AMC 8 2016 Problem 23

AMC8 2016 年第 23 题

圆★★★☆☆
Two congruent circles centered at points AA and BB each pass through the other circle's center. The line containing both AA and BB is extended to intersect the circles at points CC and DD . The circles intersect at two points, one of which is EE . What is the degree measure of ∠CED\angle CED ?
两个圆心分别在点 A 和点 B 的全等圆各自通过对方的圆心。连接点 A 和点 B 的直线和这 2 个圆交于点 C 和点 D。这 2 个圆交于 2 个点,其中之一是点 E.那么的度数是多少?
  1. A.90
  2. B.105
  3. C.120
  4. D.135
  5. E.150

正确答案 · Correct Answer

C

解析 · Solution

Solution 1

Observe that △EAB\triangle{EAB} is equilateral (all are radii of congruent circles). Therefore, m∠AEB=m∠EAB=m∠EBA=60∘m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}. Since CDCD is a straight line, we conclude that m∠EBD=180∘−60∘=120∘m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}. Since BE=BD (both are radii of the same circle), △BED\triangle{BED} is isosceles, meaning that m∠BED=m∠BDE=30∘m\angle{BED}=m\angle{BDE}=30^{\circ}. Similarly, m∠AEC=m∠ACE=30∘m\angle{AEC}=m\angle{ACE}=30^{\circ}. Now, ∠CED=m∠AEC+m∠AEB+m∠BED=30∘+60∘+30∘=120∘\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}. Therefore, the answer is 120.

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