Problem 23 · 2005 AMC 8
Medium
Geometry & Measurement
tangent-radiusisosceles-right-triangle
Isosceles right triangle ABC encloses a semicircle of area 2π. The circle has its center O on hypotenuse AB and is tangent to sides AC and BC. What is the area of triangle ABC?
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Answer: B — 8.
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Hint 1 of 2
A semicircle is half a circle — so double its area to recover a full circle, and that gives you the radius. The radius is the secret length connecting the circle to the triangle.
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Hint 2 of 2
The center O sits at the midpoint of the hypotenuse, and the distance from O to each tangent leg is the radius. Use the symmetry of the 45-45-90 to turn that radius into the leg length.
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Approach: semicircle → radius → leg via tangency
- The semicircle's area is 2π, so a full circle would be 4π = πr², giving r = 2.
- O is the midpoint of hypotenuse AB, and the perpendicular distance from O to each leg is the radius 2. In the 45-45-90, the foot of that perpendicular hits each leg at its midpoint, so the full leg is 2·2 = 4.
- Area = ½·leg·leg = ½·4·4 = 8.
- The reusable move: a tangent radius meets the side at a right angle, and a center on the hypotenuse of a 45-45-90 sits dead center — together these pin the leg to exactly twice the radius.
Another way — radius via the half-leg:
- Let each leg be s. The center O is the hypotenuse midpoint, so its horizontal and vertical distances to the right-angle vertex C are each s/2 — and those distances are the tangent radii.
- So r = s/2. With r = 2 we get s = 4 and area ½·4² = 8.
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