India competition problems

Since the triangle is isosceles, the orthocentre lies on the perpendicular $AD$ from $A$ on to $BC$. Let it cut the in-circle at $H$. Now we are given that $H$ is the orthocentre of the triangle. Let $AB = AC = b$ and $BC = 2a$. Then $BD = a$. Observe that $b > a$ since $b$ is the hypotenuse and $a$ is a leg of a right-angled triangle. Let $BH$ meet $AC$ in $E$ and $CH$ meet $AB$ in $F$. By Pythagoras theorem applied to $\triangle BDH$, we get

$$BH^2 = HD^2 + BD^2 = 4r^2 + a^2,$$
where $r$ is the in-radius of $ABC$. We want to compute $BH$ in another way. Since $A, F, H, E$ are con-cyclic, we have
$$BH \cdot BE = BF \cdot BA.$$
But $BF \cdot BA = BD \cdot BC = 2a^2$, since $A, F, D, C$ are con-cyclic. Hence $BH^2 = 4a^4/BE^2$. But
$$BE^2 = 4a^2 - CE^2 = 4a^2 - BF^2 = 4a^2 - \left(\frac{2a^2}{b}\right)^2 = \frac{4a^2(b^2 - a^2)}{b^2}.$$
This leads to
$$BH^2 = \frac{a^2b^2}{b^2 - a^2}.$$
Thus we get
$$\frac{a^2b^2}{b^2 - a^2} = a^2 + 4r^2.$$
This simplifies to $(a^4/(b^2 - a^2)) = 4r^2$. Now we relate $a, b, r$ in another way using area. We know that $[ABC] = rs$, where $s$ is the semi-perimeter of $ABC$. We have $s = (b + b + 2a)/2 = b + a$. On the other hand area can be calculated using Heron's formula:
$$[ABC]^2 = s(s - 2a)(s - b)(s - b) = (b + a)(b - a)a^2 = a^2(b^2 - a^2).$$
Hence
$$r^2 = \frac{[ABC]^2}{s^2} = \frac{a^2(b^2 - a^2)}{(b + a)^2}.$$
Using this we get
$$\frac{a^4}{b^2 - a^2} = 4 \left( \frac{a^2(b^2 - a^2)}{(b + a)^2} \right).$$
Therefore $a^2 = 4(b - a)^2$, which gives $a = 2(b - a)$ or $2b = 3a$. Finally,
$$\frac{AB}{BC} = \frac{b}{2a} = \frac{3}{4}.$$

We use the known facts $BH = 2R \cos B$ and $r = 4R \sin(A/2) \sin(B/2) \sin(C/2)$, where $R$ is the circumradius of $\triangle ABC$ and $r$ its in-radius. Therefore
$$HD = BH \sin \angle HBD = 2R \cos B \sin \left(\frac{\pi}{2} - C\right) = 2R \cos^2 B,$$
since $\angle C = \angle B$. But $\angle B = (\pi - \angle A)/2$, since $ABC$ is isosceles. Thus we obtain
$$HD = \cos^2 \left( \frac{\pi}{2} - \frac{A}{2} \right).$$
However $HD$ is also the diameter of the in circle. Therefore $HD = 2r$. Thus we get
$$2R \cos^2 \left( \frac{\pi}{2} - \frac{A}{2} \right) = 2r = 8R \sin(A/2) \sin^2\left(\frac{(\pi - A)}{4}\right).$$
This reduces to
$$\sin(A/2) = 2(1 - \sin(A/2)).$$
Therefore $\sin(A/2) = 2/3$. We also observe that $\sin(A/2) = BD/AB$. Finally
$$\frac{AB}{BC} = \frac{AB}{2BD} = \frac{1}{2\sin(A/2)} = \frac{3}{4}.$$

Let $D$ be the mid-point of $BC$. Extend $AD$ to meet the circumcircle in $L$. Then we know that $HD = DL$. But $HD = 2r$. Thus $DL = 2r$. Therefore $IL = ID + DL = r + 2r = 3r$. We also know that $LB = LI$. Therefore $LB = 3r$. This gives
$$\frac{BL}{LD} = \frac{3r}{2r} = \frac{3}{2}.$$
But $\triangle BLD$ is similar to $\triangle ABD$. So
$$\frac{AB}{BD} = \frac{BL}{LD} = \frac{3}{2}.$$
Finally,
$$\frac{AB}{BC} = \frac{AB}{2BD} = \frac{3}{4}.$$

Let $D$ be the mid-point of $BC$ and $E$ be the mid-point of $DC$. Since $DI = IH(= r)$ and $DE = EC$, the mid-point theorem implies that $IE \parallel CH$. But $CH \perp AB$. Therefore $EI \perp AB$. Let $EI$ meet $AB$ in $F$. Then $F$ is the point of tangency of the incircle of $\triangle ABC$ with $AB$. Since the incircle is also tangent to $BC$ at $D$, we have $BF = BD$. Observe that $\triangle BFE$ is similar to $\triangle BDA$. Hence
$$\frac{AB}{BD} = \frac{BE}{BF} = \frac{BE}{BD} = \frac{BD + DE}{BD} = 1 + \frac{DE}{BD} = \frac{3}{2}.$$
This gives
$$\frac{AB}{BC} = \frac{3}{4}.$$