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# IMO 2014/3

Convex quadrilateral $$ABCD$$ has $$\angle ABC = \angle CDA = 90^\circ$$. Point $$H$$ is the foot of the perpendicular from $$A$$ to $$BD$$. Points $$S$$ and $$T$$ lie on sides $$AB$$ and $$AD$$ respectively, such that $$H$$ lies inside triangle $$SCT$$ and

$\angle CHS - \angle CSB = 90 ^ \circ, \angle THC - \angle DTC = 90^ \circ .$

Prove that line $$BD$$ is tangent to the circumcirlce of triangle $$TSH$$.

Note by Calvin Lin
3 years, 6 months ago

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## Comments

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I don't have a purely synthetic solution for this as of yet. Here's my outline, I'll fill this later.

The conditions imply that the tangents to $$(CHS)$$ and $$(CHT)$$ at $$S$$ and $$T$$ are perpendicular to $$AB$$ and $$AD$$ respectively. Let $$X$$ and $$Y$$ be their centers respectively. By some tedious trig bash we obtain $$\dfrac{AX}{XH} = \dfrac{AY}{YH},$$ implying the angle bisectors of $$\angle AXH$$ and angle $$\angle AYH$$ meet at a point on $$AH,$$ or that the perpendicular bisectors of $$SH$$ and $$TH$$ meet on a point lying on $$AH.$$ Hence, the circumcircle of $$\triangle TSH$$ lies on $$AH,$$ proving the desired result.

Also, @Xuming Liang wrote a really nice synthetic solution for this problem in the AoPS thread. :)

- 3 years, 6 months ago

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I found it rather easy. :p

- 3 years, 1 month ago

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