AMC 10 · 2011 · #18
Grade 8 geometry-2dCircles A,B, and C each has radius 1. Circles A and B share one point of tangency. Circle C has a point of tangency with the midpoint of AB. What is the area inside circle C but outside circle A and circle B?
Pick an answer.
AMC 10 2011 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: Three unit circles $A$, $B$, and $C$ are placed so that $A$ and $B$ just touch each other, and $C$ touches the exact midpoint of the segment joining the centers of $A$ and $B$. Find the area of the part of circle $C$ that lies outside both circle $A$ and circle $B$.
Givens: Circles $A$, $B$, and $C$ each have radius $1$; Circles $A$ and $B$ are tangent, sharing exactly one point; Circle $C$ is tangent at the midpoint $M$ of $\overline{AB}$, the segment joining the centers of $A$ and $B$; Answer choices: (A) $3-\frac{\pi}{2}$, (B) $\frac{\pi}{2}$, (C) $2$, (D) $\frac{3\pi}{4}$, (E) $1+\frac{\pi}{2}$
Unknowns: The area of the region inside circle $C$ but outside both circle $A$ and circle $B$
Understand
Restated: Three unit circles $A$, $B$, and $C$ are placed so that $A$ and $B$ just touch each other, and $C$ touches the exact midpoint of the segment joining the centers of $A$ and $B$. Find the area of the part of circle $C$ that lies outside both circle $A$ and circle $B$.
Givens: Circles $A$, $B$, and $C$ each have radius $1$; Circles $A$ and $B$ are tangent, sharing exactly one point; Circle $C$ is tangent at the midpoint $M$ of $\overline{AB}$, the segment joining the centers of $A$ and $B$; Answer choices: (A) $3-\frac{\pi}{2}$, (B) $\frac{\pi}{2}$, (C) $2$, (D) $\frac{3\pi}{4}$, (E) $1+\frac{\pi}{2}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #17 Visualize Spatial Relationships
The wanted region is an awkward curved shape, but it is really the whole disk $C$ with two overlapping bites removed, so Tool #7 (Identify Subproblems) is the anchor: compute $\text{Area}(C)$, then the overlap $C\cap A$, then double it by symmetry. Tool #1 (Draw a Diagram) makes this exact — placing the circles on a coordinate grid fixes every center and every crossing point. Tool #17 (Visualize Spatial Relationships) is what reveals that each overlap lens is built from two congruent quarter-circle-minus-triangle segments, which is the move that makes the messy $\pi$ terms cancel.
Execute — Answer: C
8.G.B.8 Step 1 Place the circles on a grid
- Put the tangent point of $A$ and $B$ at the origin.
- Then circle $A$ has center $(-1,0)$ and circle $B$ has center $(1,0)$, so they touch exactly at the origin, which is the midpoint $M=(0,0)$ of $\overline{AB}$.
- Circle $C$ touches that same point $M$ and has radius $1$, so its center sits one unit straight up at $(0,1)$.
- The distance from $C$'s center to $A$'s center is then $\sqrt{1^2+1^2}=\sqrt{2}$.
💡 A grid pins down every center and every touch-point, turning a loose picture into exact coordinates.
7.G.B.4 Step 2 Break the target into a subtraction
- The region we want is the part of circle $C$ lying outside both $A$ and $B$.
- Start from the whole disk $C$, whose area is $\pi(1)^2=\pi$, and subtract the piece it shares with $A$ and the piece it shares with $B$.
- Circles $A$ and $B$ meet only at the single point $M$, so they enclose no shared area and nothing is subtracted twice.
- By the left-right symmetry of the picture, the overlap of $C$ with $A$ equals the overlap of $C$ with $B$.
💡 Take the full circle and peel off only the parts that poke into its two neighbors.
8.G.B.8 Step 3 Find where $C$ and $A$ cross
- Circles $C$ and $A$ meet at two points.
- One is $M=(0,0)$.
- The other is $P=(-1,1)$: it sits one unit directly above $A$'s center so it lies on $A$, and substituting into $C$ gives $(-1)^2+(1-1)^2=1$, so it lies on $C$ too.
- Notice the four points $A=(-1,0)$, $M=(0,0)$, $C=(0,1)$, $P=(-1,1)$ are the corners of a unit square.
- The overlap lens $C\cap A$ is split by the chord $MP$ into two curved pieces.
💡 The two circles cross at the tangency point and one clean lattice point that squares the whole figure up.
7.G.B.4 Step 4 The arc spans a right angle
- From $A=(-1,0)$, the direction to $M=(0,0)$ is $(1,0)$ and the direction to $P=(-1,1)$ is $(0,1)$.
- These are perpendicular, so the arc $MP$ on circle $A$ covers a right angle — one quarter of the full circle.
- A quarter of circle $A$ has area $\frac{90}{360}\pi(1)^2=\frac{\pi}{4}$.
💡 Perpendicular directions from the center mean the arc is a clean one-quarter slice of the circle.
6.G.A.1 Step 5 Peel the triangle to get the lens
- The quarter-circle contains the right triangle $AMP$ with legs $AM=1$ and $AP=1$, area $\frac12$.
- What is left over — between chord $MP$ and the arc — is a segment of area $\frac{\pi}{4}-\frac12$.
- The same chord $MP$ cuts a matching segment off circle $C$; because $A$, $M$, $C$, $P$ form a unit square, that piece is congruent, also $\frac{\pi}{4}-\frac12$.
- The two segments together make the whole lens, so $\text{Area}(C\cap A)=2\left(\frac{\pi}{4}-\frac12\right)=\frac{\pi}{2}-1$.
💡 Cutting the inscribed right triangle out of the quarter-circle leaves exactly the overlap sliver, and symmetry doubles it.
7.G.B.4 Step 6 Subtract both overlaps
- Each overlap has area $\frac{\pi}{2}-1$, and there are two of them (with $A$ and with $B$).
- Subtract them from the full disk $C$: $\pi-2\left(\frac{\pi}{2}-1\right)=\pi-\pi+2=2$.
- So the area inside circle $C$ but outside circles $A$ and $B$ is $2$, which is choice (C).
💡 The two $\pi$-pieces cancel, leaving a whole-number area.
8.G.B.8 Put the tangent point of $A$ and $B$ at the origin. Then circle $A$ has center $ 7.G.B.4 The region we want is the part of circle $C$ lying outside both $A$ and $B$. Sta 8.G.B.8 Circles $C$ and $A$ meet at two points. One is $M=(0,0)$. The other is $P=(-1,1) 7.G.B.4 From $A=(-1,0)$, the direction to $M=(0,0)$ is $(1,0)$ and the direction to $P=( 6.G.A.1 The quarter-circle contains the right triangle $AMP$ with legs $AM=1$ and $AP=1$ 7.G.B.4 Each overlap has area $\frac{\pi}{2}-1$, and there are two of them (with $A$ and Review
Reasonableness: Circle $C$ has area $\pi\approx3.14$. Each overlap is $\frac{\pi}{2}-1\approx0.57$, and two of them total about $1.14$, leaving roughly $3.14-1.14=2.0$ — matching the exact answer $2$. It is also reassuring that every $\pi$ term cancels: the leftover region can be reshaped into a straight-sided figure of whole-number area, so a $\pi$-free answer like $2$ is exactly what to expect. The $\pi$-laden options (B) $\frac{\pi}{2}$, (D) $\frac{3\pi}{4}$, and (E) $1+\frac{\pi}{2}$ are traps for stopping before the cancellation.
Alternative: Rearrange instead of subtract. Each segment that circle $C$ loses to a neighbor can be slid straight down to fill the notch left between the circles below, out to the tangent lines. After this cut-and-paste the leftover area reshapes into a square whose diagonal is the diameter $2$ of circle $C$. A square with diagonal $2$ has area $\frac12\cdot2^2=2$, confirming the answer is $2$ (C) without ever computing a segment.
CCSS standards used (min grade 8)
8.G.B.8Apply the Pythagorean Theorem to find the distance between two points in a coordinate system (Placing the circle centers on a grid and finding the center distance $\sqrt{2}$ and the crossing points $M$ and $P$.)7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Computing the disk area $\pi$ and the quarter-circle sector area $\frac{\pi}{4}$, then combining to get $2$.)6.G.A.1Find the area of triangles and other polygons by composing or decomposing shapes (Removing the right triangle $AMP$ of area $\frac12$ from the quarter-circle to isolate each circular segment.)
⭐ To find a leftover area, take the whole shape and subtract each overlapping bite — and a quarter-circle minus its triangle turns the messy $\pi$ pieces into a clean number.
⭐ To find a leftover area, take the whole shape and subtract each overlapping bite — and a quarter-circle minus its triangle turns the messy $\pi$ pieces into a clean number.
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