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Ukrainian National Mathematical Olympiad

Geometry Difficulty 4.8 AIME Prove it Ukraine

Suppose for a rhombus $ABCD$ there exists a point $T$ such that the following conditions are satisfied: $\angle ATC + \angle BTD = 180^\circ$ and the circumcircles of the triangles $ATC$ and $BTD$ are tangent to each other. Prove that the point $T$ is equidistant from the diagonals of the rhombus.

Solution

Let $P$ be the point of intersection of the rhombus diagonals, $O_1$ and $O_2$ be the centers of the described circles $\Delta ATC$ and $\Delta BTD$ respectively (Fig. 1). Hence, the following equalities are true:
$\angle A O_1 C = 2 \angle A T C = 2 \cdot (180^\circ - \angle B T D) = \angle B O_2 D,$
so the isosceles triangles $AO_1C$ and $BO_2D$ are similar. Then their altitudes are proportional to their sides, i.e. $\frac{PO_1}{PO_2} = \frac{O_1A}{O_2B} = \frac{O_1T}{O_2T}$ . Then $PT$ is a bisector (internal or external) of $\angle O_1PO_2$ , exactly what we needed to prove.

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