AMC 10 · 2011 · #15

Grade 10 geometry-3d
similar-trianglespythagorean-theoremspatial-visualization physical-representationconvert-to-algebra ↑ Prerequisites: similar-triangles
📏 Long solution 💡 3 insights
Problem
A dome sits on a pyramid's base plane and touches all four slanted faces. Find the base edge.

Pick an answer.

(A)
$3\sqrt{2}$
(B)
$\frac{13}{3}$
(C)
$4\sqrt{2}$
(D)
6
(E)
$\frac{13}{2}$
How to solve
Strategy Visualize Spatial Relationships

Tool #17 (Visualize Spatial Relationships): the whole problem collapses once you cut the solid with the right vertical plane — one chosen so that a triangular face stands square to it and therefore shows up as a single line. Tool #4 (Introduce a Variable): call half the base edge m; then every length in the slice is a formula in m and the tangency condition becomes one equation. Tool #7 (Identify Subproblems): the job splits into three separate questions — where is the hemisphere centred, what does tangency say numerically, and does the touch really land on the dome. Tool #3 (Eliminate Possibilities): the last step must show no other base edge can work, which needs the distance to move in one direction only as m grows.

1STEP 1

Symmetry pins the hemisphere's centre

Symmetry puts the centre at the middle.

dist(C,face₁) = dist(C,face₃) and dist(C,face₂) = dist(C,face₄) ⟹ C = O
2STEP 2

Name the half-edge and cut one vertical slice

One vertical slice holds the whole problem.

s = 2m, OM = m, OE = 6, ∠ EOM = 90^°
3STEP 3

Tangency means a perpendicular of length 2

Touching means a perpendicular of the radius.

OP ⊥ EM, OP = 2
4STEP 4

Similar right triangles give that distance

Similar right triangles give that distance.

△ EPO ∼ △ EOM ⟹ OP/OM = OE/EM ⟹ OP = (OM · OE)/EM = 6m/(√(m²+36))
5STEP 5

Solve for the half-edge

Solving gives the half-edge, so the edge is 3√2.

6m/(√(m²+36)) = 2 → 3m = √(m²+36) → 9m² = m²+36 → m² = 9/2 → m = 3√(2)/2, s = 3√(2)
6STEP 6

Check the touch lands on the dome, not below the floor

The touch point really lands on the dome.

EP = √(36-4) = 4√(2), EM = 9√(2)/2, MP = √(2)/2 = EM/9, z_P = 6 · 1/9 = 2/3 > 0
7STEP 7

No other base edge can work

The distance grows steadily, so no other edge works.

d(m) = 6{√(1+36/m²)} is strictly increasing; d(m) < 2 → overhang at height 2m/(√(36+m²)) > 0 → s = 3√(2) = (A)
Answer
3√(2)
Substitute back: m = 3√(2)/2 ≈ 2.1213, so √(m²+36) = √(40.5) ≈ 6.3640 and (6 × 2.1213)/6.3640 = 2.000 — the distance really is the radius. A second, independent check comes from the reciprocal-Pythagoras relation for the altitude to a hypotenuse, 1/OP² = 1/OM² + 1/OE²: here 1/4 = 2/9 + 1/36 = 8/36 + 1/36 = 9/36, which holds exactly. The size is sensible too: the hemisphere of radius 2 needs the distance from centre to wall to be at least 2, and m ≈ 2.12 clears that by a little — a snug fit, as the picture suggests. Testing the other choices gives distances 2.038 for 13/3, 2.558 for 4√(2), 2.683 for 6 and 2.858 for 13/2; none equals 2. Choice (C) 4√(2) is the trap: it is exactly EP, the distance from the apex to the touching point, not the base edge.
💡Key takeaway

Tangent means the distance from the centre equals the radius, so cut the solid with a plane that stands square to one face and the whole three-dimensional condition becomes a single altitude in a right triangle.

  • Symmetry pins the hemisphere's centre
  • Name the half-edge and cut one vertical slice
  • Tangency means a perpendicular of length 2
  • Similar right triangles give that distance
  • Solve for the half-edge
  • Check the touch lands on the dome, not below the floor
  • No other base edge can work