# APMO SOLUTIONS 

PROBLEM 1. Let $F$ be the set of all $n$-luples $\left(A_{1}, \Lambda_{2}, \ldots, A_{n}\right)$ where each $\Lambda_{i}$, $i=$ $1,2, \ldots, n$ is a subset of $\{1,2, \ldots, 1998\}$. Let $|\Lambda|$ denote the number of elements of the set $A$. Find the number

$$
\sum_{\left(A_{1}, A_{2}, \ldots, A_{n}\right)}\left|A_{1} \cup A_{2} \cup \ldots \cup A_{n}\right|
$$

## MARKING SCHEME:

Let $M$ be a subset of the set $\{1,2 \ldots, 1998\}$ and let $|M|=k$. Then the set $M$ call be obtained as the union of $t$ sets $A_{1}, A_{2}, \ldots, A_{t}$ in $\left(2^{t}-1\right)^{k}$ different ways since each clement $x \in M$ can belong to $2^{t}-1$ nonempty families of subsets $A_{1}, I_{2}, \ldots, A_{t}$.

$$
3 \text { points for describing the correct counting method }
$$

Thus we have

$$
\begin{aligned}
& \sum_{\left(A_{1}, A_{2}, \ldots, A_{t}\right) \in F}\left|A_{1} \cup A_{2} \cup \ldots \cup A_{t}\right|=\sum_{k=1}^{1998} k\binom{1998}{k}\left(2^{t}-1\right)^{k} \\
& 2 \text { points for setting up the above formula } \\
& =1998\left(2^{\prime}-1\right) \sum_{k=0}^{1997}\binom{1997}{k}\left(2^{t}-1\right)^{k}=1998\left(2^{t}-1\right) 2^{1997 t} . \\
& 2 \text { points for corrcctly carrying out the computation }
\end{aligned}
$$

PROBLEM 2. Show that for any positive integers $a$ and $b,(36 a+b)(a+36 b)$ cannot be a power of 2 .

MARKING SCCHEME:
Suppose that $(36 a+b)(a+36 b)$ is a power of 2 for some positive integers $a$ and $b$. Write $36 a+b=2^{m}=r$ and $a+36 b=2^{n}=s$. Then

$$
36 r-s=35 \times 3 \overline{7} a, a n d, 36 s-r=35 \times 37 b
$$

Hence

$$
\begin{aligned}
1 / 36<r / s= & 2^{m-n}<36, \text { or }-6<m-n<6
\end{aligned}
$$

Furthermore,

$$
4^{n}\left(4^{m-n}-1\right)=r^{2}-s^{2}=35 \times 37\left(a^{2}-b^{2}\right)
$$

Thus

$$
4^{m-n} \equiv 1(\bmod 37)
$$

## 2 points for this congruence

Observe that the 9 th power of 4 is the smallest power of 4 that is congruent 10 $1($ mod $3 \overline{1})$. Thus $9 \mid(m-n)$. Aso note that $m \neq n$. Hence $|m-n| \geq 9$, which is unt possible because $|m-n|<6$. 3 points for the final conclusion

Problem 3. Let $a, b, c$ be positive real numbers. Prove that

$$
\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right) \geq 2\left(1+\frac{a+b+c}{\sqrt[3]{a b c}}\right)
$$

## MARKING SCHEME:

Lol

$$
x=\frac{a}{\sqrt[3]{a b c}} \cdot \quad y=\frac{b}{\sqrt[3]{a b c}} . \quad z=\frac{r}{\sqrt[3]{a b c}}
$$

We need to show that

$$
\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right) \geq 2(1+x+y+z)
$$

or, since $x y z=1$,

$$
(x+y)(y+z)(z+x) \geq 2+2(x+y+z)(1)
$$

2 points for (1) by making change of variables or by assuming abe $=1$ without loss of generality
which is equivalent to

$$
(x+y+z)(x y+y z+z x-2)-x y z \geq 2(2)
$$

$\therefore$ points for reducing to (2)
Since $x y z=1$, the AM-GMS inequality implies

$$
x+y+z \geq 3 \text { and } x y+y z+z x \geq 3
$$

So. (2) follows.
3 points for surcessful application of AM-CMM inequality

## ALTERNATE SOLUTION

2 points for (1) by making change of variables or by assuming abc $=1$ without loss of generality
i.From (1), one can proceed as follows:

$$
(x+y)(y+z)(z+x)=2 x y z+x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)=2+x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{z}+\frac{1}{x}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)
$$

Thus, (1) is equivalent to

$$
x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{z}+\frac{1}{x}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right) \geq 2(x+y+z) \cdot 3
$$

$$
2 \text { points for reducing to (3) }
$$

Assume that $x \geq y \geq=$ Then

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

hence we can apply (heloyshev's incquality
1 poinls for checking the hypothesis of C'hevyshev's inequality
t.0 get.

$$
x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{z}+\frac{1}{x}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right) \geq \frac{1}{3}(x+y+z)\left(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\right)
$$

i.From the AM-GM inequality and the fact that $x y z=1$ we obtain

$$
\left(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\right) \geq 6
$$

and hence (3) follows. 2 points for successful applications of Chevyshev's and AM-GM inequalities

Problem 4. Let. $A B C^{\prime}$ be a triangle and $D$ the foot of the altitude from $A$ Lel $E$ and $F$ be on a line passing through $D$ such that $A E$ is perpendicular to $B E$. $A F$ is perpendicular to $C F$. and $E$ and $F$ are different from $D$. Let $M$ and $N$ be the midpoints of the line segments $B C$ and $E F$, respectively. Prove that.$L N$ is perpendicular to $N M$.

## MARIING SCHEME:

Let $P$ be such that $A D M P$ is a rectangle. Choose points $Q$ and $R$ on the line $A P$ such that $Q B D A$ and $A D C R$ are eectangles. Points $Q, B$ and $D$ lie on the rircle of diameter $A B$, hence $A D E Q$ is a cyclic quadrilateral. Similarly, $R, C$ and $D$ lic on the circle of diameter $A C$, hence $A D F R$ is a cyclic quadrilateral.

1 point for proving $A D E Q$ and/or $A D F R$ are ryclic
The two quadrilaterals share a side, and have the same supporting lines for other two sides. Since they are cyclic, the remaining two sides $E Q$ and $R F$ must be parallel. Thus $E, Q, R$ and $F$ are vertices of a trapezoid.

1 point for proving $E Q / / R F$
On the other hand. in rectangle $Q B C R M$ is the midpoint of $B C$. and $M P$ is parallel to $Q B$, so $P$ is the midpoint of $Q R$. Since $N$ is the midpoint of $E F$, we obtain that, in trapezoid $Q E F R, N P$ is parallel to $Q E$.

2 points for proving NP//QE
This implies that quadrilateral $A D N P$ is cyclic, having the sides parallel to the sides of $A D F R$. Moreover, $A$ lies on the circle circumscribed to this quadrilateral, because the other three vertices of the rectangle $A D M P$ lie on it. Hence the quadrilateral $A D M N$ is cyclic.

2 proints for proving $A D N P$ and $A D M N$ are cyclic
(consquently. $\angle . V M=180^{\circ}-\angle A D M=90^{\circ}$.
1 point for proving $\angle A N M=180^{\circ}-\angle A D M=90^{\circ}$

Problem 5. Determine the largest of all integers $n$ with the property that $n$ is divisible by all positive integers that are less than $\sqrt[3]{n}$.

## MARKINC SCHEME:

Observation from that $\operatorname{lcm}(2.3,4.5,6.7)=420$ is divisible by every integer less than or equal to $7=[\sqrt[3]{420}]$ and that $\operatorname{lrm}(2,3,4,5,6.7,8)=840$ is not. divisible by $9=[\sqrt[3]{840}]$, one may guess 420 is the required integer. 2 points for the correct guess

Lert $N$ be the required integer and supposic $V>120$. Put $t=[\sqrt[3]{\sqrt{2}}]$. Then

$$
\therefore \leq 1\left(1^{3}+31+3\right)(1)
$$

Sinte $t \geq 7, \quad \operatorname{cm}(2.3 .4 .5 .6 .7)=420$ should divide .1 and hence $N \geq 840$. which implies $1 \geq 9$. But then $\operatorname{lcm}(2,3,4,5,6,7.8 .9)=2520$ should divide $N$, which implies $t \geq 13=[\sqrt[3]{2520}]$.

1 points for $t \geq 13$
Observe that any four consecutive integers are divisible by 8 and that any t.wo out of four consecutive integers have ged either 1.2 , or 3 . So, we have $1(1-1)(t-2)(1-3)$ divides 6 N and in particular.

$$
\begin{aligned}
& t(t-1)(t-2)(t-3) \leq 6 \times(2) \\
& 2 \text { points for } t(t-1)(t-2)(t-3) \leq 6 N
\end{aligned}
$$

¿.From (1) and (2) follows

$$
t(t-1)(t-2)(t-3) \leq 6 t\left(t^{3}+3 t+3\right) \frac{12}{t}+\frac{7}{t^{2}}+\frac{24}{t^{3}} \geq 1
$$

Since $t \geq 13$.

$$
\frac{12}{t}+\frac{T}{t^{2}}+\frac{24}{t^{3}}<1
$$

which is a contradiction.
$\therefore$ points for the contradiction