AMC 10 · 2023 · #11

Grade 9 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

area-trapezoidpythagorean-theoremoptimizationcompleting-the-square convert-to-algebracompleting-the-squareextremal-construction ↑ Prerequisites: area-trapezoidpythagorean-theorem

📏 Long solution 💡 3 insights

Problem

Consider every isosceles trapezoid whose two legs each have length one and whose longer base is exactly twice the shorter. Those two rules do not pin down a single shape. Among the whole family, find the largest possible area.

Pick an answer.

(A)

$\frac 54$

(B)

$\frac 87$

(C)

$\frac{5\sqrt2}4$

(D)

$\frac 32$

(E)

$\frac{3\sqrt3}4$

How to solve

Strategy Extreme Principle

The question asks for a maximum, so the job is two-sided: find a ceiling the area can never break, then exhibit one trapezoid that actually sits on that ceiling. First the shape family needs a single handle. Because the legs are fixed and the long base is tied to the short one, choosing the shorter base determines everything, so one variable is enough. Dropping the two altitudes turns the trapezoid into a rectangle plus two congruent right triangles and hands the height to the Pythagorean theorem. That gives an area formula with a square root in it, which is awkward to bound directly. So I reorganize: maximize the square of the area instead of the area. Squaring clears the radical and leaves a plain quadratic shape, whose ceiling is visible either by completing the square or by AM-GM. Since area is positive, whatever maximizes the square maximizes the area itself, so nothing is lost in the trade. No calculus is needed at any point.

1STEP 1

One number controls the shape

The short base fixes everything.

shorter base = b, longer base = 2b, each leg = 1

2STEP 2

Drop the altitudes

Pythagoras gives the height.

h² + (b/2)² = 1² ⟹ h = √(1 - b²/4)

3STEP 3

Write the area as a function

The area becomes one function.

A(b) = 1/2(b + 2b) h = 3b/2√(1 - b²/4), 0 ＜ b ＜ 2

4STEP 4

Square to clear the root

Squaring clears the radical.

A² = 9b²/4(1 - b²/4) u = b²/4= 9(4u)/4(1-u) = 9u(1-u), 0 ＜ u ＜ 1

5STEP 5

Two pieces with a fixed sum

The product peaks when the pieces are equal.

u(1-u) = 1/4 - (u - 1/2)² ≤ 1/4 ⟹ A² = 9u(1-u) ≤ 9/4 ⟹ A ≤ 3/2

6STEP 6

Show the ceiling is reached

It is actually reached, so the maximum is three halves.

b = √(2), h = √(2)/2: A = 1/2(√(2) + 2√(2))·√(2)/2 = 3√(2)/2·√(2)/2 = 3/2

Answer

3/2

Test the peak by flexing the shape around b = √(2) ≈ 1.414. At b = 1.2 the height is √(1 - 0.36) = 0.8 and the area is 3(1.2)/2(0.8) = 1.44. At b = 1.6 the height is √(1 - 0.64) = 0.6 and the area is 3(1.6)/2(0.6) = 1.44 as well. Both sit below 3/2, and they match each other, which is the symmetry the substitution u = b²/4 predicts, since b = 1.2 and b = 1.6 give u values of 0.36 and 0.64, equally spaced on either side of 1/2. Pushing closer, b = 1.4 gives area ≈ 1.49970 and b = 1.45 gives ≈ 1.49803, both just under 3/2 = 1.5, with the value peaking between them exactly where b = √(2) lies. Sweeping b finely across the whole admissible interval 0 ＜ b ＜ 2 never produces anything above 1.5, and the largest value found is 1.500000 at b ≈ 1.41421, which is √(2). The endpoints behave as expected too: b → 0 and b → 2 both drive the area to 0. Finally, screen the answer list against the proven ceiling. Choice (C) is 5√(2)/4 ≈ 1.768, which is above 3/2 and therefore impossible, since A² ≤ 9/4 was proved for every legal b. Choices (A) 1.25, (B) ≈ 1.143, and (E) ≈ 1.299 are all beaten by the ordinary shape b = 1.2, which already reaches 1.44, so none of them can be the maximum. Only (D) survives. As a shape check, the maximizing trapezoid has base angles of 45°, since its right triangles have equal legs √(2)/2; a tidy angle at the optimum is the usual signature of a genuine balance point.

💡Key takeaway

When a square root blocks your view of the largest value, maximize the square of the quantity instead, and when what is left is two positive pieces adding to a constant, the product is biggest when the pieces are equal.

One number controls the shape
Drop the altitudes, get the height
Write the area as one function
Square it to clear the root
Two pieces with a fixed sum
Show the ceiling is reached