Maths Olympiad Prep

Track / Stage 4 / 330 of 340 #590 of 1964

Problem 590

AMC 12 late, AIME early
Geometry Difficulty 5.0 Find the answer

In △ABC\triangle A B C, it is known that ∠A=a,CD,BE\angle A=a, C D, B E are the altitudes on AB,ACA B, A C respectively, then DEBC=\frac{D E}{B C}=

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

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Official solution

∣cos⁡a∣ |\cos a|
3. 【Analysis and Solution】 ∵∠BDC=∠BEC,∴B,D,E,C\because \angle B D C=\angle B E C, \therefore B, D, E, C are concyclic, ∠ADE=∠ACB,△AED∼△ABC\angle A D E=\angle A C B, \triangle A E D \sim \triangle A B C, DE2BC2=S△AEDS△ABC=AD⋅AEAB⋅AC=cos⁡2a.∴DEBC=∣cos⁡a∣\frac{D E^{2}}{B C^{2}}=\frac{S_{\triangle A E D}}{S_{\triangle A B C}}=\frac{A D \cdot A E}{A B \cdot A C}=\cos ^{2} a . \therefore \frac{D E}{B C}=|\cos a|.

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