APMO/md/en-apmo1995_sol.md

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

$$\sum _{n=1}^{1995}2\sqrt{a_n-(n-1)}\ge \sum _{n=1}^{1995}{a}_n-\sum _{n=1}^{1994}(n-1)+1=\sum _{n=1}^{1995}{a}_n-\sum _{n=1}^{1995}\{(n-1)+1\}\backslash sum\_\{n=1\}^\{1995\}\; 2\; \backslash sqrt\{a\_\{n\}-(n-1)\}\; \backslash geq\; \backslash sum\_\{n=1\}^\{1995\}\; a\_\{n\}-\backslash sum\_\{n=1\}^\{1994\}(n-1)+1=\backslash sum\_\{n=1\}^\{1995\}\; a\_\{n\}-\backslash sum\_\{n=1\}^\{1995\}\backslash \{(n-1)+1\backslash \}$$

i.e.

$$0\ge \sum _{n=1}^{1995}\{a_n-(n-1)+1-2\sqrt{a_n-(n-1)}0\; \backslash geq\; \backslash sum\_\{n=1\}^\{1995\}\backslash left\backslash \{a\_\{n\}-(n-1)+1-2\; \backslash sqrt\{a\_\{n\}-(n-1)\}\backslash right.$$

[2 points] Next, note that

$${[\sqrt{a_n-(n-1)}-1]}^2=a_n-(n-1)-2\sqrt{a_n-(n-1)}+1\backslash left[\backslash sqrt\{a\_\{n\}-(n-1)\}-1\backslash right]^\{2\}=a\_\{n\}-(n-1)-2\; \backslash sqrt\{a\_\{n\}-(n-1)\}+1$$

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

$$2\sqrt{n-(n-1)}=2=n+1-(n-1),\text{ for }\mathrm{n}=1,2,\dots ,19942\; \backslash sqrt\{n-(n-1)\}=2=n+1-(n-1),\; \backslash text\; \{\; for\; \}\; \backslash mathrm\{n\}=1,2,\; \backslash ldots,\; 1994$$

and

$$2\sqrt{1995-1994}=2=1+12\; \backslash sqrt\{1995-1994\}=2=1+1$$

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

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

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

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

i.e.

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

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

By Pythagoras, we have

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

This implies,

Additionally we know that

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

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

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

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)}

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

Hence,

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

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