AMC 10 · 2017 · #16

Grade 10 geometry-2d
tangent-circlescircle-equationcoordinate-geometrysystems-of-equations convert-to-algebracoordinate-geometry ↑ Prerequisites: tangent-circlespythagorean-theorem
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A circle is wedged into the gap between two inner semicircles and one outer arc. Find its radius.

Pick an answer.

(A)
$\frac{3}{4}$
(B)
$\frac{6}{7}$
(C)
$\frac{\sqrt{3}}{2}$
(D)
$\frac{5}{8}\sqrt{2}$
(E)
$\frac{11}{12}$
How to solve
Strategy Introduce a Variable

Arcs are hard to compute with; centers are easy. The whole problem is five tangency statements, and every one of them says the same kind of thing: the distance between two centers equals a sum or a difference of radii. So tool #1 (Draw a Diagram) is used in a specific way — throw away the arcs and keep only the four centers. Tool #4 (Introduce a Variable) names the unknown radius r and the unknown position of P, which turns "squeezed into the gap" into numbers. Tool #13 (Convert to Algebra) puts coordinates on line JK so each tangency becomes a distance-formula equation. Tool #15 (Organize Information in More Ways) is the finishing move: the three equations all contain x² + y² and r², so subtracting them in pairs deletes every square at once and leaves plain linear equations.

1STEP 1

Trade every arc for a distance

Every tangency becomes a centre distance.

AB = 2 + 1 = 3, CA = R - 2, CB = R - 1
2STEP 2

Force out the big radius

Those distances force the outer radius to 3.

CA + CB = AB → (R-2) + (R-1) = 3 → 2R - 3 = 3 → R = 3
3STEP 3

Name the unknowns

Coordinates name the remaining unknowns.

C = (0,0), A = (-1,0), B = (2,0), P = (x,y); AP = 2+r, BP = 1+r, CP = 3-r
4STEP 4

Write the three equations

Three tangencies give three equations.

x² + y² &= (3-r)² = r² - 6r + 9 ; (x+1)² + y² &= (2+r)² = r² + 4r + 4 ; (x-2)² + y² &= (1+r)² = r² + 2r + 1
5STEP 5

Subtract to erase every square

Subtracting erases every square.

(2)-(1): 2x + 1 &= 10r - 5 &&→ x = 5r - 3 ; (3)-(1): -4x + 4 &= 8r - 8 &&→ x = 3 - 2r
6STEP 6

Solve and check the circle exists

Solving gives 6/7, choice (B).

5r - 3 = 3 - 2r → 7r = 6 → r = 6/7; P = (9/7, 12/7) → (B)
Answer
6/7
Check all three tangencies exactly with r = 6/7 and P = (9/7, 12/7). From A = (-1,0): AP = 1/7√(16² + 12²) = 20/7 = 2 + 6/7. From B = (2,0): BP = 1/7√(5² + 12²) = 13/7 = 1 + 6/7. From C = (0,0): CP = 1/7√(9² + 12²) = 15/7 = 3 - 6/7. All three hold on the nose, and 6/7 ≈ 0.857 is sensibly a bit smaller than the radius-1 semicircle it is wedged against. Note the trap in the answer list: 6/7 ≈ 0.857, √(3)/2 ≈ 0.866, and 5/8√(2) ≈ 0.884 are within 3% of each other, so measuring the figure cannot decide between (B), (C), and (D) — only the algebra can.
💡Key takeaway

When circles touch, forget the curves and connect the centers — every tangency becomes a distance you can put in an equation.

  • Trade every arc for a distance
  • Force out the big radius
  • Name the unknowns
  • Write the three equations
  • Subtract to erase every square
  • Solve and check the circle exists