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2021 AMC Fall 10A Problem 22

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2021 AMC Fall 10A Problem 22AMC 10 Step by Step
Name:                             Date: September 13, 2026
Solid Geometry · difficulty 4 of 5 · about 9 min · position P21-25

Inside a right circular cone with base radius 55 and height 1212 are three congruent spheres with radius rr . Each sphere is tangent to the other two spheres and also tangent to the base and side of the cone. What is rr ?

  1. A)

    32\frac{3}{2}

  2. B)

    90−40311\frac{90-40\sqrt{3}}{11}

  3. C)

    22

  4. D)

    144−25344\frac{144-25\sqrt{3}}{44}

  5. E)

    52\frac{5}{2}

Answer

B

Key insight

The three centers form an equilateral triangle of side 2r at height r, so each center is 2r/sqrt(3) from the axis; then the slant-line distance condition gives r.

Solution

Locate the centers. Each sphere touches the base, so every center is at height rr. The spheres are mutually tangent, so the centers are pairwise 2r2r apart: they form an equilateral triangle of side 2r2r in the horizontal plane at height rr, centered on the cone's axis by symmetry. The distance from a center to the axis is the circumradius of that triangle,
d=2r3=233 r.d = \frac{2r}{\sqrt3} = \frac{2\sqrt3}{3}\,r.

Slice through the axis and one center. The cross-section of the cone is a triangle with apex at height 1212 and base from −5-5 to 55; its right slant side is the line through (5,0)(5, 0) and (0,12)(0, 12), i.e. 12x+5y=6012x + 5y = 60 (a 55-1212-1313 triangle). The sphere appears as a circle of radius rr centered at (d,r)(d, r), tangent to this line, so the distance from the center to the line equals rr:
60−12d−5r13=r⟹60=18r+12d=18r+83 r.\frac{60 - 12d - 5r}{13} = r \quad\Longrightarrow\quad 60 = 18r + 12d = 18r + 8\sqrt3\, r.

Solve.
r=6018+83=309+43=30(9−43)81−48=30(9−43)33=90−40311.r = \frac{60}{18 + 8\sqrt3} = \frac{30}{9 + 4\sqrt3} = \frac{30(9 - 4\sqrt3)}{81 - 48} = \frac{30(9 - 4\sqrt3)}{33} = \frac{90 - 40\sqrt3}{11}.

The answer is (B) 90−40311\boxed{\textbf{(B)}\ \frac{90-40\sqrt{3}}{11}}.

Why this works

Three-dimensional tangency problems split into two planar pictures: a horizontal one (the centers form a regular polygon, fixing their distance from the axis) and a vertical one (a cross-section through the axis, where each sphere is a circle tangent to a line). The vertical picture is the standard "circle tangent to two lines" setup, solved either by the point-to-line distance formula or by similar triangles in the 55-1212-1313 cross-section.

Alternative approach

Avoid the distance formula: at height rr the slant side is at horizontal position 5−5r125 - \tfrac{5r}{12}. The horizontal gap from the center to the slant side is g=5−5r12−dg = 5 - \tfrac{5r}{12} - d, and the perpendicular distance is g⋅1213g \cdot \tfrac{12}{13} (the slant makes angle with the horizontal whose sine is 1213\tfrac{12}{13}). Setting this equal to rr gives 60−5r−12d=13r60 - 5r - 12d = 13r, the same equation.

The trap

Placing a sphere center at distance r (or 2r) from the axis instead of the circumradius 2r/sqrt(3) of the equilateral triangle of centers.

Common mistakes

  • Placing a sphere center at distance r (or 2r) from the axis instead of the circumradius 2r/sqrt(3) of the equilateral triangle of centers.
  • Using the horizontal gap to the slant side as the tangency distance, instead of the perpendicular distance (multiply by 1213\tfrac{12}{13}).

Techniques

Add construction lines/points (drop altitudes, extend segments, connect centers) · Place the figure on coordinates and compute