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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
2 years, 3 months ago

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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. :) Sreejato Bhattacharya · 2 years, 3 months ago

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I found it rather easy. :p Anju Pandey · 1 year, 10 months ago

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