Maths Olympiad Prep

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Geometry Difficulty 6.1 National Olympiad Prove it Bulgaria

Problem:
The tangent lines to the circumcircle kk of an isosceles △ABC\triangle ABC, AC=BCAC = BC, at the points BB and CC meet at point XX. If AXAX meets kk at point YY, find the ratio AYBY\frac{AY}{BY}.

Solution

Solution:
If ∠BAY=α\angle BAY = \alpha and ∠ABY=β\angle ABY = \beta then ∠BYX=β+α\angle BYX = \beta + \alpha. Furthermore ∠ACB=∠AYB=180∘−α−β\angle ACB = \angle AYB = 180^\circ - \alpha - \beta, implying ∠BAC=∠ABC=β+α2\angle BAC = \angle ABC = \frac{\beta + \alpha}{2}. Thus ∠AYC=β+α2\angle AYC = \frac{\beta + \alpha}{2} and ∠YCX=∠YAC=β+α2−α=β−α2\angle YCX = \angle YAC = \frac{\beta + \alpha}{2} - \alpha = \frac{\beta - \alpha}{2}. The Sine theorem for △BYX\triangle BYX and △CYX\triangle CYX gives

Figure 1

XYXB=sin⁡αsin⁡(α+β),XYXC=sin⁡β−α2sin⁡β+α2 \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=XCXB = XC we have sin⁡αsin⁡(α+β)=sin⁡β−α2sin⁡β+α2\frac{\sin \alpha}{\sin (\alpha + \beta)} = \frac{\sin \frac{\beta - \alpha}{2}}{\sin \frac{\beta + \alpha}{2}}. Hence
sin⁡α=2cos⁡β+α2sin⁡β−α2=sin⁡β−sin⁡α \sin \alpha = 2 \cos \frac{\beta + \alpha}{2} \sin \frac{\beta - \alpha}{2} = \sin \beta - \sin \alpha
and therefore AYBY=sin⁡βsin⁡α=2\frac{AY}{BY} = \frac{\sin \beta}{\sin \alpha} = 2.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.