AMC 10 · 2019 · #22

Grade 11 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
tangent-circlessimilar-trianglesequilateral-trianglepythagorean-theoremratio-proportion identify-subproblemsconvert-to-algebra ↑ Prerequisites: similar-trianglespythagorean-theorem
📏 Long solution 💡 4 insights
Problem
Two circles share a centre: a big one of radius 20 and a small one of radius 17. An equilateral triangle sits in the ring between them. One vertex touches the big circle, and the line through the opposite side just grazes the small circle. The segment from that vertex to the centre crosses the opposite side at a point splitting it three to one. Find the triangle's side length in the requested form and add up the four whole numbers in it.

Pick an answer.

(A)
42
(B)
86
(C)
92
(D)
114
(E)
130
How to solve
Strategy Draw a Diagram

Nothing here is computable until the picture has the right two extra points on it, so drawing comes first: mark S where line BC touches γ, and M the midpoint of BC. Those two points create two right angles on the same line BC — OS ⊥ BC because a radius meets a tangent squarely, and AM ⊥ BC because the altitude of an equilateral triangle hits the base at its midpoint. Two right angles on one line, plus the vertical angles at P, hand over a pair of similar triangles, and that similarity is the whole bridge between the triangle's size and the two radii. From there I name the side length s (Introduce a Variable) and break the work into small pieces (Identify Subproblems): first where P sits on BC, then the length AP, then the similarity ratio, then one linear equation in s.

1STEP 1

Mark the two right angles

The tangent and the median make right angles.

OS ⊥ BC, OS = 17 and AM ⊥ BC, BM = MC
2STEP 2

Locate the dividing point

The ratio gives its distance from the midpoint.

BP = 3s/4, CP = s/4, PM = BP - BM = 3s/4 - s/2 = s/4
3STEP 3

Distance to the vertex

Pythagoras gives that length.

AP² = AM² + PM² = 3s²/4 + s²/16 = 13s²/16 → AP = √(13)/4s, AM/AP = √(3)/2s/√(13)/4s = 2√(3)/√(13)
4STEP 4

Spot the similar triangles

Two right angles and a shared angle give similarity.

∠ AMP = ∠ OSP = 90°, ∠ APM = ∠ OPS → △ PMA ∼ △ PSO → AM/OS = PM/PS = AP/OP
5STEP 5

Turn it into one equation

The ratio becomes one equation.

AP + 17 AP/AM = 20 → AP·(AM + 17)/AM = 20 → (AM + 17)/20 = AM/AP = 2√(3)/√(13)
6STEP 6

Solve for the side

Solving gives exactly the requested form.

√(3)/2s = 40√(3)/√(13) - 17 → s = 80/√(13) - 34/√(3) = AB, m+n+p+q = 80+13+34+3 = 130
7STEP 7

Add the four numbers

Adding gives 130.

AB = 80/√(13) - 34/√(3) → (m,n,p,q) = (80,13,34,3), m+n+p+q = 130
Answer
130
Numerically 80/√(13) ≈ 22.19 and 34/√(3) ≈ 19.63, so s ≈ 2.56 — a small triangle, which is exactly right: the ring between radius 17 and radius 20 is only 3 wide, so an equilateral triangle squeezed into it cannot be large. Checking the picture, AM ≈ 2.22, so AP ≈ 2.31 and OP ≈ 17.69; then PS = √(OP² - OS²) ≈ 4.91, and the similarity ratio PM/PS ≈ 0.64/4.91 ≈ 0.130 matches AM/OS ≈ 2.22/17 ≈ 0.130. Building the triangle in coordinates confirms every condition: all three sides equal, |OA| = 20, the distance from O to line BC is exactly 17, and BP/CP = 3. The nearest vertex to O is C at about 17.53, safely outside γ — the tangency point S lands beyond C, off the segment itself, which is how the triangle can stay outside the small circle while its extended side line still touches it. Answer (E) stands.
💡Key takeaway

Drop the two perpendiculars onto line BC — one from A, one from O — and the bow-tie of similar triangles at P ties the triangle's size straight to the two radii.

  • Mark the two right angles
  • Locate P from the ratio
  • Pythagoras gives AP
  • Spot the similar triangles
  • Turn similarity into one equation
  • Solve for the side length
  • Read off the requested sum

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