AMC 8 · 2005 · #23

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid

area-circlesarea-trianglesisosceles-trianglereflection-symmetry identify-subproblemsreflection-unfolding ↑ Prerequisites: area-circlesarea-triangles

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

Isosceles right triangle ABC has its right angle at C and encloses a semicircle whose area is 2π. The center O of the semicircle lies on the hypotenuse AB, and the semicircle is tangent to the two legs AC and BC. Find the area of triangle ABC.

Pick an answer.

(A)

(B)

(C)

$3\pi$

(D)

(E)

$4\pi$

AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The figure already shows the triangle and semicircle; the missing ingredient is the two radii from O to the points of tangency on AC and BC. Tool #1 (Draw a Diagram) means adding those two radii: each is perpendicular to its leg, so together with the right angle at C they form a small square inside the triangle. Once the square appears, the radius (found from the area 2π) and a 45-45-90 triangle at corner A together give the leg length. Tool #13 (Use Symmetry) is the alternative angle: reflect the triangle across hypotenuse AB and the half-disk becomes a full circle inside a square. The triangle is half that square, so the triangle's area is half the square's area.

1STEP 1

Halve the disk formula: from the semicircle's area $\frac{1}{2}$ π r² = 2π solve r² = 4, so the radius is r = 2.

\frac{1}{2}

π r² = 2π → π r² = 4π → r² = 4 → r = 2

2STEP 2

Draw the two radii to the touch points D and E: three right angles plus OD = OE make ODCE a square, so CD = CE = 2.

OD = OE = 2, ∠ ODC = ∠ OEC = ∠ DCE = 90° → ODCE is a square; CD = CE = 2

3STEP 3

At corner A the angles 90° and 45° make △ ADO a 45-45-90 triangle, so its legs match: AD = 2.

∠ ODA = 90°, ∠ DAO = 45° → △ ADO is 45-45-90 → AD = OD = 2

4STEP 4

Each leg is AD + DC = 4, so the isosceles right triangle's area is $\frac{1}{2}$ · 4 · 4 = 8.

AC = 2 + 2 = 4 = BC; [△ ABC] =

\frac{1}{2}

· AC · BC =

\frac{1}{2}

· 4 · 4 = 8 → (B)

Answer

Sanity check the size. The semicircle has area 2π ≈ 6.28, and it fits inside the triangle, so the triangle's area must be larger than 6.28. That rules out (A) 6. Choice (B) 8 is just a bit larger than 6.28, which fits the picture: the half-disk fills most of the triangle but leaves two small corner regions near A and B. Choices (D) 10 and (E) 4π ≈ 12.57 would leave too much empty room; (C) 3π ≈ 9.42 has no reason to involve π once the semicircle's area cancels. A direct check: with leg 4, the leg-to-tangent-point distance is AD = 2, the radius OD = 2, and the right triangle's area

\frac{1}{2}

· 4 · 4 = 8 matches answer (B) exactly.

💡Key takeaway

When a circle is tangent to lines, the first move is almost always to draw the radii to the touch points. Here those two radii build a square inside the triangle, and the 45-45-90 corners fill in the rest of each leg — giving leg 4 and area 8.

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