APMO/md/en-apmo2004_sol.md

APMO 2004 - Problems and Solutions

Problem 1

Determine all finite nonempty sets $S$ of positive integers satisfying

$$\frac{i+j}{(i,j)}\text{ is an element of }S\text{ for all }i,j\text{ in }S\text{, }\backslash frac\{i+j\}\{(i,\; j)\}\; \backslash text\; \{\; is\; an\; element\; of\; \}\; S\; \backslash text\; \{\; for\; all\; \}\; i,\; j\; \backslash text\; \{\; in\; \}\; S\; \backslash text\; \{,\; \}$$

where $(i, j)$ is the greatest common divisor of $i$ and $j$. Answer: $S={2}$.

Solution

Let $k \in S$. Then $\frac{k+k}{(k, k)}=2$ is in $S$ as well. Suppose for the sake of contradiction that there is an odd number in $S$, and let $k$ be the largest such odd number. Since $(k, 2)=1, \frac{k+2}{(k, 2)}=k+2>k$ is in $S$ as well, a contradiction. Hence $S$ has no odd numbers. Now suppose that $\ell>2$ is the second smallest number in $S$. Then $\ell$ is even and $\frac{\ell+2}{(\ell, 2)}=\frac{\ell}{2}+1$ is in $S$. Since $\ell>2 \Longrightarrow \frac{\ell}{2}+1>2, \frac{\ell}{2}+1 \geq \ell \Longleftrightarrow \ell \leq 2$, a contradiction again. Therefore $S$ can only contain 2 , and $S={2}$ is the only solution.

Problem 2

Let $O$ be the circumcentre and $H$ the orthocentre of an acute triangle $A B C$. Prove that the area of one of the triangles $\mathrm{AOH}, \mathrm{BOH}$ and COH is equal to the sum of the areas of the other two.

Solution 1

Suppose, without loss of generality, that $B$ and $C$ lies in the same side of line $O H$. Such line is the Euler line of $A B C$, so the centroid $G$ lies in this line.

Let $M$ be the midpoint of $B C$. Then the distance between $M$ and the line $O H$ is the average of the distances from $B$ and C to OH , and the sum of the areas of triangles BOH and COH is

$$[BOH]+[COH]=\frac{OH\cdot d(B,OH)}{2}+\frac{OH\cdot d(C,OH)}{2}=\frac{OH\cdot 2d(M,OH)}{2}\mathrm{.}[B\; O\; H]+[C\; O\; H]=\backslash frac\{O\; H\; \backslash cdot\; d(B,\; O\; H)\}\{2\}+\backslash frac\{O\; H\; \backslash cdot\; d(C,\; O\; H)\}\{2\}=\backslash frac\{O\; H\; \backslash cdot\; 2\; d(M,\; O\; H)\}\{2\}\; .$$

Since $A G=2 G M, d(A, O H)=2 d(M, O H)$. Hence

$$[BOH]+[COH]=\frac{OH\cdot d(A,OH)}{2}=[AOH][B\; O\; H]+[C\; O\; H]=\backslash frac\{O\; H\; \backslash cdot\; d(A,\; O\; H)\}\{2\}=[A\; O\; H]$$

and the result follows.

Solution 2

One can use barycentric coordinates: it is well known that

$$\begin{array}{c}{\displaystyle A=(1:0:0),B=(0:1:0),C=(0:0:1),}\\ {\displaystyle O=(\sin 2A:\sin 2B:\sin 2C)\text{ and }H=(\tan A:\tan B:\tan C)\mathrm{.}}\end{array}\backslash begin\{gathered\}\; A=(1:\; 0:\; 0),\; \backslash quad\; B=(0:\; 1:\; 0),\; \backslash quad\; C=(0:\; 0:\; 1),\; \backslash \backslash \; O=(\backslash sin\; 2\; A:\; \backslash sin\; 2\; B:\; \backslash sin\; 2\; C)\; \backslash quad\; \backslash text\; \{\; and\; \}\; \backslash quad\; H=(\backslash tan\; A:\; \backslash tan\; B:\; \backslash tan\; C)\; .\; \backslash end\{gathered\}$$

Then the (signed) area of $A O H$ is proportional to

Adding all three expressions we find that the sum of the signed sums of the areas is a constant times

By multilinearity of the determinant, this sum equals

which contains, in its rows, the coordinates of the centroid, the circumcenter, and the orthocenter. Since these three points lie in the Euler line of $A B C$, the signed sum of the areas is 0 , which means that one of the areas of $A O H, B O H, C O H$ is the sum of the other two areas. Comment: Both solutions can be adapted to prove a stronger result: if the centroid $G$ of triangle $A B C$ belongs to line $X Y$ then one of the areas of triangles $A X Y, B X Y$, and $C X Y$ is equal to the sum of the other two.

Problem 3

Let a set $S$ of 2004 points in the plane be given, no three of which are collinear. Let $\mathcal{L}$ denote the set of all lines (extended indefinitely in both directions) determined by pairs of points from the set. Show that it is possible to colour the points of $S$ with at most two colours, such that for any points $p, q$ of $S$, the number of lines in $\mathcal{L}$ which separate $p$ from $q$ is odd if and only if $p$ and $q$ have the same colour. Note: A line $\ell$ separates two points $p$ and $q$ if $p$ and $q$ lie on opposite sides of $\ell$ with neither point on $\ell$.

Solution

Choose any point $p$ from $S$ and color it, say, blue. Let $n(q, r)$ be the number of lines from $\mathcal{L}$ that separates $q$ and $r$. Then color any other point $q$ blue if $n(p, q)$ is odd and red if $n(p, q)$ is even. Now it remains to show that $q$ and $r$ have the same color if and only if $n(q, r)$ is odd for all $q \neq p$ and $r \neq p$, which is equivalent to proving that $n(p, q)+n(p, r)+n(q, r)$ is always odd. For this purpose, consider the seven numbered regions defined by lines $p q$, $p r$, and $q r$ :

Any line that do not pass through any of points $p, q, r$ meets the sides $p q, q r, p r$ of triangle $p q r$ in an even number of points (two sides or no sides), so these lines do not affect the parity of $n(p, q)+n(p, r)+n(q, r)$. Hence the only lines that need to be considered are the ones that pass through one of vertices $p, q, r$ and cuts the opposite side in the triangle $p q r$. Let $n_{i}$ be the number of points in region $i, p, q$, and $r$ excluded, as depicted in the diagram. Then the lines through $p$ that separate $q$ and $r$ are the lines passing through $p$ and points from regions 1,4 , and 7 . The same applies for $p, q$ and regions 2,5 , and 7 ; and $p, r$ and regions 3,6 , and 7. Therefore

and the result follows. Comment: The problem statement is also true if 2004 is replaced by any even number and is not true if 2004 is replaced by any odd number greater than 1.

Problem 4

For a real number $x$, let $\lfloor x\rfloor$ stand for the largest integer that is less than or equal to $x$. Prove that

$$\lfloor \frac{(n-1)!}{n(n+1)}\rfloor \backslash left\backslash lfloor\backslash frac\{(n-1)!\}\{n(n+1)\}\backslash right\backslash rfloor$$

is even for every positive integer $n$.

Solution

Consider four cases:

• $n \leq 5$. Then $\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor=0$ is an even number.
• $n$ and $n+1$ are both composite (in particular, $n \geq 8$ ). Then $n=a b$ and $n+1=c d$ for $a, b, c, d \geq 2$. Moreover, since $n$ and $n+1$ are coprime, $a, b, c, d$ are all distinct and smaller than $n$, and one can choose $a, b, c, d$ such that exactly one of these four numbers is even. Hence $\frac{(n-1)!}{n(n+1)}$ is an integer. As $n \geq 8>6,(n-1)!$ has at least three even factors, so $\frac{(n-1)!}{n(n+1)}$ is an even integer.
• $n \geq 7$ is an odd prime. By Wilson's theorem, $(n-1)!\equiv-1(\bmod n)$, that is, $\frac{(n-1)!+1}{n}$ is an integer, as $\frac{(n-1)!+n+1}{n}=\frac{(n-1)!+1}{n}+1$ is. As before, $\frac{(n-1)!}{n+1}$ is an even integer; therefore $\frac{(n-1)!+n+1}{n+1}=\frac{(n-1)!}{n+1}+1$ is an odd integer. Also, $n$ and $n+1$ are coprime and $n$ divides the odd integer $\frac{(n-1)!+n+1}{n+1}$, so $\frac{(n-1)!+n+1}{n(n+1)}$ is also an odd integer. Then

$$\lfloor \frac{(n-1)!}{n(n+1)}\rfloor =\frac{(n-1)!+n+1}{n(n+1)}-1\backslash left\backslash lfloor\backslash frac\{(n-1)!\}\{n(n+1)\}\backslash right\backslash rfloor=\backslash frac\{(n-1)!+n+1\}\{n(n+1)\}-1$$

is even.

• $n+1 \geq 7$ is an odd prime. Again, since $n$ is composite, $\frac{(n-1)!}{n}$ is an even integer, and $\frac{(n-1)!+n}{n}$ is an odd integer. By Wilson's theorem, $n!\equiv-1(\bmod n+1) \Longleftrightarrow(n-1)!\equiv 1$ $(\bmod n+1)$. This means that $n+1$ divides $(n-1)!+n$, and since $n$ and $n+1$ are coprime, $n+1$ also divides $\frac{(n-1)!+n}{n}$. Then $\frac{(n-1)!+n}{n(n+1)}$ is also an odd integer and

$$\lfloor \frac{(n-1)!}{n(n+1)}\rfloor =\frac{(n-1)!+n}{n(n+1)}-1\backslash left\backslash lfloor\backslash frac\{(n-1)!\}\{n(n+1)\}\backslash right\backslash rfloor=\backslash frac\{(n-1)!+n\}\{n(n+1)\}-1$$

is even.

Problem 5

Prove that

$$(a^2+2)(b^2+2)(c^2+2)\ge 9(ab+bc+ca)\backslash left(a^\{2\}+2\backslash right)\backslash left(b^\{2\}+2\backslash right)\backslash left(c^\{2\}+2\backslash right)\; \backslash geq\; 9(a\; b+b\; c+c\; a)$$

for all real numbers $a, b, c>0$.

Solution 1

Let $p=a+b+c, q=a b+b c+c a$, and $r=a b c$. The inequality simplifies to

$$a^2{b}^2{c}^2+2(a^2{b}^2+b^2{c}^2+c^2{a}^2)+4(a^2+b^2+c^2)+8-9(ab+bc+ca)\ge 0a^\{2\}\; b^\{2\}\; c^\{2\}+2\backslash left(a^\{2\}\; b^\{2\}+b^\{2\}\; c^\{2\}+c^\{2\}\; a^\{2\}\backslash right)+4\backslash left(a^\{2\}+b^\{2\}+c^\{2\}\backslash right)+8-9(a\; b+b\; c+c\; a)\; \backslash geq\; 0$$

Since $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=q^{2}-2 p r$ and $a^{2}+b^{2}+c^{2}=p^{2}-2 q$,

$$r^2+2q^2-4pr+4p^2-8q+8-9q\ge 0r^\{2\}+2\; q^\{2\}-4\; p\; r+4\; p^\{2\}-8\; q+8-9\; q\; \backslash geq\; 0$$

which simplifies to

$$r^2+2q^2+4p^2-17q-4pr+8\ge 0r^\{2\}+2\; q^\{2\}+4\; p^\{2\}-17\; q-4\; p\; r+8\; \backslash geq\; 0$$

Bearing in mind that equality occurs for $a=b=c=1$, which means that, for instance, $p=3 r$, one can rewrite $(I)$ as

$${(r-\frac{p}{3})}^2-\frac{10}{3}pr+\frac{35}{9}{p}^2+2q^2-17q+8\ge 0\backslash left(r-\backslash frac\{p\}\{3\}\backslash right)^\{2\}-\backslash frac\{10\}\{3\}\; p\; r+\backslash frac\{35\}\{9\}\; p^\{2\}+2\; q^\{2\}-17\; q+8\; \backslash geq\; 0$$

Since $(a b-b c)^{2}+(b c-c a)^{2}+(c a-a b)^{2} \geq 0$ is equivalent to $q^{2} \geq 3 p r$, rewrite $(I I)$ as

$${(r-\frac{p}{3})}^2+\frac{10}{9}(q^2-3pr)+\frac{35}{9}{p}^2+\frac{8}{9}{q}^2-17q+8\ge 0\backslash left(r-\backslash frac\{p\}\{3\}\backslash right)^\{2\}+\backslash frac\{10\}\{9\}\backslash left(q^\{2\}-3\; p\; r\backslash right)+\backslash frac\{35\}\{9\}\; p^\{2\}+\backslash frac\{8\}\{9\}\; q^\{2\}-17\; q+8\; \backslash geq\; 0$$

Finally, $a=b=c=1$ implies $q=3$; then rewrite (III) as

$${(r-\frac{p}{3})}^2+\frac{10}{9}(q^2-3pr)+\frac{35}{9}(p^2-3q)+\frac{8}{9}(q-3{)}^2\ge 0\backslash left(r-\backslash frac\{p\}\{3\}\backslash right)^\{2\}+\backslash frac\{10\}\{9\}\backslash left(q^\{2\}-3\; p\; r\backslash right)+\backslash frac\{35\}\{9\}\backslash left(p^\{2\}-3\; q\backslash right)+\backslash frac\{8\}\{9\}(q-3)^\{2\}\; \backslash geq\; 0$$

This final inequality is true because $q^{2} \geq 3 p r$ and $p^{2}-3 q=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \geq 0$.

Solution 2

We prove the stronger inequality

$$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c{)}^2\backslash left(a^\{2\}+2\backslash right)\backslash left(b^\{2\}+2\backslash right)\backslash left(c^\{2\}+2\backslash right)\; \backslash geq\; 3(a+b+c)^\{2\}$$

which implies the proposed inequality because $(a+b+c)^{2} \geq 3(a b+b c+c a)$ is equivalent to $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0$, which is immediate. The inequality $(*)$ is equivalent to

$$((b^2+2)(c^2+2)-3)a^2-6(b+c)a+2(b^2+2)(c^2+2)-3(b+c{)}^2\ge 0\backslash left(\backslash left(b^\{2\}+2\backslash right)\backslash left(c^\{2\}+2\backslash right)-3\backslash right)\; a^\{2\}-6(b+c)\; a+2\backslash left(b^\{2\}+2\backslash right)\backslash left(c^\{2\}+2\backslash right)-3(b+c)^\{2\}\; \backslash geq\; 0$$

Seeing this inequality as a quadratic inequality in $a$ with positive leading coefficient $\left(b^{2}+2\right)\left(c^{2}+\right.$ 2) $-3=b^{2} c^{2}+2 b^{2}+2 c^{2}+1$, it suffices to prove that its discriminant is non-positive, which is equivalent to

$$(3(b+c){)}^2-((b^2+2)(c^2+2)-3)(2(b^2+2)(c^2+2)-3(b+c{)}^2)\le 0(3(b+c))^\{2\}-\backslash left(\backslash left(b^\{2\}+2\backslash right)\backslash left(c^\{2\}+2\backslash right)-3\backslash right)\backslash left(2\backslash left(b^\{2\}+2\backslash right)\backslash left(c^\{2\}+2\backslash right)-3(b+c)^\{2\}\backslash right)\; \backslash leq\; 0$$

This simplifies to

$$-2(b^2+2)(c^2+2)+3(b+c{)}^2+6\le 0-2\backslash left(b^\{2\}+2\backslash right)\backslash left(c^\{2\}+2\backslash right)+3(b+c)^\{2\}+6\; \backslash leq\; 0$$

Now we look $(* *)$ as a quadratic inequality in $b$ with negative leading coefficient $-2 c^{2}-1$ :

$$(-2c^2-1)b^2+6cb-c^2-2\le 0\backslash left(-2\; c^\{2\}-1\backslash right)\; b^\{2\}+6\; c\; b-c^\{2\}-2\; \backslash leq\; 0$$

If suffices to show that the discriminant of $(* *)$ is non-positive, which is equivalent to

$$9c^2-(2c^2+1)(c^2+2)\le 09\; c^\{2\}-\backslash left(2\; c^\{2\}+1\backslash right)\backslash left(c^\{2\}+2\backslash right)\; \backslash leq\; 0$$

It simplifies to $-2\left(c^{2}-1\right)^{2} \leq 0$, which is true. The equality occurs for $c^{2}=1$, that is, $c=1$, for which $b=\frac{6 c}{2\left(2 c^{2}+1\right)}=1$, and $a=\frac{6(b+c)}{2\left(\left(b^{2}+2\right)\left(c^{2}+2\right)-3\right)}=1$.

Solution 3

Let $A, B, C$ angles in $(0, \pi / 2)$ such that $a=\sqrt{2} \tan A, b=\sqrt{2} \tan B$, and $c=\sqrt{2} \tan C$. Then the inequality is equivalent to

$$4{\sec }^{2}A{\sec }^{2}B{\sec }^{2}C\ge 9(\tan A\tan B+\tan B\tan C+\tan C\tan A)4\; \backslash sec\; ^\{2\}\; A\; \backslash sec\; ^\{2\}\; B\; \backslash sec\; ^\{2\}\; C\; \backslash geq\; 9(\backslash tan\; A\; \backslash tan\; B+\backslash tan\; B\; \backslash tan\; C+\backslash tan\; C\; \backslash tan\; A)$$

Substituting $\sec x=\frac{1}{\cos x}$ for $x \in{A, B, C}$ and clearing denominators, the inequality is equivalent to

$$\cos A\cos B\cos C(\sin A\sin B\cos C+\cos A\sin B\sin C+\sin A\cos B\sin C)\le \frac{4}{9}\backslash cos\; A\; \backslash cos\; B\; \backslash cos\; C(\backslash sin\; A\; \backslash sin\; B\; \backslash cos\; C+\backslash cos\; A\; \backslash sin\; B\; \backslash sin\; C+\backslash sin\; A\; \backslash cos\; B\; \backslash sin\; C)\; \backslash leq\; \backslash frac\{4\}\{9\}$$

Since

we rewrite our inequality as

$$\cos A\cos B\cos C(\cos A\cos B\cos C-\cos (A+B+C))\le \frac{4}{9}\backslash cos\; A\; \backslash cos\; B\; \backslash cos\; C(\backslash cos\; A\; \backslash cos\; B\; \backslash cos\; C-\backslash cos\; (A+B+C))\; \backslash leq\; \backslash frac\{4\}\{9\}$$

The cosine function is concave down on $(0, \pi / 2)$. Therefore, if $\theta=\frac{A+B+C}{3}$, by the AM-GM inequality and Jensen's inequality,

$$\cos A\cos B\cos C\le {\left(\frac{\cos A+\cos B+\cos C}{3}\right)}^3\le {\cos }^3\frac{A+B+C}{3}={\cos }^3\theta \backslash cos\; A\; \backslash cos\; B\; \backslash cos\; C\; \backslash leq\backslash left(\backslash frac\{\backslash cos\; A+\backslash cos\; B+\backslash cos\; C\}\{3\}\backslash right)^\{3\}\; \backslash leq\; \backslash cos\; ^\{3\}\; \backslash frac\{A+B+C\}\{3\}=\backslash cos\; ^\{3\}\; \backslash theta$$

Therefore, since $\cos A \cos B \cos C-\cos (A+B+C)=\sin A \sin B \cos C+\cos A \sin B \sin C+$ $\sin A \cos B \sin C>0$, and recalling that $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$, $\cos A \cos B \cos C(\cos A \cos B \cos C-\cos (A+B+C)) \leq \cos ^{3} \theta\left(\cos ^{3} \theta-\cos 3 \theta\right)=3 \cos ^{4} \theta\left(1-\cos ^{2} \theta\right)$. Finally, by AM-GM (notice that $1-\cos ^{2} \theta=\sin ^{2} \theta>0$ ), $3 \cos ^{4} \theta\left(1-\cos ^{2} \theta\right)=\frac{3}{2} \cos ^{2} \theta \cdot \cos ^{2} \theta\left(2-2 \cos ^{2} \theta\right) \leq \frac{3}{2}\left(\frac{\cos ^{2} \theta+\cos ^{2} \theta+\left(2-2 \cos ^{2} \theta\right)}{3}\right)^{3}=\frac{4}{9}$, and the result follows.