AMC 10 · 2006 · #16
Grade 8 geometry-2dA circle of radius 1 is tangent to a circle of radius 2. The sides of △ABC are tangent to the circles as shown, and the sides AB and AC are congruent. What is the area of △ABC?
Pick an answer.
AMC 10 2006 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: An isosceles triangle $ABC$ (with $AB = AC$) has two circles stacked inside it along its line of symmetry: a circle of radius $2$ sitting on the base and a circle of radius $1$ above it. Each circle touches both equal sides, the two circles touch each other, and the big circle also touches the base. We want the area of $\triangle ABC$.
Givens: $\triangle ABC$ is isosceles with $AB = AC$; A large circle has radius $2$; a small circle has radius $1$; Both circles are tangent to the two equal sides $\overline{AB}$ and $\overline{AC}$; The large circle is tangent to base $\overline{BC}$; the two circles are tangent to each other; Answer choices: (A) $\tfrac{35}{2}$, (B) $15\sqrt{2}$, (C) $\tfrac{64}{3}$, (D) $16\sqrt{2}$, (E) $24$
Unknowns: The area of $\triangle ABC$
Understand
Restated: An isosceles triangle $ABC$ (with $AB = AC$) has two circles stacked inside it along its line of symmetry: a circle of radius $2$ sitting on the base and a circle of radius $1$ above it. Each circle touches both equal sides, the two circles touch each other, and the big circle also touches the base. We want the area of $\triangle ABC$.
Givens: $\triangle ABC$ is isosceles with $AB = AC$; A large circle has radius $2$; a small circle has radius $1$; Both circles are tangent to the two equal sides $\overline{AB}$ and $\overline{AC}$; The large circle is tangent to base $\overline{BC}$; the two circles are tangent to each other; Answer choices: (A) $\tfrac{35}{2}$, (B) $15\sqrt{2}$, (C) $\tfrac{64}{3}$, (D) $16\sqrt{2}$, (E) $24$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #13 Convert to Algebra, #7 Identify Subproblems, #8 Analyze the Units
The figure hides its structure until you draw the altitude from $A$: it is the axis of symmetry, so both circle centers sit on it and every measurement lines up vertically. Tool #1 (Draw a Diagram) exposes the family of similar right triangles formed by the apex $A$, each center, and each point where a circle grazes a side. Tool #4 (Introduce a Variable) names the two apex-to-center distances so the shared shape of those triangles becomes a clean ratio. Tool #13 (Convert to Algebra) turns 'the circles touch' into an equation that pins those distances down and hands us the height for free. Tool #7 (Identify Subproblems) splits the leftover work into finding the base, and Tool #8 (Analyze the Units) closes with the area formula. The whole solution is one geometric picture read carefully, not a formula hunt.
Execute — Answer: D
8.G.A.5 Step 1 Draw the axis of symmetry
- Since $AB = AC$, the altitude from apex $A$ down to base $BC$ is also the triangle's mirror line.
- By symmetry each inscribed circle is centered on this line: call the small center $S$ (radius $1$) and the large center $L$ (radius $2$), with $L$ below $S$.
- Where a circle touches a side, the radius to that point is perpendicular to the side.
- So $A$, a center, and its touch-point on side $AB$ form a right triangle with the right angle at the touch-point and the radius as the short leg.
💡 Equal sides fold onto each other across the altitude, so anything balanced inside the triangle must sit right on that fold line.
8.G.A.4 Step 2 The two circles make similar triangles
- Look at the two right triangles: apex $A$ to small center $S$ to its touch-point, and apex $A$ to large center $L$ to its touch-point.
- Both contain the same apex angle at $A$ and both have a right angle at the touch-point, so by AA they are similar.
- In similar right triangles the ratio (radius)$\div$(distance from $A$ to the center) is the same.
- Let $AS = d$ and $AL = D$.
- Then $\dfrac{1}{d} = \dfrac{2}{D}$, which means $D = 2d$: the big center is exactly twice as far from $A$ as the small center.
💡 Both circles hug the same wedge at $A$, so the far circle must be scaled up in exactly the same proportion as its distance.
8.EE.C.8 Step 3 Use the circles touching to find distances
- The two circles are externally tangent, so the gap between their centers equals the sum of the radii: $SL = 1 + 2 = 3$.
- Both centers are on the axis with $L$ farther from $A$, so $SL = D - d = 3$.
- Combine with $D = 2d$: substituting gives $2d - d = d = 3$, hence $d = 3$ and $D = 6$.
- Now the height: the large circle rests on the base, so its center $L$ is a distance equal to its radius, $2$, above the base.
- The height from $A$ to the base is therefore $AL + 2 = 6 + 2 = 8$.
💡 'Touching' turns a picture into a length equation, and once the far center is nailed down, dropping one more radius lands you on the base.
8.G.B.7 Step 4 Find the base with the Pythagorean theorem
- In the small circle's right triangle ($AS = 3$, radius $= 1$), the third side along the axis is $\sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}$, so its along-axis leg to hypotenuse ratio is $\dfrac{2\sqrt{2}}{3}$.
- The full half-triangle $A$-foot-$B$ is similar to it (same apex angle, right angle at the foot), with the altitude $8$ playing the along-axis leg.
- Matching ratios, $\dfrac{8}{AB} = \dfrac{2\sqrt{2}}{3}$, so $AB = \dfrac{24}{2\sqrt{2}} = 6\sqrt{2}$.
- Then by the Pythagorean theorem the half-base is $\sqrt{AB^2 - 8^2} = \sqrt{72 - 64} = \sqrt{8} = 2\sqrt{2}$, so the whole base $BC = 4\sqrt{2}$.
💡 The little circle's triangle is a shrunk copy of the whole half-triangle, so its side ratio unlocks the real one.
7.G.B.6 Step 5 Compute the area
- The triangle has base $BC = 4\sqrt{2}$ and height $8$.
- Its area is $\dfrac{1}{2} \cdot 4\sqrt{2} \cdot 8 = 16\sqrt{2}$.
- This is choice $\textbf{(D)}$.
💡 With base and height in hand, the area is just half their product.
8.G.A.5 Since $AB = AC$, the altitude from apex $A$ down to base $BC$ is also the triang 8.G.A.4 Look at the two right triangles: apex $A$ to small center $S$ to its touch-point 8.EE.C.8 The two circles are externally tangent, so the gap between their centers equals 8.G.B.7 In the small circle's right triangle ($AS = 3$, radius $= 1$), the third side al 7.G.B.6 The triangle has base $BC = 4\sqrt{2}$ and height $8$. Its area is $\dfrac{1}{2} Review
Reasonableness: Check the size: $16\sqrt{2} \approx 22.6$, which sits sensibly between the base $4\sqrt{2}\approx 5.7$ and height $8$ triangle you can picture. The numbers all cross-check: with $AB = 6\sqrt{2}$, half-base $2\sqrt{2}$, and height $8$, we get $(2\sqrt{2})^2 + 8^2 = 8 + 64 = 72 = (6\sqrt{2})^2$, so the right triangle closes exactly. Both circles also fit: the small center is $3$ from $A$ and the large center $6$ from $A$, a gap of $3 = 1+2$, and the large center is $8 - 6 = 2$ above the base, matching its radius. The trap answers come from slips: $\textbf{(E) } 24$ is $\tfrac12\cdot 6 \cdot 8$ (using $6$ instead of $4\sqrt2$ for the base), and $\textbf{(A) }\tfrac{35}{2}$/$\textbf{(C) }\tfrac{64}{3}$ do not survive the tangency ratio $D = 2d$.
Alternative: Use the inradius formula. The large circle is the triangle's incircle (it touches the base and both equal sides), so its radius $r = 2$ equals Area$\,/\,$semiperimeter: $2 = \dfrac{K}{s}$. With base $4\sqrt2$ and equal sides $6\sqrt2$, the semiperimeter is $s = \dfrac{4\sqrt2 + 6\sqrt2 + 6\sqrt2}{2} = 8\sqrt2$, giving $K = 2 \cdot 8\sqrt2 = 16\sqrt2$, the same area. This confirms the big circle really is the incircle and the answer is consistent.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Arguing from symmetry that both centers lie on the altitude and that each radius meets a side at a right angle, setting up the right triangles.)8.G.A.4Understand that a two-dimensional figure is similar to another using transformations (Recognizing the two apex right triangles as similar (AA) so that radius over apex-distance is a shared ratio, giving $D = 2d$.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving $D - d = 3$ together with $D = 2d$ to get the apex-to-center distances $d = 3$ and $D = 6$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the along-axis leg $2\sqrt2$, the side $AB = 6\sqrt2$, and the half-base $2\sqrt2$ to recover the full base.)7.G.B.6Solve real-world problems involving area, surface area, and volume (Applying the triangle area formula $\tfrac12 \cdot \text{base} \cdot \text{height}$ to finish.)
⭐ Draw the mirror line of the isosceles triangle: both circles sit on it, their apex triangles are similar so the far center is twice as far as the near one, and 'the circles touch' fixes every length — leading to base $4\sqrt2$, height $8$, and area $16\sqrt2$.
⭐ Draw the mirror line of the isosceles triangle: both circles sit on it, their apex triangles are similar so the far center is twice as far as the near one, and 'the circles touch' fixes every length — leading to base $4\sqrt2$, height $8$, and area $16\sqrt2$.
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