AMC 10 · 2019 · #21
Grade 8 geometry-3dProblem
A sphere with center O has radius 6. A triangle with sides of length 15,15, and 24 is situated in space so that each of its sides is tangent to the sphere. What is the distance between O and the plane determined by the triangle?
Pick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A sphere of radius $6$ centered at $O$ rests against a triangle with sides $15, 15, 24$ — each side of the triangle is tangent to the sphere. How far is $O$ from the plane that holds the triangle?
Givens: Sphere center $O$, radius $R = 6$; Triangle in space with side lengths $15, 15, 24$; Each of the three sides is tangent to the sphere; Answer choices: (A) $2\sqrt{3}$, (B) $4$, (C) $3\sqrt{2}$, (D) $2\sqrt{5}$, (E) $5$
Unknowns: $d$, the perpendicular distance from $O$ to the plane of the triangle
Understand
Restated: A sphere of radius $6$ centered at $O$ rests against a triangle with sides $15, 15, 24$ — each side of the triangle is tangent to the sphere. How far is $O$ from the plane that holds the triangle?
Givens: Sphere center $O$, radius $R = 6$; Triangle in space with side lengths $15, 15, 24$; Each of the three sides is tangent to the sphere; Answer choices: (A) $2\sqrt{3}$, (B) $4$, (C) $3\sqrt{2}$, (D) $2\sqrt{5}$, (E) $5$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #9 Solve an Easier Related Problem
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 $R^2 = r^2 + d^2$ 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.
Execute — Answer: D
8.G.B.7 Step 1 - Draw the 15-15-24 isosceles triangle with the base $24$ horizontal.
- Drop the altitude from the apex to the base; by symmetry it bisects the base into two $12$-segments.
- Each half is a right triangle with legs $12$ and $h$, hypotenuse $15$.
- So $h^2 = 15^2 - 12^2 = 225 - 144 = 81$, giving $h = 9$ (a scaled 3-4-5).
💡 The Pythagorean theorem turns the isosceles split into a familiar 9-12-15 right triangle.
6.G.A.1 Step 2 - The triangle's area is half base times height: $A = \tfrac{1}{2} \cdot 24 \cdot 9 = 108$.
- Its semiperimeter is $s = \tfrac{15+15+24}{2} = 27$.
💡 Area uses the standard triangle formula; semiperimeter is half the perimeter.
7.G.B.4 Step 3 - The three points where the sphere touches the triangle's sides are all in the plane of the triangle AND on the sphere — so they lie on the circle where the sphere meets the plane.
- A circle tangent to all three sides of a triangle from the inside IS the incircle.
- So the sphere-plane cross-section is the incircle, with radius $r = A/s = 108/27 = 4$.
💡 Inradius equals area over semiperimeter — and the sphere's slice IS the incircle.
8.G.B.7 Step 4 - Now look at the 3D picture from the side.
- The center $O$, the foot of the perpendicular $F$ (center of the incircle), and any tangent point $T$ on the sphere form a right triangle: $OT = R = 6$ is a radius of the sphere, $FT = r = 4$ is the incircle radius, and $\angle OFT = 90^\circ$ because $OF$ is perpendicular to the triangle's plane.
💡 A point on the small circle, the plane's foot, and the sphere's center form a right triangle in 3D.
8.G.B.7 Step 5 - Apply the Pythagorean theorem to triangle $OFT$: $d^2 + r^2 = R^2$, so $d^2 = 36 - 16 = 20$, giving $d = \sqrt{20} = 2\sqrt{5}$.
- The answer is (D).
💡 Pythagorean theorem nails the distance: $R^2 = r^2 + d^2$.
8.G.B.7 Draw the 15-15-24 isosceles triangle with the base $24$ horizontal. Drop the alt 6.G.A.1 The triangle's area is half base times height: $A = \tfrac{1}{2} \cdot 24 \cdot 7.G.B.4 The three points where the sphere touches the triangle's sides are all in the pl 8.G.B.7 Now look at the 3D picture from the side. The center $O$, the foot of the perpen 8.G.B.7 Apply the Pythagorean theorem to triangle $OFT$: $d^2 + r^2 = R^2$, so $d^2 = 36 Review
Reasonableness: Sanity check the magnitudes. $2\sqrt{5} \approx 4.47$, which is less than the sphere's radius $6$ — it must be, otherwise the plane would not intersect the sphere at all (and there would be no tangencies). Also $r = 4 < R = 6$, so the cross-sectional circle is smaller than a great circle, consistent with the plane being a non-equatorial slice. Finally, $r^2 + d^2 = 16 + 20 = 36 = R^2$ checks the right-triangle relationship directly.
Alternative: Tool #13 (Convert to Algebra): set up coordinates with $O$ at the origin and the triangle's plane at height $z = d$. Place an incircle of radius $r$ centered at $(0, 0, d)$ in that plane and demand each side of the 15-15-24 triangle (parametrized in that plane) be tangent to the sphere $x^2+y^2+z^2=36$. The tangency condition collapses to $r^2 + d^2 = 36$, with $r$ determined by the triangle's geometry — same answer, more algebra.
CCSS standards used (min grade 8)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the triangle's area via $\tfrac{1}{2} \cdot \text{base} \cdot \text{height}$ and getting the semiperimeter.)7.G.B.4Know the formulas for area and circumference of a circle (Using the inradius formula $r = A/s$ to find the radius of the sphere-plane cross-section circle.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Splitting the 15-15-24 triangle into two 9-12-15 right triangles to get the altitude, and applying $R^2 = r^2 + d^2$ in 3D to extract the distance $d$.)
⭐ This 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 $r = A/s$ you already know — the answer is $d = \sqrt{6^2 - 4^2} = 2\sqrt{5}$.
⭐ This 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 $r = A/s$ you already know — the answer is $d = \sqrt{6^2 - 4^2} = 2\sqrt{5}$.
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