Competition · AMC preparation · step 4 of 4

AMC 8 · 2016 · #23

Grade 7 geometry-2d

Archetype Angle Chase in a Synthetic Figure └ Sub-archetype Isosceles Equal-Angle Chase

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 the figure

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

Solve the middle triangle

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

Solve the left triangle

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

Solve the right triangle

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

Add the three subangles

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

∠ CED = 30° + 60° + 30° = 120° → (C)

Treating ∠ CED as the sum of adjacent non-overlapping pieces is the Grade 4 "angle measure is additive" idea.

The angle ∠ CED at the crossing point measures 30° + 60° + 30° = 120°.

▸ Why?

At E the rays EC, EA, EB, ED fan out in that order, so ∠ CED is cut into the three adjacent pieces ∠ CEA, ∠ AEB, ∠ BED with no gaps or overlaps, and the whole angle is exactly those three measures added together.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

The middle piece ∠ AEB is 60° because triangle ABE is equilateral, so all three of its angles are equal 60° angles.

▸ Why?

The sides AB, AE, and BE each run from a circle's center out to its edge, so every one of them equals the radius r and the three sides are equal.

A circle's radius is constantEvery point on a circle sits exactly one radius from the center, so all radii of a circle are equal.

▸ Why?

With all three sides equal, flipping the triangle lays it back exactly onto itself, which forces its three angles to be equal.

Rigid motion preserves length and angleSliding, turning, or flipping a shape lays it exactly onto a copy, so every length and angle stays the same (folding is the flip case).

▸ Why?

Three equal angles that must total 180° each have to be 60°.

A triangle's angles sum to 180 degreesThe three corners of any flat triangle always add up to one straight angle, 180 degrees.

▸ Why?

Each outer piece, ∠ CEA on the left and its mirror image ∠ BED on the right, is 30°, because each is a base angle of an isosceles triangle whose top angle is 120°.

▸ Why?

The top angle ∠ CAE and the equilateral's angle ∠ EAB = 60° sit side by side along the straight line through C, A, B, so ∠ CAE takes the leftover and is 120°.

A straight angle is 180 degreesAngles sitting side by side along a straight line always fill up 180 degrees.

▸ Why?

The triangle has two equal sides, so it is mirror-symmetric and its two base angles are equal.

▸ Why?

Sides CA and AE are both radii of circle A, so they are the same length.

A circle's radius is constantEvery point on a circle sits exactly one radius from the center, so all radii of a circle are equal.

▸ Why?

Because those two sides are equal, flipping the triangle across the line between them swaps the two base angles onto each other, so they must be equal.

Rigid motion preserves length and angleSliding, turning, or flipping a shape lays it exactly onto a copy, so every length and angle stays the same (folding is the flip case).

▸ Why?

The three angles add to 180°, so after the 120° top angle the two equal base angles split the remaining 60° and each is 30°.

A triangle's angles sum to 180 degreesThe three corners of any flat triangle always add up to one straight angle, 180 degrees.

4.MD.C.7Identify Subproblems

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!

Draw the figure
Solve the middle triangle
Solve the left triangle
Solve the right triangle
Add the three subangles

A parent dashboard for the family lives at sensimlab.com.

Draw the figure

Solve the middle triangle

Solve the left triangle

Solve the right triangle

Add the three subangles

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