AMC 8 · 2016 · #23

Grade 7 geometry-2d

Archetype Casework by Position / Vertex Type

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

📏 Long solution 💡 4 insights

Problem

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 $\angle CED$ ?

$\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150$

Pick an answer.

(A)

(B)

105

(C)

120

(D)

135

(E)

150

View mode:

Toolkit + CCSS Solution

Understand

Restated: 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 $\angle CED$.

Givens: Two circles of the same radius $r$, centered at $A$ and $B$; Circle $A$ passes through $B$ and circle $B$ passes through $A$, so $AB = r$; $C$ is on circle $A$ and on line $AB$ with $A$ between $C$ and $B$, so $CA = r$ and $CB = 2r$ (a diameter of circle $A$); $D$ is on circle $B$ and on line $AB$ with $B$ between $A$ and $D$, so $BD = r$ and $AD = 2r$ (a diameter of circle $B$); $E$ is one of the two points where the circles cross, so $AE = BE = r$; Answer choices: (A) $90$, (B) $105$, (C) $120$, (D) $135$, (E) $150$ (degrees)

Unknowns: The degree measure of $\angle CED$, the angle at $E$ formed by rays $EC$ and $ED$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

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 $\angle CED$ into three friendlier pieces — $\angle CEA$, $\angle AEB$, $\angle BED$ — and find each from a single triangle ($\triangle CAE$, $\triangle ABE$, $\triangle BDE$). Each piece is a one-triangle calculation; adding them gives the answer.

Execute — Answer: C

#1 Draw a Diagram 4.G.A.1 Step 1

Draw the figure (Tool #1).
Sketch circle $A$ and circle $B$ with the same radius; place them so each passes through the other's center.
Mark the line through $A$ and $B$ and label its far hits with the circles as $C$ (on circle $A$, with $A$ between $C$ and $B$) and $D$ (on circle $B$, with $B$ between $A$ and $D$).
Pick one of the two crossing points and call it $E$.
Now draw the three radii $AE$, $BE$, and the segment $AB$.
The picture shows a small triangle $\triangle ABE$ in the middle and two larger triangles $\triangle CAE$ and $\triangle BDE$ on either side, all sharing the vertex $E$.

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

💡 Sketching the points, lines, and angles of the problem is exactly the Grade 4 "draw points, lines, rays, and angles" skill.

#7 Identify Subproblems 5.G.B.4 Step 2

Solve the middle triangle $\triangle ABE$.
Its three sides $AB, AE, BE$ are all equal to $r$, so $\triangle ABE$ is equilateral and every angle in it is $60^\circ$.
In particular $\angle AEB = 60^\circ$ and $\angle EAB = 60^\circ$.

$$AB = AE = BE = r \;\Rightarrow\; \triangle ABE \text{ equilateral} \;\Rightarrow\; \angle AEB = \angle EAB = 60^\circ$$

💡 Recognizing a triangle as equilateral from three equal sides is the Grade 5 "classify two-dimensional figures by properties" skill.

#7 Identify Subproblems 7.G.B.5 Step 3

Solve the left triangle $\triangle CAE$.
Since $C, A, B$ lie on a straight line, $\angle CAE$ and $\angle EAB$ together make a straight angle of $180^\circ$, so $\angle CAE = 180^\circ - 60^\circ = 120^\circ$.
Also $CA = AE = r$ (both radii of circle $A$), so $\triangle CAE$ is isosceles with equal base angles $\angle ACE = \angle AEC$.
The three angles of the triangle add to $180^\circ$.

$$2 \cdot \angle AEC + 120^\circ = 180^\circ \;\Rightarrow\; \angle AEC = 30^\circ$$

💡 Using a straight line to get a supplementary angle and then the triangle-angle sum is the Grade 7 "angle facts to solve for an unknown angle" move.

#7 Identify Subproblems 7.G.B.5 Step 4

Solve the right triangle $\triangle BDE$ by mirror image.
Since $A, B, D$ are collinear, $\angle EBD = 180^\circ - \angle EBA = 180^\circ - 60^\circ = 120^\circ$.
And $BD = BE = r$ (both radii of circle $B$), so $\triangle BDE$ is isosceles with $\angle BDE = \angle BED$.

$$2 \cdot \angle BED + 120^\circ = 180^\circ \;\Rightarrow\; \angle BED = 30^\circ$$

💡 Repeating the exact same supplementary-angle + isosceles-triangle reasoning on the symmetric side reinforces the subproblem habit.

#7 Identify Subproblems 4.MD.C.7 Step 5

Add the three subangles to assemble $\angle CED$.
From $E$ the three rays $EC, EA, EB, ED$ fan out in order, so $\angle CED = \angle CEA + \angle AEB + \angle BED = 30^\circ + 60^\circ + 30^\circ = 120^\circ$, which is choice (C).

$$\angle CED = 30^\circ + 60^\circ + 30^\circ = 120^\circ \;\Rightarrow\; \textbf{(C)}$$

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

[1] #1 4.G.A.1 Draw the figure (Tool #1). Sketch circle $A$ and circle $B$ with the same radius

[2] #7 5.G.B.4 Solve the middle triangle $\triangle ABE$. Its three sides $AB, AE, BE$ are all

[3] #7 7.G.B.5 Solve the left triangle $\triangle CAE$. Since $C, A, B$ lie on a straight line,

[4] #7 7.G.B.5 Solve the right triangle $\triangle BDE$ by mirror image. Since $A, B, D$ are co

[5] #7 4.MD.C.7 Add the three subangles to assemble $\angle CED$. From $E$ the three rays $EC, E

Review

Reasonableness: The answer $120^\circ$ is between $90^\circ$ and $180^\circ$, 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 $\angle CEB = 90^\circ$. Adding the already-found $\angle BED = 30^\circ$ gives $\angle CED = 90^\circ + 30^\circ = 120^\circ$, confirming (C) by a second route.

Alternative: Tool #13 (Convert to Algebra) with coordinates: place $A = (0,0)$ and $B = (1,0)$ with $r = 1$, so $C = (-1, 0)$ and $D = (2, 0)$. The two intersection points satisfy $x^2 + y^2 = 1$ and $(x-1)^2 + y^2 = 1$, giving $x = \tfrac{1}{2}$, so $E = (\tfrac{1}{2}, \tfrac{\sqrt{3}}{2})$. Vectors $\vec{EC} = (-\tfrac{3}{2}, -\tfrac{\sqrt{3}}{2})$ and $\vec{ED} = (\tfrac{3}{2}, -\tfrac{\sqrt{3}}{2})$ have $\vec{EC} \cdot \vec{ED} = -\tfrac{9}{4} + \tfrac{3}{4} = -\tfrac{3}{2}$ and $|\vec{EC}| = |\vec{ED}| = \sqrt{3}$, so $\cos(\angle CED) = -\tfrac{3/2}{3} = -\tfrac{1}{2}$, giving $\angle CED = 120^\circ$ — same answer (C).

CCSS standards used (min grade 7)

4.G.A.1 Draw points, lines, line segments, rays, angles, and perpendicular and parallel lines (Sketching the two circles, the centers $A, B$, the shared line and its extensions to $C, D$, and the radii $AE, BE$ that build the three triangles.)
4.MD.C.7 Recognize angle measure as additive; decompose angles into non-overlapping parts (Writing $\angle CED = \angle CEA + \angle AEB + \angle BED$ and adding $30^\circ + 60^\circ + 30^\circ = 120^\circ$.)
5.G.B.4 Classify two-dimensional figures in a hierarchy based on properties (Identifying $\triangle ABE$ as equilateral (three equal sides) and $\triangle CAE, \triangle BDE$ as isosceles (two equal radii).)
7.G.B.5 Use facts about supplementary, complementary, vertical, and adjacent angles to solve for unknown angles (Turning each $60^\circ$ angle at $A$ (or $B$) into the supplementary $120^\circ$ angle inside $\triangle CAE$ (or $\triangle BDE$), then using the triangle-angle sum to find the $30^\circ$ base angles at $E$.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 7)

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