Join the USAMO Guide Discord Server. Regular updates are posted there!

Let me Join!
Dismiss

HMMT 2014 #0jj8

HMMT 2014Hard

From module Angle Chasing Mastery

Community difficultyNo ratings yetBe the first to rate this problem.
Understand the scale
Scale 0.5: Easiest math competition problems, often solvable even for early middle school students without experience. 1: Problems strictly for beginners, often at a moderate middle school level. 1.5: Problems for stronger beginner students, on the level of the middling problems in most middle school contests. 2: For motivated beginners, harder questions from the previous categories. 2.5: More advanced beginner problems, hardest questions from previous categories. 3: Early intermediate problems that require more creative thinking. 3.5: Tougher early intermediate problems that consistently stretch into higher-level creative thinking or conceptual knowledge. 4: Intermediate-level problems. 4.5: Upper intermediate problems approaching the upper bound for non-invitational sprint math competitions. 5: More difficult AIME problems or simple proof-based Olympiad-style problems. 6: High-level AIME-styled questions or introductory Olympiad-level questions. 7: Tougher Olympiad-level questions, may require more technical knowledge. 8: High-level Olympiad-level questions. 9: Expert Olympiad-level questions. 9.5: The hardest problems appearing on Olympiads which the strongest students could reasonably solve. 10: Historically notorious problems, generally unsuitable for even very hard competitions because they are exceedingly tedious, long, and difficult.
Angle chasingDistance chasing

Problem

Problem:

Let ABCDABCD be a trapezoid with AB∥CDAB \parallel CD and ∠D=90∘\angle D = 90^{\circ}. Suppose that there is a point EE on CDCD such that AE=BEAE = BE and that triangles AEDAED and CEBCEB are similar, but not congruent. Given that CDAB=2014\frac{CD}{AB} = 2014, find BCAD\frac{BC}{AD}.

← Back to all problems