Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME Prove it Soviet Union

Problem:

a. The convex hexagon $ABCDEF$ has all angles equal. Prove that $AB - DE = EF - BC = CD - FA$ .

b. Given six lengths $a_{1}$ , $a_{2}$ , $a_{3}$ , $a_{4}$ , $a_{5}$ , $a_{6}$ satisfying $a_{1} - a_{4} = a_{5} - a_{2} = a_{3} - a_{6}$ , show that you can construct a hexagon with sides $a_{1}$ , $a_{2}$ , $a_{3}$ , $a_{4}$ , $a_{5}$ , $a_{6}$ and equal angles.

Solution

Solution:

a. Extend $AB$ , $CD$ , $EF$ . We get an equilateral triangle with sides $AF + AB + BC$ , $BC + CD + DE$ , $ED + EF + FA$ . Hence $AB - DE = CD - FA = EF - BC$ , as required.

b. Take an equilateral triangle with sides $s$ , $t$ , $u$ of lengths $a_{2} + a_{3} + a_{4}$ , $a_{4} + a_{5} + a_{6}$ , and $a_{6} + a_{1} + a_{2}$ respectively. Construct $BC$ of length $a_{2}$ parallel to $t$ with $B$ on $u$ and $C$ on $s$ . Construct $DE$ of length $a_{4}$ parallel to $u$ with $D$ on $s$ and $E$ on $t$ . Construct $FA$ of length $a_{6}$ parallel to $s$ with $F$ on $t$ and $A$ on $u$ . Then $ABCDEF$ is the required hexagon, with $AB = a_{1}$ , $BC = a_{2}$ , etc.

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