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Geometry Difficulty 4.7 AIME Prove it United States

Problem:

Let $ABCD$ be a quadrilateral with an inscribed circle $\omega$ and let $P$ be the intersection of its diagonals $AC$ and $BD$ . Let $R_{1}, R_{2}, R_{3}, R_{4}$ be the circumradii of triangles $APB$ , $BPC$ , $CPD$ , $DPA$ respectively. If $R_{1}=31$ and $R_{2}=24$ and $R_{3}=12$ , find $R_{4}$ .

Solution

Solution:

Note that $\angle APB = 180^{\circ} - \angle BPC = \angle CPD = 180^{\circ} - \angle DPA$ so $\sin APB = \sin BPC = \sin CPD = \sin DPA$ . Now let $\omega$ touch sides $AB$ , $BC$ , $CD$ , $DA$ at $E$ , $F$ , $G$ , $H$ respectively. Then $AB + CD = AE + BF + CG + DH = BC + DA$ so
$\frac{AB}{\sin APB} + \frac{CD}{\sin CPD} = \frac{BC}{\sin BPC} + \frac{DA}{\sin DPA}$
and by the Extended Law of Sines this implies
$2R_{1} + 2R_{3} = 2R_{2} + 2R_{4}$
which immediately yields $R_{4} = R_{1} + R_{3} - R_{2} = 19$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.