APMO/md/en-apmo2002_sol.md

# XIV APMO: Solutions and Marking Schemes 

1. Let $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ be a sequence of non-negative integers, where $n$ is a positive integer.

Let

$$
A_{n}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}
$$

Prove that

$$
a_{1}!a_{2}!\ldots a_{n}!\geq\left(\left\lfloor A_{n}\right\rfloor!\right)^{n}
$$

where $\left\lfloor A_{n}\right\rfloor$ is the greatest integer less than or equal to $A_{n}$, and $a!=1 \times 2 \times \cdots \times a$ for $a \geq 1$ (and $0!=1$ ). When does equality hold?

## Solution 1.

Assume without loss of generality that $a_{1} \geq a_{2} \geq \cdots \geq a_{n} \geq 0$, and let $s=\left\lfloor A_{n}\right\rfloor$. Let $k$ be any (fixed) index for which $a_{k} \geq s \geq a_{k+1}$.

Our inequality is equivalent to proving that

$$
\frac{a_{1}!}{s!} \cdot \frac{a_{2}!}{s!} \cdot \ldots \cdot \frac{a_{k}!}{s!} \geq \frac{s!}{a_{k+1}!} \cdot \frac{s!}{a_{k+2}!} \cdot \ldots \cdot \frac{s!}{a_{n}!}
$$

Now for $i=1,2, \ldots, k, a_{i}!/ s!$ is the product of $a_{i}-s$ factors. For example, 9!/5! $=9 \cdot 8 \cdot 7 \cdot 6$. The left side of inequality (1) therefore is the product of $A=a_{1}+a_{2}+\cdots+a_{k}-k s$ factors, all of which are greater than $s$. Similarly, the right side of (1) is the product of $B=(n-k) s-$ $\left(a_{k+1}+a_{k+2}+\cdots+a_{n}\right)$ factors, all of which are at most $s$. Since $\sum_{i=1}^{n} a_{i}=n A_{n} \geq n s, A \geq B$. This proves the inequality. [ 5 marks to here.]

Equality in (1) holds if and only if either:
(i) $A=B=0$, that is, both sides of (1) are the empty product, which occurs if and only if $a_{1}=a_{2}=\cdots=a_{n}$; or
(ii) $a_{1}=1$ and $s=0$, that is, the only factors on either side of (1) are 1 's, which occurs if and only if $a_{i} \in\{0,1\}$ for all $i$. [2 marks for both (i) and (ii), no marks for (i) only.]

Solution 2.
Assume without loss of generality that $0 \leq a_{1} \leq a_{2} \leq \cdots \leq a_{n}$. Let $d=a_{n}-a_{1}$ and $m=\left|\left\{i: a_{i}=a_{1}\right\}\right|$. Our proof is by induction on $d$.

We first do the case $d=a_{n}-a_{1}=0$ or 1 separately. Then $a_{1}=a_{2}=\cdots=a_{m}=a$ and $a_{m+1}=\cdots=a_{n}=a+1$ for some $1 \leq m \leq n$ and $a \geq 0$. In this case we have $\left\lfloor A_{n}\right\rfloor=a$, so the inequality to be proven is just $a_{1}!a_{2}!\ldots a_{n}!\geq(a!)^{n}$, which is obvious. Equality holds if and only if either $m=n$, that is, $a_{1}=a_{2}=\cdots=a_{n}=a$; or if $a=0$, that is, $a_{1}=\cdots=a_{m}=0$ and $a_{m+1}=\cdots=a_{n}=1$. [ 2 marks to here.]

So assume that $d=a_{n}-a_{1} \geq 2$ and that the inequality holds for all sequences with smaller values of $d$, or with the same value of $d$ and smaller values of $m$. Then the sequence

$$
a_{1}+1, a_{2}, a_{3}, \ldots, a_{n-1}, a_{n}-1
$$

though not necessarily in non-decreasing order any more, does have either a smaller value of $d$, or the same value of $d$ and a smaller value of $m$, but in any case has the same value of $A_{n}$. Thus, by induction and since $a_{n}>a_{1}+1$,

$$
\begin{aligned}
a_{1}!a_{2}!\ldots a_{n}! & =\left(a_{1}+1\right)!a_{2}!\ldots a_{n-1}!\left(a_{n}-1\right)!\cdot \frac{a_{n}}{a_{1}+1} \\
& \geq\left(\left\lfloor A_{n}\right\rfloor!\right)^{n} \cdot \frac{a_{n}}{a_{1}+1} \\
& >\left(\left\lfloor A_{n}\right\rfloor!\right)^{n}
\end{aligned}
$$

which completes the proof. Equality cannot hold in this case.
2. Find all positive integers $a$ and $b$ such that

$$
\frac{a^{2}+b}{b^{2}-a} \text { and } \frac{b^{2}+a}{a^{2}-b}
$$

are both integers.
Solution.
By the symmetry of the problem, we may suppose that $a \leq b$. Notice that $b^{2}-a \geq 0$, so that if $\frac{a^{2}+b}{b^{2}-a}$ is a positive integer, then $a^{2}+b \geq b^{2}-a$. Rearranging this inequality and factorizing, we find that $(a+b)(a-b+1) \geq 0$. Since $a, b>0$, we must have $a \geq b-1$. [3 marks to here.] We therefore have two cases:

Case 1: $a=b$. Substituting, we have

$$
\frac{a^{2}+a}{a^{2}-a}=\frac{a+1}{a-1}=1+\frac{2}{a-1},
$$

which is an integer if and only if $(a-1) \mid 2$. As $a>0$, the only possible values are $a-1=1$ or 2. Hence, $(a, b)=(2,2)$ or $(3,3)$. [1 mark.]

Case 2: $a=b-1$. Substituting, we have

$$
\frac{b^{2}+a}{a^{2}-b}=\frac{(a+1)^{2}+a}{a^{2}-(a+1)}=\frac{a^{2}+3 a+1}{a^{2}-a-1}=1+\frac{4 a+2}{a^{2}-a-1} .
$$

Once again, notice that $4 a+2>0$, and hence, for $\frac{4 a+2}{a^{2}-a-1}$ to be an integer, we must have $4 a+2 \geq a^{2}-a-1$, that is, $a^{2}-5 a-3 \leq 0$. Hence, since $a$ is an integer, we can bound $a$ by $1 \leq \bar{a} \leq 5$. Checking all the ordered pairs $(a, b)=(1,2),(2,3), \ldots,(5,6)$, we find that only $(1,2)$ and $(2,3)$ satisfy the given conditions. [3 marks.]

Thus, the ordered pairs that work are

$$
(2,2),(3,3),(1,2),(2,3),(2,1),(3,2)
$$

where the last two pairs follow by symmetry. [2 marks if these solutions are found without proof that there are no others.]
3. Let $A B C$ be an equilateral triangle. Let $P$ be a point on the side $A C$ and $Q$ be a point on the side $A B$ so that both triangles $A B P$ and $A C Q$ are acute. Let $R$ be the orthocentre of triangle $A B P$ and $S$ be the orthocentre of triangle $A C Q$. Let $T$ be the point common to the segments $B P$ and $C Q$. Find all possible values of $\angle C B P$ and $\angle B C Q$ such that triangle $T R S$ is equilateral.

## Solution.

We are going to show that this can only happen when

$$
\angle C B P=\angle B C Q=15^{\circ} .
$$

Lemma. If $\angle C B P>\angle B C Q$, then $R T>S T$.
Proof. Let $A D, B E$ and $C F$ be the altitudes of triangle $A B C$ concurrent at its centre $G$. Then $P$ lies on $C E, Q$ lies on $B F$, and thus $T$ lies in triangle $B D G$.
![](https://cdn.mathpix.com/cropped/2024_11_22_f9f048d50025eb9b7941g-3.jpg?height=590&width=679&top_left_y=968&top_left_x=636)

Note that -

$$
\angle F A S=\angle F C Q=30^{\circ}-\angle B C Q>30^{\circ}-\angle C B P=\angle E B P=\angle E A R
$$

Since $A F=A E$, we have $F S>E R$ so that

$$
G S=G F-F S<G E-E R=G R .
$$

Let $T_{x}$ be the projection of $T$ onto $B C$ and $T_{y}$ be the projection of $T$ onto $A D$, and similarly for $R$ and $S$. We have

$$
R_{x} T_{x}=D R_{x}+D T_{x}>\left|D S_{x}-D T_{x}\right|=S_{x} T_{x}
$$

and

$$
R_{y} T_{y}=G R_{y}+G T_{y}>G S_{y}+G T_{y}=S_{y} T_{y}
$$

It follows that $R T>S T$.
[1 mark for stating the Lemma, 3 marks for proving it.]
Thus, if $\triangle T R S$ is equilateral, we must have $\angle C B P=\angle B C Q$.
![](https://cdn.mathpix.com/cropped/2024_11_22_f9f048d50025eb9b7941g-4.jpg?height=589&width=663&top_left_y=291&top_left_x=639)

It is clear from the symmetry of the figure that $T R=T S$, so $\triangle T R S$ is equilateral if and only if $\angle R T A=30^{\circ}$. Now, as $B R$ is an altitude of the triangle $A B C, \angle R B A=30^{\circ}$. So $\triangle T R S$ is equilateral if and only if $R T B A$ is a cyclic quadrilateral. Therefore, $\triangle T R S$ is equilateral if and only if $\angle T B R=\angle T A R$. But

$$
\begin{aligned}
90^{\circ} & =\angle T B A+\angle B A R \\
& =(\angle T B R+\angle R B A)+(\angle B A T+\angle T A R) \\
& =\left(\angle T B R+30^{\circ}\right)+\left(30^{\circ}+\angle T A R\right)
\end{aligned}
$$

and so

$$
30^{\circ}=\angle T A R+\angle T B R
$$

But these angles must be equal, so $\angle T A R=\angle T B R=15^{\circ}$. Therefore $\angle C B P=\angle B C Q=15^{\circ}$. [3 marks for finishing the proof with the assumption that $\angle C B P=\angle B C Q$.]
4. Let $x, y, z$ be positive numbers such that

$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1
$$

Show that

$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z}
$$

Solution 1.

$$
\begin{aligned}
\sum_{\text {cyclic }} \sqrt{x+y z} & =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\frac{1}{x}+\frac{1}{y z}} \\
& =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\frac{1}{x}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{1}{y z}} \quad[1 \text { mark. }] \\
& =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)} \quad[1 \text { mark.] }
\end{aligned}
$$

$$
\begin{aligned}
& =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\left(\frac{1}{x}+\frac{1}{\sqrt{y z}}\right)^{2}+\frac{(\sqrt{y}-\sqrt{z})^{2}}{x y z}} \quad[2 \text { marks. }] \\
& \geq \sqrt{x y z} \sum_{\text {cyclic }}\left(\frac{1}{x}+\frac{1}{\sqrt{y z}}\right) \quad[1 \text { mark. }] \\
& =\sqrt{x y z}\left(1+\sum_{\text {cyclic }} \frac{1}{\sqrt{y z}}\right) \quad[1 \text { mark. }] \\
& =\sqrt{x y z}+\sum_{\text {cyclic }} \sqrt{x} \quad[1 \text { mark. }]
\end{aligned}
$$

Note. It is easy to check that equality holds if and only if $x=y=z=3$.
Solution 2.
Squaring both sides of the given inequality, we obtain

$$
\begin{aligned}
& \sum_{\text {cyclic }} x+\sum_{\text {cyclic }} y z+2 \sum_{\text {cyclic }} \sqrt{x+y z} \sqrt{y+z x} \\
& \quad \geq x y z+2 \sqrt{x y z} \sum_{\text {cyclic }} \sqrt{x}+\sum_{\text {cyclic }} x+2 \sum_{\text {cyclic }} \sqrt{x y} \quad \text { [1 mark.] }
\end{aligned}
$$

It follows from the given condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ that $x y z=\sum_{\text {cyclic }} x y$. Therefore, the given inequality is equivalent to

$$
\sum_{\text {cyclic }} \sqrt{x+y z} \sqrt{y+z x} \geq \sqrt{x y z} \sum_{\text {cyclic }} \sqrt{x}+\sum_{\text {cyclic }} \sqrt{x y} . \quad[2 \text { marks.] }
$$

Using the Cauchy-Schwarz inequality [or just $x^{2}+y^{2} \geq 2 x y$ ], we see that

$$
(x+y z)(y+z x) \geq\left(\sqrt{x y}+\sqrt{x y z^{2}}\right)^{2}, \quad[1 \text { mark. }]
$$

or

$$
\sqrt{x+y z} \sqrt{y+z x} \geq \sqrt{x y}+\sqrt{z} \sqrt{x y z} . \quad[1 \text { mark. }]
$$

Taking the cyclic sum of this inequality over $x, y$ and $z$, we get the desired inequality. [2 marks.]

Solution 3.
This is another way of presenting the idea in the first solution.
Using the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ and the AM-GM inequality, we have

$$
\begin{aligned}
x+y z-\left(\sqrt{\frac{y z}{x}}+\sqrt{x}\right)^{2} & =y z\left(1-\frac{1}{x}\right)-2 \sqrt{y z} \\
& =y z\left(\frac{1}{y}+\frac{1}{z}\right)-2 \sqrt{y z}=y+z-2 \sqrt{y z} \geq 0
\end{aligned}
$$

which gives

$$
\sqrt{x+y z} \geq \sqrt{\frac{y z}{x}}+\sqrt{x} . \quad[3 \text { marks. }]
$$

Similarly, we have

$$
\sqrt{y+z x} \geq \sqrt{\frac{z x}{y}}+\sqrt{y} \text { and } \sqrt{z+x y} \geq \sqrt{\frac{x y}{z}}+\sqrt{z}
$$

Addition yields

$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}+\sqrt{x}+\sqrt{y}+\sqrt{z}
$$

[2 marks.] Using the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ again, we have

$$
\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}=\sqrt{x y z}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\sqrt{x y z}, \quad[1 \text { mark. }]
$$

and thus

$$
\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z} . \quad[1 \text { mark. }]
$$

## Solution 4.

This is also another way of presenting the idea in the first solution.
We make the substitution $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$. Then it is enough to show that

$$
\sqrt{\frac{1}{a}+\frac{1}{b c}}+\sqrt{\frac{1}{b}+\frac{1}{c a}}+\sqrt{\frac{1}{c}+\frac{1}{a b}} \geq \sqrt{\frac{1}{a b c}}+\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}},
$$

where $a+b+c=1$. Multiplying this inequality by $\sqrt{a b c}$, we find that it can be written

$$
\sqrt{a+b c}+\sqrt{b+c a}+\sqrt{c+a b} \geq 1+\sqrt{b c}+\sqrt{c a}+\sqrt{a b} . \quad[1 \text { mark. }]
$$

This is equivalent to

$$
\begin{aligned}
& \sqrt{a(a+b+c)+b c}+\sqrt{b(a+b+c)+c a}+\sqrt{c(a+b+c)+a b} \\
& \geq a+b+c+\sqrt{b c}+\sqrt{c a}+\sqrt{a b}, \quad[1 \text { mark. }]
\end{aligned}
$$

which in turn is equivalent to

$$
\sqrt{(a+b)(a+c)}+\sqrt{(b+c)(b+a)}+\sqrt{(c+a)(c+b)} \geq a+b+c+\sqrt{b c}+\sqrt{c a}+\sqrt{a b}
$$

[1 mark.] (This is a homogeneous version of the original inequality.) By the Cauchy-Schwarz inequality (or since $b+c \geq 2 \sqrt{b c}$ ), we have

$$
\left[(\sqrt{a})^{2}+(\sqrt{b})^{2}\right]\left[(\sqrt{a})^{2}+(\sqrt{c})^{2}\right] \geq(\sqrt{a} \sqrt{a}+\sqrt{b} \sqrt{c})^{2}
$$

or

$$
\sqrt{(a+b)(a+c)} \geq a+\sqrt{b c} . \quad[2 \text { marks. }]
$$

Taking the cyclic sum of this inequality over $a, b, c$, we get the desired inequality. [2 marks.]
5. Let R denote the set of all real numbers. Find all functions $f$ from R to R satisfying:
(i) there are only finitely many $s$ in R such that $f(s)=0$, and
(ii) $f\left(x^{4}+y\right)=x^{3} f(x)+f(f(y))$ for all $x, y$ in $\mathbf{R}$.

## Solution 1.

The only such function is the identity function on $R$.
Setting $(x, y)=(1,0)$ in the given functional equation (ii), we have $f(f(0))=0$. Setting $x=0$ in (ii), we find

$$
f(y)=f(f(y))
$$

[1 mark.] and thus $f(0)=f(f(0))=0$ [1 mark.]. It follows from (ii) that $f\left(x^{4}+y\right)=$ $x^{3} f(x)+f(y)$ for all $x, y \in \mathbf{R}$. Set $y=0$ to obtain

$$
f\left(x^{4}\right)=x^{3} f(x)
$$

for all $x \in \mathrm{R}$, and so

$$
f\left(x^{4}+y\right)=f\left(x^{4}\right)+f(y)
$$

for all $x, y \in \mathbf{R}$. The functional equation (3) suggests that $f$ is additive, that is, $f(a+b)=$ $f(a)+f(b)$ for all $a, b \in \mathbf{R}$. [1 mark.] We now show this.

First assume that $a \geq 0$ and $b \in R$. It follows from (3) that

$$
f(a+b)=f\left(\left(a^{1 / 4}\right)^{4}+b\right)=f\left(\left(a^{1 / 4}\right)^{4}\right)+f(b)=f(a)+f(b)
$$

We next note that $f$ is an odd function, since from (2)

$$
f(-x)=\frac{f\left(x^{4}\right)}{(-x)^{3}}=\frac{f\left(x^{4}\right)}{-x^{3}}=-f(x), \quad x \neq 0
$$

Since $f$ is odd, we have that, for $a<0$ and $b \in R$,

$$
\begin{aligned}
f(a+b) & =-f((-a)+(-b))=-(f(-a)+f(-b)) \\
& =-(-f(a)-f(b))=f(a)+f(b)
\end{aligned}
$$

Therefore, we conclude that $f(a+b)=f(a)+f(b)$ for all $a, b \in R$. [2 mers.].]
We now show that $\{s \in \mathrm{R} \mid f(s)=0\}=\{0\}$. Recall that $f(0)=0$. Assume that there is a nonzero $h \in \mathrm{R}$ such that $f(h)=0$. Then, using the fact that $f$ is additive, we inductively have $f(n h)=0$ or $n h \in\{s \in R \mid f(s)=0\}$ for all $n \in \mathbf{N}$. However, this is a contradiction to the given condition (i). [1 mark.]

It's now easy to check that $f$ is one-to-one. Assume that $f(a)=f^{\prime}(b)$ for some $a, b \in \mathbb{P}$. Then, we have $f(b)=f(a)=f(a-b)+f(b)$ or $f(a-b)=0$. This implies that $a-b \in\{s \in$ $\mathbf{R} \mid f(s)=0\}=\{0\}$ or $a=b$, as desired. From (1) and the fact that $f$ is one-to-one, we deduce that $f(x)=x$ for all $x \in \mathbf{R}$. [1 mark.] This completes the proof.

## Solution 2.

Again, the only such function is the identity function on R .
As in Solution 1, we first show that $f(f(y))=f(y), f(0)=0$, and $f\left(x^{4}\right)=x^{3} f(x)$. [2 marks.] From the latter follows

$$
f(x)=0 \Longrightarrow f\left(x^{4}\right)=0
$$

and from condition (i) we get that $f(x)=0$ only possibly for $x \in\{0,1,-1\}$. [1 mark.]
Next we prove

$$
f(a)=b \Longrightarrow f(\sqrt[4]{|a-b|})=0
$$

This is clear if $a=b$. If $a>b$ then

$$
\begin{aligned}
f(a) & =f((a-b)+b)=(a-b)^{3 / 4} f(\sqrt[4]{a-b})+f(f(b)) \\
& =(a-b)^{3 / 4} f(\sqrt[4]{a-b})+f(b) \\
& =(a-b)^{3 / 4} f(\sqrt[4]{a-b})+f(f(a)) \\
& =(a-b)^{3 / 4} f(\sqrt[4]{a-b})+f(a),
\end{aligned}
$$

so $(a-b)^{3 / 4} f(\sqrt[4]{a-b})=0$ which means $f(\sqrt[4]{|a-b|})=0$. If $a<b$ we get similarly

$$
\begin{aligned}
f(b) & =f((b-a)+a)=(b-a)^{3 / 4} f(\sqrt[4]{b-a})+f(f(a)) \\
& =(b-a)^{3 / 4} f(\sqrt[4]{b-a})+f(b)
\end{aligned}
$$

and again $f(\sqrt[4]{|a-b|})=0$. [2 marks.]
Thus $f(a)=b \Longrightarrow|a-b| \in\{0,1\}$. Suppose that $f(x)=x+b$ for some $x$, where $|b|=1$. Then from $f\left(x^{4}\right)=x^{3} f(x)$ and $f\left(x^{4}\right)=x^{4}+a$ for some $|a| \leq 1$ we get $x^{3}=a / b$, so $|x| \leq 1$. Thus $f(x)=x$ for all $x$ except possibly $x= \pm 1$. [ 1 mark.] But for example,

$$
f(1)=f\left(2^{4}-15\right)=2^{3} f(2)+f(f(-15))=2^{3} \cdot 2-15=1
$$

and

$$
f(-1)=f\left(2^{4}-17\right)=2^{3} f(2)+f(f(-17))=2^{3} \cdot 2-17=-1
$$

[1 mark.] This finishes the proof.