AMC 10 · 2005 · #16

Grade 9 geometry-2d

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

tangent-circlescoordinate-geometrypythagorean-theorem easier-related-problemconvert-to-algebra ↑ Prerequisites: coordinate-geometrypythagorean-theoremquadratic-equations

📏 Long solution 💡 3 insights 📊 Diagram

Problem

Three equal small circles sit in a corner, one touching both axes and the other two each touching it and one axis. A larger circle touches both axes and the two outer small circles. Find how many times larger its radius is.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Introduce a Variable

The question asks only for a ratio, and every condition in the problem is about tangency, which survives scaling. So the picture can be shrunk until s=1, leaving a single unknown r. Coordinates then do all the work: a circle tangent to both axes in the first quadrant has its centre at (ρ,ρ), and external tangency between two circles becomes one distance equation. That converts the whole configuration into a single quadratic in r.

1STEP 1

Scale the picture so s = 1

Scaling is free, so set the small radius to one.

scale by 1/s: s↦ 1, r↦ r/s; the ratio r/s is unchanged

2STEP 2

A circle touching both axes

Touching both axes puts a centre on the diagonal.

tangent to both axes, radius ρ → centre=(ρ,ρ); O₁=(1,1), O=(r,r)

3STEP 3

Locate the second and third circles

External tangency and one axis locate the two outer centres.

O₂=(3,1), O₃=(1,3)

4STEP 4

Write the tangency equation

Centre distance equals radius sum, giving one equation.

(r-3)²+(r-1)²=(r+1)²

5STEP 5

Solve the quadratic

It factors, and the degenerate root is excluded.

2r²-8r+10=r²+2r+1 → r²-10r+9=0 → (r-1)(r-9)=0

6STEP 6

Verify and read off the ratio

Checking every condition confirms the ratio 9, choice (D).

√((9-3)²+(9-1)²)=√(36+64)=10=9+1 ✓, r/s=9

Answer

With s=1, the big circle has to clear the second circle sitting out at (3,1), so its centre must be well past x=3; a radius near the small choices 5 or 6 would be too tight, and r=9 places the centre at (9,9) with the 6-8-10 triangle closing exactly. Note the big circle is tangent only to the second and third circles, not to the first: the distance from (9,9) to (1,1) is 8√(2)≈ 11.31, not 10, which agrees with the problem, since it never claims tangency to the first circle. The figure in the problem is drawn with a radius-9 circle and unit circles at (1,1), (3,1), (1,3), matching the result.

💡Key takeaway

A circle touching both axes has its centre at (radius, radius), and two circles touching on the outside are the sum of their radii apart, so one distance equation settles the whole picture.

Scale the picture so s = 1
A circle touching both axes
Locate the second and third circles
Write the tangency equation
Solve the quadratic
Verify and read off the ratio