
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

![](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

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