AMC 10 · 2012 · #16
Grade 8 geometry-2dThree circles with radius 2 are mutually tangent. What is the total area of the circles and the region bounded by them, as shown in the figure?
Pick an answer.
AMC 10 2012 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 circles, each with radius $2$, are placed so that every circle just touches the other two. The picture shades all three circles together with the curved patch trapped in the middle between them. Find the total shaded area — the three circles plus the middle region.
Givens: There are $3$ circles, each of radius $2$; The circles are mutually tangent (each one touches both of the others at a single point); The shaded figure is the three full circles together with the curved region bounded between them; Answer choices: (A) $10\pi+4\sqrt{3}$, (B) $13\pi-\sqrt{3}$, (C) $12\pi+\sqrt{3}$, (D) $10\pi+9$, (E) $13\pi$
Unknowns: The total area of the three circles plus the curved middle region
Understand
Restated: Three circles, each with radius $2$, are placed so that every circle just touches the other two. The picture shades all three circles together with the curved patch trapped in the middle between them. Find the total shaded area — the three circles plus the middle region.
Givens: There are $3$ circles, each of radius $2$; The circles are mutually tangent (each one touches both of the others at a single point); The shaded figure is the three full circles together with the curved region bounded between them; Answer choices: (A) $10\pi+4\sqrt{3}$, (B) $13\pi-\sqrt{3}$, (C) $12\pi+\sqrt{3}$, (D) $10\pi+9$, (E) $13\pi$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #16 Change Focus / Count the Complement
The curved shape has no single area formula, so Tool #7 (Identify Subproblems) splits it into two things we can measure: the three whole circles, and the curved patch in the middle. Tool #1 (Draw a Diagram) adds the key hidden line — joining the three centers makes an equilateral triangle that pins down the middle patch. Tool #16 (Change Focus / Count the Complement) handles that patch by measuring the triangle and subtracting the circle slices that stick into it, instead of trying to integrate a curved region directly.
Execute — Answer: A
7.G.B.6 Step 1 Split into circles plus middle patch
- The whole shaded figure is two separate things added together: the three full circles, and the little curved region trapped between them.
- Find each area on its own, then add.
- This turns one strange curved shape into two familiar subproblems.
💡 A shape with no formula becomes easy once you cut it into pieces you already know.
7.G.B.4 Step 2 Add up the three circles
- Each circle has radius $2$, so its area is $\pi r^2=\pi\cdot2^2=4\pi$.
- There are three identical circles, so together they cover $3\times4\pi=12\pi$.
- The tangency doesn't change this — tangent circles meet at a single point and don't overlap in area.
💡 Three whole circles are just three times one circle's area, since they only touch, never overlap.
8.G.B.7 Step 3 Frame the middle with a triangle
- Draw the hidden line segments joining the three centers.
- Because each pair of circles is tangent, each pair of centers is $2+2=4$ apart, so these segments form an equilateral triangle of side $4$.
- To get its area, drop a height: it splits the base $4$ into halves of $2$, and by the Pythagorean theorem the height is $\sqrt{4^2-2^2}=\sqrt{12}=2\sqrt{3}$.
- So the triangle's area is $\tfrac12\cdot4\cdot2\sqrt{3}=4\sqrt{3}$.
💡 Joining tangent centers makes an equilateral triangle, and its height comes straight from the Pythagorean theorem.
7.G.B.4 Step 4 Subtract the circle slices
- That triangle isn't the middle patch yet: at each corner a slice of a circle pokes into the triangle.
- Each corner angle of an equilateral triangle is $60^\circ$, so each slice is a $60^\circ$ sector.
- Three of them add to $60^\circ\times3=180^\circ$, which is exactly half of a full circle.
- Half a circle of radius $2$ has area $\tfrac12\cdot4\pi=2\pi$.
- The middle patch is the triangle with those slices removed: $4\sqrt{3}-2\pi$.
💡 Three $60^\circ$ corners make a straight $180^\circ$, so the pieces sticking in are exactly half a circle.
7.G.B.6 Step 5 Add the two parts
- Add the three circles and the middle patch: $12\pi+(4\sqrt{3}-2\pi)$.
- Combine the $\pi$ terms: $12\pi-2\pi=10\pi$, and keep the $4\sqrt{3}$.
- The total is $10\pi+4\sqrt{3}$, which is choice (A).
💡 Recombining the measured pieces gives the whole figure's area.
7.G.B.6 The whole shaded figure is two separate things added together: the three full ci 7.G.B.4 Each circle has radius $2$, so its area is $\pi r^2=\pi\cdot2^2=4\pi$. There are 8.G.B.7 Draw the hidden line segments joining the three centers. Because each pair of ci 7.G.B.4 That triangle isn't the middle patch yet: at each corner a slice of a circle pok 7.G.B.6 Add the three circles and the middle patch: $12\pi+(4\sqrt{3}-2\pi)$. Combine th Review
Reasonableness: The answer must be a bit more than the three circles alone ($12\pi$) but the middle patch is small — it is $4\sqrt{3}-2\pi\approx6.93-6.28\approx0.65$, a sliver, so the total should be just above $12\pi\approx37.7$. Choice (A) gives $10\pi+4\sqrt{3}\approx31.4+6.9\approx38.3$, which fits. Watch the trap of forgetting to subtract the sectors: that would give $12\pi+4\sqrt{3}$ (choice C uses $+\sqrt{3}$, a similar bait), which overcounts. Removing the half-circle of overlap is what turns $12\pi$ into $10\pi$.
Alternative: Think of it as three circles with a $60^\circ$ wedge of each folded inward to fill the gap, plus the triangle. The three $60^\circ$ wedges total half a circle $=2\pi$; the circles supply $12\pi$ and the equilateral triangle supplies $4\sqrt{3}$, but the wedges are counted once (inside the triangle) rather than twice, so subtract $2\pi$: $12\pi+4\sqrt{3}-2\pi=10\pi+4\sqrt{3}$, again (A).
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for area and circumference of a circle (Computing each circle's area $\pi(2)^2=4\pi$ and the half-circle of sectors $2\pi$.)7.G.B.6Solve real-world problems involving area, surface area, and volume (Decomposing the shaded figure into circles plus a middle patch and recombining the parts.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the equilateral triangle's height $2\sqrt{3}$ to get its area $4\sqrt{3}$.)
⭐ Cut the odd shape into three whole circles plus a middle patch, and the patch is just the center triangle minus half a circle.
⭐ Cut the odd shape into three whole circles plus a middle patch, and the patch is just the center triangle minus half a circle.
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