AMC 10 · 2019 · #18
Grade 8 geometry-3dPick an answer.
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.
Find the triangle's height
Being isosceles gives the height at once.
The Pythagorean theorem turns the isosceles split into a familiar 9-12-15 right triangle.
8.G.B.7Draw A DiagramArea and semiperimeter
Compute the area and the semiperimeter.
Area uses the standard triangle formula; semiperimeter is half the perimeter.
6.G.A.1Identify SubproblemsThe inradius
Their ratio is the inradius.
Inradius equals area over semiperimeter — and the sphere's slice IS the incircle.
The inscribed radius is the area divided by the semiperimeter, and the sphere's slice is exactly that circle.
▸ Why?
Joining the centre to the corners cuts the triangle into three pieces, each half a side times the same height.
▸ Why?
That shared height is the distance to each side, which a tangent circle meets square on.
Build the right triangle
The sphere's radius is the hypotenuse.
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 DiagramCompute the distance
Pythagoras gives two root five.
Pythagorean theorem nails the distance: R² = r² + d².
8.G.B.7Identify SubproblemsThis AMC 12 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 = A/s you already know — the answer is d = √(6² - 4²) = 2√(5).