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, 2022

Geometry Difficulty 5.4 AIME, harder Prove it Japan

There is a convex quadrilateral $ABCD$ and a point $P$ inside such that the lines $AP$ and $AD$ are orthogonal and the lines $BP$ and $CD$ are orthogonal. When $AB = 7$ , $AP = 3$ , $BP = 6$ , $AD = 5$ , $CD = 10$ , find the area of triangle $ABC$ .

Solution

We can assume $A$ , $B$ , $C$ , $D$ are in counter-clockwise order. Since the lines $AP$ and $AD$ , the lines $BP$ and $CD$ are orthogonal respectively, if we rotate the triangle $APB$ counter-clockwise by $90^\circ$ then $AP$ and $AD$ are parallel and $BP$ and $CD$ are parallel. The points $P$ , $A$ , $B$ and $D$ , $A$ , $C$ are in counter-clockwise order respectively hence we get $\angle APB = \angle ADC$ .

We have $AP : BP = 1 : 2 = AD : CD$ thus the triangle $APB$ and $ADC$ are similar.
Therefore $AC = \frac{AB \cdot AD}{AP} = \frac{35}{3}$ is obtained. Since $APB$ and $ADC$ are similar, we also have $\angle BAC = \angle BAP + \angle PAC = \angle CAD + \angle PAC = \angle PAD = 90^\circ$ . Hence the area of the triangle $ABC$ is $\frac{1}{2} \cdot AB \cdot AC = \frac{245}{6}$ .

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