APMO/md/en-apmo1998_sol.md

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

$$\sum _{(A_1,A_2,\dots ,A_n)}\mid A_1\cup A_2\cup \dots \cup A_n\mid \backslash sum\_\{\backslash left(A\_\{1\},\; A\_\{2\},\; \backslash ldots,\; A\_\{n\}\backslash right)\}\backslash left|A\_\{1\}\; \backslash cup\; A\_\{2\}\; \backslash cup\; \backslash ldots\; \backslash cup\; A\_\{n\}\backslash right|$$

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 $A_{1}, I_{2}, \ldots, A_{t}$.

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

Thus we have

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

$$36r-s=35\times 3\bar{7}a,and,36s-r=35\times 37b36\; r-s=35\; \backslash times\; 3\; \backslash overline\{7\}\; a,\; a\; n\; d,\; 36\; s-r=35\; \backslash times\; 37\; b$$

Hence

Furthermore,

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

Thus

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

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

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

MARKING SCHEME:

Lol

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

We need to show that

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

or, since $x y z=1$,

$$(x+y)(y+z)(z+x)\ge 2+2(x+y+z)(1)(x+y)(y+z)(z+x)\; \backslash geq\; 2+2(x+y+z)(1)$$

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

$$(x+y+z)(xy+yz+zx-2)-xyz\ge 2(2)(x+y+z)(x\; y+y\; z+z\; x-2)-x\; y\; z\; \backslash geq\; 2(2)$$

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

$$x+y+z\ge 3\text{ and }xy+yz+zx\ge 3x+y+z\; \backslash geq\; 3\; \backslash text\; \{\; and\; \}\; x\; y+y\; z+z\; x\; \backslash geq\; 3$$

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:

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

Thus, (1) is equivalent to

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

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

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

$$\frac{1}{y}+\frac{1}{z}\ge \frac{1}{z}+\frac{1}{x}\ge \frac{1}{x}+\frac{1}{y}\backslash frac\{1\}\{y\}+\backslash frac\{1\}\{z\}\; \backslash geq\; \backslash frac\{1\}\{z\}+\backslash frac\{1\}\{x\}\; \backslash geq\; \backslash frac\{1\}\{x\}+\backslash frac\{1\}\{y\}$$

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

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

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

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

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

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

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.

¿.From (1) and (2) follows

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

Since $t \geq 13$.

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