exam: INMO

problem: 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

![](https://cdn.mathpix.com/cropped/2024_06_05_8eb8bd67d3f7fc489f33g-1.jpg?height=328&width=496&top_left_y=600&top_left_x=1313)

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

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

$$
B H \cdot B E=B F \cdot B A
$$

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

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

This leads to

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

Thus we get

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

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::

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

Hence

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

Using this we get

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

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

$$
\frac{A B}{B C}=\frac{b}{2 a}=\frac{3}{4}
$$

## 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

$$
H D=B H \sin \angle H B D=2 R \cos B \sin \left(\frac{\pi}{2}-C\right)=2 R \cos ^{2} B
$$

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

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

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

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

This reduces to

$$
\sin (A / 2)=2(1-\sin (A / 2))
$$

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

$$
\frac{A B}{B C}=\frac{A B}{2 B D}=\frac{1}{2 \sin (A / 2)}=\frac{3}{4}
$$

## 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

$$
\frac{B L}{L D}=\frac{3 r}{2 r}=\frac{3}{2}
$$

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

$$
\frac{A B}{B D}=\frac{B L}{L D}=\frac{3}{2}
$$

Finally,

$$
\frac{A B}{B C}=\frac{A B}{2 B D}=\frac{3}{4}
$$

## 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

$$
\frac{A B}{B D}=\frac{B E}{B F}=\frac{B E}{B D}=\frac{B D+D E}{B D}=1+\frac{D E}{B D}=\frac{3}{2}
$$

This gives

$$
\frac{A B}{B C}=\frac{3}{4}
$$

solution_match: Solution:


problem: 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

$$
a b(a+b)=c^{3}, \quad b c(b+c)=a^{3}
$$

Therefore

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

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

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

Similarly, we get other two cases:

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

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

$$
\begin{aligned}
\frac{a^{3}}{c}+\frac{b^{3}}{c} & =\frac{c^{3}}{a}+a^{2} \\
\frac{b^{3}}{a}+\frac{c^{3}}{a} & =\frac{a^{3}}{b}+b^{2} \\
\frac{c^{3}}{b}+\frac{a^{3}}{b} & =\frac{b^{3}}{c}+c^{2}
\end{aligned}
$$

Adding, we get

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

Using C-S inequality, we have

$$
\begin{aligned}
\left(a^{2}+b^{2}+c^{2}\right)^{2} & =\left(\frac{\sqrt{a^{3}}}{\sqrt{c}} \cdot \sqrt{a c}+\frac{\sqrt{b^{3}}}{\sqrt{a}} \cdot \sqrt{b a}+\frac{\sqrt{c^{3}}}{\sqrt{b}} \cdot \sqrt{c b}\right)^{2} \\
& \leq\left(\frac{a^{3}}{c}+\frac{b^{3}}{a}+\frac{c^{3}}{b}\right)(a c+b a+c b) \\
& =\left(a^{2}+b^{2}+c^{2}\right)(a b+b c+c a)
\end{aligned}
$$

Thus we obtain

$$
a^{2}+b^{2}+c^{2} \leq a b+b c+c a
$$

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


problem:
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 $n \in \mathbb{N}$.

solution_match: ## Solution:


problem: 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.


problem:
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

$$
\frac{1}{r^{\prime}}=\frac{1}{r}+\frac{1}{B D}
$$

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

$$
r^{\prime}=P D \tan (\theta / 2)=Q D \cot (\theta / 2)
$$

Hence

$$
P Q=P D+Q D=r^{\prime}\left(\cot \frac{\theta}{2}+\tan \frac{\theta}{2}\right)=\frac{2 r^{\prime}}{\sin \theta}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_8eb8bd67d3f7fc489f33g-5.jpg?height=336&width=613&top_left_y=1252&top_left_x=1271)

But we observe that

$$
D P=\frac{B D+D A-A B}{2}, \quad D Q=\frac{B D+D C-B C}{2}
$$

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

$$
\begin{aligned}
\frac{a c}{2}=[A B C]=[A B D]+[C B D]=r^{\prime} \frac{(A B+B D+D A)}{2} & +r^{\prime} \frac{(C B+B D+D C)}{2} \\
= & r^{\prime} \frac{(c+a+b+2 B D)}{2}=r^{\prime}(s+B D)
\end{aligned}
$$

But

$$
r^{\prime}=\frac{P Q \sin \theta}{2}=\frac{P Q \cdot h}{2 B D}
$$

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

$$
a c=2 \times r^{\prime}(s+B D)=2 \times \frac{P Q \cdot h}{2 \times B D}(s+B D)=\frac{(b-c-a+2 B D) c a(s+B D)}{2 \times B D \times b}
$$

Thus we get

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

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

$$
r s=[A B C]=r^{\prime}(s+B D)
$$

Hence

$$
\frac{1}{r^{\prime}}=\frac{1}{r}+\frac{B D}{r s}=\frac{1}{r}+\frac{1}{B D}
$$

## Alternate Solution 1: Observe that

$$
\frac{r^{\prime}}{r}=\frac{A P}{A X}=\frac{C Q}{C X}=\frac{A P+C Q}{A C}
$$

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

$$
\frac{r^{\prime}}{r}=\frac{\left(s_{1}-B D\right)+\left(s_{2}-B D\right)}{A C}=\frac{s_{1}+s_{2}-2 B D}{b}
$$

But

$$
s_{1}+s_{2}=\frac{A D+B D+c}{2}+\frac{C D+B D+a}{2}=\frac{(a+b+c)+2 B D}{2}=\frac{s+B D}{2}
$$

This gives

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

We also have

$$
r^{\prime}=\frac{[A B D]}{s_{1}}=\frac{[C B D]}{s_{2}}=\frac{[A B D]+[C B D]}{s_{1}+s_{2}}=\frac{[A B C]}{s+B D}=\frac{r s}{s+B D}
$$

This implies that

$$
\frac{r^{\prime}}{r}=\frac{s}{s+B D}
$$

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

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

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

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

Therefore

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

This gives

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

But

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

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

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

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

$$
1=\tan 45^{\circ}=\tan \left(\beta_{1}+\beta_{2}\right)=\frac{\tan \beta_{1}+\tan \beta_{2}}{1-\tan \beta_{1} \tan \beta_{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_8eb8bd67d3f7fc489f33g-7.jpg?height=342&width=569&top_left_y=661&top_left_x=1277)

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

$$
1=\frac{\left(r^{\prime} / y_{2}\right)+\left(r^{\prime} / x_{2}\right)}{1-\left(r^{\prime}\right)^{2} / x_{2} y_{2}}
$$

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

$$
r^{\prime}=\frac{x_{2} y_{2}-x_{1} y_{1}}{x_{2}+y_{2}}
$$

We also know

$$
r=\frac{A B+B C-A C}{2}=\frac{x_{2}+y_{2}-\left(x_{1}+y_{1}\right)}{2}=\frac{\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)}{2}
$$

Finally,

$$
\begin{aligned}
\frac{1}{r}+\frac{1}{B D} & =\frac{2}{\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)}+\frac{1}{x_{1}+x_{2}} \\
& =\frac{2 x_{1}+2 x_{2}+\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)}{\left(x_{1}+x_{2}\right)\left(\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)\right)}
\end{aligned}
$$

But we can write

$$
2 x_{1}+2 x_{2}+\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)=\left(x_{1}+x_{2}+x_{2}-x_{1}\right)+\left(y_{1}+y_{2}+y_{2}-y_{1}\right)=2\left(x_{2}+y_{2}\right)
$$

and

$$
\begin{aligned}
\left(x_{1}+x_{2}\right)\left(\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)\right)=2\left(x_{1}+x_{2}\right) & \left(x_{2}-y_{1}\right) \\
& =2\left(x_{2}\left(x_{2}+x_{1}-y_{1}\right)-x_{1} y_{1}\right)=2\left(x_{2} y_{2}-x_{1} y_{1}\right)
\end{aligned}
$$

Therefore

$$
\frac{1}{r}+\frac{1}{B D}=\frac{2\left(x_{2}+y_{2}\right)}{2\left(x_{2} y_{2}-x_{1} y_{1}\right)}=\frac{1}{r^{\prime}}
$$

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


problem: 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

$$
a^{2}=a+k d, \quad(a+p d)^{2}=a+l d, \quad(a+q d)^{2}=a+m d
$$

Thus we have

$$
a+l d=(a+p d)^{2}=a^{2}+2 p a d+p^{2} d^{2}=a+k d+2 p a d+p^{2} d^{2}
$$

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 $a, d$ :

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

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

$$
p^{2} a^{2}=p^{2} a+k p^{2} d
$$

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

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

This shows that pa satisfies the equation

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

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

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

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:

$$
a^{2}=a+k d, \quad p^{2} d+2 p a=n_{1}, \quad q^{2} d+2 q a=n_{2}
$$

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

$$
\begin{aligned}
u^{2} y & =u v y+k x v^{2} \\
2 p u y+p^{2} v x & =v y n_{1} \\
2 q u y+q^{2} v x & =v y n_{2}
\end{aligned}
$$

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

$$
u^{2}=u v+k x v
$$

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.