AMC 10 · 2011 · #17

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

tangent-circlescoordinate-geometryarea-triangles physical-representationidentify-subproblems ↑ Prerequisites: tangent-circles

📏 Long solution 💡 3 insights

Problem

Three circles of different sizes each touch the other two from outside. Find the area of the triangle on the touching points.

Pick an answer.

(A)

$\frac{3}{5}$

(B)

$\frac{4}{5}$

(C)

(D)

$\frac{6}{5}$

(E)

$\frac{4}{3}$

How to solve

Strategy Draw a Diagram

The problem hands over three numbers and asks for an area, so the first job is Tool #1 (Draw a Diagram) — but a careful one. The whole problem turns on a fact that is easy to assume and worth proving: when two circles are externally tangent, the touching point is not floating anywhere, it lies on the segment joining the two centers, exactly one radius from each center. Once that is nailed down, the three centers are forced to be 3, 4, and 5 apart, the picture is rigid, and the answer exists at all. Tool #15 (Organize Information in More Ways) then re-records the rigid picture as coordinates, which turns "where is the third touch point?" into arithmetic. Tool #7 (Identify Subproblems) finishes: a slanted triangle is hard to measure, but a rectangle around it minus three right-triangle corners is easy.

1STEP 1

Pin down where two circles touch

Each touching point sits on the line of centres.

OO' = r + s, OT = r, TO' = s

2STEP 2

The centers form a 3-4-5 triangle

The centres form a right triangle.

O₁O₂ = 3, O₁O₃ = 4, O₂O₃ = 5, 3² + 4² = 9 + 16 = 25 = 5²

3STEP 3

Put the right angle on the axes

Putting the right angle on the axes fixes two points.

O₁ = (0,0), O₂ = (3,0), O₃ = (0,4); P = (1,0), Q = (0,1)

4STEP 4

Walk two-fifths along the hypotenuse

The third sits partway along the hypotenuse.

R = (3 - 6/5, 8/5) = (9/5, 8/5), O₃R = √(81/25 + 144/25) = 3

5STEP 5

Box it in, name the corners

Boxing the triangle leaves three corner pieces.

rectangle = 72/25; 1/2, 1/2·4/5·8/5 = 16/25, 1/2·3/5·9/5 = 27/50

6STEP 6

Subtract the three corners

Subtracting them gives 6/5, choice (D).

72/25 - 42/25 = 30/25 = 6/5 → (D)

Answer

6/5

First a size check that needs no computation. All three touch points are exactly 1 away from the point (1,1): (1,0) and (0,1) obviously, and (9/5, 8/5) because (4/5)² + (3/5)² = 1. So the triangle is inscribed in a circle of radius 1, and the largest triangle that fits inside such a circle is the equilateral one, with area 3√(3)/4 ≈ 1.299. That rules out choice (E) 4/3 ≈ 1.333 outright, and it says the answer should be a bit under 1.299 — 6/5 = 1.2 fits, since the triangle is close to, but not exactly, equilateral. Second, a whole-minus-parts cross-check: the center triangle has area 1/2 · 3 · 4 = 6, and the three corner pieces cut off at O₁, O₂, O₃ have areas 1/2 · 1 · 1 · sin 90° = 1/2, 1/2 · 2 · 2 · 4/5 = 8/5, and 1/2 · 3 · 3 · 3/5 = 27/10, totalling 24/5; then 6 - 24/5 = 6/5. Same answer, confirming (D).

💡Key takeaway

Circles that touch on the outside meet on the line joining their centers, so the three radii alone lock the whole picture in place — centers 3, 4, 5 apart, touch points at (1,0), (0,1), (9/5, 8/5), and a rectangle minus three corners gives 6/5.

Pin down where two circles touch
The centers form a 3-4-5 triangle
Put the right angle on the axes
Walk two-fifths along the hypotenuse
Box it in, name the corners
Subtract the three corners