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Problem 1302

AIME late

Geometry Difficulty 5.4 Prove it HMMT February · United States · 2024

Let $ABTCD$ be a convex pentagon with area $22$ such that $AB = CD$ and the circumcircles of triangles $TAB$ and $TCD$ are internally tangent. Given that $\angle ATD = 90^{\circ}$ , $\angle BTC = 120^{\circ}$ , $BT = 4$ , and $CT = 5$ , compute the area of triangle $TAD$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

Paste $\triangle TCD$ outside the pentagon to get $\triangle ABX \cong \triangle DCT$ . From the tangent circles condition, we get
$\begin{aligned} \angle XBT & = 360^{\circ} - \angle XBA - \angle ABT \\ & = 360^{\circ} - \angle DCT - \angle ABT \\ & = 360^{\circ} - 270^{\circ} = 90^{\circ} \\ \angle XAT & = 90^{\circ} - \angle BXA - \angle ATB \\ & = 90^{\circ} - \angle CTD - \angle ATB \\ & = 90^{\circ} - (120^{\circ} - 90^{\circ}) = 60^{\circ} . \end{aligned}$
Moreover, if $x = AT$ and $y = TD$ , then notice that
$\begin{aligned} [ABTCD] & = [ABT] + [CDT] + [ATD] \\ & = [XAT] - [XBT] + [ATD] \\ & = \frac{1}{2} x y \sin 60^{\circ} - \frac{1}{2} \cdot 4 \cdot 5 + \frac{1}{2} x y \\ & = \frac{2 + \sqrt{3}}{4} x y - 10 \end{aligned}$
so we have
$x y = 32 \cdot \frac{4}{2 + \sqrt{3}} = 128(2 - \sqrt{3}) \Longrightarrow [ATD] = \frac{1}{2} x y = 64(2 - \sqrt{3}) .$

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