APMO/md/en-apmo2003_sol.md

XV APMO: Solutions and Marking Schemes

1. Let $a, b, c, d, e, f$ be real numbers such that the polynomial

$$p(x)=x^8-4x^7+7x^6+a{x}^5+b{x}^4+c{x}^3+d{x}^2+ex+fp(x)=x^\{8\}-4\; x^\{7\}+7\; x^\{6\}+a\; x^\{5\}+b\; x^\{4\}+c\; x^\{3\}+d\; x^\{2\}+e\; x+f$$

factorises into eight linear factors $x-x_{i}$, with $x_{i}>0$ for $i=1,2, \ldots, 8$. Determine all possible values of $f$.

Solution.

From

$$x^8-4x^7+7x^6+a{x}^5+b{x}^4+c{x}^3+d{x}^2+ex+f=(x-x_1)(x-x_2)\dots (x-x_8)x^\{8\}-4\; x^\{7\}+7\; x^\{6\}+a\; x^\{5\}+b\; x^\{4\}+c\; x^\{3\}+d\; x^\{2\}+e\; x+f=\backslash left(x-x\_\{1\}\backslash right)\backslash left(x-x\_\{2\}\backslash right)\; \backslash ldots\backslash left(x-x\_\{8\}\backslash right)$$

we have

$$\sum _{i=1}^8{x}_i=4\text{ and }\sum x_i{x}_j=7\backslash sum\_\{i=1\}^\{8\}\; x\_\{i\}=4\; \backslash quad\; \backslash text\; \{\; and\; \}\; \backslash quad\; \backslash sum\; x\_\{i\}\; x\_\{j\}=7$$

where the second sum is over all pairs $(i, j)$ of integers where $1 \leq i<j \leq 8$. Since this sum can also be written

$$\frac{1}{2}[{\left(\sum _{i=1}^8{x}_i\right)}^2-\sum _{i=1}^8{x}_i^2]\backslash frac\{1\}\{2\}\backslash left[\backslash left(\backslash sum\_\{i=1\}^\{8\}\; x\_\{i\}\backslash right)^\{2\}-\backslash sum\_\{i=1\}^\{8\}\; x\_\{i\}^\{2\}\backslash right]$$

we get

$$14={\left(\sum _{i=1}^8{x}_i\right)}^2-\sum _{i=1}^8{x}_i^2=16-\sum _{i=1}^8{x}_i^{2}14=\backslash left(\backslash sum\_\{i=1\}^\{8\}\; x\_\{i\}\backslash right)^\{2\}-\backslash sum\_\{i=1\}^\{8\}\; x\_\{i\}^\{2\}=16-\backslash sum\_\{i=1\}^\{8\}\; x\_\{i\}^\{2\}$$

so

$$\sum _{i=1}^8{x}_i^2=2\text{ while }\sum _{i=1}^8{x}_i=4.[3\text{ marks }]\backslash sum\_\{i=1\}^\{8\}\; x\_\{i\}^\{2\}=2\; \backslash quad\; \backslash text\; \{\; while\; \}\; \backslash quad\; \backslash sum\_\{i=1\}^\{8\}\; x\_\{i\}=4\; .\; \backslash quad[3\; \backslash text\; \{\; marks\; \}]$$

Now

$$\sum _{i=1}^8{(2x_i-1)}^2=4\sum _{i=1}^8{x}_i^2-4\sum _{i=1}^8{x}_i+8=4(2)-4(4)+8=0\backslash sum\_\{i=1\}^\{8\}\backslash left(2\; x\_\{i\}-1\backslash right)^\{2\}=4\; \backslash sum\_\{i=1\}^\{8\}\; x\_\{i\}^\{2\}-4\; \backslash sum\_\{i=1\}^\{8\}\; x\_\{i\}+8=4(2)-4(4)+8=0$$

which forces $x_{i}=1 / 2$ for all $i$. [3 marks] Therefore

$$f=\prod _{i=1}^8{x}_i={\left(\frac{1}{2}\right)}^8=\frac{1}{256}\mathrm{.}[1mx]f=\backslash prod\_\{i=1\}^\{8\}\; x\_\{i\}=\backslash left(\backslash frac\{1\}\{2\}\backslash right)^\{8\}=\backslash frac\{1\}\{256\}\; .\; \backslash quad[1\; m\; x]$$

Alternate solution: After obtaining (1) [3 marks], use Cauchy's inequality to get

$$16={(x_1\cdot 1+x_2\cdot 1+\cdots +x_8\cdot 1)}^2\le (x_1^{2-}+x_2^2+\cdots +x_8^2)(1^2+1^2+\cdots +1^2)=8\cdot 2=1616=\backslash left(x\_\{1\}\; \backslash cdot\; 1+x\_\{2\}\; \backslash cdot\; 1+\backslash cdots+x\_\{8\}\; \backslash cdot\; 1\backslash right)^\{2\}\; \backslash leq\backslash left(x\_\{1\}^\{2-\}+x\_\{2\}^\{2\}+\backslash cdots+x\_\{8\}^\{2\}\backslash right)\backslash left(1^\{2\}+1^\{2\}+\backslash cdots+1^\{2\}\backslash right)=8\; \backslash cdot\; 2=16$$

or the power mean inequality to get

$$\frac{1}{2}=\frac{1}{8}\sum _{i=1}^8{x}_i\le {\left(\frac{1}{8}\sum _{i=1}^8{x}_i^2\right)}^{1\mathrm{/}2}=\frac{1}{2}\mathrm{.}[2\text{ marks }]\backslash frac\{1\}\{2\}=\backslash frac\{1\}\{8\}\; \backslash sum\_\{i=1\}^\{8\}\; x\_\{i\}\; \backslash leq\backslash left(\backslash frac\{1\}\{8\}\; \backslash sum\_\{i=1\}^\{8\}\; x\_\{i\}^\{2\}\backslash right)^\{1\; /\; 2\}=\backslash frac\{1\}\{2\}\; .\; \backslash quad[2\; \backslash text\; \{\; marks\; \}]$$

Either way, equality must hold, which can only happen if all the terms $x_{i}$ are equal, that is, if $x_{i}=1 / 2$ for all $i$. [1 mark] Thus $f=1 / 256$ as above. [ 1 mark] 2. Suppose $A B C D$ is a square piece of cardboard with side length $a$. On a plane are two parallel lines $\ell_{1}$ and $\ell_{2}$, which are also $a$ units apart. The square $A B C D$ is placed on the plane so that sides $A B$ and $A D$ intersect $\ell_{1}$ at $E$ and $F$ respectively. Also, sides $C B$ and $C D$ intersect $\ell_{2}$ at $G$ and $H$ respectively. Let the perimeters of $\triangle A E F$ and $\triangle C G H$ be $m_{1}$ and $m_{2}$ respectively. Prove that no matter how the square was placed, $m_{1}+m_{2}$ remains constant.

Solution 1. Let $E H$ intersect $F G$ at $O$. The distance from $G$ to line $F D$ and line $E F$ are both $a$. So $F G$ bisects $\angle E F D$. Similarly, $E H$ bisects $\angle B E F$. So $O$ is an excentre of $\triangle A E F$. Similarly, $O$ is an excentre of $\triangle C G H$. [2 marks] Construct these excircles with centre $O$. Let $M, N, P, Q$ be on sides $A B, B C, C D, D A$ respectively, where these excircles touch the square. Then $O M \perp A B, O N \perp B C, O P \perp C D$, and $O Q \perp D A$. Since $A B | C D$ and $A D | B C, M, O, P$ are collinear and $N, O, Q$ are collinear. Now $M P=N Q=a$. [2 marks] Using the fact that the two tangents from a point to a circle have the same length, we get $E F=E M+F Q$ and $G H=G N+H P$. [1 mark] Then

$$m_1=AE+AF+EF=AE+AF+(EM+FQ)=AM+AQ=OQ+OMm\_\{1\}=A\; E+A\; F+E\; F=A\; E+A\; F+(E\; M+F\; Q)=A\; M+A\; Q=O\; Q+O\; M$$

and

$$m_2=CG+CH+GH=CG+CH+(GN+HP)=CN+CP=OP+ON\mathrm{.}[1\text{ mark }]m\_\{2\}=C\; G+C\; H+G\; H=C\; G+C\; H+(G\; N+H\; P)=C\; N+C\; P=O\; P+O\; N\; .\; \backslash quad[1\; \backslash text\; \{\; mark\; \}]$$

Therefore

$$m_1+m_2=(OQ+OM)+(OP+ON)=MP+NQ=2a\mathrm{.}[1\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{k}]m\_\{1\}+m\_\{2\}=(O\; Q+O\; M)+(O\; P+O\; N)=M\; P+N\; Q=2\; a\; .\; \backslash quad[1\; \backslash mathrm\{mark\}]$$

Solution 2.

Extend $A B$ to $I$ and $D C$ to $J$ so that $A E=B I=C J$. Let $\ell_{2}$ intersect $I J$ at $M$, and let $K$ lie on $I J$ so that $G K \perp I J$. Then, since $A E=G K, \triangle A E F$ and $\triangle K G M$ are congruent. [1 mark] Thus, since $G K=C J$ and $G C=K J$,

$$m_1+m_2=\mathrm{perimeter}(KGM)+\mathrm{perimeter}(CGH)=\mathrm{perimeter}(HMJ)\mathrm{.}[\mathbf{2}\text{ marks }]m\_\{1\}+m\_\{2\}=\backslash operatorname\{perimeter\}(K\; G\; M)+\backslash operatorname\{perimeter\}(C\; G\; H)=\backslash operatorname\{perimeter\}(H\; M\; J)\; .\; \backslash quad[\backslash mathbf\{2\}\; \backslash text\; \{\; marks\; \}]$$

Let $L$ lie on $C D$ so that $E L \perp C D$. Then a circle with centre $E$ and radius $a$ will touch $D C$ at $L, I J$ at $I$, and the interior of $H M$ at some point $N$, so

$$\mathrm{perimeter}(HMJ)=JH+(HN+NM)+JM=(JH+HL)+(MI+JM)=JL+IJ=a+a=2a\backslash operatorname\{perimeter\}(H\; M\; J)=J\; H+(H\; N+N\; M)+J\; M=(J\; H+H\; L)+(M\; I+J\; M)=J\; L+I\; J=a+a=2\; a$$

[4 marks] Thus $m_{1}+m_{2}=2 a$. Solution 3. Without loss of generality, assume the square has side $a=1$. Let $\theta$ be the acute angle between $\ell_{1}$ (or $\ell_{2}$ ) and the sides $A B$ and $C D$ of the square. Then, letting $E F=x$ and $G H=y$, we have

$$EA=x\cos \theta ,AF=x\sin \theta ,CH=y\cos \theta ,CG=y\sin \theta E\; A=x\; \backslash cos\; \backslash theta,\; \backslash quad\; A\; F=x\; \backslash sin\; \backslash theta,\; \backslash quad\; C\; H=y\; \backslash cos\; \backslash theta,\; \backslash quad\; C\; G=y\; \backslash sin\; \backslash theta$$

Thus

$$m_1+m_2=(x+y)(\sin \theta +\cos \theta +1)\mathrm{.}[2\text{ marks }]m\_\{1\}+m\_\{2\}=(x+y)(\backslash sin\; \backslash theta+\backslash cos\; \backslash theta+1)\; .\; \backslash quad[2\; \backslash text\; \{\; marks\; \}]$$

Draw lines parallel to $\ell_{1}, \ell_{2}$ through $A$ and $C$ respectively. The distance between these lines is $\sin \theta+\cos \theta$ [1 mark], as can be seen by drawing a mutual perpendicular to these lines through $B$, say. Also, the altitudes from $A$ to $E F$ and from $C$ to $G H$ have lengths $x \sin \theta \cos \theta$ and $y \sin \theta \cos \theta$ respectively [ 1 mark]. Therefore the distance between $\ell_{1}$ and $\ell_{2}$ must be

$$(\sin \theta +\cos \theta )-x\sin \theta \cos \theta -y\sin \theta \cos \theta (\backslash sin\; \backslash theta+\backslash cos\; \backslash theta)-x\; \backslash sin\; \backslash theta\; \backslash cos\; \backslash theta-y\; \backslash sin\; \backslash theta\; \backslash cos\; \backslash theta$$

But we are given that this distance is $a=1$, so

$$(x+y)\sin \theta \cos \theta +1=\sin \theta +\cos \theta (x+y)\; \backslash sin\; \backslash theta\; \backslash cos\; \backslash theta+1=\backslash sin\; \backslash theta+\backslash cos\; \backslash theta$$

or

$$x+y=\frac{\sin \theta +\cos \theta -1}{\sin \theta \cos \theta }\cdot [1\text{ mark }]x+y=\backslash frac\{\backslash sin\; \backslash theta+\backslash cos\; \backslash theta-1\}\{\backslash sin\; \backslash theta\; \backslash cos\; \backslash theta\}\; \backslash cdot\; \backslash quad[1\; \backslash text\; \{\; mark\; \}]$$

Therefore, by (1),

3. Let $k \geq 14$ be an integer, and let $p_{k}$ be the largest prime number which is strictly less than $k$. You may assume that $p_{k} \geq 3 k / 4$. Let $n$ be a composite integer. Prove: (a) if $n=2 p_{k}$, then $n$ does not divide $(n-k)$ !; (b) if $n>2 p_{k}$, then $n$ divides $(n-k)$ !.

Solution. (a) Note that $n-k=2 p_{k}-k<2 p_{k}-p_{k}=p_{k}$, so $p_{k} \nmid(n-k)$ !, so $2 p_{k} \nless(n-k)$ !. [1 mark] (b) Note that $n>2 p_{k} \geq 3 k / 2$ implies $k<2 n / 3$, so $n-k>n / 3$. So if we can find integers $a, b \geq 3$ such that $n=a b$ and $a \neq b$, then both $a$ and $b$ will appear separately in the product $(n-k)!=1 \times 2 \times \cdots \times(n-k)$, which means $n \mid(n-k)!$. Observe that $k \geq 14$ implies $p_{k} \geq 13$, so that $n>2 p_{k} \geq 26$.

If $n=2^{\alpha}$ for some integer $\alpha \geq 5$, then take $a=2^{2}, b=2^{\alpha-2}$. [ 1 mark] Otherwise, since $n \geq 26>16$, we can take $a$ to be an odd prime factor of $n$ and $b=n / a$ [1 mark], unless $b<3$ or $b=a$.

Case (i): $b<3$. Since $n$ is composite, this means $b=2$, so that $2 a=n>2 p_{k}$. As $a$ is a prime number and $p_{k}$ is the largest prime number which is strictly less than $k$, it follows that $a \geq k$. From $n-k=2 a-k \geq$ $2 a-a=a>2$ we see that $n=2 a$ divides into $(n-k)$. [ 2 marks]

Case (ii): $b=a$. Then $n=a^{2}$ and $a>6$ since $n \geq 26$. Thus $n-k>n / 3=a^{2} / 3>2 a$, so that both $a$ and $2 a$ appear among ${1,2, \ldots, n-k}$. Hence $n=a^{2}$ divides into $(n-k)!$. [2 marks] 4. Let $a, b, c$ be the sides of a triangle, with $a+b+c=1$, and let $n \geq 2$ be an integer. Show that

Solution.

Without loss of generality, assume $a \leq b \leq c$. As $a+b>c$, we have

$$\frac{\sqrt[n]{2}}{2}=\frac{\sqrt[n]{2}}{2}(a+b+c)>\frac{\sqrt[n]{2}}{2}(c+c)=\sqrt[n]{2c^n}\ge \sqrt[n]{b^n+c^n}[2\text{ marks }]\backslash frac\{\backslash sqrt[n]\{2\}\}\{2\}=\backslash frac\{\backslash sqrt[n]\{2\}\}\{2\}(a+b+c)>\backslash frac\{\backslash sqrt[n]\{2\}\}\{2\}(c+c)=\backslash sqrt[n]\{2\; c^\{n\}\}\; \backslash geq\; \backslash sqrt[n]\{b^\{n\}+c^\{n\}\}\; \backslash quad\; \backslash quad[2\; \backslash text\; \{\; marks\; \}]$$

As $a \leq c$ and $n \geq 2$, we have

Thus

Likewise

Adding (1), (2) and (3), we get

5. Given two positive integers $m$ and $n$, find the smallest positive integer $k$ such that among any $k$ people, either there are $2 m$ of them who form $m$ pairs of mutually acquainted people or there are $2 n$ of them forming $n$ pairs of mutually unacquainted people.

Solution.

Let the smallest positive integer $k$ satisfying the condition of the problem be denoted $r(m, n)$. We shall show that

$$r(m,n)=2(m+n)-\min \{m,n\}-1r(m,\; n)=2(m+n)-\backslash min\; \backslash \{m,\; n\backslash \}-1$$

Observe that, by symmetry, $r(m, n)=r(n, m)$. Therefore it suffices to consider the case where $m \geq n$, and to prove that

$$r(m,n)=2m+n-1.[1\text{ mark }]r(m,\; n)=2\; m+n-1\; .\; \backslash quad[1\; \backslash text\; \{\; mark\; \}]$$

First we prove that

$$r(m,n)\ge 2m+n-1r(m,\; n)\; \backslash geq\; 2\; m+n-1$$

by an example. Call a group of $k$ people, every two of whom are mutually acquainted, a $k$-clique. Consider a set of $2 m+n-2$ people consisting of a $(2 m-1)$-clique together with an additional $n-1$ people none of whom know anyone else. (Call such people isolated.) Then there are not $2 m$ people forming $m$ mutually acquainted pairs, and there also are not $2 n$ people forming $n$ mutually unacquainted pairs. Thus $r(m, n) \geq$ $(2 m-1)+(n-1)+1=2 m+n-1$ by the definition of $r(m, n)$. [1 mark]

To establish (1), we need to prove that $r(m, n) \leq 2 m+n-1$. To do this, we now show that

$$r(m,n)\le r(m-1,n-1)+3\text{ for all }m\ge n\ge 2r(m,\; n)\; \backslash leq\; r(m-1,\; n-1)+3\; \backslash quad\; \backslash text\; \{\; for\; all\; \}\; m\; \backslash geq\; n\; \backslash geq\; 2$$

Let $G$ be a group of $t=r(m-1, n-1)+3$ people. Notice that

$$t\ge 2(m-1)+(n-1)-1+3=2m+n-1\ge 2m\ge 2nt\; \backslash geq\; 2(m-1)+(n-1)-1+3=2\; m+n-1\; \backslash geq\; 2\; m\; \backslash geq\; 2\; n$$

If $G$ is a $t$-clique, then $G$ contains $2 m$ people forming $m$ mutually acquainted pairs, and if $G$ has only isolated people, then $G$ contains $2 n$ people forming $n$ mutually unacquainted pairs. Otherwise, there are three people in $G$, say $a, b$ and $c$, such that $a, b$ are acquainted but $a, c$ are not. Now consider the group $A$ obtained byremoving $a, b$ and $c$ from $G$. A has $t-3=r(m-1, n-1)$ people, so by the definition of $r(m-1, n-1)$, A either contains $2(m-1)$ people forming $m-1$ mutually acquainted pairs, or else contains $2(n-1)$ people forming $n-1$ mutually unacquainted pairs. In the former case, we add the acquainted pair $a, b$ to $A$ to form $m$ mutually acquainted pairs in $G$. In the latter case, we add the unacquainted pair $a, c$ to $A$ to form $n$ mutually unacquainted pairs in $G$. This proves (2). [3 marks]

Trivially, $r(s, 1)=2 s$ for all $s[\mathbf{1}$ mark], so $r(m, n) \leq 2 m+n-1$ holds whenever $n=1$. Proceeding by induction on $n$, by (2) we obtain

$$r(m,n)\le r(m-1,n-1)+3\le 2(m-1)+(n-1)-1+3=2m+n-1r(m,\; n)\; \backslash leq\; r(m-1,\; n-1)+3\; \backslash leq\; 2(m-1)+(n-1)-1+3=2\; m+n-1$$

which completes the proof. [1 mark] Note. Give an additional 1 mark to any student who gets at most 5 marks by the above marking scheme, but in addition gives a valid argument that $r(2,2)=5$.