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Bulgarian Mathematical Olympiad

Geometry Difficulty 6.1 National Olympiad Prove it Bulgaria

Problem:
The tangent lines to the circumcircle $k$ of an isosceles $\triangle ABC$ , $AC = BC$ , at the points $B$ and $C$ meet at point $X$ . If $AX$ meets $k$ at point $Y$ , find the ratio $\frac{AY}{BY}$ .

Solution

Solution:
If $\angle BAY = \alpha$ and $\angle ABY = \beta$ then $\angle BYX = \beta + \alpha$ . Furthermore $\angle ACB = \angle AYB = 180^\circ - \alpha - \beta$ , implying $\angle BAC = \angle ABC = \frac{\beta + \alpha}{2}$ . Thus $\angle AYC = \frac{\beta + \alpha}{2}$ and $\angle YCX = \angle YAC = \frac{\beta + \alpha}{2} - \alpha = \frac{\beta - \alpha}{2}$ . The Sine theorem for $\triangle BYX$ and $\triangle CYX$ gives

$\frac{XY}{XB} = \frac{\sin \alpha}{\sin (\alpha + \beta)}, \quad \frac{XY}{XC} = \frac{\sin \frac{\beta - \alpha}{2}}{\sin \frac{\beta + \alpha}{2}}$
and since $XB = XC$ we have $\frac{\sin \alpha}{\sin (\alpha + \beta)} = \frac{\sin \frac{\beta - \alpha}{2}}{\sin \frac{\beta + \alpha}{2}}$ . Hence
$\sin \alpha = 2 \cos \frac{\beta + \alpha}{2} \sin \frac{\beta - \alpha}{2} = \sin \beta - \sin \alpha$
and therefore $\frac{AY}{BY} = \frac{\sin \beta}{\sin \alpha} = 2$ .

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