INMO/md/en-sol-inmo16.md

INMO-2016 problems and solutions

1. Let $A B C$ be triangle in which $A B=A C$. Suppose the orthocentre of the triangle lies on the in-circle. Find the ratio $A B / B C$.

Solution: Since the triangle is isosceles, the orthocentre lies on the perpendicular $A D$ from $A$ on to $B C$. Let it cut the in-circle at $H$. Now we are given that $H$ is the orthocentre of the triangle. Let $A B=A C=b$ and $B C=2 a$. Then $B D=a$. Observe that $b>a$ since $b$ is the hypotenuse and $a$ is a leg of a right-angled triangle. Let $B H$ meet $A C$ in $E$ and $C H$ meet $A B$ in $F$. By Pythagoras theorem applied to $\triangle B D H$, we get

$$B{H}^2=H{D}^2+B{D}^2=4r^2+a^{2}B\; H^\{2\}=H\; D^\{2\}+B\; D^\{2\}=4\; r^\{2\}+a^\{2\}$$

where $r$ is the in-radius of $A B C$. We want to compute $B H$ in another way. Since $A, F, H, E$ are con-cyclic, we have

$$BH\cdot BE=BF\cdot BAB\; H\; \backslash cdot\; B\; E=B\; F\; \backslash cdot\; B\; A$$

But $B F \cdot B A=B D \cdot B C=2 a^{2}$, since $A, F, D, C$ are con-cyclic. Hence $B H^{2}=4 a^{4} / B E^{2}$. But

$$B{E}^2=4a^2-C{E}^2=4a^2-B{F}^2=4a^2-{\left(\frac{2a^2}{b}\right)}^2=\frac{4a^2(b^2-a^2)}{b^2}B\; E^\{2\}=4\; a^\{2\}-C\; E^\{2\}=4\; a^\{2\}-B\; F^\{2\}=4\; a^\{2\}-\backslash left(\backslash frac\{2\; a^\{2\}\}\{b\}\backslash right)^\{2\}=\backslash frac\{4\; a^\{2\}\backslash left(b^\{2\}-a^\{2\}\backslash right)\}\{b^\{2\}\}$$

This leads to

$$B{H}^2=\frac{a^2{b}^2}{b^2-a^2}B\; H^\{2\}=\backslash frac\{a^\{2\}\; b^\{2\}\}\{b^\{2\}-a^\{2\}\}$$

Thus we get

$$\frac{a^2{b}^2}{b^2-a^2}=a^2+4r^2\backslash frac\{a^\{2\}\; b^\{2\}\}\{b^\{2\}-a^\{2\}\}=a^\{2\}+4\; r^\{2\}$$

This simplifies to $\left(a^{4} /\left(b^{2}-a^{2}\right)\right)=4 r^{2}$. Now we relate $a, b, r$ in another way using area. We know that $[A B C]=r s$, where $s$ is the semi-perimeter of $A B C$. We have $s=(b+b+2 a) / 2=b+a$. On the other hand area can be calculated using Heron's formula::

$$[ABC{]}^2=s(s-2a)(s-b)(s-b)=(b+a)(b-a)a^2=a^2(b^2-a^2)[A\; B\; C]^\{2\}=s(s-2\; a)(s-b)(s-b)=(b+a)(b-a)\; a^\{2\}=a^\{2\}\backslash left(b^\{2\}-a^\{2\}\backslash right)$$

Hence

$$r^2=\frac{[ABC{]}^2}{s^2}=\frac{a^2(b^2-a^2)}{(b+a{)}^2}r^\{2\}=\backslash frac\{[A\; B\; C]^\{2\}\}\{s^\{2\}\}=\backslash frac\{a^\{2\}\backslash left(b^\{2\}-a^\{2\}\backslash right)\}\{(b+a)^\{2\}\}$$

Using this we get

$$\frac{a^4}{b^2-a^2}=4\left(\frac{a^2(b^2-a^2)}{(b+a{)}^2}\right)\backslash frac\{a^\{4\}\}\{b^\{2\}-a^\{2\}\}=4\backslash left(\backslash frac\{a^\{2\}\backslash left(b^\{2\}-a^\{2\}\backslash right)\}\{(b+a)^\{2\}\}\backslash right)$$

Therefore $a^{2}=4(b-a)^{2}$, which gives $a=2(b-a)$ or $2 b=3 a$. Finally,

$$\frac{AB}{BC}=\frac{b}{2a}=\frac{3}{4}\backslash frac\{A\; B\}\{B\; C\}=\backslash frac\{b\}\{2\; a\}=\backslash frac\{3\}\{4\}$$

Alternate Solution 1:

We use the known facts $B H=2 R \cos B$ and $r=4 R \sin (A / 2) \sin (B / 2) \sin (C / 2)$, where $R$ is the circumradius of $\triangle A B C$ and $r$ its in-radius. Therefore

$$HD=BH\sin \mathrm{\angle }HBD=2R\cos B\sin (\frac{\pi }{2}-C)=2R{\cos }^{2}BH\; D=B\; H\; \backslash sin\; \backslash angle\; H\; B\; D=2\; R\; \backslash cos\; B\; \backslash sin\; \backslash left(\backslash frac\{\backslash pi\}\{2\}-C\backslash right)=2\; R\; \backslash cos\; ^\{2\}\; B$$

since $\angle C=\angle B$. But $\angle B=(\pi-\angle A) / 2$, since $A B C$ is isosceles. Thus we obtain

$$HD=2R{\cos }^2(\frac{\pi }{2}-\frac{A}{2})H\; D=2\; R\; \backslash cos\; ^\{2\}\backslash left(\backslash frac\{\backslash pi\}\{2\}-\backslash frac\{A\}\{2\}\backslash right)$$

However $H D$ is also the diameter of the in circle. Therefore $H D=2 r$. Thus we get

$$2R{\cos }^2(\frac{\pi }{2}-\frac{A}{2})=2r=8R\sin (A\mathrm{/}2){\sin }^2((\pi -A)\mathrm{/}4)2\; R\; \backslash cos\; ^\{2\}\backslash left(\backslash frac\{\backslash pi\}\{2\}-\backslash frac\{A\}\{2\}\backslash right)=2\; r=8\; R\; \backslash sin\; (A\; /\; 2)\; \backslash sin\; ^\{2\}((\backslash pi-A)\; /\; 4)$$

This reduces to

$$\sin (A\mathrm{/}2)=2(1-\sin (A\mathrm{/}2))\backslash sin\; (A\; /\; 2)=2(1-\backslash sin\; (A\; /\; 2))$$

Therefore $\sin (A / 2)=2 / 3$. We also observe that $\sin (A / 2)=B D / A B$. Finally

$$\frac{AB}{BC}=\frac{AB}{2BD}=\frac{1}{2\sin (A\mathrm{/}2)}=\frac{3}{4}\backslash frac\{A\; B\}\{B\; C\}=\backslash frac\{A\; B\}\{2\; B\; D\}=\backslash frac\{1\}\{2\; \backslash sin\; (A\; /\; 2)\}=\backslash frac\{3\}\{4\}$$

Alternate Solution 2:

Let $D$ be the mid-point of $B C$. Extend $A D$ to meet the circumcircle in $L$. Then we know that $H D=D L$. But $H D=2 r$. Thus $D L=2 r$. Therefore $I L=I D+D L=r+2 r=3 r$. We also know that $L B=L I$. Therefore $L B=3 r$. This gives

$$\frac{BL}{LD}=\frac{3r}{2r}=\frac{3}{2}\backslash frac\{B\; L\}\{L\; D\}=\backslash frac\{3\; r\}\{2\; r\}=\backslash frac\{3\}\{2\}$$

But $\triangle B L D$ is similar to $\triangle A B D$. So

$$\frac{AB}{BD}=\frac{BL}{LD}=\frac{3}{2}\backslash frac\{A\; B\}\{B\; D\}=\backslash frac\{B\; L\}\{L\; D\}=\backslash frac\{3\}\{2\}$$

Finally,

$$\frac{AB}{BC}=\frac{AB}{2BD}=\frac{3}{4}\backslash frac\{A\; B\}\{B\; C\}=\backslash frac\{A\; B\}\{2\; B\; D\}=\backslash frac\{3\}\{4\}$$

Alternate Solution 3:

Let $D$ be the mid-point of $B C$ and $E$ be the mid-point of $D C$. Since $D I=I H(=r)$ and $D E=E C$, the mid-point theorem implies that $I E | C H$. But $C H \perp A B$. Therefore $E I \perp A B$. Let $E I$ meet $A B$ in $F$. Then $F$ is the point of tangency of the incircle of $\triangle A B C$ with $A B$. Since the incircle is also tangent to $B C$ at $D$, we have $B F=B D$. Observe that $\triangle B F E$ is similar to $\triangle B D A$. Hence

$$\frac{AB}{BD}=\frac{BE}{BF}=\frac{BE}{BD}=\frac{BD+DE}{BD}=1+\frac{DE}{BD}=\frac{3}{2}\backslash frac\{A\; B\}\{B\; D\}=\backslash frac\{B\; E\}\{B\; F\}=\backslash frac\{B\; E\}\{B\; D\}=\backslash frac\{B\; D+D\; E\}\{B\; D\}=1+\backslash frac\{D\; E\}\{B\; D\}=\backslash frac\{3\}\{2\}$$

This gives

$$\frac{AB}{BC}=\frac{3}{4}\backslash frac\{A\; B\}\{B\; C\}=\backslash frac\{3\}\{4\}$$

2. For positive real numbers $a, b, c$, which of the following statements necessarily implies $a=b=c$ : (I) $a\left(b^{3}+c^{3}\right)=b\left(c^{3}+a^{3}\right)=c\left(a^{3}+b^{3}\right)$, (II) $a\left(a^{3}+b^{3}\right)=b\left(b^{3}+c^{3}\right)=c\left(c^{3}+a^{3}\right)$ ? Justify your answer.

Solution: We show that (I) need not imply that $a=b=c$ where as (II) always implies $a=b=c$.

Observe that $a\left(b^{3}+c^{3}\right)=b\left(c^{3}+a^{3}\right)$ gives $c^{3}(a-b)=a b\left(a^{2}-b^{2}\right)$. This gives either $a=b$ or $a b(a+b)=c^{3}$. Similarly, $b=c$ or $b c(b+c)=a^{3}$. If $a \neq b$ and $b \neq c$, we obtain

$$ab(a+b)=c^3,bc(b+c)=a^{3}a\; b(a+b)=c^\{3\},\; \backslash quad\; b\; c(b+c)=a^\{3\}$$

Therefore

$$b(a^2-c^2)+b^2(a-c)=c^3-a^{3}b\backslash left(a^\{2\}-c^\{2\}\backslash right)+b^\{2\}(a-c)=c^\{3\}-a^\{3\}$$

This gives $(a-c)\left(a^{2}+b^{2}+c^{2}+a b+b c+c a\right)=0$. Since $a, b, c$ are positive, the only possibility is $a=c$. We have therefore 4 possibilities: $a=b=c ; a \neq b, b \neq c$ and $c=a ; b \neq c, c \neq a$ and $a=b$; $c \neq a, a \neq b$ and $b=c$.

Suppose $a=b$ and $b, a \neq c$. Then $b\left(c^{3}+a^{3}\right)=c\left(a^{3}+b^{3}\right)$ gives $a c^{3}+a^{4}=2 c a^{3}$. This implies that $a(a-c)\left(a^{2}-a c-c^{2}\right)=0$. Therefore $a^{2}-a c-c^{2}=0$. Putting $a / c=x$, we get the quadratic equation $x^{2}-x-1=0$. Hence $x=(1+\sqrt{5}) / 2$. Thus we get

$$a=b=\left(\frac{1+\sqrt{5}}{2}\right)c,c\text{ arbitrary positive real number. }a=b=\backslash left(\backslash frac\{1+\backslash sqrt\{5\}\}\{2\}\backslash right)\; c,\; \backslash quad\; c\; \backslash text\; \{\; arbitrary\; positive\; real\; number.\; \}$$

Similarly, we get other two cases:

And $a=b=c$ is the fourth possibility.

Consider (II): $a\left(a^{3}+b^{3}\right)=b\left(b^{3}+c^{3}\right)=c\left(c^{3}+a^{3}\right)$. Suppose $a, b, c$ are mutually distinct. We may assume $a=\max {a, b, c}$. Hence $a>b$ and $a>c$. Using $a>b$, we get from the first relation that $a^{3}+b^{3}<b^{3}+c^{3}$. Therefore $a^{3}<c^{3}$ forcing $a<c$. This contradicts $a>c$. We conclude that $a, b, c$ cannot be mutually distinct. This means some two must be equal. If $a=b$, the equality of the first two expressions give $a^{3}+b^{3}=b^{3}+c^{3}$ so that $a=c$. Similarly, we can show that $b=c$ implies $b=a$ and $c=a$ gives $c=b$.

Alternate for (II) by a contestant: We can write

Adding, we get

$$\frac{a^3}{c}+\frac{b^3}{a}+\frac{c^3}{b}=a^2+b^2+c^2\backslash frac\{a^\{3\}\}\{c\}+\backslash frac\{b^\{3\}\}\{a\}+\backslash frac\{c^\{3\}\}\{b\}=a^\{2\}+b^\{2\}+c^\{2\}$$

Using C-S inequality, we have

Thus we obtain

$$a^2+b^2+c^2\le ab+bc+caa^\{2\}+b^\{2\}+c^\{2\}\; \backslash leq\; a\; b+b\; c+c\; a$$

However this implies $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \leq 0$ and hence $a=b=c$.

3. Let $\mathbb{N}$ denote the set of all natural numbers. Define a function $T: \mathbb{N} \rightarrow \mathbb{N}$ by $T(2 k)=k$ and $T(2 k+1)=2 k+2$. We write $T^{2}(n)=T(T(n))$ and in general $T^{k}(n)=T^{k-1}(T(n))$ for any $k>1$.

(i) Show that for each $n \in \mathbb{N}$, there exists $k$ such that $T^{k}(n)=1$.

(ii) For $k \in \mathbb{N}$, let $c_{k}$ denote the number of elements in the set $\left{n: T^{k}(n)=1\right}$. Prove that $c_{k+2}=c_{k+1}+c_{k}$, for $k \geq 1$.

Solution:

(i) For $n=1$, we have $T(1)=2$ and $T^{2}(1)=T(2)=1$. Hence we may assume that $n>1$.

Suppose $n>1$ is even. Then $T(n)=n / 2$. We observe that $(n / 2) \leq n-1$ for $n>1$.

Suppose $n>1$ is odd so that $n \geq 3$. Then $T(n)=n+1$ and $T^{2}(n)=(n+1) / 2$. Again we see that $(n+1) / 2 \leq(n-1)$ for $n \geq 3$.

Thus we see that in at most $2(n-1)$ steps $T$ sends $n$ to 1 . Hence $k \leq 2(n-1)$. (Here $2(n-1)$ is only a bound. In reality, less number of steps will do.)

(ii) We show that $c_{n}=f_{n+1}$, where $f_{n}$ is the $n$-th Fibonacci number.

Let $n \in \mathbb{N}$ and let $k \in \mathbb{N}$ be such that $T^{k}(n)=1$. Here $n$ can be odd or even. If $n$ is even, it can be either of the form $4 d+2$ or of the form $4 d$.

If $n$ is odd, then $1=T^{k}(n)=T^{k-1}(n+1)$. (Observe that $k>1$; otherwise we get $n+1=1$ which is impossible since $n \in \mathbb{N}$.) Here $n+1$ is even.

If $n=4 d+2$, then again $1=T^{k}(4 d+2)=T^{k-1}(2 d+1)$. Here $2 d+1=n / 2$ is odd.

Thus each solution of $T^{k-1}(m)=1$ produces exactly one solution of $T^{k}(n)=1$ and $n$ is either odd or of the form $4 d+2$.

If $n=4 d$, we see that $1=T^{k}(4 d)=T^{k-1}(2 d)=T^{k-2}(d)$. This shows that each solution of $T^{k-2}(m)=1$ produces exactly one solution of $T^{k}(n)=1$ of the form $4 d$.

Thus the number of solutions of $T^{k}(n)=1$ is equal to the number of solutions of $T^{k-1}(m)=1$ and the number of solutions of $T^{k-2}(l)=1$ for $k>2$. This shows that $c_{k}=c_{k-1}+c_{k-2}$ for $k>2$. We also observe that 2 is the only number which goes to 1 in one step and 4 is the only number which goes to 1 in two steps. Hence $c_{1}=1$ and $c_{2}=2$. This proves that $c_{n}=f_{n+1}$ for all $n \in \mathbb{N}$.

4. Suppose 2016 points of the circumference of a circle are coloured red and the remaining points are coloured blue. Given any natural number $n \geq 3$, prove that there is a regular $n$-sided polygon all of whose vertices are blue.

Solution: Let $A_{1}, A_{2}, \ldots, A_{2016}$ be 2016 points on the circle which are coloured red and the remain- ing blue. Let $n \geq 3$ and let $B_{1}, B_{2}, \ldots, B_{n}$ be a regular $n$-sided polygon inscribed in this circle with the vertices chosen in anti-clock-wise direction. We place $B_{1}$ at $A_{1}$. (It is possible, in this position, some other $B$ 's also coincide with some other $A$ 's.) Rotate the polygon in anti-clock-wise direction gradually till some $B$ 's coincide with (an equal number of) $A$ 's second time. We again rotate the polygon in the same direction till some $B$ 's coincide with an equal number of $A$ 's third time, and so on until we return to the original position, i.e., $B_{1}$ at $A_{1}$. We see that the number of rotations will not be more than $2016 \times n$, that is, at most these many times some $B$ 's would have coincided with an equal number of $A$ 's. Since the interval $\left(0,360^{\circ}\right)$ has infinitely many points, we can find a value $\alpha^{\circ} \in\left(0,360^{\circ}\right)$ through which the polygon can be rotated from its initial position such that no $B$ coincides with any $A$. This gives a $n$-sided regular polygon having only blue vertices.

Alternate Solution: Consider a regular $2017 \times n$-gon on the circle; say, $A_{1} A_{2} A_{3} \cdots A_{2017 n}$. For each $j, 1 \leq j \leq 2017$, consider the points $\left{A_{k}: k \equiv j(\bmod 2017)\right}$. These are the vertices of a regular $n$-gon, say $S_{j}$. We get 2017 regular $n$-gons; $S_{1}, S_{2}, \ldots, S_{2017}$. Since there are only 2016 red points, by pigeon-hole principle there must be some $n$-gon among these 2017 which does not contain any red point. But then it is a blue $n$-gon.

5. Let $A B C$ be a right-angled triangle with $\angle B=90^{\circ}$. Let $D$ be a point on $A C$ such that the in-radii of the triangles $A B D$ and $C B D$ are equal. If this common value is $r^{\prime}$ and if $r$ is the in-radius of triangle $A B C$, prove that

$$\frac{1}{r^{\mathrm{\prime }}}=\frac{1}{r}+\frac{1}{BD}\backslash frac\{1\}\{r^\{\backslash prime\}\}=\backslash frac\{1\}\{r\}+\backslash frac\{1\}\{B\; D\}$$

Solution: Let $E$ and $F$ be the incentres of triangles $A B D$ and $C B D$ respectively. Let the incircles of triangles $A B D$ and $C B D$ touch $A C$ in $P$ and $Q$ respectively. If $\angle B D A=\theta$, we see that

$$r^{\mathrm{\prime }}=PD\tan (\theta \mathrm{/}2)=QD\cot (\theta \mathrm{/}2)r^\{\backslash prime\}=P\; D\; \backslash tan\; (\backslash theta\; /\; 2)=Q\; D\; \backslash cot\; (\backslash theta\; /\; 2)$$

Hence

$$PQ=PD+QD=r^{\mathrm{\prime }}(\cot \frac{\theta }{2}+\tan \frac{\theta }{2})=\frac{2r^{\mathrm{\prime }}}{\sin \theta }P\; Q=P\; D+Q\; D=r^\{\backslash prime\}\backslash left(\backslash cot\; \backslash frac\{\backslash theta\}\{2\}+\backslash tan\; \backslash frac\{\backslash theta\}\{2\}\backslash right)=\backslash frac\{2\; r^\{\backslash prime\}\}\{\backslash sin\; \backslash theta\}$$

But we observe that

$$DP=\frac{BD+DA-AB}{2},DQ=\frac{BD+DC-BC}{2}D\; P=\backslash frac\{B\; D+D\; A-A\; B\}\{2\},\; \backslash quad\; D\; Q=\backslash frac\{B\; D+D\; C-B\; C\}\{2\}$$

Thus $P Q=(b-c-a+2 B D) / 2$. We also have

But

$$r^{\mathrm{\prime }}=\frac{PQ\sin \theta }{2}=\frac{PQ\cdot h}{2BD}r^\{\backslash prime\}=\backslash frac\{P\; Q\; \backslash sin\; \backslash theta\}\{2\}=\backslash frac\{P\; Q\; \backslash cdot\; h\}\{2\; B\; D\}$$

where $h$ is the altitude from $B$ on to $A C$. But we know that $h=a c / b$. Thus we get

$$ac=2\times r^{\mathrm{\prime }}(s+BD)=2\times \frac{PQ\cdot h}{2\times BD}(s+BD)=\frac{(b-c-a+2BD)ca(s+BD)}{2\times BD\times b}a\; c=2\; \backslash times\; r^\{\backslash prime\}(s+B\; D)=2\; \backslash times\; \backslash frac\{P\; Q\; \backslash cdot\; h\}\{2\; \backslash times\; B\; D\}(s+B\; D)=\backslash frac\{(b-c-a+2\; B\; D)\; c\; a(s+B\; D)\}\{2\; \backslash times\; B\; D\; \backslash times\; b\}$$

Thus we get

$$2\times BD\times b=2\times (BD-(s-b))(s+BD)2\; \backslash times\; B\; D\; \backslash times\; b=2\; \backslash times(B\; D-(s-b))(s+B\; D)$$

This gives $B D^{2}=s(s-b)$. Since $A B C$ is a right-angled triangle $r=s-b$. Thus we get $B D^{2}=r s$. On the other hand, we also have $[A B C]=r^{\prime}(s+B D)$. Thus we get

$$rs=[ABC]=r^{\mathrm{\prime }}(s+BD)r\; s=[A\; B\; C]=r^\{\backslash prime\}(s+B\; D)$$

Hence

$$\frac{1}{r^{\mathrm{\prime }}}=\frac{1}{r}+\frac{BD}{rs}=\frac{1}{r}+\frac{1}{BD}\backslash frac\{1\}\{r^\{\backslash prime\}\}=\backslash frac\{1\}\{r\}+\backslash frac\{B\; D\}\{r\; s\}=\backslash frac\{1\}\{r\}+\backslash frac\{1\}\{B\; D\}$$

Alternate Solution 1: Observe that

$$\frac{r^{\mathrm{\prime }}}{r}=\frac{AP}{AX}=\frac{CQ}{CX}=\frac{AP+CQ}{AC}\backslash frac\{r^\{\backslash prime\}\}\{r\}=\backslash frac\{A\; P\}\{A\; X\}=\backslash frac\{C\; Q\}\{C\; X\}=\backslash frac\{A\; P+C\; Q\}\{A\; C\}$$

where $X$ is the point at which the incircle of $A B C$ touches the side $A C$. If $s_{1}$ and $s_{2}$ are respectively the semi-perimeters of triangles $A B D$ and $C B D$, we know $A P=s_{1}-B D$ and $C Q=s_{2}-B D$. Therefore

$$\frac{r^{\mathrm{\prime }}}{r}=\frac{(s_1-BD)+(s_2-BD)}{AC}=\frac{s_1+s_2-2BD}{b}\backslash frac\{r^\{\backslash prime\}\}\{r\}=\backslash frac\{\backslash left(s\_\{1\}-B\; D\backslash right)+\backslash left(s\_\{2\}-B\; D\backslash right)\}\{A\; C\}=\backslash frac\{s\_\{1\}+s\_\{2\}-2\; B\; D\}\{b\}$$

But

$$s_1+s_2=\frac{AD+BD+c}{2}+\frac{CD+BD+a}{2}=\frac{(a+b+c)+2BD}{2}=\frac{s+BD}{2}s\_\{1\}+s\_\{2\}=\backslash frac\{A\; D+B\; D+c\}\{2\}+\backslash frac\{C\; D+B\; D+a\}\{2\}=\backslash frac\{(a+b+c)+2\; B\; D\}\{2\}=\backslash frac\{s+B\; D\}\{2\}$$

This gives

$$\frac{r^{\mathrm{\prime }}}{r}=\frac{s+BD-2BD}{b}=\frac{s-BD}{b}\backslash frac\{r^\{\backslash prime\}\}\{r\}=\backslash frac\{s+B\; D-2\; B\; D\}\{b\}=\backslash frac\{s-B\; D\}\{b\}$$

We also have

$$r^{\mathrm{\prime }}=\frac{[ABD]}{s_1}=\frac{[CBD]}{s_2}=\frac{[ABD]+[CBD]}{s_1+s_2}=\frac{[ABC]}{s+BD}=\frac{rs}{s+BD}r^\{\backslash prime\}=\backslash frac\{[A\; B\; D]\}\{s\_\{1\}\}=\backslash frac\{[C\; B\; D]\}\{s\_\{2\}\}=\backslash frac\{[A\; B\; D]+[C\; B\; D]\}\{s\_\{1\}+s\_\{2\}\}=\backslash frac\{[A\; B\; C]\}\{s+B\; D\}=\backslash frac\{r\; s\}\{s+B\; D\}$$

This implies that

$$\frac{r^{\mathrm{\prime }}}{r}=\frac{s}{s+BD}\backslash frac\{r^\{\backslash prime\}\}\{r\}=\backslash frac\{s\}\{s+B\; D\}$$

Comparing the two expressions for $r^{\prime} / r$, we see that

$$\frac{s-BD}{b}=\frac{s}{s+BD}\backslash frac\{s-B\; D\}\{b\}=\backslash frac\{s\}\{s+B\; D\}$$

Therefore $s^{2}-B D^{2}=b s$, or $B D^{2}=s(s-b)$. Thus we get $B D=\sqrt{s(s-b)}$.

We know now that

$$\frac{r^{\mathrm{\prime }}}{r}=\frac{s}{s+BD}=\frac{s-BD}{b}=\frac{BD}{(s-b)+BD}=\frac{\sqrt{s(s-b)}}{(s-b)+\sqrt{s(s-b)}}=\frac{\sqrt{s}}{\sqrt{s-b}+\sqrt{s}}\backslash frac\{r^\{\backslash prime\}\}\{r\}=\backslash frac\{s\}\{s+B\; D\}=\backslash frac\{s-B\; D\}\{b\}=\backslash frac\{B\; D\}\{(s-b)+B\; D\}=\backslash frac\{\backslash sqrt\{s(s-b)\}\}\{(s-b)+\backslash sqrt\{s(s-b)\}\}=\backslash frac\{\backslash sqrt\{s\}\}\{\backslash sqrt\{s-b\}+\backslash sqrt\{s\}\}$$

Therefore

$$\frac{r}{r^{\mathrm{\prime }}}=1+\sqrt{\frac{s-b}{s}}\backslash frac\{r\}\{r^\{\backslash prime\}\}=1+\backslash sqrt\{\backslash frac\{s-b\}\{s\}\}$$

This gives

$$\frac{1}{r^{\mathrm{\prime }}}=\frac{1}{r}+\left(\sqrt{\frac{s-b}{s}}\right)\frac{1}{r}\backslash frac\{1\}\{r^\{\backslash prime\}\}=\backslash frac\{1\}\{r\}+\backslash left(\backslash sqrt\{\backslash frac\{s-b\}\{s\}\}\backslash right)\; \backslash frac\{1\}\{r\}$$

But

$$\left(\sqrt{\frac{s-b}{s}}\right)\frac{1}{r}=\left(\frac{s-b}{\sqrt{s(s-b)}}\right)\frac{1}{r}=\left(\frac{s-b}{BD}\right)\frac{1}{r}\backslash left(\backslash sqrt\{\backslash frac\{s-b\}\{s\}\}\backslash right)\; \backslash frac\{1\}\{r\}=\backslash left(\backslash frac\{s-b\}\{\backslash sqrt\{s(s-b)\}\}\backslash right)\; \backslash frac\{1\}\{r\}=\backslash left(\backslash frac\{s-b\}\{B\; D\}\backslash right)\; \backslash frac\{1\}\{r\}$$

If $\angle B=90^{\circ}$, we know that $r=s-b$. Therfore we get

$$\frac{1}{r^{\mathrm{\prime }}}=\frac{1}{r}+\left(\frac{s-b}{BD}\right)\frac{1}{r}=\frac{1}{r}+\frac{1}{BD}\backslash frac\{1\}\{r^\{\backslash prime\}\}=\backslash frac\{1\}\{r\}+\backslash left(\backslash frac\{s-b\}\{B\; D\}\backslash right)\; \backslash frac\{1\}\{r\}=\backslash frac\{1\}\{r\}+\backslash frac\{1\}\{B\; D\}$$

Alternate Solution 2 by a contestant: Observe that $\angle E D F=90^{\circ}$. Hence $\triangle E D P$ is similar to $\triangle D F Q$. Therefore $D P \cdot D Q=E P \cdot F Q$. Taking $D P=y_{1}$ and $D Q=x_{1}$, we get $x_{1} y_{1}=\left(r^{\prime}\right)^{2}$. We also observe that $B D=x_{1}+x_{2}=y_{1}+y_{2}$. Since $\angle E B F=45^{\circ}$, we get

$$1=\tan {45}^{\circ }=\tan ({\beta }_1+{\beta }_2)=\frac{\tan {\beta }_1+\tan {\beta }_2}{1-\tan {\beta }_1\tan {\beta }_2}1=\backslash tan\; 45^\{\backslash circ\}=\backslash tan\; \backslash left(\backslash beta\_\{1\}+\backslash beta\_\{2\}\backslash right)=\backslash frac\{\backslash tan\; \backslash beta\_\{1\}+\backslash tan\; \backslash beta\_\{2\}\}\{1-\backslash tan\; \backslash beta\_\{1\}\; \backslash tan\; \backslash beta\_\{2\}\}$$

But $\tan \beta_{1}=r^{\prime} / y_{2}$ and $\tan \beta_{2}=r^{\prime} / x_{2}$. Hence we obtain

$$1=\frac{\left(r^{\mathrm{\prime }}\mathrm{/}{y}_2\right)+\left(r^{\mathrm{\prime }}\mathrm{/}{x}_2\right)}{1-{\left(r^{\mathrm{\prime }}\right)}^2\mathrm{/}{x}_2{y}_2}1=\backslash frac\{\backslash left(r^\{\backslash prime\}\; /\; y\_\{2\}\backslash right)+\backslash left(r^\{\backslash prime\}\; /\; x\_\{2\}\backslash right)\}\{1-\backslash left(r^\{\backslash prime\}\backslash right)^\{2\}\; /\; x\_\{2\}\; y\_\{2\}\}$$

Solving for $r^{\prime}$, we get

$$r^{\mathrm{\prime }}=\frac{x_2{y}_2-x_1{y}_1}{x_2+y_2}r^\{\backslash prime\}=\backslash frac\{x\_\{2\}\; y\_\{2\}-x\_\{1\}\; y\_\{1\}\}\{x\_\{2\}+y\_\{2\}\}$$

We also know

$$r=\frac{AB+BC-AC}{2}=\frac{x_2+y_2-(x_1+y_1)}{2}=\frac{(x_2-x_1)+(y_2-y_1)}{2}r=\backslash frac\{A\; B+B\; C-A\; C\}\{2\}=\backslash frac\{x\_\{2\}+y\_\{2\}-\backslash left(x\_\{1\}+y\_\{1\}\backslash right)\}\{2\}=\backslash frac\{\backslash left(x\_\{2\}-x\_\{1\}\backslash right)+\backslash left(y\_\{2\}-y\_\{1\}\backslash right)\}\{2\}$$

Finally,

But we can write

$$2x_1+2x_2+(x_2-x_1)+(y_2-y_1)=(x_1+x_2+x_2-x_1)+(y_1+y_2+y_2-y_1)=2(x_2+y_2)2\; x\_\{1\}+2\; x\_\{2\}+\backslash left(x\_\{2\}-x\_\{1\}\backslash right)+\backslash left(y\_\{2\}-y\_\{1\}\backslash right)=\backslash left(x\_\{1\}+x\_\{2\}+x\_\{2\}-x\_\{1\}\backslash right)+\backslash left(y\_\{1\}+y\_\{2\}+y\_\{2\}-y\_\{1\}\backslash right)=2\backslash left(x\_\{2\}+y\_\{2\}\backslash right)$$

and

Therefore

$$\frac{1}{r}+\frac{1}{BD}=\frac{2(x_2+y_2)}{2(x_2{y}_2-x_1{y}_1)}=\frac{1}{r^{\mathrm{\prime }}}\backslash frac\{1\}\{r\}+\backslash frac\{1\}\{B\; D\}=\backslash frac\{2\backslash left(x\_\{2\}+y\_\{2\}\backslash right)\}\{2\backslash left(x\_\{2\}\; y\_\{2\}-x\_\{1\}\; y\_\{1\}\backslash right)\}=\backslash frac\{1\}\{r^\{\backslash prime\}\}$$

Remark: One can also choose $B=(0,0), A=(0, a)$ and $C=(1,0)$ and the coordinate geometry proof gets reduced considerbly.

6. Consider a non-constant arithmetic progression $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ Suppose there exist relatively prime positive integers $p>1$ and $q>1$ such that $a_{1}^{2}, a_{p+1}^{2}$ and $a_{q+1}^{2}$ are also the terms of the same arithmetic progression. Prove that the terms of the arithmetic progression are all integers.

Solution: Let us take $a_{1}=a$. We have

$$a^2=a+kd,(a+pd{)}^2=a+ld,(a+qd{)}^2=a+mda^\{2\}=a+k\; d,\; \backslash quad(a+p\; d)^\{2\}=a+l\; d,\; \backslash quad(a+q\; d)^\{2\}=a+m\; d$$

Thus we have

$$a+ld=(a+pd{)}^2=a^2+2pad+p^2{d}^2=a+kd+2pad+p^2{d}^{2}a+l\; d=(a+p\; d)^\{2\}=a^\{2\}+2\; p\; a\; d+p^\{2\}\; d^\{2\}=a+k\; d+2\; p\; a\; d+p^\{2\}\; d^\{2\}$$

Since we have non-constant AP, we see that $d \neq 0$. Hence we obtain $2 p a+p^{2} d=l-k$. Similarly, we get $2 q a+q^{2} d=m-k$. Observe that $p^{2} q-p q^{2} \neq 0$. Otherwise $p=q$ and $\operatorname{gcd}(p, q)=p>1$ which is a contradiction to the given hypothesis that $\operatorname{gcd}(p, q)=1$. Hence we can solve the two equations for $a, d$ :

$$a=\frac{p^2(m-k)-q^2(l-k)}{2(p^{2}q-p{q}^2)},d=\frac{q(l-k)-p(m-k)}{p^{2}q-p{q}^2}a=\backslash frac\{p^\{2\}(m-k)-q^\{2\}(l-k)\}\{2\backslash left(p^\{2\}\; q-p\; q^\{2\}\backslash right)\},\; \backslash quad\; d=\backslash frac\{q(l-k)-p(m-k)\}\{p^\{2\}\; q-p\; q^\{2\}\}$$

It follows that $a, d$ are rational numbers. We also have

$$p^2{a}^2=p^{2}a+k{p}^{2}dp^\{2\}\; a^\{2\}=p^\{2\}\; a+k\; p^\{2\}\; d$$

But $p^{2} d=l-k-2 p a$. Thus we get

$$p^2{a}^2=p^{2}a+k(l-k-2pa)=(p-2k)pa+k(l-k)p^\{2\}\; a^\{2\}=p^\{2\}\; a+k(l-k-2\; p\; a)=(p-2\; k)\; p\; a+k(l-k)$$

This shows that pa satisfies the equation

$$x^2-(p-2k)x-k(l-k)=0x^\{2\}-(p-2\; k)\; x-k(l-k)=0$$

Since $a$ is rational, we see that $p a$ is rational. Write $p a=w / z$, where $w$ is an integer and $z$ is a natural numbers such that $\operatorname{gcd}(w, z)=1$. Substituting in the equation, we obtain

$$w^2-(p-2k)wz-k(l-k)z^2=0w^\{2\}-(p-2\; k)\; w\; z-k(l-k)\; z^\{2\}=0$$

This shows $z$ divides $w$. Since $\operatorname{gcd}(w, z)=1$, it follows that $z=1$ and $p a=w$ an integer. (In fact any rational solution of a monic polynomial with integer coefficients is necessarily an integer.) Similarly, we can prove that $q a$ is an integer. Since $\operatorname{gcd}(p, q)=1$, there are integers $u$ and $v$ such that $p u+q v=1$. Therefore $a=(p a) u+(q a) v$. It follows that $a$ is an integer.

But $p^{2} d=l-k-2 p a$. Hence $p^{2} d$ is an integer. Similarly, $q^{2} d$ is also an integer. Since $\operatorname{gcd}\left(p^{2}, q^{2}\right)=1$, it follows that $d$ is an integer. Combining these two, we see that all the terms of the AP are integers.

Alternatively, we can prove that $a$ and $d$ are integers in another way. We have seen that $a$ and $d$ are rationals; and we have three relations:

$$a^2=a+kd,p^{2}d+2pa=n_1,q^{2}d+2qa=n_{2}a^\{2\}=a+k\; d,\; \backslash quad\; p^\{2\}\; d+2\; p\; a=n\_\{1\},\; \backslash quad\; q^\{2\}\; d+2\; q\; a=n\_\{2\}$$

where $n_{1}=l-k$ and $n_{2}=m-k$. Let $a=u / v$ and $d=x / y$ where $u, x$ are integers and $v, y$ are natural numbers, and $\operatorname{gcd}(u, v)=1, \operatorname{gcd}(x, y)=1$. Putting this in these relations, we obtain

Now (1) shows that $v \mid u^{2} y$. Since $\operatorname{gcd}(u, v)=1$, it follows that $v \mid y$. Similarly (2) shows that $y \mid p^{2} v x$. Using $\operatorname{gcd}(y, x)=1$, we get that $y \mid p^{2} v$. Similarly, (3) shows that $y \mid q^{2} v$. Therefore $y$ divides $\operatorname{gcd}\left(p^{2} v, q^{2} v\right)=v$. The two results $v \mid y$ and $y \mid v$ imply $v=y$, since both $v, y$ are positive.

Substitute this in (1) to get

$$u^2=uv+kxvu^\{2\}=u\; v+k\; x\; v$$

This shows that $v \mid u^{2}$. Since $\operatorname{gcd}(u, v)=1$, it follows that $v=1$. This gives $v=y=1$. Finally $a=u$ and $d=x$ which are integers.