Problem 23 · 2016 AMC 8
Hard
Geometry & Measurement
spatial-reasoningarea-decomposition
Two congruent circles centered at points A and B each pass through the other circle's center. The line containing both A and B is extended to intersect the circles at points C and D. The circles intersect at two points, one of which is E. What is the degree measure of ∠CED?
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Answer: C — 120°.
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Hint 1 of 3
Mark every segment that is a radius. Each circle passes through the other's center, so AB, AE, and BE are ALL radii of equal length — that makes ▵AEB equilateral, handing you a free 60° at E.
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Hint 2 of 3
C and D are the far ends of the line through both centers, so CA and BD are full DIAMETERS. The key fact: an angle inscribed in a semicircle (standing on a diameter) is a right angle — so ∠CEA = 90° and ∠BED = 90°.
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Hint 3 of 3
Now ∠CED isn't just one of those angles — the two right angles OVERLAP in the middle (∠AEB), so combine them carefully rather than adding blindly.
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Approach: equilateral triangle for the middle angle + Thales' right angles on the diameters
- AE = EB = AB (all radii of the two congruent circles, each through the other's center), so ▵AEB is equilateral and ∠AEB = 60°.
- CA is a diameter of A's circle, so by Thales' theorem (angle in a semicircle) ∠CEA = 90°; likewise BD is a diameter of B's circle, so ∠BED = 90°.
- Going from ray EC to ray ED, the two 90° angles share the overlap ∠AEB, so ∠CED = ∠CEA + ∠BED − ∠AEB = 90 + 90 − 60 = 120°.
- Why this transfers: two facts unlock most circle-angle problems — equal radii build equilateral/isosceles triangles, and any angle subtending a DIAMETER is exactly 90° (Thales). Spot the diameter, get a right angle for free.
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