AMC 10 · 2006 · #16
Grade 8 geometry-2d
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Draw the axis of symmetry
Since AB = AC, the altitude from A is the mirror line, and both centers sit on it: S (radius 1) above L (radius 2).
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.5Draw A DiagramThe two circles make similar triangles
Apex A, a center, and its touch-point form right triangles sharing the angle at A, so they are similar by AA and AL = 2·AS.
Both circles hug the same wedge at A, so the far circle must be scaled up in exactly the same proportion as its distance.
Both circles hug the same wedge, so the far one is scaled up in exactly the same proportion as its distance.
▸ Why?
The radius to a touch point meets the side square on, so each circle sits in a right triangle at that corner.
▸ Why?
Those triangles share every angle, so all their matching sides sit in one fixed ratio.
Use the circles touching to find distances
Externally tangent circles put the centers 1 + 2 = 3 apart, so AS = 3 and AL = 6; adding the bottom radius 2 gives height 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.EE.C.8Convert To AlgebraFind the base with the Pythagorean theorem
The small triangle (AS = 3, radius 1) has axis leg 2√(2); scaled up to height 8 the half-base is 2√(2), so BC = 4√(2).
The little circle's triangle is a shrunk copy of the whole half-triangle, so its side ratio unlocks the real one.
8.G.B.7Identify SubproblemsCompute the area
Base 4√(2) times height 8, halved, is 16√(2) — choice (D).
With base and height in hand, the area is just half their product.
7.G.B.6Analyze The UnitsDraw 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√2, height 8, and area 16√2.
- Draw the axis of symmetry
- The two circles make similar triangles
- Use the circles touching to find distances
- Find the base with the Pythagorean theorem
- Compute the area
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