AMC 8 · 2016 · #23

Grade 7 geometry-2d

Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role

angle-sum-triangleline-symmetryspatial-visualization identify-subproblemscasework ↑ Prerequisites: angle-sum-triangleline-symmetry

📏 Long solution 💡 4 insights

Problem

Two congruent circles centered at A and B are placed so each one passes through the other's center, making them overlap. The line through A and B is extended until it hits the circles at the two far points C (past A) and D (past B). The circles cross at two points; call one of them E. Find the size of ∠ CED.

Pick an answer.

(A)

(B)

105

(C)

120

(D)

135

(E)

150

AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

There is no figure printed with the problem, so step one is Tool #1 (Draw a Diagram): sketch the two overlapping circles, mark the centers A, B, the line they share, the far points C and D, and one intersection point E. Drawing in the segments AE, BE, AB immediately reveals that they are all equal to the radius r. That picture has a lot going on at once, so apply Tool #7 (Identify Subproblems) to chop ∠ CED into three friendlier pieces — ∠ CEA, ∠ AEB, ∠ BED — and find each from a single triangle (△ CAE, △ ABE, △ BDE). Each piece is a one-triangle calculation; adding them gives the answer.

1STEP 1

Draw a diagram: sketch the two overlapping circles, mark C, D, E, then join AE, BE, AB — three triangles △CAE, △ABE, △BDE all meet at E.

AB = r, AE = r, BE = r, CA = r, BD = r

2STEP 2

The middle triangle △ABE has all three sides equal to r, so it is equilateral with every angle 60°; in particular ∠AEB = 60°.

AB = AE = BE = r → △ ABE equilateral → ∠ AEB = ∠ EAB = 60°

3STEP 3

C, A, B are collinear, so ∠CAE = 180° − 60° = 120°; with CA = AE = r the isosceles base angles give ∠AEC = 30°.

2 · ∠ AEC + 120° = 180° → ∠ AEC = 30°

4STEP 4

By mirror symmetry on the right, A, B, D collinear gives ∠EBD = 120°, and BD = BE = r makes △BDE isosceles, so ∠BED = 30°.

2 · ∠ BED + 120° = 180° → ∠ BED = 30°

5STEP 5

Rays EC, EA, EB, ED fan out in order, so ∠CED = 30° + 60° + 30° = 120° — choice (C).

∠ CED = 30° + 60° + 30° = 120° → (C)

Answer

120

The answer 120° is between 90° and 180°, which matches the picture: E sits above the line CD, and the rays EC and ED open wider than a right angle but stay less than a straight angle. A clean cross-check: CB is a full diameter of circle A (length 2r), and E is on circle A, so by Thales' theorem ∠ CEB = 90°. Adding the already-found ∠ BED = 30° gives ∠ CED = 90° + 30° = 120°, confirming (C) by a second route.

💡Key takeaway

This AMC 8 problem only needs the Grade 7 idea of "angles on a straight line add to 180°, and angles in a triangle add to 180°" you already know!

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