AMC 10 · 2019 · #21
Grade 8 geometry-3dPick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #7 (Subproblems): split into (a) find the inradius of the 15-15-24 triangle inside the plane and (b) use a right triangle in 3D to get d. Tool #1 (Diagram): a 2D picture for the triangle plus a side-view sphere-plane cross-section makes the relationship R² = r² + d² visible. Tool #9 (Easier Problem): the 15-15-24 triangle splits along its axis of symmetry into two 9-12-15 right triangles — a scaled 3-4-5 — so area is immediate.
Drop the altitude of the 15-15-24 isosceles triangle; it splits the base into two 12s, so the height is h = 9 (a 9-12-15 triangle).
The Pythagorean theorem turns the isosceles split into a familiar 9-12-15 right triangle.
8.G.B.7Draw A DiagramArea = ½·24·9 = 108 and semiperimeter s = = 27.
Area uses the standard triangle formula; semiperimeter is half the perimeter.
6.G.A.1Identify SubproblemsThe three tangent points lie on the sphere's cross-section circle, which is exactly the incircle, so r = = = 4.
Inradius equals area over semiperimeter — and the sphere's slice IS the incircle.
7.G.B.4Identify SubproblemsSide-view: O, the plane's foot F, and a tangent point T form a right triangle: OT = R = 6, FT = r = 4, ∠OFT = 90°.
A point on the small circle, the plane's foot, and the sphere's center form a right triangle in 3D.
8.G.B.7Draw A DiagramPythagoras on OFT: d² = R² - r² = 36 - 16 = 20, so d = √(20) = 2√(5) — choice (D).
Pythagorean theorem nails the distance: R² = r² + d².
8.G.B.7Identify SubproblemsThis AMC 10 problem only needs Grade 8 Pythagorean theorem (twice — once to get the triangle's height, once to relate the sphere radius, incircle radius, and distance) plus the inradius rule r = you already know — the answer is d = √(6² - 4²) = 2√(5).