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Geometry Difficulty 4.8 AIME Find the answer

2. ABCDA B C D is a square. BDEFB D E F is a rhombus with AA, EE and FF collinear. Find ADE\angle A D E.

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

Solution

SolUtion
Let XX be the foot of the perpendicular from AA to BDB D and let YY be the foot of the perpendicular from EE to BDB D.

Let BDB D have length 2 .
The length of AXA X is 1 because it is half the diagonal ACA C, which is the same as BDB D. This is also the same length of EYE Y because AXYEA X Y E is a rectangle.

Triangle EYDE Y D is half of an equilateral triangle (since DE=DB=2D E=D B=2 ), so EDY=30\angle E D Y=30^{\circ}.
Since BDA=45,ADE=4530=15\angle B D A=45^{\circ}, \angle A D E=45^{\circ}-30^{\circ}=15^{\circ}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.