AMC 10 · 2015 · #24

Grade 10 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

perpendicular-bisectorpower-of-a-pointpythagorean-theoremquadratic-equations convert-to-algebracasework ↑ Prerequisites: perpendicular-bisectorpythagorean-theorem

📏 Long solution 💡 4 insights

Problem

Four circles share two points, with radii in a fixed ratio and equal centre gaps. Add the four centre distances to the chord's midpoint.

Pick an answer.

(A)

180

(B)

184

(C)

188

(D)

192

(E)

196

How to solve

Strategy Introduce a Variable

A picture (Tool #1) shows the key collapse: every center is equidistant from P and Q, so all four centers and R sit on one line. That makes a single signed coordinate per center enough to encode both its position and its radius (Tool #4, Introduce a Variable), and the two clues about a pair become one quadratic. The problem then splits into two separate jobs (Tool #7): first list every pair shape the clues permit (Tool #2, and the degree of the quadratic is what proves the list is complete), then use "no two circles congruent" to eliminate the pairings that are not allowed (Tool #3). The second job is the real content — without it we would only know a value is possible, not that it is forced.

1STEP 1

Trap every center on one line

Every centre sits on one line.

AP = AQ, BP = BQ, CP = CQ, DP = DQ ⟹ A, B, C, D, R lie on the perpendicular bisector of PQ

2STEP 2

One coordinate per center

One coordinate describes each circle.

r² = t² + 24² = t² + 576

3STEP 3

Write the pair conditions

The pair conditions become two equations.

64(p² + 576) = 25(q² + 576), (q - p)² = 39²

4STEP 4

Reduce to a single quadratic

Eliminating leaves one quadratic.

64(p²+576) = 25((p+39)²+576) ⟹ 39p² - 1950p - 15561 = 0 ⟹ p² - 50p - 399 = 0

5STEP 5

List the two pair shapes

It yields exactly two pair shapes.

Shape I: {57, 96}, radii 15√(17), 24√(17), sum 153 Shape II: {7, 32}, radii 25, 40, sum 39

6STEP 6

Prove the list is complete

The degree proves the list is complete.

deg(p²-50p-399) = 2 ⟹ at most two p ; both are realized ⟹ exactly two pair shapes

7STEP 7

Non-congruence forces one of each

Distinct sizes force one of each shape.

d ↦ √(d²+576) is strictly increasing ⟹ congruent ⇔ equal distance to R

8STEP 8

Add the four distances

Adding the four gives 192, choice (D).

AR+BR+CR+DR = (57+96) + (7+32) = 153 + 39 = 192 → (D)

Answer

192

Every circle carries a chord of length 48, so every radius must exceed 24; the four radii 25, 40, 15√(17) ≈ 61.8, 24√(17) ≈ 99.0 all clear that bar and are pairwise different, honoring "no two congruent". Each declared pair really has the required ratio: 25/40 = 5/8 and 15√(17)/24√(17) = 5/8. Each declared centre distance is right: 96 - 57 = 39 and 7 + 32 = 39. Two further checks confirm the argument closed both halves. First, the forbidden pairings give 153+153 = 306 and 39+39 = 78, and neither is on the answer list, so the mixed configuration is the only candidate the problem could have meant. Second, the quadratic had exactly two roots, both used, so no fifth or sixth circle position was overlooked. The forced total 192 is choice (D).

💡Key takeaway

Every center sits on one line through the middle of the shared chord, so a single number per circle turns the picture into one quadratic; its two roots are the only two shapes, and "no two circles alike" is what forces one of each.

Trap every center on one line
One coordinate per center
Write the pair conditions
Reduce to a single quadratic
List the two pair shapes
Prove the list is complete
Non-congruence forces one of each
Add the four distances