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 $(5,8,11)$.


$\mathrm{P}$ 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 $\mathrm{x}=\sqrt{ } 129$.


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 $r_{1}=1 / 2, r_{2}$ $=1, \mathrm{r}_{3}=5 / 4, \mathrm{r}_{4}=2$.


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 $(\mathrm{yz}-$ $\left.x^{2}\right) /(1-x)=(x+y+z)$


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 $f(1985)$ $=\mathrm{f}(5)+\mathrm{f}(397)=0$.


$\mathrm{O}$ 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 $1 / A D+1 / B E+1 / C F=2 / R$.


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 $f(x)=4\left(x^{2}(1+x) /(1-x)\right)^{1 / 3}$.


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 $\mathrm{BM}>\mathrm{CN}$.


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 $\mathrm{a}_{2 \mathrm{k}-1}=\mathrm{d}_{\mathrm{k}}=\mathrm{c}_{\mathrm{k}}^{2}+1$.


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 $\mathrm{a}+\mathrm{b}+\mathrm{c}$ $=3 \pi / 4$


$\mathrm{ABCD}$ 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.


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.


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 $p=a+c$.


$\mathrm{P}$ 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.


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 $\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}\right)$.


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.


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 $x_{1}<x_{2}<\ldots<x_{n+1}$ with no three in AP and $x_{n+1}>a_{n+1}$. 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.


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 $\mathrm{y}=\mathrm{z}=-1$.


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.


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 $k<A$ it is sufficient to show that $20 h^{2} k+20 h k^{2} \leq(2 k+h)(2 k+h+k)(2 k+2 h+k)$
$=18 \mathrm{k}^{3}+27 \mathrm{hk}^{2}+13 \mathrm{~h}^{2} \mathrm{k}+2 \mathrm{~h}^{3}$ or $18 \mathrm{k}^{3}+7 \mathrm{hk}^{2}-7 \mathrm{~h}^{2} \mathrm{k}+2 \mathrm{~h}^{3} \geq 0$. But $7 \mathrm{k}^{2}-7 \mathrm{hk}+2 \mathrm{~h}^{2}=7(\mathrm{k}-$ $\mathrm{h} / 2)^{2}+\mathrm{h}^{2} / 4 \geq 0$ and $h$ and $\mathrm{k}$ are non-negative, so $18 \mathrm{k}^{3}+\mathrm{h}\left(7 \mathrm{k}^{2}-7 \mathrm{hk}+2 \mathrm{~h}^{2}\right) \geq 0$.

Thus we have established that $0<=\mathrm{X}<1 / 20$, which shows that $\mathrm{f}(\mathrm{a}, \mathrm{b}, \mathrm{c})<1 / 20$, which is slightly stronger than the required result.


The incircle of the triangle $\mathrm{ABC}$ touches $\mathrm{AC}$ at $\mathrm{M}$ and $\mathrm{BC}$ at $\mathrm{N}$ and has center $\mathrm{O}$. $\mathrm{AO}$ meets $\mathrm{MN}$ at $\mathrm{P}$ and BO meets MN at $\mathrm{Q}$. Show that MP.OA $=\mathrm{BC} . \mathrm{OQ}$.

Solution:
![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-09.jpg?height=442&width=594&top_left_y=887&top_left_x=248)

The key to getting started is to notice that angle $\mathrm{AQB}=90^{\circ}$.

Angle $\mathrm{BAQ}=90^{\circ}-\mathrm{B} / 2$, so angle $\mathrm{OAQ}=90^{\circ}-\mathrm{B} / 2-\mathrm{A} / 2=\mathrm{C} / 2$. So $\mathrm{OQ}=\mathrm{AO} \sin \mathrm{C} / 2$. Thus we have to show that $\mathrm{MP}=\mathrm{BC} \sin \mathrm{C} / 2$.

Let the incircle touch $\mathrm{AB}$ at $\mathrm{L}$ and let $\mathrm{Y}$ be the midpoint of ML (also the intersection of ML with AO). Angle NMC $=90^{\circ}$ - C/2. It is also A/2 + angle MPY, so angle MPY $=90-\mathrm{C} / 2-$ $\mathrm{A} / 2=\mathrm{B} / 2$. Hence $\mathrm{MP}=\mathrm{MY} / \sin \mathrm{B} / 2$. We have $\mathrm{MY}=\mathrm{MO} \sin \mathrm{MOA}=\mathrm{r} \cos \mathrm{A} / 2$ (where $\mathrm{r}$ is the inradius, as usual). So $\mathrm{MP}=(\mathrm{r} \cos \mathrm{A} / 2) / \sin \mathrm{B} / 2$. We have $\mathrm{BC}=\mathrm{BN}+\mathrm{NC}=\mathrm{r}(\cot \mathrm{B} / 2+$ $\cot \mathrm{C} / 2)$, so $\mathrm{MP} / \mathrm{BC}=(\cos \mathrm{A} / 2) /(\sin \mathrm{B} / 2(\cot \mathrm{B} / 2+\cot \mathrm{C} / 2))$. Hence $\mathrm{MP} /(\mathrm{BC} \sin \mathrm{C} / 2)=($ $\cos \mathrm{A} / 2) /(\cos \mathrm{B} / 2 \sin \mathrm{C} / 2+\sin \mathrm{B} / 2 \cos \mathrm{C} / 2)=\cos \mathrm{A} / 2 / \sin (\mathrm{B} / 2+\mathrm{C} / 2)=1$.


The function $\mathrm{f}$ on the positive integers satisfies $\mathrm{f}(1)=1, \mathrm{f}(2 \mathrm{n}+1)=\mathrm{f}(2 \mathrm{n})+1$ and $\mathrm{f}(2 \mathrm{n})=3$ $f(n)$. Find the set of all $m$ such that $m=f(n)$ for some $n$.

Solution: We show that to obtain $\mathrm{f}(\mathrm{n})$, one writes $\mathrm{n}$ in base 2 and then reads it in base 3 . For example 12 $=1100_{2}$, so $f(12)=1100_{3}=36$. Let $g(n)$ be defined in this way. Then certainly $g(1)=1$. Now $2 n+1$ has the same binary expansion as $2 n$ except for a final 1 , so $g(2 n+1)=g(2 n)+1$. Similarly, $2 \mathrm{n}$ has the same binary expansion as $\mathrm{n}$ with the addition of a final zero. Hence $g(2 n)=3 g(n)$. So $g$ is the same as f. Hence the set of all $m$ such that $m=f(n)$ for some $n$ is the the set of all $\mathrm{m}$ which can be written in base 3 without a digit 2 .


Show that there are infinitely many solutions in positive integers to $2 a^{2}-3 a+1=3 b^{2}+b$.

Solution:
Put $\mathrm{A}=\mathrm{a}-1$ and the equation becomes $\mathrm{A}(2 \mathrm{~A}+1)=\mathrm{b}(3 \mathrm{~b}+1)$. Let $\mathrm{d}$ be the greatest common divisor of A and $b$. Put $A=d x, b=d y$. Then $x(2 d x+1)=y(3 d y+1)$. Since $x$ and $y$ are coprime, $\mathrm{x}$ must divide $3 \mathrm{dy}+1$. So put $3 \mathrm{dy}+1=\mathrm{nx}$. Then $2 \mathrm{dx}+1=\mathrm{ny}$. Solving for $\mathrm{x}$ and $\mathrm{y}$ in terms of $n$ and $d$ we get $x=(n+3 d) /\left(n^{2}-6 d^{2}\right), y=(n+2 d) /\left(n^{2}-6 d^{2}\right)$.

So we would certainly be home if we could show that there were infinitely many solutions to $\mathrm{n}^{2}-6 \mathrm{~d}^{2}=1$. It is not hard to find the first few: $1^{2}-6.0^{2}=1,5^{2}-6 \cdot 2^{2}=1,49^{2}-6 \cdot 20^{2}=1$. We notice that $49^{2}=2.5^{2}-1$, so we wonder whether $\mathrm{n}=2.49^{2}-1$ might be another solution and indeed we find it gives $d=1960=2 \cdot 49.20$. This suggests we try $\left(2 n^{2}-1\right)^{2}-6(2 n d)^{2}=4 n^{4}-$ $4 n^{2}+1-24 n^{2} d^{2}=4 n^{2}\left(n^{2}-6 d^{2}-1\right)+1=1$. So there are indeed infinitely many solutions to $n^{2}$ $-6 \mathrm{~d}^{2}=1$ and we are done.


The function $\mathrm{f}$ is defined on the non-negative integers. $\mathrm{f}\left(2^{\mathrm{n}}-1\right)=0$ for $\mathrm{n}=0,1,2, \ldots$. If $\mathrm{m}$ is not of the form $2^{n}-1$, then $f(m)=f(m+1)+1$. Show that $f(n)+n=2^{k}-1$ for some $k$, and find $\mathrm{f}\left(2^{1990}\right)$

Solution: We claim that if $2^{\mathrm{m}}<=\mathrm{n}<2^{\mathrm{m}+1}$, then $\mathrm{f}(\mathrm{n})=2^{\mathrm{m}+1}-\mathrm{n}-1$. Put $\mathrm{r}=2^{\mathrm{m}+1}-\mathrm{n}$. Then the claim follows by induction on $\mathrm{r}$. Hence $\mathrm{f}\left(2^{1990}\right)=2^{1990}-1$.


I is the incenter of the triangle $\mathrm{ABC}$ and the incircle touches $\mathrm{BC}, \mathrm{CA}, \mathrm{AB}$ at $\mathrm{D}, \mathrm{E}, \mathrm{F}$ respectively. AD meets the incircle again at P. M is the midpoint of EF. Show that PMID is cyclic (or the points are collinear).

Solution:
![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-10.jpg?height=352&width=545&top_left_y=2057&top_left_x=253)

$\angle \mathrm{AEI}=\angle \mathrm{AME}=90^{\circ}$, so $\mathrm{AEI}$ and $\mathrm{AME}$ are similar. Hence $\mathrm{AM} / \mathrm{AE}=\mathrm{AE} / \mathrm{AI}$ or $\mathrm{AM} \cdot \mathrm{AI}=$ $\mathrm{AE}^{2}$. $\mathrm{AE}$ is tangent to the incircle, so $\mathrm{AE} 2=\mathrm{AP} \cdot \mathrm{AD}$. Hence $\mathrm{AM} \cdot \mathrm{AI}=\mathrm{AP} \cdot \mathrm{AD}$, so if $\mathrm{P}, \mathrm{M}, \mathrm{I}, \mathrm{D}$ are not collinear, then they are cyclic.
$\mathrm{f}(\mathrm{x})=(\mathrm{x}+\mathrm{b})^{2}+\mathrm{c}$, where $\mathrm{b}$ and $\mathrm{c}$ are integers. If the prime $\mathrm{p}$ divides $\mathrm{c}$, but $\mathrm{p}^{2}$ does not divide $c$, show that $f(n)$ is not divisible by $p^{2}$ for any integer $n$. If an odd prime $q$ does not divide $c$, but divides $\mathrm{f}(\mathrm{n})$ for some $\mathrm{n}$, show that for any $\mathrm{r}$, we can find $\mathrm{N}$ such that $\mathrm{q}^{\mathrm{r}}$ divides $\mathrm{f}(\mathrm{N})$.


Solution:
The first part is trivial. If $\mathrm{p}$ does not divide $(\mathrm{x}+\mathrm{b})$, then it does not divide $(\mathrm{x}+\mathrm{b})^{2}$, so it does not divide $(\mathrm{x}+\mathrm{b})^{2}+\mathrm{c}$. On the other hand, if $\mathrm{p}$ does divide $\mathrm{x}+\mathrm{b}$, then $\mathrm{p}^{2}$ divides $(\mathrm{x}+\mathrm{b})^{2}$, so $\mathrm{p}^{2}$ does not divide $(\mathrm{x}+\mathrm{b})^{2}+\mathrm{c}$.

For the second part, we use induction on $\mathrm{r}$. For $\mathrm{r}=1$, we are given that $\mathrm{q}$ divides $\mathrm{f}(\mathrm{n})$. Now suppose that $q^{r}$ divides $f(N)$ for some $N$. If $q^{r+1}$ divides $f(N)$, then we are done. So suppose $q^{r+1}$ does not divide $\mathrm{f}(\mathrm{N})$, so $\mathrm{f}(\mathrm{N})=\mathrm{q}^{\mathrm{r}} \mathrm{h}$ where $\mathrm{q}$ does not divide h. We have $\mathrm{f}\left(\mathrm{N}+\mathrm{kq}^{\mathrm{r}}\right)=\mathrm{f}(\mathrm{N})+$ $q^{r}(2 N+2 b) k=q^{r} h+q^{r}(2 N+2 b) k$. Now $q$ divides $(N+b)^{2}+c$, and does not divide $c$, so it does not divide $(\mathrm{N}+\mathrm{b})^{2}$ and hence does not divide $\mathrm{N}+\mathrm{b}$. It is odd, so it does not divide $2 \mathrm{~N}+2 \mathrm{~b}$. Hence we can find $k$ such that $k(2 N+2 b)=-h$ mod $q$. Then we have $\mathrm{q}^{\mathrm{r}+1}$ divides $\mathrm{f}\left(\mathrm{N}+\mathrm{kq}^{\mathrm{r}}\right)$, which completes the induction.


The circle $\mathrm{C}$ has diameter $\mathrm{AB}$. The tangent at $\mathrm{B}$ is $\mathrm{T}$. For each point $\mathrm{M}$ (not equal to $\mathrm{A}$ ) on $\mathrm{C}$ there is a circle $\mathrm{C}^{\prime}$ which touches $\mathrm{T}$ and touches $\mathrm{C}$ at M. Find the point at which $\mathrm{C}^{\prime}$ touches $\mathrm{T}$ and find the locus of the center of $\mathrm{C}^{\prime}$ as $\mathrm{M}$ varies. Show that there is a circle orthogonal to all the circles $\mathrm{C}^{\prime}$.

## Answer

$\mathrm{C}^{\prime}$ touches $\mathrm{T}$ at the intersection of $\mathrm{T}$ and the line $\mathrm{AM}$

the locus of the center is a parabola vertex B

the circle center A radius $\mathrm{AB}$ is orthogonal to all circles $\mathrm{C}^{\prime}$

Solution:
![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-11.jpg?height=366&width=323&top_left_y=1847&top_left_x=250)

Let $\mathrm{O}$ be the center of $\mathrm{C}$. Let the line $\mathrm{AM}$ meet $\mathrm{T}$ at $\mathrm{N}$. Let the perpendicular to $\mathrm{T}$ at $\mathrm{N}$ meet the line $\mathrm{OM}$ at $\mathrm{O}^{\prime}$. Then $\angle \mathrm{O}^{\prime} \mathrm{NM}=\angle \mathrm{MAB}$ ( $\mathrm{O}^{\prime} \mathrm{N}$ parallel to $\mathrm{AB}$, because both perpendicular to $\mathrm{T})=\angle \mathrm{OMA}(\mathrm{OM}=\mathrm{OA})=\angle \mathrm{O}^{\prime} \mathrm{MN}$. So $\mathrm{O}^{\prime} \mathrm{M}=\mathrm{O}^{\prime} \mathrm{N}$. Hence $\mathrm{O}^{\prime}$ is the center of $\mathrm{C}^{\prime}$.

Take B to be the origin and $\mathrm{A}$ to be the point $(2 \mathrm{a}, 0)$, so $\mathrm{O}$ is $(\mathrm{a}, 0)$ and $\mathrm{C}$ has radius a. If $\mathrm{O}^{\prime}$ is $(x, y)$, then we require that $O^{\prime} O=x+a$ or $(x-a)^{2}+y^{2}=(x+a)^{2}$, or $y^{2}=4 a x$, which is a parabola with vertex $B$ and axis the $x$-axis.

Triangles $\mathrm{AMB}, \mathrm{ABN}$ are similar $\left(\angle \mathrm{AMB}=\angle \mathrm{ABN}=90^{\circ}\right)$, so $\mathrm{AM} / \mathrm{AB}=\mathrm{AB} / \mathrm{AN}$ and hence $\mathrm{AM} \cdot \mathrm{AN}=\mathrm{AB}^{2}$. Now consider the circle center $\mathrm{A}$ radius $\mathrm{AB}$. It must meet the circle $\mathrm{C}^{\prime}$, because it contains the point $M$. Suppose it meets at $X$. Then $A X^{2}=A B^{2}=A M \cdot A N$, so $A X$ is tangent to $\mathrm{C}^{\prime}$ and hence the circles are orthogonal.


$A$ and $B$ are opposite corners of an $n \times n$ board, divided into $n^{2}$ squares by lines parallel to the sides. In each square the diagonal parallel to $\mathrm{AB}$ is drawn, so that the board is divided into $2 \mathrm{n}^{2}$ small triangles. The board has $(n+1)^{2}$ nodes and large number of line segments, each of length 1 or $\sqrt{ } 2$. A piece moves from $\mathrm{A}$ to $\mathrm{B}$ along the line segments. It never moves along the same segment twice and its path includes exactly two sides of every small triangle on the board. For which $\mathrm{n}$ is this possible?

## Answer

$\mathrm{n}=2$ only

Solution:
The diagram above shows that $\mathrm{n}=2$ is possible (the path is AHEFGHCDIHB). Now suppose $\mathrm{n}$ $>2$.

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.


$\mathrm{f}(\mathrm{x})$ 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.


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.


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.


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.


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 $\mathrm{H}$.


$a_{n}$ 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 |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |


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.


$\mathrm{ABC}$ 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 $\mathrm{PQC}=(\sqrt{3}) / 4$.


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, $\mathrm{A}=\mathrm{B}= \pm 4$.


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 $\mathrm{ABC}$.

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.]


Find all functions $\mathrm{f}$ on the positive integers with positive integer values such that (1) if $\mathrm{x}<\mathrm{y}$, then $f(x)<f(y)$, and $(2) f(y f(x))=x^{2} f(x y)$.

Solution:
Answer: $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$.

Note that ( 1 ) implies $f$ is $(1,1)$.

Put $y=1$. Then $f(f(x))=x^{2} f(x)$.

Put $y=f(z)$, then $f(f(z) f(x))=x^{2} f(x f(z))=x^{2} z^{2} f(x z)=f(f(x z))$. But $f$ is $(1,1)$ so $f(x z)=$ $\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{z})$.

Now suppose $\mathrm{f}(\mathrm{m})>\mathrm{m}^{2}$ 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 $\mathrm{f}(\mathrm{m})=\mathrm{m}^{2}$.


$\mathrm{ABC}$ 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 .


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.


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 $B$.


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 $1<b<1992$ such that if 1993 is written in base $\mathrm{b}$ then all its digits are the same.

Solution:
Any even number $2 \mathrm{n}$ can be written as 22 in base $\mathrm{n}-1$. In particular $1994=22996$.

We have to show that we cannot write $1993=$ aaa ... $\mathrm{a}_{\mathrm{b}}$. If the number has $\mathrm{n}$ digits, then 1993 $=\mathrm{a}\left(1+\mathrm{b}+\ldots+\mathrm{b}^{\mathrm{n}-1}\right)=\mathrm{a}\left(\mathrm{b}^{\mathrm{n}}-1\right) /(\mathrm{b}-1)$. But 1993 is prime, so a must be 1 . Hence $\mathrm{b}^{\mathrm{n}-1}+\ldots+\mathrm{b}$ $-1992=0$. So b must divide $1992=2^{3} 3.83$. We cannot have $\mathrm{n}=2$, for then $\mathrm{b}=1992$ and we require $\mathrm{b}<1992$. So $\mathrm{n}>2$. 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:

```
\(1+2+2^{2}+\ldots+2^{9}=1023,1+\ldots+2^{10}=2047\).
\(1+3+\ldots+3^{6}=1093,1+\ldots+3^{\wedge} 7=3280\)
\(1+4+\ldots+4^{5}=1365,1+\ldots+4^{6}=5461\)
```

```
\(1+6+\ldots 6^{4}=1555,1+\ldots+6^{5}=9331\)
\(1+8+8^{2}+8^{3}=585,1+\ldots+8^{4}=4681\)
\(1+12+12^{2}+12^{3}=1885,1+\ldots+12^{4}=22621\)
\(1+24+24^{2}=601,1+\ldots+24^{3}=14425\)
```


$\mathrm{ABCD}$ 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 $1 \mathrm{~s}$.


$\mathrm{ABC}$ 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.


$\mathrm{n}$ 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 $0 \leq m<n$ can be written as a sum of one element from each of the $r$ subsets.

## Answer

smallest integer such that $\mathrm{k}^{\mathrm{r}} \geq \mathrm{n}$.

Solution:
We can form at most $\mathrm{k}^{r}$ distinct sums, so $\mathrm{k}^{r}$ must be $\geq n$.

Now consider $\mathrm{A}_{1}=\{0,1,2, \ldots, \mathrm{k}-1\}, \mathrm{A}_{2}=\{0, \mathrm{k}, 2 \mathrm{k}, \ldots,(\mathrm{k}-1) \mathrm{k}\}, \mathrm{A}_{3}=\left\{0, \mathrm{k}^{2}, 2 \mathrm{k}^{2}, \ldots,(\mathrm{k}-\right.$ 1) $\left.\mathrm{k}^{2}\right\}, \ldots, \mathrm{A}_{\mathrm{r}}=\left\{0, \mathrm{k}^{\mathrm{r}-1}, 2 \mathrm{k}^{r-1}, \ldots,(\mathrm{k}-1) \mathrm{k}^{r^{r-1}}\right\}$. Then for any non-negative integer $\mathrm{m}<\mathrm{k}^{\mathrm{r}}$, we can write $\mathrm{m}$ with $\mathrm{r}$ digits in base $\mathrm{k}$ (using leading zeros as necessary) and hence as a sum of one element from each $A_{i}$. This subset works for $(k-1) k^{r-1}<n \leq k^{r}$. For smaller $n$ above $(k-1)^{r}$ we cannot use all the elements given above, but we do not need them, so we just replace the elements which are too large by arbitrary elements under n.

For example, suppose $\mathrm{n}=17, \mathrm{r}=4$. We need $\mathrm{k}=3$. So we form $\mathrm{A}_{1}=\{0,1,2\}, \mathrm{A}_{2}=\{0,3,6\}$, $\mathrm{A}_{3}=\{0,9,18\}, \mathrm{A}_{4}=\{0,27,54\}$. Now $18,27,54$ are unnecessary, so we pad out $\mathrm{A}_{3}$ and $\mathrm{A}_{4}$ with other elements. We could take $\mathrm{A}_{3}=\{0,1,9\}, \mathrm{A}_{4}=\{0,1,2\}$.


Find all possible values for the sum of the digits of a square.

Solution:
Answer: any non-negative integer $=0,1,4$ or $7 \bmod 9$.

$0^{2}=0,( \pm 1)^{2}=1,( \pm 2)^{2}=4,( \pm 3)^{2}=0,( \pm 4)^{2}=7 \bmod 9$, so the condition is necessary.

We exhibit squares which give these values.

$0 \bmod 9$. Obviously $0^{2}=0$. We have $9^{2}=81,99^{2}=9801$ and in general $9 \ldots 9^{2}=\left(10^{\mathrm{n}}-1\right)^{2}=$ $10^{2 \mathrm{n}}-2.10^{\mathrm{n}}+1=9 \ldots 980 \ldots 01$, with digit sum $9 n$.

1 mod 9 . Obviously $1^{2}=1$ with digit sum 1 , and $8^{2}=64$ with digit sum 10 . We also have $98^{2}$ $=9604,998^{2}=996004$, and in general $9 \ldots 98^{2}=\left(10^{\mathrm{n}}-2\right)^{2}=10^{2 \mathrm{n}}-4.10^{\mathrm{n}}+4=9 \ldots 960 \ldots 04$, with digit sum $9 n+1$.

$4 \bmod 9$ Obviously $2^{2}=4$ with digit sum 4 , and $7^{2}=49$ with digit sum 13 . Also $97^{2}=9409$ with digit sum $22,997^{2}=994009$ with digit sum 31 , and in general $9 \ldots 97^{2}=\left(10^{\mathrm{n}}-3\right)^{2}=10^{2 \mathrm{n}}-$ $6.10^{\mathrm{n}}+9=9 \ldots 940 \ldots 09$, with digit sum $9 n+4$.

$7 \bmod 9$ Obviously $4^{2}=16$, with digit sum 7. Also $95^{2}=9025$, digit sum $16,995^{2}=990025$ with digit sum 25 , and in general $9 \ldots 95^{2}=\left(10^{\mathrm{n}}-5\right)^{2}=10^{2 \mathrm{n}}-10^{\mathrm{n}+1}+25=9 \ldots 90 \ldots 025$, with digit sum $9 n-2$.


Find all solutions in real numbers $x_{1}, x_{2}, \ldots, x_{n+1}$ all at least 1 such that: (1) $x_{1}{ }^{1 / 2}+x_{2}{ }^{1 / 3}+x_{3}{ }^{1 / 4}$ $+\ldots+\mathrm{x}_{\mathrm{n}}^{1 /(\mathrm{n}+1)}=\mathrm{n}_{\mathrm{n}+1}{ }^{1 / 2} ;$ and (2) $\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots+\mathrm{x}_{\mathrm{n}}\right) / \mathrm{n}=\mathrm{x}_{\mathrm{n}+1}$.

## Answer

The only solution is the obvious, all $\mathrm{x}_{\mathrm{i}}=1$.

Solution: By Cauchy-Schwartz, $\left(\sum \mathrm{x}_{\mathrm{i}}^{1 / 2}\right)^{2} \leq\left(\sum 1\right)\left(\&\right.$ sum $\left.\mathrm{x}_{\mathrm{i}}\right)$, with equality iff all $\mathrm{x}_{\mathrm{i}}$ equal. In other words, if we put $x_{n+1}=\left(x_{1}+x_{2}+\ldots+x_{n}\right) / n$, then $\sum x_{i}^{1 / 2} \leq n x_{n+1}^{1 / 2}$. But since all $x_{i} \geq 1$, we have $\mathrm{x}_{1}^{1 / 2}+\mathrm{x}_{2}{ }^{1 / 3}+\mathrm{x}_{3}{ }^{1 / 4}+\ldots+\mathrm{x}_{\mathrm{n}}{ }^{1 /(n+1)} \leq \sum \mathrm{x}_{\mathrm{i}}^{1 / 2}$ with equality iff $\mathrm{x}_{2}=\mathrm{x}_{3}=\ldots=\mathrm{x}_{\mathrm{n}}=1$. Hence $\mathrm{x}_{1}{ }^{1 / 2}+\mathrm{x}_{2}{ }^{1 / 3}+\mathrm{x}_{3}{ }^{1 / 4}+\ldots+\mathrm{x}_{\mathrm{n}}{ }^{1 /(n+1)} \leq \mathrm{x}_{\mathrm{n}+1}{ }^{1 / 2}$ with equality iff all $\mathrm{x}_{\mathrm{i}}=1$.


$\mathrm{ABCD}$ is an $\mathrm{n} \mathrm{x}$ board. We call a diagonal row of cells a positive diagonal if it is parallel to AC. How many coins must be placed on an $\mathrm{n} x \mathrm{n}$ board such that every cell either has a coin or is in the same row, column or positive diagonal as a coin?

## Answer

smallest integer $\geq(2 n-1) / 3$

[so $2 \mathrm{~m}-1$ for $\mathrm{n}=3 \mathrm{~m}-1,2 \mathrm{~m}$ for $\mathrm{n}=3 \mathrm{~m}, 2 \mathrm{~m}+1$ for $\mathrm{n}=3 \mathrm{~m}+1$ ]

Solution:
There must be at least $\mathrm{n}-\mathrm{k}$ rows without a coin and at least $\mathrm{n}-\mathrm{k}$ columns without a coin. Let $\mathrm{r}_{1}$, $\mathrm{r}_{2}, \ldots, \mathrm{r}_{\mathrm{n}-\mathrm{k}}$ be cells in the top row without a coin which are also in a column without a coin. Let $\mathrm{r}_{1}, \mathrm{c}_{2}, \mathrm{c}_{3}, \ldots, \mathrm{c}_{\mathrm{n}-\mathrm{k}}$ be cells in the first column without a coin which are also in a row without a coin. Each of the $2 \mathrm{n}-2 \mathrm{k}-1 \mathrm{r}_{\mathrm{i}}$ and $\mathrm{c}_{\mathrm{j}}$ are on a different positive diagonal, so we must have $\mathrm{k} \geq$ $2 n-2 k-1$ and hence $k \geq(2 n-1) / 3$.

Let (i,j) denote the cell in row $\mathrm{i}$, col $\mathrm{j}$. For $\mathrm{n}=3 \mathrm{~m}-1$, put coins in (m,1), $(\mathrm{m}-1,2),(m-2,3), \ldots$, $(1, m)$ and in $(2 m-1, m+1),(2 m-2, m+2), \ldots,(m+1,2 m-1)$. It is easy to check that this works. For $\mathrm{n}=3 \mathrm{~m}$, put an additional coin in ( $2 \mathrm{~m}, 2 \mathrm{~m}$ ), it is easy to check that works. For $\mathrm{n}=3 \mathrm{~m}+1$ we can use the same arrangement as for $3 m+2$.

![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-24.jpg?height=382&width=374&top_left_y=1945&top_left_x=264)


$\mathrm{f}$ is a function defined on the positive integers with positive integer values. Use $\mathrm{f}^{\mathrm{m}}(\mathrm{n})$ to mean $\mathrm{f}(\mathrm{f}(\ldots \mathrm{f}(\mathrm{n}) \ldots))=.\mathrm{n}$ where $\mathrm{f}$ is taken $\mathrm{m}$ times, so that $\mathrm{f}^{2}(\mathrm{n})=\mathrm{f}(\mathrm{f}(\mathrm{n}))$, for example. Find the largest possible $0<k<1$ such that for some function $\mathrm{f}$, we have $\mathrm{f}^{\mathrm{m}}(\mathrm{n}) \neq \mathrm{n}$ for $\mathrm{m}=1,2, \ldots$, $[\mathrm{kn}]$, but $\mathrm{f}^{\mathrm{m}}(\mathrm{n})=\mathrm{n}$ for some $\mathrm{m}$ (which may depend on $\left.\mathrm{n}\right)$.

## Answer

we can get $\mathrm{k}$ arbitrarily close to 1

Solution:
The basic idea is to take a block of integers $m+1, m+2, \ldots, M$ and to define $f(m+1)=m+2$, $f(m+2)=m+3, \ldots, f(M-1)=M, f(M)=m+1$. Then for any integer $h$ in the block we have $f^{n}(h)$ $\neq \mathrm{h}$ for $\mathrm{n}=1,2, \ldots, \mathrm{M}-\mathrm{m}-1$ and $\mathrm{f}^{\mathrm{M}-\mathrm{m}}(\mathrm{h})=\mathrm{h}$. Note that the ratio $(\mathrm{M}-\mathrm{m}) / \mathrm{h}$ is worst (smallest) for $\mathrm{h}=\mathrm{M}$.

For example, take the first block to be $1,2, \ldots, \mathrm{N}$, the second block to be $\mathrm{N}+1, \ldots, \mathrm{N}^{2}$, the third block, $N^{2}+1, \ldots, N^{3}$ and so on. Then for any integer $n$ we have $f^{\mathrm{m}}(\mathrm{n}) \neq \mathrm{n}$ for $\mathrm{m}<\mathrm{kn}$ where $\mathrm{k}=1-1 / \mathrm{N}$.


Find the smallest positive integer $\mathrm{n}$ so that a cube with side $\mathrm{n}$ can be divided into 1996 cubes each with side a positive integer.

Solution:
Answer: 13.

Divide all the cubes into unit cubes. Then the 1996 cubes must each contain at least one unit cube, so the large cube contains at least 1996 unit cubes. But $12^{3}=1728<1996<2197=13^{3}$, so it is certainly not possible for $\mathrm{n}<13$.

It can be achieved with 13 by $1.5^{3}+11.2^{3}+1984.1^{3}=13^{3}$ (actually packing the cubes together to form a $13 \times 13 \times 13$ cube is trivial since there are so many unit cubes).

## Problem 2

$\mathrm{M}$ is the midpoint of the median $\mathrm{AD}$ of the triangle $\mathrm{ABC}$. The ray $\mathrm{BM}$ meets $\mathrm{AC}$ at $N$. Show that $\mathrm{AB}$ is tangent to the circumcircle of $\mathrm{NBC}$ iff $\mathrm{BM} / \mathrm{BN}=(\mathrm{BC} / \mathrm{BN})^{2}$.


Solution:
![](https://cdn.mathpix.com/cropped/2024_05_09_1d234a78607cd781abe6g-26.jpg?height=603&width=551&top_left_y=264&top_left_x=250)

Applying Menelaus to the triangle $\mathrm{ADC}$, we have ( $\mathrm{AM} / \mathrm{MD})(\mathrm{BD} / \mathrm{DC})(\mathrm{CN} / \mathrm{NA})=1$, so $(\mathrm{CN} / \mathrm{NA})=2$. Hence $\mathrm{AN} / \mathrm{AC}=1 / 3$. Applying Menelaus to the triangle $\mathrm{BNC}$, we have $(\mathrm{BM} / \mathrm{MN})(\mathrm{AN} / \mathrm{AC})(\mathrm{CD} / \mathrm{DB})=1$, so $\mathrm{BM} / \mathrm{MN}=3$. That is true irrespective of whether $\mathrm{AB}$ is tangent to the circle NBC.

If $\mathrm{AB}$ is tangent, then $\mathrm{AB}^{2}=\mathrm{AN} \cdot \mathrm{AC}=1 / 3 \mathrm{AC}$. Also angle $\mathrm{ABN}=$ angle $\mathrm{BCN}$, so triangles $\mathrm{ANB}$ and $\mathrm{ABC}$ are similar. Hence $\mathrm{BC} / \mathrm{BN}=\mathrm{AC} / \mathrm{AB}$. Hence $(\mathrm{BC} / \mathrm{BN})^{2}=3=\mathrm{BM} / \mathrm{BN}$.

Conversely, if $(\mathrm{BC} / \mathrm{BN})^{2}=\mathrm{BM} / \mathrm{BN}$, then $(\mathrm{BC} / \mathrm{BN})^{2}=3$.

Now applying the cosine formula to $\mathrm{AMN}$ and $\mathrm{AMB}$ and using $\cos \mathrm{AMN}+\cos \mathrm{AMB}=0$, we have $\left(3 \mathrm{AN}^{2}-3 \mathrm{AM}^{2}-3 \mathrm{MN}^{2}\right)+\left(\mathrm{AB}^{2}-\mathrm{AM}^{2}-\mathrm{BM}^{2}\right)=0$, so $A \mathrm{~B}^{2}+\mathrm{AC}^{2} / 3=\mathrm{AD}^{2}+3 / 4 \mathrm{BN}^{2}$. Similarly from triangles $\mathrm{ADC}$ and $\mathrm{ADB}$ we get $\mathrm{AB}^{2}+\mathrm{AC}^{2}=2 \mathrm{AD}^{2}+\mathrm{BC}^{2} / 2$. So using $\mathrm{BN}^{2}=$ $\mathrm{BC}^{2} / 3$ we get $2 \mathrm{AB}^{2}+2 / 3 \mathrm{AC}^{2}=A B^{2}+\mathrm{AC}^{2}$ and hence $(\mathrm{AC} / \mathrm{AB})^{2}=3=(\mathrm{BC} / \mathrm{BN})^{2}$. So $\mathrm{AC} / \mathrm{AB}$ $=\mathrm{BC} / \mathrm{BN}$. Note that is not enough to conclude that triangles $\mathrm{ABC}$ and $\mathrm{BNC}$ are similar, because the common angle $\mathrm{C}$ is not between $\mathrm{AC}$ and $\mathrm{AB}$. However, we have $\mathrm{AN} / \mathrm{AB}=(1 / 3)$ $\mathrm{AC} / \mathrm{AB}=\mathrm{AB} / \mathrm{AC}$, so $\mathrm{AB}^{2}=\mathrm{AN} \cdot \mathrm{AC}$, so $\mathrm{AB}$ is tangent to the circle NBC.


$\mathrm{n}=\mathrm{k}^{2}-\mathrm{k}+1$, where $\mathrm{k}$ is a prime plus one. Show that we can color some squares of an $\mathrm{n} \mathrm{x}$ board black so that each row and column has exactly $\mathrm{k}$ black squares, but there is no rectangle with sides parallel to the sides of the board which has its four corner squares black.

Solution:
We can regard the rows as lines and the columns as points. Black squares denote incidence. So line 3 contains point 4 iff square $(3,4)$ is black. The condition about rectangles then means that there is at most one line through two distinct points.

Suppose we take the points to be ( $\mathrm{a}, \mathrm{b}, \mathrm{c})$, where $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are residues $\bmod \mathrm{p}$, not all zero, and the coordinates are homogeneous, so that we regard (a, b, c), $(2 a, 2 b, 2 c), \ldots,((p-1) a,(p-1) b$, $(\mathrm{p}-1) \mathrm{c})$ as the same point. That gives $\left(\mathrm{p}^{3}-1\right) /(\mathrm{p}-1)=\mathrm{p}^{2}+\mathrm{p}+1$ points, which is the correct number.

We can take lines to be $l \mathrm{x}+m \mathrm{y}+n \mathrm{z}=0$, where the point is $(\mathrm{x}, \mathrm{y}, \mathrm{z})$. In other words, the lines are also triples $(l, m, n)$, with $l, m, n$ residues mod p, not all zero and $(l, m, n),(2 l, 2 m, 2 n), \ldots$, $((p-1) l,(p-1) m,(p-1) n)$ representing the same line.

One way of writing the points is $\mathrm{p}^{2}$ of the form $(\mathrm{a}, \mathrm{b}, 1), \mathrm{p}$ of the form $(\mathrm{a}, 1,0)$ and lastly $(1,0$, $0)$ ) Similarly for the lines. We must show that (1) each point is on $p+1$ lines (so each column has $\mathrm{p}+1$ black squares), (2) each line has $\mathrm{p}+1$ points (so each row has $\mathrm{p}+1$ black squares, (3) two lines meet in just one point (so no rectangles).

(1): Consider the point $\mathrm{P}(\mathrm{a}, \mathrm{b}, 1)$ with a non-zero. Then for any $m$, there is a unique $l$ such that $l \mathrm{a}+m \mathrm{~b}+1.1=0$, so there are $\mathrm{p}$ lines of the form $(l, m, 1)$ which contain P. Similarly, there is a unique $l$ such that $l \mathrm{a}+1 \mathrm{~b}+0.1=0$, so one line of the form $(l, 1,0)$ contains $\mathrm{P}$. The line $(1,0,0)$ does not contain $\mathrm{P}$. So P lies on just $\mathrm{p}+1$ lines. Similarly for $(\mathrm{a}, \mathrm{b}, 1)$ with $\mathrm{b}$ nonzero. The point $(0,0,1)$ does not lie on any lines $(l, m, 1)$, but lies on $(l, 1,0)$ and $(1,0,0)$, so again it lies on $\mathrm{p}+1$ lines.

Consider the point $\mathrm{Q}(\mathrm{a}, 1,0)$ with a non-zero. For any $\mathrm{m}$, there is a unique $l$ such that $\mathrm{Q}$ lies on $(l, m, 0)$. There is also a unique $l$ such that $\mathrm{Q}$ lies on $(l, 1,0)$. $\mathrm{Q}$ does not lie on $(1,0,0)$, so it lies on just p+1 lines. Similarly, the point $(0,1,0)$ lies on the p lines $(l, 0,0)$ and on $(1,0$, $0)$, but no others.

Finally, the point $(1,0,0)$ lies on the $\mathrm{p}$ lines $(0, m, 1)$, the line $(0,1,0)$ and no others. Thus in all cases a point lies on just $\mathrm{p}+1$ lines. The proof of (2) is identical.

(3). Suppose the lines are $(l, m, n)$ and ( $L, M, N$ ). If $l$ and $L$ are non-zero, then we can take the lines as $\left(1, m^{\prime}, n^{\prime}\right)$ and $\left(1, M^{\prime}, N^{\prime}\right)$. So any point (x, y, z) on both satisfies $\mathrm{x}+m^{\prime} \mathrm{y}+n^{\prime} \mathrm{z}=0(*)$ and $\mathrm{x}+M^{\prime} \mathrm{y}+N^{\prime} \mathrm{z}=0$. Subtracting, $\left(m^{\prime}-M^{\prime}\right) \mathrm{y}+\left(n^{\prime}-N^{\prime}\right) \mathrm{z}=0$. The coefficients cannot both be zero, since the lines are distinct. So the ratio $\mathrm{y}: \mathrm{z}$ is fixed. Then *) $^{*}$ gives the ratio $\mathrm{x}: \mathrm{y}$. So the point is uniquely determined. If just one of $l, L$ is non-zero, then we can take the lines as ( 0 , $\left.m^{\prime}, n^{\prime}\right),\left(1, M^{\prime}, N^{\prime}\right)$. We cannot have both $m^{\prime}$ and $n^{\prime}$ zero, so the ratio $\mathrm{y}: \mathrm{z}$ is determined, then the other line determines the ratio $\mathrm{x}: \mathrm{y}$. So again the point is uniquely determined. Finally, suppose $l$ and $L$ are both zero. Then since the lines are distinct y and $\mathrm{z}$ must both be zero. So the unique point on both lines is $(1,0,0)$.


$\mathrm{n}>2$ 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 $1 / a(n+1)+1 /((n+1-a)(n+1))=(n+1-a+a) /(a(n+1-$ $a)(n+1))=1 /(a(n+1-a))$.


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 $\mathrm{n}$.


$A_{1}, A_{2}, \ldots, A_{n}$ 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 $1 / a+1 / b+1 / c+$ $1 / \mathrm{d}=0$.


$k>=1$ 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.


$n>1$ 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.


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 $o(n) / e(n)=\left(2^{n}-1\right) /\left(2^{n}+1\right)$.


$\mathrm{ABC}$ 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.


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 $\left(x_{1}{ }^{2}+x_{2}{ }^{2}+\ldots+x_{1997}{ }^{2}\right) / 4 \leq 9 / 4$.


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 $3 \bmod 4$.


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 $\angle \mathrm{A}=\angle \mathrm{B}$.


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).


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.