INMO/md/en-inmosol-15.md

Problems and Solutions: INMO-2015

1. Let $A B C$ be a right-angled triangle with $\angle B=90^{\circ}$. Let $B D$ be the altitude from $B$ on to $A C$. Let $P, Q$ and $I$ be the incentres of triangles $A B D, C B D$ and $A B C$ respectively. Show that the circumcentre of of the triangle $P I Q$ lies on the hypotenuse $A C$.

Solution: We begin with the following lemma:

Lemma: Let $X Y Z$ be a triangle with $\angle X Y Z=90+\alpha$. Construct an isosceles triangle $X E Z$, externally on the side $X Z$, with base angle $\alpha$. Then $E$ is the circumcentre of $\triangle X Y Z$.

Proof of the Lemma: Draw $E D \perp$ $X Z$. Then $D E$ is the perpendicular bisector of $X Z$. We also observe that $\angle X E D=\angle Z E D=90-\alpha$. Observe that $E$ is on the perpendicular bisector of $X Z$. Construct the circumcircle of $X Y Z$. Draw perpendicular bisector of $X Y$ and let it meet $D E$ in $F$. Then $F$ is the circumcentre of $\triangle X Y Z$. Join $X F$. Then $\angle X F D=90-\alpha$. But we know

that $\angle X E D=90-\alpha$. Hence $E=F$. Let $r_{1}, r_{2}$ and $r$ be the inradii of the triangles $A B D, C B D$ and $A B C$ respectively. Join $P D$ and $D Q$. Observe that $\angle P D Q=90^{\circ}$. Hence

$$P{Q}^2=P{D}^2+D{Q}^2=2r_1^2+2r_2^{2}P\; Q^\{2\}=P\; D^\{2\}+D\; Q^\{2\}=2\; r\_\{1\}^\{2\}+2\; r\_\{2\}^\{2\}$$

Let $s_{1}=(A B+B D+D A) / 2$. Observe that $B D=c a / b$ and $A D=$ $\sqrt{A B^{2}-B D^{2}}=\sqrt{c^{2}-\left(\frac{c a}{b}\right)^{2}}=c^{2} / b$. This gives $s_{1}=c s / b$. But $r_{1}=$ $s_{1}-c=(c / b)(s-b)=c r / b$. Similarly, $r_{2}=a r / b$. Hence

$$P{Q}^2=2r^2\left(\frac{c^2+a^2}{b^2}\right)=2r^{2}P\; Q^\{2\}=2\; r^\{2\}\backslash left(\backslash frac\{c^\{2\}+a^\{2\}\}\{b^\{2\}\}\backslash right)=2\; r^\{2\}$$

Consider $\triangle P I Q$. Observe that $\angle P I Q=90+(B / 2)=135$. Hence $P Q$ subtends $90^{\circ}$ on the circumference of the circumcircle of $\triangle P I Q$. But we have seen that $\angle P D Q=90^{\circ}$. Now construct a circle with $P Q$ as diameter. Let it cut $A C$ again in $K$. It follows that $\angle P K Q=90^{\circ}$ and the points $P, D, K, Q$ are concyclic. We also notice $\angle K P Q=\angle K D Q=45^{\circ}$

and $\angle P Q K=\angle P D A=45^{\circ}$.

Thus $P K Q$ is an isosceles right-angled triangle with $K P=K Q$. Therfore $K P^{2}+K Q^{2}=P Q^{2}=2 r^{2}$ and hence $K P=K Q=r$.

Now $\angle P I Q=90+45$ and $\angle P K Q=2 \times 45^{\circ}=90^{\circ}$ with $K P=K Q=r$.

Hence $K$ is the circumcentre of $\triangle P I Q$.

(Incidentally, This also shows that $K I=r$ and hence $K$ is the point of contact of the incircle of $\triangle A B C$ with $A C$.)

Solution 2: Here we use computation to prove that the point of contact $K$ of the incircle with $A C$ is the circumcentre of $\triangle P I Q$. We show that $K P=K Q=r$. Let $r_{1}$ and $r_{2}$ be the inradii of triangles $A B D$ and $C B D$ respectively. Draw $P L \perp A C$ and $Q M \perp A C$. If $s_{1}$ is the semiperimeter of $\triangle A B D$, then $A L=s_{1}-B D$.

But

$$s_1=\frac{AB+BD+DA}{2},BD=\frac{ca}{b},AD=\frac{c^2}{b}s\_\{1\}=\backslash frac\{A\; B+B\; D+D\; A\}\{2\},\; \backslash quad\; B\; D=\backslash frac\{c\; a\}\{b\},\; \backslash quad\; A\; D=\backslash frac\{c^\{2\}\}\{b\}$$

Hence $s_{1}=c s / b$. This gives $r_{1}=s_{1}-c=c r / b, A L=s_{1}-B D=c(s-a) / b$. Hence $K L=A K-A L=(s-a)-\frac{c(s-a)}{b}=\frac{(b-c)(s-a)}{b}$. We observe that $2 r^{2}=\frac{(c+a-b)^{2}}{2}=\frac{c^{2}+a^{2}+b^{2}-2 b c-2 a b+2 c a}{2}=\left(b^{2}-b a-b c+a c\right)=(b-c)(b-a)$.

This gives

Thus $K L=r a / b$. Finally,

$$K{P}^2=K{L}^2+L{P}^2=\frac{r^2{a}^2}{b^2}+\frac{r^2+c^2}{b^2}=r^{2}K\; P^\{2\}=K\; L^\{2\}+L\; P^\{2\}=\backslash frac\{r^\{2\}\; a^\{2\}\}\{b^\{2\}\}+\backslash frac\{r^\{2\}+c^\{2\}\}\{b^\{2\}\}=r^\{2\}$$

Thus $K P=r$. Similarly, $K Q=r$. This gives $K P=K I=K Q=r$ and therefore $K$ is the circumcentre of $\triangle K I Q$.

(Incidentally, this also shows that $K L=c a / b=r_{2}$ and $K M=r_{1}$.)

2. For any natural number $n>1$, write the infinite decimal expansion of $1 / n$ (for example, we write $1 / 2=0.4 \overline{9}$ as its infinite decimal expansion, not 0.5 ). Determine the length of the non-periodic part of the (infinite) decimal expansion of $1 / n$.

Solution: For any prime $p$, let $\nu_{p}(n)$ be the maximum power of $p$ dividing $n$; ie $p^{\nu_{p}(n)}$ divides $n$ but not higher power. Let $r$ be the length of the non-periodic part of the infinite decimal expansion of $1 / n$.

Write

$$\frac{1}{n}=0.a_1{a}_2\cdots a_r\overline{b_1{b}_2\cdots b_s}\backslash frac\{1\}\{n\}=0\; .\; a\_\{1\}\; a\_\{2\}\; \backslash cdots\; a\_\{r\}\; \backslash overline\{b\_\{1\}\; b\_\{2\}\; \backslash cdots\; b\_\{s\}\}$$

We show that $r=\max \left(\nu_{2}(n), \nu_{5}(n)\right)$.

Let $a$ and $b$ be the numbers $a_{1} a_{2} \cdots a_{r}$ and $b=b_{1} b_{2} \cdots b_{s}$ respectively. (Here $a_{1}$ and $b_{1}$ can be both 0.) Then

$$\frac{1}{n}=\frac{1}{{10}^r}(a+\sum _{k\ge 1}\frac{b}{{\left({10}^s\right)}^k})=\frac{1}{{10}^r}(a+\frac{b}{{10}^s-1})\backslash frac\{1\}\{n\}=\backslash frac\{1\}\{10^\{r\}\}\backslash left(a+\backslash sum\_\{k\; \backslash geq\; 1\}\; \backslash frac\{b\}\{\backslash left(10^\{s\}\backslash right)^\{k\}\}\backslash right)=\backslash frac\{1\}\{10^\{r\}\}\backslash left(a+\backslash frac\{b\}\{10^\{s\}-1\}\backslash right)$$

Thus we get $10^{r}\left(10^{s}-1\right)=n\left(\left(10^{s}-1\right) a+b\right)$. It shows that $r \geq$ $\max \left(\nu_{2}(n), \nu_{5}(n)\right)$. Suppose $r>\max \left(\nu_{2}(n), \nu_{5}(n)\right)$. Then 10 divides $b-a$. Hence the last digits of $a$ and $b$ are equal: $a_{r}=b_{s}$. This means

$$\frac{1}{n}=0.a_1{a}_2\cdots a_{r-1}\overline{b_s{b}_1{b}_2\cdots b_{s-1}}\backslash frac\{1\}\{n\}=0\; .\; a\_\{1\}\; a\_\{2\}\; \backslash cdots\; a\_\{r-1\}\; \backslash overline\{b\_\{s\}\; b\_\{1\}\; b\_\{2\}\; \backslash cdots\; b\_\{s-1\}\}$$

This contradicts the definition of $r$. Therefore $r=\max \left(\nu_{2}(n), \nu_{5}(n)\right)$.

3. Find all real functions $f$ from $\mathbb{R} \rightarrow \mathbb{R}$ satisfying the relation

$$f(x^2+yf(x))=xf(x+y)f\backslash left(x^\{2\}+y\; f(x)\backslash right)=x\; f(x+y)$$

Solution: Put $x=0$ and we get $f(y f(0))=0$. If $f(0) \neq 0$, then $y f(0)$ takes all real values when $y$ varies over real line. We get $f(x) \equiv 0$. Suppose $f(0)=0$. Taking $y=-x$, we get $f\left(x^{2}-x f(x)\right)=0$ for all real $x$.

Suppose there exists $x_{0} \neq 0$ in $\mathbb{R}$ such that $f\left(x_{0}\right)=0$. Putting $x=x_{0}$ in the given relation we get

$$f\left(x_0^2\right)=x_{0}f(x_0+y)f\backslash left(x\_\{0\}^\{2\}\backslash right)=x\_\{0\}\; f\backslash left(x\_\{0\}+y\backslash right)$$

for all $y \in \mathbb{R}$. Now the left side is a constant and hence it follows that $f$ is a constant function. But the only constant function which satisfies the equation is identically zero function, which is already obtained. Hence we may consider the case where $f(x) \neq 0$ for all $x \neq 0$.

Since $f\left(x^{2}-x f(x)\right)=0$, we conclude that $x^{2}-x f(x)=0$ for all $x \neq 0$. This implies that $f(x)=x$ for all $x \neq 0$. Since $f(0)=0$, we conclude that $f(x)=x$ for all $x \in R$.

Thus we have two functions: $f(x) \equiv 0$ and $f(x)=x$ for all $x \in \mathbb{R}$.

4. There are four basket-ball players $A, B, C, D$. Initially, the ball is with $A$. The ball is always passed from one person to a different person. In how many ways can the ball come back to $A$ after seven passes? (For example $A \rightarrow C \rightarrow B \rightarrow D \rightarrow A \rightarrow B \rightarrow C \rightarrow A$ and $A \rightarrow D \rightarrow A \rightarrow D \rightarrow C \rightarrow A \rightarrow B \rightarrow A$ are two ways in which the ball can come back to $A$ after seven passes.)

Solution: Let $x_{n}$ be the number of ways in which $A$ can get back the ball after $n$ passes. Let $y_{n}$ be the number of ways in which the ball goes back to a fixed person other than $A$ after $n$ passes. Then

$$x_n=3y_{n-1}\text{, }x\_\{n\}=3\; y\_\{n-1\}\; \backslash text\; \{,\; \}$$

and

$$y_n=x_{n-1}+2y_{n-1}y\_\{n\}=x\_\{n-1\}+2\; y\_\{n-1\}$$

We also have $x_{1}=0, x_{2}=3, y_{1}=1$ and $y_{2}=2$.

Eliminating $y_{n}$ and $y_{n-1}$, we get $x_{n+1}=3 x_{n-1}+2 x_{n}$. Thus

Alternate solution: Since the ball goes back to one of the other 3 persons, we have

$$x_n+3y_n=3^{n}x\_\{n\}+3\; y\_\{n\}=3^\{n\}$$

since there are $3^{n}$ ways of passing the ball in $n$ passes. Using $x_{n}=$ $3 y_{n-1}$, we obtain

$$x_{n-1}+x_n=3^{n-1}x\_\{n-1\}+x\_\{n\}=3^\{n-1\}$$

with $x_{1}=0$. Thus

$$\begin{array}{r}{x}_7=3^6-x_6=3^6-3^5+x_5=3^6-3^5+3^4-x_4=3^6-3^5+3^4-3^3+x_3\\ =3^6-3^5+3^4-3^3+3^2-x_2=3^6-3^5+3^4-3^3+3^2-3\\ =(2\times 3^5)+(2\times 3^3)+(2\times 3)=486+54+6=546\end{array}\backslash begin\{array\}\{r\}\; x\_\{7\}=3^\{6\}-x\_\{6\}=3^\{6\}-3^\{5\}+x\_\{5\}=3^\{6\}-3^\{5\}+3^\{4\}-x\_\{4\}=3^\{6\}-3^\{5\}+3^\{4\}-3^\{3\}+x\_\{3\}\; \backslash \backslash \; =3^\{6\}-3^\{5\}+3^\{4\}-3^\{3\}+3^\{2\}-x\_\{2\}=3^\{6\}-3^\{5\}+3^\{4\}-3^\{3\}+3^\{2\}-3\; \backslash \backslash \; =\backslash left(2\; \backslash times\; 3^\{5\}\backslash right)+\backslash left(2\; \backslash times\; 3^\{3\}\backslash right)+(2\; \backslash times\; 3)=486+54+6=546\; \backslash end\{array\}$$

5. Let $A B C D$ be a convex quadrilateral. Let the diagonals $A C$ and $B D$ intersect in $P$. Let $P E, P F, P G$ and $P H$ be the altitudes from $P$ on to the sides $A B, B C, C D$ and $D A$ respectively. Show that $A B C D$ has an incircle if and only if

$$\frac{1}{PE}+\frac{1}{PG}=\frac{1}{PF}+\frac{1}{PH}\backslash frac\{1\}\{P\; E\}+\backslash frac\{1\}\{P\; G\}=\backslash frac\{1\}\{P\; F\}+\backslash frac\{1\}\{P\; H\}$$

Solution: Let $A P=p, B P=q, C P=r, D P=s ; A B=a, B C=b$, $C D=c$ and $D A=d$. Let $\angle A P B=\angle C P D=\theta$. Then $\angle B P C=\angle D P A=$ $\pi-\theta$. Let us also write $P E=h_{1}, P F=h_{2}, P G=h_{3}$ and $P H=h_{4}$.

Observe that

$$h_{1}a=pq\sin \theta ,h_{2}b=qr\sin \theta ,h_{3}c=rs\sin \theta ,h_{4}d=sp\sin \theta h\_\{1\}\; a=p\; q\; \backslash sin\; \backslash theta,\; \backslash quad\; h\_\{2\}\; b=q\; r\; \backslash sin\; \backslash theta,\; \backslash quad\; h\_\{3\}\; c=r\; s\; \backslash sin\; \backslash theta,\; \backslash quad\; h\_\{4\}\; d=s\; p\; \backslash sin\; \backslash theta$$

Hence

$$\frac{1}{h_1}+\frac{1}{h_3}=\frac{1}{h_2}+\frac{1}{h_4}\backslash frac\{1\}\{h\_\{1\}\}+\backslash frac\{1\}\{h\_\{3\}\}=\backslash frac\{1\}\{h\_\{2\}\}+\backslash frac\{1\}\{h\_\{4\}\}$$

is equivalent to

$$\frac{a}{pq}+\frac{c}{rs}=\frac{b}{qr}+\frac{d}{sp}\backslash frac\{a\}\{p\; q\}+\backslash frac\{c\}\{r\; s\}=\backslash frac\{b\}\{q\; r\}+\backslash frac\{d\}\{s\; p\}$$

This is the same as

$$ars+cpq=bsp+dqra\; r\; s+c\; p\; q=b\; s\; p+d\; q\; r$$

Thus we have to prove that $a+c=b+d$ if and only if $a r s+c p q=b s p+d q r$. Now we can write $a+c=b+d$ as

$$a^2+c^2+2ac=b^2+d^2+2bda^\{2\}+c^\{2\}+2\; a\; c=b^\{2\}+d^\{2\}+2\; b\; d$$

But we know that

Hence $a+c=b+d$ is equivalent to

$$-pq\cos \theta +-rs\cos \theta +ac=ps\cos \theta +qr\cos \theta +bd-p\; q\; \backslash cos\; \backslash theta+-r\; s\; \backslash cos\; \backslash theta+a\; c=p\; s\; \backslash cos\; \backslash theta+q\; r\; \backslash cos\; \backslash theta+b\; d$$

Similarly, by squaring ars $+c p q=b s p+d q r$ we can show that it is equivalent to

$$-pq\cos \theta +-rs\cos \theta +ac=ps\cos \theta +qr\cos \theta +bd-p\; q\; \backslash cos\; \backslash theta+-r\; s\; \backslash cos\; \backslash theta+a\; c=p\; s\; \backslash cos\; \backslash theta+q\; r\; \backslash cos\; \backslash theta+b\; d$$

We conclude that $a+c=b+d$ is equivalent to $c p q+a r s=b p s+d q r$. Hence $A B C D$ has an in circle if and only if

$$\frac{1}{h_1}+\frac{1}{h_3}=\frac{1}{h_2}+\frac{1}{h_4}\backslash frac\{1\}\{h\_\{1\}\}+\backslash frac\{1\}\{h\_\{3\}\}=\backslash frac\{1\}\{h\_\{2\}\}+\backslash frac\{1\}\{h\_\{4\}\}$$

6. From a set of 11 square integers, show that one can choose 6 numbers $a^{2}, b^{2}, c^{2}, d^{2}, e^{2}, f^{2}$ such that

$$a^2+b^2+c^2\equiv d^2+e^2+f^2(12)a^\{2\}+b^\{2\}+c^\{2\}\; \backslash equiv\; d^\{2\}+e^\{2\}+f^\{2\}\; \backslash quad(\backslash bmod\; 12)$$

Solution: The first observation is that we can find 5 pairs of squares such that the two numbers in a pair have the same parity. We can see this as follows:

Odd numbers	Even numbers	Odd pairs	Even pairs	Total pairs
0	11	0	5	5
1	10	0	5	5
2	9	1	4	5
3	8	1	4	5
4	7	2	3	5
5	6	2	3	5
6	5	3	2	5
7	4	3	2	5
8	3	4	1	5
9	2	4	1	5
10	1	5	0	5
11	0	5	0	5

Let us take such 5 pairs: say $\left(x_{1}^{2}, y_{1}^{2}\right),\left(x_{2}^{2}, y_{2}^{2}\right), \ldots,\left(x_{5}^{2}, y_{5}^{2}\right)$. Then $x_{j}^{2}-y_{j}^{2}$ is divisible by 4 for $1 \leq j \leq 5$. Let $r_{j}$ be the remainder when $x_{j}^{2}-y_{j}^{2}$ is divisible by $3,1 \leq j \leq 3$. We have 5 remainders $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$. But these can be 0,1 or 2 . Hence either one of the remainders occur 3 times or each of the remainders occur once. If, for example $r_{1}=r_{2}=r_{3}$, then 3 divides $r_{1}+r_{2}+r_{3}$; if $r_{1}=0, r_{2}=1$ and $r_{3}=2$, then again 3 divides $r_{1}+r_{2}+r_{3}$. Thus we can always find three remainders whose sum is divisible by 3 . This means we can find 3 pairs, say, $\left(x_{1}^{2}, y_{1}^{2}\right),\left(x_{2}^{2}, y_{2}^{2}\right),\left(x_{3}^{2}, y_{3}^{2}\right)$ such that 3 divides $\left(x_{1}^{2}-y_{1}^{2}\right)+\left(x_{2}^{2}-y_{2}^{2}\right)+\left(x_{3}^{2}-y_{3}^{2}\right)$. Since each difference is divisible by 4 , we conclude that we can find 6 numbers $a^{2}, b^{2}, c^{2}, d^{2}, e^{2}, f^{2}$ such that

$$a^2+b^2+c^2\equiv d^2+e^2+f^2(12)a^\{2\}+b^\{2\}+c^\{2\}\; \backslash equiv\; d^\{2\}+e^\{2\}+f^\{2\}\; \backslash quad(\backslash bmod\; 12)$$