AMC 10 · 2006 · #13

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
tangent-circlessystems-of-equationsarea-circles convert-to-algebra ↑ Prerequisites: tangent-circlessystems-of-equations
📏 Medium solution 💡 2 insights 📊 Diagram
Problem
Three circles are centred at the vertices of a right triangle with sides 3, 4 and 5. Every two of them touch from the outside. Find the total area of the three circles.

Pick an answer.

(A)
$12\pi$
(B)
$\frac{25\pi}{2}$
(C)
$13\pi$
(D)
$\frac{27\pi}{2}$
(E)
$14\pi$
How to solve
Strategy Introduce a Variable

The only quantities the question asks about are the radii, and the picture never states them, so Tool #4 (Introduce a Variable) names one radius per vertex. The real work is converting the phrase "mutually externally tangent" into equations, and that conversion has to be an if-and-only-if: it is not enough that tangency forces PQ=r_P+r_Q, because the answer is only a real answer if some genuine set of three circles exists. Tool #15 (Organize Information in More Ways) then handles the resulting system — instead of substituting one equation into another, add all three at once, which exploits the symmetry and produces the total of the radii in a single line. Tool #14 (Extreme Principle) closes the existence gap by looking at the smallest radius, the one that would go negative first, and showing that the triangle inequality is exactly what keeps it positive. Tool #7 (Identify Subproblems) finishes: factor π out and the whole problem collapses to one sum of three squares.

1STEP 1

Name the radii, translate tangency

Touching from outside means each centre distance is a sum of radii.

r_A+r_B=AB, r_B+r_C=BC, r_C+r_A=CA
2STEP 2

Add all three, then halve

Adding all three and halving gives the semiperimeter.

2(r_A+r_B+r_C)=3+4+5=12 → r_A+r_B+r_C=6
3STEP 3

Read off each radius

One subtraction each reads off 1, 2 and 3.

r_A=6-5=1, r_B=6-4=2, r_C=6-3=3
4STEP 4

Check the three circles exist

The triangle inequality makes every radius positive, so the circles exist.

2(s-a)=b+c-a ＞ 0⇔ b+c ＞ a; 3+4-5=2 ＞ 0
5STEP 5

Add the three areas

Adding the areas gives 14π, choice (E).

π r_A²+π r_B²+π r_C²=π(1²+2²+3²)=π(1+4+9)=14π → (E)
Answer
14π
First put the radii straight back into the three tangency conditions: 1+2=3, 1+3=4, 2+3=5 — all three sides matched, all radii positive, so the picture is real. Next, a structural check that kills a choice without computing anything. The radii sum to 6, and for any three positive numbers with a fixed sum, the sum of their squares is smallest when they are all equal: r_A²+r_B²+r_C² ≥ 6²/3=12, with equality only if r_A=r_B=r_C, which by r=s-(opposite side) happens only for an equilateral triangle. A 3-4-5 triangle is scalene, so the total area is strictly more than 12π — choice (A) is impossible for shape reasons alone. Choice (B) is the designed trap: attach a radius to each side by halving it, getting 3/2,2,5/2, and the areas total π·(9+16+25)/4=25π/2. It fails the test in Step 1 — 3/2+2=7/2 is not a side of the triangle — because radii belong to vertices, not sides. Finally a size check on the picture: the three circular sectors cut off inside the triangle do not overlap (the circles only touch), so their total area must be under the triangle's area of 6. The angles are 90°, about 53°, and about 37°, sitting at the radii 1, 2, 3, giving sector areas about 0.79, 1.85, and 2.90, total about 5.54. Just under 6, and snugly so — exactly what mutually touching circles packed into the corners should look like.
💡Key takeaway

Two circles touch on the outside exactly when the gap between their centres is one radius plus the other, so each side of the triangle becomes an equation — and adding all three equations at once counts every radius twice, handing you the total in a single line.

  • Name the radii, translate tangency
  • Add all three, then halve
  • Read off each radius
  • Check the three circles exist
  • Add the three areas

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