AMC 10 · 2003 · #11

Grade 8 geometry-2d
pythagorean-theoremarea-circlesequilateral-trianglecircumradius convert-to-algebraidentify-subproblems ↑ Prerequisites: pythagorean-theorem
📏 Long solution 💡 3 insights
Problem
A square and an equilateral triangle are built so that their perimeters are equal. One circle is drawn through the four corners of the square and its area is called A; another circle is drawn through the three corners of the triangle and its area is called B. Find the value of A/B.

Pick an answer.

(A)
$\frac{9}{16}$
(B)
$\frac{3}{4}$
(C)
$\frac{27}{32}$
(D)
$\frac{3\sqrt{6}}{8}$
(E)
1
How to solve
Strategy Introduce a Variable

Nothing in the problem has a number attached, so tool #4 (Introduce a Variable) supplies the missing handle: name the shared perimeter P and every length in the picture becomes an expression in P. Tool #7 (Identify Subproblems) then splits the work into two independent radius hunts, one for the square and one for the triangle, with the ratio assembled only at the very end. Tool #1 (Draw a Diagram) is what makes each radius findable: drawing the center and dropping the right segments turns both radius questions into right triangles, and right triangles are something the Pythagorean theorem can finish. Because only a ratio is wanted, P and π are expected to cancel, which is also a built-in check that the setup was consistent.

1STEP 1

Name the perimeter and get the sides

With shared perimeter P, the sides are P/4 and P/3 — different, since the shapes have different side counts.

s_square=P/4, s_triangle=P/3
2STEP 2

Find the square's circle radius

A square's diagonal is a full diameter, giving radius squared P²/32.

d=P/4√(2)=P√(2)/4, r_A=d/2=P√(2)/8, r_A²=2P²/64=P²/32
3STEP 3

Find the triangle's altitude

For the triangle, dropping the altitude gives height s√(3)/2.

h²=s²-(s/2)²=3s²/4, h=s√(3)/2
4STEP 4

Find the triangle's circle radius

Placing the centre on that altitude gives radius squared P²/27.

r²=s²/4+(h-r)² → 2hr=s² → r_B=s/√(3)=P/3√(3), r_B²=P²/27
5STEP 5

Divide the two areas

Dividing cancels pi and the perimeter, leaving 27/32, choice (C).

A/B=(π P²/32)/(π P²/27)=1/32 · 27=27/32 → (C)
Answer
27/32
The triangle is the pointier shape, so for the same perimeter its corners reach further from its center and it needs the larger circle. That forces A/B < 1, which rules out (E) and says the answer should be a little under 1; 27/32=0.84375 fits. A numerical run confirms it: take P=12, so the square has side 3 and radius 3√(2)/2≈ 2.121 with r_A²=4.5, while the triangle has side 4 and radius 4/√(3)≈ 2.309 with r_B²≈ 5.333, giving 4.5/5.333=0.84375. Choice (D) is worth naming as the trap: 3√(6)/8=√(27/32)≈ 0.919 is the ratio of the radii, not of the areas, so it is exactly what a solver gets by forgetting to square.
💡Key takeaway

When two shapes share a perimeter, give that perimeter a letter, use right triangles inside each shape to find its circle's radius, and let the shared letters cancel when you divide.

  • Name the perimeter and get the sides
  • Find the square's circle radius
  • Find the triangle's altitude
  • Find the triangle's circle radius
  • Divide the two areas