APMO/md/en-apmo1995_sol.md

# SOLUTIONS 

Note: On the left side of the page the maximum number of points that may be awarded for every part of the solution is indicated in brackets.

Question 1. Suppose that $\left(a_{1}, a_{2}, \ldots, a_{1995}\right)$ is a solution of the given system of incqualities. Then

$$
\sum_{n=1}^{1995} 2 \sqrt{a_{n}-(n-1)} \geq \sum_{n=1}^{1995} a_{n}-\sum_{n=1}^{1994}(n-1)+1=\sum_{n=1}^{1995} a_{n}-\sum_{n=1}^{1995}\{(n-1)+1\}
$$

i.e.

$$
0 \geq \sum_{n=1}^{1995}\left\{a_{n}-(n-1)+1-2 \sqrt{a_{n}-(n-1)}\right.
$$

[2 points]
Next, note that

$$
\left[\sqrt{a_{n}-(n-1)}-1\right]^{2}=a_{n}-(n-1)-2 \sqrt{a_{n}-(n-1)}+1
$$

for $n=1,2, \ldots, 1995$.
[1 point]
Hence,
$0 \geq \sum_{n=1}^{1995}\left[a_{n}-(n-1)-2 \sqrt{a_{n}-(n-1)}+1\right]=\sum_{n=1}^{1995}\left[\sqrt{a_{n}-(n-1)}-1\right]^{2} \geq 0$.
Therefore, $\sqrt{a_{n}-(n-1)}=1$, for $\mathrm{n}=1,2, \ldots, 1995$. It follows that $a_{n}=\mathrm{n}$ for $\mathrm{n}=1,2, \ldots, 1995$.

## [2 points]

Conversely, if $\sqrt{a_{n}-(n-1)}=1$, for $\mathrm{n}=1,2, \ldots, 1995$, then

$$
2 \sqrt{n-(n-1)}=2=n+1-(n-1), \text { for } \mathrm{n}=1,2, \ldots, 1994
$$

and

$$
2 \sqrt{1995-1994}=2=1+1
$$

which shows that $a_{\mathrm{n}}=\mathrm{n}$, for $\mathrm{n}=1,2, \ldots, 1995$, is indeed a solution of the given system of inequalities.
[2 points]
Question 2. The answer is 14.
[1 point]
Denote the required number by M . We observe that the sequence $2 \cdot 101,3 \cdot 97$, $5 \cdot 89,7 \cdot 83,11 \cdot 79,13 \cdot 73,17 \cdot 71,19 \cdot 67,23 \cdot 61=1403,29 \cdot 59=1711,31 \cdot 53=1643$, $37 \cdot 47=1739,41 \cdot 43=1763$ satisfies conditions i) and ii) and contains no prime number. Hence, $M>13$.

## [3 points]

Now we show that a sequence with 14 elements that satisfies conditions i) and ii) will contain a prime number. We proceed by contradiction. Suppose the elements are $a_{1}, a_{2}, \ldots, a_{14}$. Since none of them is a prime number, each element will contain at least two prime factors. We take any two prime factors from each $a_{i}$, and list them in ascending order $\mathrm{p}_{1}<\mathrm{p}_{2}<\ldots<\mathrm{p}_{26}<\mathrm{p}_{27}<\mathrm{p}_{28}$. As the 14 th prime is 43 , this means $43 \leq p_{14}, 47 \leq p_{15}$ and so on. Now $43 \cdot 47=2021>1995$. This means that $p_{14}$ must pair up with one of the $p_{1}, p_{2} \ldots p_{13}$ to form a certain $a_{i}$. Likewise $p_{15}$ must pair up with one of the $p_{1}, p_{2} \ldots p_{13}$ to form another $a_{i}$, and so on (without repetition). Hence there exist $p_{i}, p_{j}, 13<i<j$, that must pair up together to form some $a_{i}$. But then $a_{i} \geq p_{i} p_{j} \geq 43 \cdot 47>1995$, a contradiction.
[3 points]
Question 3. Let T be the intersection of PQ and $\mathrm{RS}, \mathrm{T}$ lies outside C , the circle PQRS.
i) Clearly any point on C belongs to the set A .
ii) Let $\mathrm{r}=\sqrt{T P \cdot T Q}=\sqrt{T R \cdot T S}$, and consider the circle with center T and radius r. Let $V$ a point on this circle. Since $T V^{2}=T P \cdot T Q=T R \cdot T S$, TV is tangent to the circles PQV and RSV. Therefore, PQV is tangent to RSV. That means, V is in the set A :
[4 points]
Conversely, assume V is in A , i.e. PQV is tangent to RSV. If the circles PQV and RSV are the same, then $\mathrm{PQV}=\mathrm{RSV}=\mathrm{PQRS}$. Otherwise, let the line TV intersect $P Q V$ in $V_{1}$, and RSV in $V_{2}$. Then
$\mathrm{TP} \cdot \mathrm{TQ}=\mathrm{TV} \cdot \mathrm{TV}_{1}$
$\mathrm{TR} \cdot \mathrm{TS}=\mathrm{TV} \cdot \mathrm{TV}_{2}$.

Due to the fact that $P Q R$ and $S$ are on a circle, we have $T P \cdot T Q=T R \cdot T S$, thus $\mathrm{TV} \cdot \mathrm{TV}_{1}=\mathrm{TV} \cdot \mathrm{TV}_{2}$. Moreover, since T does not lie on $\mathrm{C}, \mathrm{T} \neq \mathrm{V}$, which implies $T V_{1}=T V_{2}$, i.e., $V_{1}=V_{2}=V$.
All this means that TV is tangent to the circles PQV and RSV, therefore V lies on the circle with center T and radius $\mathrm{r}=\sqrt{T P \cdot T Q}=\sqrt{T R \cdot T S}$.
[3 points]
Question 4. First, we will show that MS is perpendicular to A'B'. Since SAMB, SBN'A', SA'M'B' and SB'NA are rectangles, it follows that MNM'N' is a rectangle with its sides parallel to $\mathrm{AA}^{\prime}$ and $\mathrm{BB}^{\prime}$.
Moreover, the perpendicular bisectors of $\mathrm{AA}^{\prime}$ and $\mathrm{BB}^{\prime}$ pass through O , and they coincide with those of $\mathrm{MN}^{\prime}$ and $\mathrm{NM}^{\prime}$. Therefore, O is the center of the rectangle. Let I and H be the intersections of MS with AB and $\mathrm{A}^{\prime} \mathrm{B}^{\prime}$. We then have $\angle \mathrm{HSA}^{\prime}=\angle \mathrm{ASI}$, $\angle \mathrm{ASI}=\angle \mathrm{SAI}$, $\angle \mathrm{SAI}=\angle \mathrm{A}^{\prime} \mathrm{AB}=\angle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{B}$.
In the triangle $\mathrm{SA}^{\prime} \mathrm{B}^{\prime}, \angle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{B}$ or $\angle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{S}$ is the complementary angle of $\angle \mathrm{SA}^{\prime} \mathrm{B}^{\prime}$. The angles $\mathrm{HSA}^{\prime}$ and $\mathrm{SA}^{\prime} \mathrm{B}$ are complementary angles and the triangle $\mathrm{SA}^{\prime} \mathrm{H}$ is a right-angled triangle with right angle at H . Therefore, $\mathrm{MS} \perp \mathrm{A}^{\prime} \mathrm{B}$ '.
[1 point]
Next, we will show that $A B^{2}+A^{\prime} B^{12}=4 R^{2}$ and that $M N^{\prime 2}+N^{\prime} M^{12}$ is constant.
Let D be the second intersection of $\mathrm{MN}^{\prime}$ with the circle, then $\mathrm{AD}=\mathrm{AB}$, since they subtend equal angles. This implies

$$
\mathrm{AB}^{2}+\mathrm{AB}^{\prime} \mathrm{B}^{\prime 2}=\mathrm{A}^{\prime} \mathrm{D}^{2}+\mathrm{A}^{\prime} \mathrm{B}^{12}
$$

But, we know $\mathrm{DA}^{\prime} \| \mathrm{MH}$, since $\angle \mathrm{BDA}^{\prime}=\angle \mathrm{BAA}^{\prime}=\angle \mathrm{BMH}$, that means $\angle \mathrm{DA}^{\prime} \mathrm{B}^{\prime}$ $=90^{\circ}$ and it is inscribed in the circle, therefore D and $\mathrm{B}^{\prime}$ are diametrically opposed, what finally implies

$$
\mathrm{AB}^{2}+\mathrm{A}^{\prime} \mathrm{B}^{12}=\mathrm{A}^{\prime} \mathrm{D}^{2}+\mathrm{A}^{\prime} \mathrm{B}^{12}=\mathrm{DB}^{12}=(2 \mathrm{R})^{2}=4 \mathrm{R}^{2}
$$

i.e.

$$
\mathrm{AB}^{2}+\mathrm{A}^{\prime} \mathrm{B}^{\prime 2}=4 \mathrm{R}^{2}
$$

[2 points]
To see that $\mathrm{MN}^{12}+\mathrm{N}^{\prime} \mathrm{M}^{12}$ is constant consider the following equalities

$$
\begin{aligned}
\mathrm{MN}^{\prime 2}=\left(\mathrm{MB}+\mathrm{BN} N^{\prime}\right)^{2} & =\mathrm{MB}^{2}+\mathrm{BN}^{\prime 2}+2 \mathrm{MB} \cdot \mathrm{BN}^{\prime} \\
& =\mathrm{SA}^{2}+\mathrm{SA}^{\prime 2}+2 \mathrm{SA} \cdot \mathrm{SA}^{\prime 2} \\
\mathrm{M}^{\prime} \mathrm{N}^{\prime 2}=\left(\mathrm{N}^{\prime} \mathrm{A}^{\prime}+\mathrm{A}^{\prime} \mathrm{M}^{\prime}\right)^{2} & =\mathrm{N}^{\prime} \mathrm{A}^{\prime 2}+\mathrm{A}^{\prime} \mathrm{M}^{\prime 2}+2 \mathrm{~N}^{\prime} \mathrm{A}^{\prime} \cdot \mathrm{A}^{\prime} \mathrm{M}^{\prime} \\
& =\mathrm{SB}^{2}+\mathrm{SB}^{\prime 2}+2 \mathrm{SB} \cdot \mathrm{SB}^{\prime}
\end{aligned}
$$

By Pythagoras, we have

$$
\mathrm{AB}^{2}+\mathrm{A}^{\prime} \mathrm{B}^{12}=\left(\mathrm{SA}^{2}+\mathrm{SB}^{2}\right)+\left(\mathrm{SA}^{12}+\mathrm{SB}^{12}\right)
$$

This implies,

$$
\begin{aligned}
\mathrm{MN}^{\prime 2}+\mathrm{M}^{\prime} \mathrm{N}^{\prime 2}= & \mathrm{SA}^{2}+\mathrm{SB}^{2}+\mathrm{SA}^{12}+\mathrm{SB}^{12}+2 \mathrm{SA} \cdot \mathrm{SA}^{\prime}+2 \mathrm{SBSB}^{\prime} \\
& =\mathrm{AB}^{2}+\mathrm{A}^{\prime} \mathrm{B}^{12}+4 \mathrm{SA} \cdot \mathrm{SA}^{\prime} \\
& =8 \mathrm{R}^{2}-40 \mathrm{~S}^{2}
\end{aligned}
$$

Additionally we know that

$$
\mathrm{MN}^{\prime 2}+\mathrm{M}^{\prime} \mathrm{N}^{\prime 2}=\mathrm{MM}^{12}=4 \mathrm{OM}^{2}
$$

[2 points]
But, $40 M^{2}=8 R^{2}-40 S^{2}$.
Therefore,

$$
\mathrm{MN}^{12}+\mathrm{M}^{\prime} \mathrm{N}^{\prime 2}=4 \mathrm{OM}^{2}
$$

This last quantity is clearly a constant.

## [1 point]

Finally, it is clear that the vertices of the rectangle $M N N^{\prime} N^{\prime}$ lie on the circle with center O and radius $\mathrm{OM}=\sqrt{2 R^{2}-O S^{2}}$. Therefore, the set of points consists of a circle.

## [1 point]

Question 5. The minimum value of $k$ is $\mathrm{k}^{*}=4$.

## [1 point]

First, we define a function $f$ from $Z$ to $\{1,2,3,4\}$ recursively as follows: $f(0)=1$. For any positive integer $i, f(i)$ is defined to be the minimum positive integer not in $A_{i}:=\{\mathrm{f}(\mathrm{j}): \mathrm{i}-\mathrm{j} \in\{5,7,12\}$ and $-\mathrm{i}<\mathrm{j}<\mathrm{i}\}$, and $\mathrm{f}(-\mathrm{i})$ the minimum positive integer not in $B_{i}:=\{f(j): j+i \in\{5,7,12\}$ and $-i<j<i\}$. Note that $\left|A_{i}\right| \leq 3$ and $\left|B_{i}\right| \leq 3$ for any $i$. So, f is a function from Z to $\{1,2,3,4\}$ such that $\mathrm{f}(\mathrm{x}) \neq f(\mathrm{y})$ whenever $\mid \mathrm{x}$ $-\mathrm{y} \mid \in\{5,7,12\}$. This gives that $\mathrm{k}^{*} \leq 4$.
[3 points]
Next, we claim that $k^{*} \geq 4$. Suppose it is not the case. Then there exists a function f from Z to $\{1,2,3\}$ with the property that $\mathrm{f}(\mathrm{x}) \neq \mathrm{f}(\mathrm{y})$ whenever $|\mathrm{x}-\mathrm{y}| \in$ $\{5,7,12\}$. For any integer $x$, consider the values $f(x), f(x-5)$, and $f(x+7)$. These three values are different. Now consider $f(x+2)$. Since $f(x+2) \notin\{f(x-5), f(x+$ 7)\}

$$
f(x)=f(x+2) \text { for any integer } x
$$

Hence,

$$
\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{x}+2)=\mathrm{f}(\mathrm{x}+4)=\ldots=\mathrm{f}(\mathrm{x}+12)
$$

which is impossible. Thus $\mathrm{k}^{*}>4$.
[3 points]