AMC 10 · 2004 · #25
Grade 8 geometry-2dA circle of radius 1 is internally tangent to two circles of radius 2 at points A and B, where AB is a diameter of the smaller circle. What is the area of the region, shaded in the picture, that is outside the smaller circle and inside each of the two larger circles?
Pick an answer.
AMC 10 2004 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: A small circle of radius $1$ is internally tangent to two larger circles of radius $2$, touching them at the two ends $A$ and $B$ of one of its diameters. Find the area of the region that lies outside the small circle but inside both large circles at once.
Givens: The small circle has radius $1$; its diameter $AB$ has length $2$.; Each large circle has radius $2$ and is internally tangent to the small circle — one at $A$, the other at $B$.; From the picture, one large circle is centered at $A$ and the other at $B$ (each passes through the opposite tangent point).; Answer choices: (A) $\frac{5}{3}\pi - 3\sqrt{2}$, (B) $\frac{5}{3}\pi - 2\sqrt{3}$, (C) $\frac{8}{3}\pi - 3\sqrt{3}$, (D) $\frac{8}{3}\pi - 3\sqrt{2}$, (E) $\frac{8}{3}\pi - 2\sqrt{3}$.
Unknowns: The area of the shaded region: outside the small circle and inside both large circles.
Understand
Restated: A small circle of radius $1$ is internally tangent to two larger circles of radius $2$, touching them at the two ends $A$ and $B$ of one of its diameters. Find the area of the region that lies outside the small circle but inside both large circles at once.
Givens: The small circle has radius $1$; its diameter $AB$ has length $2$.; Each large circle has radius $2$ and is internally tangent to the small circle — one at $A$, the other at $B$.; From the picture, one large circle is centered at $A$ and the other at $B$ (each passes through the opposite tangent point).; Answer choices: (A) $\frac{5}{3}\pi - 3\sqrt{2}$, (B) $\frac{5}{3}\pi - 2\sqrt{3}$, (C) $\frac{8}{3}\pi - 3\sqrt{3}$, (D) $\frac{8}{3}\pi - 3\sqrt{2}$, (E) $\frac{8}{3}\pi - 2\sqrt{3}$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems
Everything turns on placing the four key points, so Tool #1 (Draw a Diagram) leads: set coordinates with the small circle at the origin, $A=(0,1)$, $B=(0,-1)$, large circles centered at $A$ and $B$. Two facts fall out immediately — the large circles cross at $(\pm\sqrt3,0)$, and the small disk sits entirely inside the overlap. That second fact powers Tool #16 (Change Focus / Count the Complement): instead of chasing the awkward shaded ring directly, compute the whole lens where the two big disks overlap, then simply subtract the small disk's area $\pi$ at the very end. Computing the lens itself is a clean Tool #7 (Identify Subproblems) job: the chord joining the two crossing points splits the lens into two identical circular segments, and each segment is one sector minus one triangle. The tell-tale $60^\circ$ angles (from an equilateral triangle of side $2$) produce the $\sqrt3$, and remembering to subtract $\pi$ is exactly what separates choice (B) from the decoys (C)/(D)/(E).
Execute — Answer: B
8.G.B.7 Step 1 Set coordinates and find where the big circles cross
- Put the small circle at the origin $O=(0,0)$ so its diameter endpoints are $A=(0,1)$ and $B=(0,-1)$.
- The large circle tangent at $A$ is centered at $A$ with radius $2$ (it passes through $B$, since $AB=2$); likewise the other is centered at $B$ with radius $2$.
- Their equations are $x^2+(y-1)^2=4$ and $x^2+(y+1)^2=4$.
- Subtracting one from the other kills the $x^2$ and gives $(y-1)^2-(y+1)^2=0$, i.e.
- $-4y=0$, so $y=0$.
- Then $x^2+1=4$, so $x=\pm\sqrt3$.
- The circles cross at $P=(\sqrt3,0)$ and $Q=(-\sqrt3,0)$.
💡 By symmetry the two crossing points must sit level with the center, and the horizontal reach $\sqrt3$ is just the leftover leg once you use up $1$ of the radius $2$ going vertically.
7.G.B.6 Step 2 Reframe: add the small disk back, subtract it at the end
- The shaded region is (inside both big circles) with the small disk carved out.
- First check the small disk really sits inside the overlap.
- Any point of the small circle is $(\cos\theta,\sin\theta)$; its distance to center $A=(0,1)$ is $\sqrt{\cos^2\theta+(\sin\theta-1)^2}=\sqrt{2-2\sin\theta}\le 2$, always.
- The same holds for $B$, so the whole small disk lies inside both big circles.
- Therefore the shaded area equals the area of the full overlap lens minus the area of the small disk, $\pi\cdot 1^2=\pi$.
- This is the load-bearing move: compute the easy lens, then subtract $\pi$ once.
💡 Rather than measure a lumpy ring, measure the whole filled lens — a shape we know how to cut up — and take the small circle out only at the finish.
8.G.A.5 Step 3 Spot the equilateral triangle to get the angle
- Look at triangle $ABP$ with $A=(0,1)$, $B=(0,-1)$, $P=(\sqrt3,0)$.
- Its sides are $AB=2$, $AP=\sqrt{3+1}=2$, and $BP=\sqrt{3+1}=2$ — all equal, so $ABP$ is equilateral and every angle is $60^\circ$.
- The same is true of $ABQ$.
- In particular, at center $B$ the chord $PQ$ opens an angle $\angle PBQ=\angle PBA+\angle ABQ=60^\circ+60^\circ=120^\circ$; by symmetry $\angle PAQ=120^\circ$ at center $A$ as well.
- These $120^\circ$ central angles are what let us size the two arcs of the lens.
💡 Three equal sides force three $60^\circ$ angles, and stacking two of them across the diameter makes the $120^\circ$ wedge that each arc of the lens rides on.
7.G.B.4 Step 4 Cut the lens into two segments and add them
- The chord $PQ$ (the $x$-axis) splits the overlap lens into an upper piece bounded by the lower circle (center $B$) and a lower piece bounded by the upper circle (center $A$); by symmetry the two pieces are congruent.
- Take the upper piece: it is the $120^\circ$ circular sector of the circle centered at $B$ (bounded by $BP$, $BQ$, and the arc through $A$) minus the triangle $PBQ$.
- The sector is $\frac{120^\circ}{360^\circ}=\frac13$ of that circle: $\frac13\cdot \pi\cdot 2^2=\frac{4\pi}{3}$.
- Triangle $PBQ$ has base $PQ=2\sqrt3$ and height $1$ (from $B$ up to the $x$-axis), so its area is $\frac12\cdot 2\sqrt3\cdot 1=\sqrt3$.
- One segment $=\frac{4\pi}{3}-\sqrt3$, and the lens is two of them: $\text{area(lens)}=2\left(\frac{4\pi}{3}-\sqrt3\right)=\frac{8\pi}{3}-2\sqrt3$.
💡 A curved lens is hard, but a pie-slice minus a straight triangle is easy — and the lens is just two of those slices back to back.
7.G.B.6 Step 5 Subtract the small circle for the final area
- From Step 2, the shaded region is the lens with the small disk removed: $\left(\frac{8\pi}{3}-2\sqrt3\right)-\pi$.
- Writing $\pi=\frac{3\pi}{3}$ gives $\frac{8\pi}{3}-\frac{3\pi}{3}-2\sqrt3=\frac{5\pi}{3}-2\sqrt3$.
- That is choice (B).
- The decoys (C), (D), (E) all keep the full $\frac{8\pi}{3}$ — they are exactly what you get if you forget to remove the small circle — so subtracting the $\pi$ is the whole ballgame.
💡 The lens was the whole overlap; peeling the small circle back off it turns $\frac{8\pi}{3}$ into $\frac{5\pi}{3}$ and lands the answer.
8.G.B.7 Put the small circle at the origin $O=(0,0)$ so its diameter endpoints are $A=(0 7.G.B.6 The shaded region is (inside both big circles) with the small disk carved out. F 8.G.A.5 Look at triangle $ABP$ with $A=(0,1)$, $B=(0,-1)$, $P=(\sqrt3,0)$. Its sides are 7.G.B.4 The chord $PQ$ (the $x$-axis) splits the overlap lens into an upper piece bounde 7.G.B.6 From Step 2, the shaded region is the lens with the small disk removed: $\left(\ Review
Reasonableness: Numerically the lens is $\frac{8\pi}{3}-2\sqrt3\approx 8.378-3.464=4.914$, and subtracting $\pi\approx3.142$ leaves $\approx1.772$, matching $\frac{5\pi}{3}-2\sqrt3\approx5.236-3.464=1.772$. That is positive and comfortably smaller than the lens, as a carved-out sub-region must be — a good sanity check. The $\sqrt3$ is the right radical: it comes from a $30$-$60$-$90$ / equilateral-triangle setup (angles of $60^\circ$ and $120^\circ$), so a $\sqrt2$ (which would need a $45^\circ$ angle) or a bare $3\sqrt3$ (no leftover circle subtracted) has no place here, ruling out (A), (C), (D). Only (B) has both the correct $\frac{5\pi}{3}$ (small circle removed) and the correct $2\sqrt3$.
Alternative: Use the two-circle overlap formula directly. For two circles of radius $R$ whose centers are distance $d$ apart, the overlap area is $2R^2\cos^{-1}\!\left(\frac{d}{2R}\right)-\frac{d}{2}\sqrt{4R^2-d^2}$. Here $R=2$, $d=2$, so it is $2(4)\cos^{-1}\!\left(\frac12\right)-1\cdot\sqrt{16-4}=8\cdot\frac{\pi}{3}-\sqrt{12}=\frac{8\pi}{3}-2\sqrt3$. Subtracting the small disk $\pi$ again gives $\frac{5\pi}{3}-2\sqrt3$, confirming (B) without decomposing by hand.
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Solving the two circle equations to locate the crossing points $(\pm\sqrt3,0)$ and confirming the side lengths $AP=BP=2$.)8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Recognizing $\triangle ABP$ as equilateral to get $60^\circ$ angles and adding them to the $120^\circ$ central angle of each lens arc.)7.G.B.4Know the formulas for area and circumference of a circle (Computing a $120^\circ$ sector as one-third of the full circle of radius $2$ (area $\frac{4\pi}{3}$).)7.G.B.6Solve real-world problems involving area, surface area, and volume (Assembling the lens as sector-minus-triangle segments and subtracting the small disk to get the composite shaded area.)
⭐ Instead of measuring the awkward shaded ring, fill in the small circle to get the tidy lens where both big circles overlap; cut that lens into a pie-slice minus a triangle to find it is $\frac{8\pi}{3}-2\sqrt3$, then take the small circle $\pi$ back out to get $\frac{5\pi}{3}-2\sqrt3$.
⭐ Instead of measuring the awkward shaded ring, fill in the small circle to get the tidy lens where both big circles overlap; cut that lens into a pie-slice minus a triangle to find it is $\frac{8\pi}{3}-2\sqrt3$, then take the small circle $\pi$ back out to get $\frac{5\pi}{3}-2\sqrt3$.
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