Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it Soviet Union

Problem:

$ABCD$ is any convex quadrilateral. Construct a new quadrilateral as follows. Take $A'$ so that $A$ is the midpoint of $DA'$ ; similarly, $B'$ so that $B$ is the midpoint of $AB'$ ; $C'$ so that $C$ is the midpoint of $BC'$ ; and $D'$ so that $D$ is the midpoint of $CD'$ . Show that the area of $A'B'C'D'$ is five times the area of $ABCD$ .

Solution

Solution:

Compare the triangles $A'B'A$ and $ADB$ . The base of $A'B'A$ can be taken as $A'A$ , which is the same length as $AD$ . The height of $A'B'A$ is $AB'$ times $\sin B'A'A'$ , which is twice $AB$ times $\sin BAD$ . So $\text{area } A'B'A = 2 \times \text{area } ADB$ .

Similarly, $\text{area } B'C'B = 2 \times \text{area } BAC$ , $\text{area } C'D'C = 2 \times \text{area } CBD$ , and $\text{area } D'A'D = 2 \times \text{area } DCA$ .

So adding, the $\text{area } A'B'A + \text{area } C'D'C = 2 \times \text{area } ABCD$ , and $\text{area } B'C'B + \text{area } D'A'D = 2 \times \text{area } ABCD$ .

But $ABCD = A'B'A + B'C'B + C'D'C + D'A'D + ABCD$ . Hence result.

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