AMC 10 · 2003 · #17

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
equilateral-trianglearea-circlesthirty-sixty-ninety-triangle convert-to-algebra ↑ Prerequisites: equilateral-triangle
📏 Long solution 💡 3 insights
Problem
An equilateral triangle sits inside a circle so that all three corners touch the circle. The number giving the triangle's perimeter (in inches) is the same as the number giving the circle's area (in square inches). Find the circle's radius.

Pick an answer.

(A)
$\frac{3\sqrt{2}}{\pi}$
(B)
$\frac{3\sqrt{3}}{\pi}$
(C)
$\sqrt{3}$
(D)
$\frac{6}{\pi}$
(E)
$\sqrt{3}\pi$

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The radius R is what we want, and everything else in the problem can be written in terms of it, so Tool #4 (Introduce a Variable) makes R the single unknown to chase. Tool #1 (Draw a Diagram) is the key that unlocks the geometry: sketching the circle with its inscribed triangle and dropping a line from the center to one side turns the picture into a 30-60-90 right triangle, which pins the side length to R. Once both the perimeter and the area are expressions in R, Tool #13 (Convert to Algebra) lets us set them equal and solve one clean equation. The plan is to express two things (3 × side and π R²) in the same variable, then let algebra finish.

1STEP 1

Draw the picture and name the radius

Sketch the circle with the triangle's three corners on it. Call the radius RR, the side ss; the radii cut the center into 120∘120^\circ angles.

R = radius, s = side, central angle for one side = 120°
2STEP 2

Find the side from the radius with a 30-60-90 triangle

A perpendicular from the center halves the 120∘120^\circ angle into a 30-60-90 triangle with hypotenuse RR; the 1:3:21:\sqrt{3}:2 ratio gives s=3Rs=\sqrt{3}R.

s/2 = √(3)/2R → s = √(3) R
3STEP 3

Write the perimeter and the area in terms of R

The perimeter is 3s=33R3s=3\sqrt{3}R and the area is πR2\pi R^{2}, so 'the two numbers are equal' becomes 33R=πR23\sqrt{3}R=\pi R^{2}.

P = 3s = 3√(3) R, A = π R², 3√(3) R = π R²
4STEP 4

Solve the equation for the radius

Since R≠0R \neq 0, cancel one RR to get 33=πR3\sqrt{3}=\pi R, then divide by π\pi: R=33πR=\frac{3\sqrt{3}}{\pi}, choice (B).

3√(3) R = π R² → 3√(3) = π R → R = 3√(3)/π → (B)
Answer
3√(3)/π
Plug R = 3√(3)/π back in. The area is π R² = π·27/π² = 27/π. The perimeter is 3√(3) R = 3√(3)·3√(3)/π = (9 · 3)/π = 27/π. The two match, so the radius is correct. Numerically R ≈ 5.196/3.1416≈ 1.65 inches, a sensible size for a small circle whose perimeter number and area number happen to coincide.
💡Key takeaway

Write both the perimeter and the area using the same radius R, set them equal, and cancel the shared R to turn a scary squared equation into an easy one.

  • Draw the picture and name the radius
  • Find the side from the radius with a 30-60-90 triangle
  • Write the perimeter and the area in terms of R
  • Solve the equation for the radius

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