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 $TSH$ .

#Geometry #IMO

Easy Math Editor

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Comments

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 - 6 years, 11 months ago

I found it rather easy. :p

Anju Pandey - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$