AMC 10 · 2006 · #21

Grade 11 geometry-2d
area-ellipseellipse-eccentricitypythagorean-theoremsystems-of-equations convert-to-algebra ↑ Prerequisites: area-ellipsepythagorean-theorem
📏 Long solution 💡 4 insights
Problem
An ellipse has a rectangle's opposite corners as its foci and passes through the other two corners. The rectangle's area and the ellipse's area are both given. Find the rectangle's perimeter.

Pick an answer.

(A)
$\frac {16\sqrt {2006}}{\pi}$
(B)
$\frac {1003}4$
(C)
$8\sqrt {1003}$
(D)
$6\sqrt {2006}$
(E)
$\frac {32\sqrt {1003}}\pi$
How to solve
Strategy Introduce a Variable

Four lengths matter here — the rectangle's two sides and the ellipse's two semi-axes — so Tool #4 (Introduce a Variable) names them l, w, a, b and lets the two area facts become two equations. Tool #1 (Draw a Diagram) is what reveals the bridge: the corner A sits at the end of both sides AB and AD, and those are exactly the two distances from A to the foci, while BD is both the rectangle's diagonal and the focus-to-focus distance. Tool #15 (Organize Information in More Ways) is the decisive move — the perimeter needs only l+w, never l and w separately, so the whole problem should be rewritten in terms of the sum l+w and the product lw. Tool #13 (Convert to Algebra) then grinds those relations together. Finally Tool #6 (Guess and Check) is used honestly at the end: the equations alone only say what the perimeter must be if such a rectangle exists, so an actual pair of side lengths has to be produced.

1STEP 1

Name the four lengths

The two given areas become two products.

lw = 2006, abπ = 2006π → ab = 2006
2STEP 2

Corner A gives half the perimeter

The defining property makes the perimeter a multiple of one axis.

AB+AD = 2a → l+w = 2a → P = 2(l+w) = 4a
3STEP 3

The diagonal is the focal distance

The diagonal is exactly the focal distance.

2c = BD = √(l²+w²) → c² = (l²+w²)/4
4STEP 4

The crux: b² is half the rectangle's area

The ellipse relation makes the other axis squared half the rectangle's area.

b² = a²-c² = (l+w)²-(l²+w²)/4 = 2lw/4 = lw/2 = 1003 → b=√(1003)
5STEP 5

The ellipse's area now pins down a

The given ellipse area then pins the first axis.

a = 2006/√(1003) = 2√(1003), a² = 4012, b² = 1003, c² = 3009 > 0
6STEP 6

Check the rectangle exists, then read off the perimeter

A real rectangle exists, so the perimeter is 8√(1003), choice (C).

t²-4√(1003) t+2006=0 → t = 2√(1003)±√(2006) > 0; P = 4a = 8√(1003) → (C)
Answer
8√(1003)
Everything checks numerically. With √(1003)≈ 31.670 and √(2006)≈ 44.788, the sides are l≈ 108.129 and w≈ 18.552: their product is 4012-2006=2006 exactly, and their sum is 4√(1003)≈ 126.68. The diagonal is √(l²+w²)=√(16048-4012)=√(12036)≈ 109.71, so c²=3009 and a²-c²=4012-3009=1003=b² as required, and the ellipse's area is π ab=π · 2√(1003)·√(1003)=2006π exactly. Two independent smell tests also pass. First, π cancels the moment abπ=2006π is written, so a perimeter containing π is impossible — that rules out (A) and (E) before any work. Second, among all rectangles of area 2006 the square has the smallest perimeter, 4√(2006)≈ 179.2; the answer 8√(1003)≈ 253.4 comfortably exceeds it, consistent with the very elongated rectangle the algebra produced.
💡Key takeaway

The two sides meeting at a corner are that corner's distances to the two foci, so the semi-perimeter is the ellipse's 2a; and because (l+w)²-(l²+w²)=2lw, the short semi-axis knows only the rectangle's area, which pins the perimeter at 8√(1003).

  • Name the four lengths
  • Corner A gives half the perimeter
  • The diagonal is the focal distance
  • The crux: b² is half the rectangle's area
  • The ellipse's area now pins down a
  • Check the rectangle exists, then read off the perimeter