Maths Olympiad Prep

Track / Stage 5 / 12 of 400 #612 of 1964

Problem 612

AIME late
Geometry Difficulty 5.0 Find the answer

Inside square ABCDA B C D, a point EE is chosen so that triangle DECD E C is equilateral. Find the measure of ∠AEB\angle A E B.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Answer: 150∘150^{\circ}

Solution: Since △DEC\triangle D E C is an equilateral triangle, then ∣DE∣=∣CE∣|D E|=|C E| each angle has a measure of 60∘60^{\circ}. This implies that ∠ADE\angle A D E has measure of 30∘30^{\circ}. Since ∣AD∣=∣DE∣|A D|=|D E|, then △ADE\triangle A D E is an isosceles triangle. Thus, ∠DAE=∠DEA=75∘\angle D A E=\angle D E A=75^{\circ}. The same argument on triangle BECB E C will give us ∠CBE=\angle C B E= ∠CEB=75∘\angle C E B=75^{\circ}. Thus, ∠AEB=150∘\angle A E B=150^{\circ}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.