AMC 8 · 2016 · #23
Grade 7 geometry-2dProblem
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?
Pick an answer.
AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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)
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
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 $E$.
💡 Sketching the points, lines, and angles of the problem is exactly the Grade 4 "draw points, lines, rays, and angles" skill.
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 $\angle EAB = 60^\circ$.
💡 Recognizing a triangle as equilateral from three equal sides is the Grade 5 "classify two-dimensional figures by properties" skill.
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 $180^\circ$.
💡 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.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 $\angle BDE = \angle BED$.
💡 Repeating the exact same supplementary-angle + isosceles-triangle reasoning on the symmetric side reinforces the subproblem habit.
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).
💡 Treating $\angle CED$ as the sum of adjacent non-overlapping pieces is the Grade 4 "angle measure is additive" idea.
4.G.A.1 Draw the figure (Tool #1). Sketch circle $A$ and circle $B$ with the same radius 5.G.B.4 Solve the middle triangle $\triangle ABE$. Its three sides $AB, AE, BE$ are all 7.G.B.5 Solve the left triangle $\triangle CAE$. Since $C, A, B$ lie on a straight line, 7.G.B.5 Solve the right triangle $\triangle BDE$ by mirror image. Since $A, B, D$ are co 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.1Draw 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.7Recognize 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.4Classify 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.5Use 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!
⭐ 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!
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