AMC 10 · 2007 · #11

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

isosceles-trianglecoordinate-geometrypythagorean-theoremarea-circles convert-to-algebra ↑ Prerequisites: pythagorean-theorem

📏 Medium solution 💡 3 insights

Problem

An isosceles triangle has two equal sides of length 3 and a base of length 2. A circle passes through all three vertices of this triangle. Find the area of that circle.

Pick an answer.

(A)

$2\pi$

(B)

$\frac{5}{2}\pi$

(C)

$\frac{81}{32}\pi$

(D)

$3\pi$

(E)

$\frac{7}{2}\pi$

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The area of the circle is π R², so everything hinges on the radius R. Drawing the triangle with coordinates (Tool #1) and putting the base flat with its midpoint at the origin exposes the triangle's mirror symmetry, which forces the circle's center onto the vertical axis. Naming the center's height k (Tool #4) turns 'the center is equidistant from all three vertices' into equations. The job then splits into small subproblems (Tool #7): first the triangle's height, then the center's position, then the radius, then the area.

1STEP 1

Set coordinates and find the height

Put the base on the x-axis with B=(-1,0), C=(1,0) and apex A=(0,h). Then 1²+h²=3², so the height is h=2√2.

1²+h²=3² → h²=8 → h=2√2

2STEP 2

Put the center on the axis of symmetry

Symmetry puts the center at O=(0,k). Out to a base corner R²=1+k²; straight up to the apex R²=(2√2-k)².

R²=1+k² and R²=(2√2-k)²

3STEP 3

Solve for the center's height

Set them equal: 1+k²=(2√2-k)². The k² terms cancel, leaving 1=8-4√2 k, so k=7/(4√2).

1+k²=8-4√2 k+k² → 4√2 k=7 → k=7/4√2

4STEP 4

Compute the radius squared and the area

Squaring gives k²=49/32, so R²=1+49/32=81/32 and the area is π R²=81/32π, choice (C).

R²=1+49/32=81/32 → Area=π R²=81/32π

Answer

81/32π

Sanity-check the size: R²=81/32≈2.53, so R≈1.59. The circle must be big enough to reach every vertex; the apex is 2√2≈2.83 above the base, and the center at height k=7/4√2≈1.24 leaves 2.83-1.24≈1.59 up to the apex and √(1+1.24²)≈1.59 out to a base corner — the two distances agree, confirming a single consistent radius. The area 81/32π≈2.53π lands neatly between choices 2π and 5/2π, matching only (C).

💡Key takeaway

Put the shape on a grid, use symmetry to pin the circle's center on the mirror line, and remember the area only needs R², not R itself.

Set coordinates and find the height
Put the center on the axis of symmetry
Solve for the center's height
Compute the radius squared and the area

Set coordinates and find the height

Put the center on the axis of symmetry

Solve for the center's height

Compute the radius squared and the area

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