ - -$a_{1237}<40-400 / 40=30$ - -$a_{1537}<30-300 / 30=20$ - -$\mathrm{a}_{1737}<20-200 / 20=10$ - -$\mathrm{a}_{1837}<10-100 / 10=0$. - -Contradiction. So we must have $\mathrm{a}_{\mathrm{n}}<0$ for some $\mathrm{n}<2002$. - -## Problem B3 - -A game is played on a 2001 x 2001 board as follows. The first player's piece is the policeman, the second player's piece is the robber. Each piece can move one square south, one square east or one square northwest. In addition, the policeman (but not the robber) can move from the bottom right to the top left square in a single move. The policeman starts in the central square, and the robber starts one square diagonally northeast of the policeman. If the policeman moves onto the same square as the robber, then the robber is captured and the first player wins. However, the robber may move onto the same square as the policeman without being captured (and play continues). Show that the robber can avoid capture for at least 10000 moves, but that the policeman can ultimately capture the robber. - -## Solution - -Color the squares with three colors as follows: - -``` -0\mathrm{k}$. Suppose $2 \mathrm{x}^{\mathrm{k}-1}-\mathrm{x}^{k}=\mathrm{h}>0$. Then we can increase $\mathrm{x}_{\mathrm{k}}$ by $\mathrm{h}, \mathrm{x}_{\mathrm{k}+1}$ by $2 \mathrm{~h}, \mathrm{x}_{\mathrm{k}+2}$ by $4 \mathrm{~h}$, ... So the sum will be increased by $\mathrm{h}\left( \pm 1 \pm 2 \pm \ldots \pm 2^{\mathrm{m}-1}+2^{\mathrm{m}}\right)$ for some $\mathrm{m} \geq 0$. But $\pm 1 \pm 2 \pm$ $\ldots \pm 2^{\mathrm{m}-1} \geq-\left(1+2+\ldots+2^{\mathrm{m}-1}\right)=-2^{\mathrm{m}}+1$, so the overall sum will be increased by at least 1 . So the set of $x_{i}$ was not maximal. - -## Problem B1 - -A $\square\{1,2,3, \ldots, 49\}$ does not contain six consecutive integers. Find the largest possible value of |A|. How many such subsets are there (of the maximum size)? - -## Answer - -$\max =41$; no. ways 495 - -## Solution - -We must exclude at least one element of each of the 8 sets $\{1,2, \ldots, 6\},\{7, \ldots, 12\},\{13, \ldots$, $18\}, \ldots,\{43, \ldots, 48\}$. So $|\mathrm{A}| \leq 41$. But a value of 41 is certainly possible, for example, exclude $2,8,14, \ldots, 44$. - -The largest excluded element must be at least 44 (or we have the 6 consecutive elements 44, 45, 46, 47, 48, 49). The smallest excluded element must be at most 6 . If we exclude 2 and 44, then the difference between them is 7.6 and so the other 6 excluded elements are fixed. But if we exclude 3 and 44, for example, then there are several possible choices for the other elements. - -There are 5 ways of choosing the smallest and largest excluded element to get a difference of 7. 6 between them ( 2 and 44, 3 and 45, 4 and 46, 5 and 47,6 and 48). There are 4 ways to get a difference of 7. $6-1$ ( 3 and 44, 4 and 45, 5 and 46, 6 and 47). There are 3 ways to get a difference of 7. $6-2$ ( 4 and 44, 5 and 45, 6 and 46), 2 ways to get a difference of 7. $6-3$ (5 and 44, 6 and 45), and 1 way to get a difference of 7. $6-4$ (6 and 44). - -If the difference is $7.6-1$, then we can shorten any of the 7 gaps, so there are 7 possibilities. For example, with 3 and 44, we could shorten the first gap, so excluding 3, 8, 14, 20, 26, 32, 38 and 44 , or the second gap, so excluding $3,9,14,20,26,32,38$ and 44 , and so on. - -If the difference is $7 \cdot 6-2$, then we can shorten one gap by two ( 7 possibilities) or two gaps by one (21 possibilities), total 28 . If the difference is $7 \cdot 6-3$, then we can shorten on gap by three (7), one by two and one by one (42) or three by one (35), total 84. Finally, if the difference is 7. 6 - 4, we can shorten one by four (7), one by three and one by 1 (42), two by two (21), one by two and two by one (105), or four by one (35), total 210. - -So the total number of possibilities is $5 \cdot 1+4 \cdot 7+3 \cdot 28+2 \cdot 84+1 \cdot 210=495$. - -## Problem B2 - -$\mathrm{ABCD}$ is a square. $\mathrm{P}, \mathrm{Q}$ are points on the sides $\mathrm{BC}, \mathrm{CD}$ respectively, distinct from the endpoints such that $\mathrm{BP}=\mathrm{CQ}$. $\mathrm{X}, \mathrm{Y}$ are points on $\mathrm{AP}, \mathrm{AQ}$ respectively. Show that there is a triangle with side lengths $B X, X Y, Y D$. - -## Solution - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-52.jpg?height=439&width=328&top_left_y=1508&top_left_x=247) - -We have $\mathrm{DY}<\mathrm{BY} \leq \mathrm{BX}+\mathrm{XY}$ (this is almost obvious, but to prove formally use the cosine formula for BAY and DAY and notice that $\angle \mathrm{BAY}>\angle \mathrm{DAY}$ ). Similarly, $\mathrm{BX}<\mathrm{DX} \leq \mathrm{DY}+$ YX. So it remains to show that $X Y\angle \mathrm{PAQ}$, so it follows by the same observation as above that $\mathrm{XY}^{\prime}>\mathrm{XY}$. But the claim is almost obvious. Note that $\mathrm{PQ}^{\prime}=\mathrm{AB}$ - -![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-53.jpg?height=334&width=371&top_left_y=256&top_left_x=266) - -So take $\mathrm{P}^{\prime}$ on $\mathrm{AD}$ with $\angle \mathrm{P}^{\prime} \mathrm{PQ} \mathrm{P}^{\prime}=90^{\circ}$. Then $\mathrm{A}$ lies inside the circle $\mathrm{P}^{\prime} \mathrm{PQ}^{\prime}$, so extend $\mathrm{PA}$ to meet it again at $\mathrm{A}^{\prime}$. Then $\angle \mathrm{PA}^{\prime} \mathrm{Q}^{\prime}=\angle \mathrm{PP}^{\prime} \mathrm{Q}^{\prime}=45^{\circ}$, so $\angle \mathrm{PAQ}^{\prime}=\angle \mathrm{PA}^{\prime} \mathrm{Q}^{\prime}+\angle \mathrm{AQ}^{\prime} \mathrm{Q}^{\prime}>45^{\circ}$. But $\angle \mathrm{PAQ}^{\prime}+\angle \mathrm{PAQ}=90^{\circ}$, so $\angle \mathrm{PAQ}^{\prime}>\angle \mathrm{PAQ}$ as claimed. - -## Problem B3 - -The sequences $\mathrm{a}_{0}, \mathrm{a}_{1}, \mathrm{a}_{2}, \ldots$ and $\mathrm{b}_{0}, \mathrm{~b}_{1}, \mathrm{~b}_{2}, \ldots$ are defined by $\mathrm{a}_{0}=1, \mathrm{~b}_{0}=4, \mathrm{a}_{\mathrm{n}+1}=\mathrm{a}_{\mathrm{n}}{ }^{2001}+\mathrm{b}_{\mathrm{n}}$, $b_{n+1}=b_{n}^{2001}+a_{n}$. Show that no member of either sequence is divisible by 2003. - -## Solution - -2003 is prime, so $\mathrm{a}^{2002}=1 \bmod 2003$ for any a not divisible by 2003. Thus $a_{n+1}=a_{n}^{-1}+b_{n}$ $\bmod 2003, b_{n+1}=b_{n}^{-1}+a_{n} \bmod 2003$. Put $c_{n}=a_{n} b_{n}$. Then $c_{n+1}=c_{n}+1 / c_{n}+2=\left(c_{n}+1\right)_{2} / c_{n}$ $\bmod 2003$. So if $\mathrm{c}_{\mathrm{n}} \neq 0 \bmod 2003$, then $\mathrm{c}_{\mathrm{n}+1} \neq 0 \bmod 2003$ unless $\mathrm{c}_{\mathrm{n}}=-1 \bmod 2003$. Then if $c_{n+1}=-1 \bmod 2003$, we must have $\left(c_{n}{ }^{2}+3 c_{n}+1\right) / c_{n}=0 \bmod 2003$, so $c_{n}{ }^{2}+3 c_{n}+1=0 \bmod$ 2003. Note that $c_{0}=4$. So it is sufficient to show that there are no solutions to $x^{2}+3 x+1=0$ $\bmod 2003$, or equivalently to $(x-1000)^{2}=1000^{2}-1=502 \bmod 2003$. In other words, we have to show that 502 is a quadratic non-residue mod 2003. - -The easiest way to do that is to use the law of quadratic reciprocity, but that is almost certainly outside the syllabus. We note that 4. $502=5 \bmod 2003$, so 502 is a square iff 5 is a square. It is sufficient to show that $5^{1001}=-1 \bmod 2003$, for then if we had $x^{2}=5$, we would have $\mathrm{x}^{2002}=-1 \bmod 2003$, whereas we know that $\mathrm{x}^{2002}=1 \bmod 2003$. We note that $1001=$ 7. 11. 13. We start by showing that $5^{7}=8 \bmod 2003$. We have $5^{5}=3125=1122 \bmod 2003$, so $5^{6}=5610=1604 \bmod 2003$, so $5^{7}=8020=8 \bmod 2003$. - -We calculate successively $2^{11}=2048=45 \bmod 2003$, so $2^{22}=2025=22 \bmod 2003$. Multiplying by 22 is relatively easy, so $2^{44}=484,2^{66}=10648=633,2^{88}=13926=-95,2^{110}$ $=-2090=-87,2^{132}=-1914=89,2^{143}=4005=-1$ all $\bmod 2003$. Hence $8^{11 \cdot 13}=-1 \bmod 2003$, so $5^{1001}=-1 \bmod 2003$, as required, and we are done. -