IMO 2014 problem 3, first day

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 $$\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}.$$ Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.

This is the third problem from IMO 2014, South Africa. How to do this? Thanks.

• – Jaakko Seppälä Jul 10 '14 at 19:37
• Oh thank you I see there is a solution. Sorry I don't follow that site too much, so I didn't check. Mainly I follow Math SE, not the artofproblemsolving site, not that I mean to say it is bad, of course. – shadow10 Jul 11 '14 at 11:43

Let $M$ be the midpoint of $CH$. Furthermore, let $P$ be a point on (the extension of) $CH$ such that $SP$ is perpendicular to $CH$. Because $\angle CHS - \angle CSB = 90^{\circ}$, it is straightforward to show that the triangles $\triangle SBC$ and $\triangle SPH$ are similar. Consider spiral homotheties $$B(90^\circ \text{counter clockwise},\rho:=\frac{BS}{BC}=\frac{PS}{PH})$$ and $$P(90^\circ \text{counter clockwise},\frac{1}{\rho}).$$ The combination of these two transformations is a $180^{\circ}$ rotation that maps $C$ to $H$. Therefore, $M$ (i.e., the midpoint of $CH$) is the center of this rotation and is invariant under the transformation. Using this fact, it is straightforward to show that $MB=MP$. It is also easy to show that $M$ lies on the perpendicular bisector of $BD$, meaning that $MB=MD$. Therefore, we can conclude that $S$, $P$, and $T$ are collinear. In other words, $CH$ is perpendicular to $ST$.
To show that $BD$ is tangent to the circumcircle of $\triangle HST$ it suffices to show that
$$\angle DHT = \angle HST.\tag{*}$$
Note that, $CPSB$ is cyclic and therefore $\angle HPB = \angle CPB = \angle CSB = \angle HSP$. Hence, $BP$ is perpendicular to $HS$. Similarly, we have $\angle HPD = \angle HTP$ and $DP$ is perpendicular to $HT$. Let $X$ denote the intersection of $HS$ and $BP$. Similarly, let $Y$ be the intersection of $HT$ and $DP$. We have $\triangle XBS \sim \triangle PCS$, $\triangle YDT \sim \triangle PCT$, $\triangle XPH \sim \triangle PSH$, and $\triangle YPH \sim \triangle PTH$. Therefore, we can deduce
where the last equation holds because $XPYH$ is cyclic. Now, (*) follows immediately using the fact that $\angle XPH = \angle HST$.