2023 AMC 12B Problems/Problem 11

Problem

What is the maximum area of an isosceles trapezoid that has legs of length $1$ and one base twice as long as the other?

$\textbf{(A) }\frac 54 \qquad \textbf{(B) } \frac 87 \qquad \textbf{(C)} \frac{5\sqrt2}4 \qquad \textbf{(D) } \frac 32 \qquad \textbf{(E) } \frac{3\sqrt3}4$

Solution 1

Let the trapezoid be $ABCD$ with $AD = BC = 1, \; AB = x, CD = 2x$ . Extend $AD$ and $BC$ to meet at point $E$ . Then, notice $\triangle ABE \sim \triangle DCE$ with side length ratio $1:2$ and $AE = BE = 1$ . Thus, $[DCE] = 4 \cdot [ABE]$ and $[ABCD] = [DCE] - [ABE] = \frac{3}{4} \cdot [DCE]$ .

The problem reduces to maximizing the area of $[DCE]$ , an isosceles triangle with legs of length $2$ . Analyzing the sine area formula, this is clearly maximized when $\angle DEC = 90^{\circ}$ , so $[DCE] = 2$ and $[ABCD] = \frac{3}{4} \cdot 2 = \boxed{\frac{3}{2}}.$

-PIDay

Solution 2

Denote by $x$ the length of the shorter base. Thus, the height of the trapezoid is $\begin{align*} \sqrt{1^2 - \left( \frac{x}{2} \right)^2} . \end{align*}$

Thus, the area of the trapezoid is $\begin{align*} \frac{1}{2} \left( x + 2 x \right) \sqrt{1^2 - \left( \frac{x}{2} \right)^2} & = \frac{3}{4} \sqrt{x^2 \left( 4 - x^2 \right)} \\ & \leq \frac{3}{4} \frac{x^2 + \left( 4 - x^2 \right)}{2} \\ & = \boxed{\textbf{(D) } \frac{3}{2}} , \end{align*}$

where the inequality follows from the AM-GM inequality and it is binding if and only if $x^2 = 4 - x^2$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3 (Calculus)

Derive the expression for area $A = \frac{3}{4}x\sqrt{4-x^2}$ as in the solution above. To find the minimum, we can take the derivative with respect to $x$ : $\frac{dA}{dx} = \frac{3}{4}\sqrt{4-x^2}+\left(\frac{3x}{4}\right)\frac{-2x}{2\sqrt{4-x^2}} = \frac{6-3x^2}{2\sqrt{4-x^2}}.$ This expression is equal to zero when $x=\pm\sqrt{2}$ , so $A$ has two critical points at $\pm\sqrt{2}$ . But given the bounds of the problem, we can conclude $x = \sqrt{2}$ maximizes $A$ (alternatively you can do first derivative test). Plugging that value back in, we get $A_{\text{max}} = \boxed{(\text{D})\ \frac{3}{2}}$ .

~cantalon

(Slightly Simpler)

Or rewrite the expression for area to be

$A = \frac{3}{4}\sqrt{4x^2-x^4}$

Now to find the minimum, we can just find the minimum of what's inside the square root (since the square root function is increasing). Take the derivative of $4x^2-x^4$ , $f'(x)=8x-4x^3.$ This is equal to zero at $x=0,\pm\sqrt{2}$ but the solution must be positive so $x=\sqrt{2}$ .

Solution 4 (Trigonometry)

Let the length of the shorter base of the trapezoid be $2x$ and the height of the trapezoid be $y$ .

$[asy] unitsize(100); pair A=(-1, 0), B=(1, 0), C=(0.5, 0.5), D=(-0.5, 0.5); draw(A--B--C--D--cycle, black); label("$2x$",(0,0.58),(0,0)); label("$2x$",(0,-0.08),(0,0)); label("$x$",(-0.75,-0.08),(0,0)); label("$x$",(0.75,-0.08),(0,0)); draw(D--(-0.5,0),black); draw(C--(0.5,0),black); label("$y$",(0.58,0.25)); label("$y$",(-0.42,0.25)); [/asy]$

Each leg has length $1$ if and only if $x^2+y^2=1$ , where $x$ and $y$ are positive real numbers. The general solution to this equation is $(x,y)=(\cos t,\sin t)$ for any number $0<t<\frac{\pi}{2}$ so that $x$ and $y$ are positive. The area to maximize is $\frac{1}{2}(2x+4x)y=3xy$ Hence, we maximize $3\sin t\cos t$ for $0<t<\frac{\pi}{2}$ . $\begin{align*} 3xy &= 3\sin t\cos t \\ &= \frac{3}{2}(2\sin t\cos t) \\ &= \frac{3}{2}\sin(2t) \end{align*}$ The maximum of $\sin(2t)$ is $1$ , thus the maximum of $3xy$ is $\boxed{\text{(D) }\frac{3}{2}}$ which occurs at $t=\frac{\pi}{4}$ , satisfying the inequality $0<t<\frac{\pi}{2}$ .

~Robabob1

Solution 5

Denote $x$ and $2x$ as the two bases. We can straightforwardly find that the height of the trapezoid is $\sqrt{1-\frac{x^2}{4}}$ by the Pythagorean theorem.

The area of this trapezoid is then given by the expression $\frac{x+2x}{2}\cdot\sqrt{1-\frac{x^2}{4}}.$

Let $m$ be the maximum value that this expression achieves. Since both $x$ and $m$ are positive, we can perform the following operations to maximize $16m^2$ , thus $m^2$ , and thus $m$ . We have

$\frac{x+2x}{2}\cdot\sqrt{1-\frac{x^2}{4}} = m$ $\frac{9x^2}{4}\cdot\frac{4-x^2}{4} = m^2$ $-9x^4 + 36x^2 = 16m^2.$

The maximum vaue of this quartic occurs at $x^2 = \frac{-36}{2(-9)} = 2$ , or when $x = \sqrt{2}.$

It follows that the area of the trapezoid is equal to

$\frac{3\sqrt{2}}{2}\cdot\frac{1}{\sqrt{2}} = \boxed{\text{(D) }\frac{3}{2}}.$

-Benedict T (countmath1)

Video Solution 1 by OmegaLearn

https://youtu.be/WlXBbaHl-z4

Video Solution

https://youtu.be/e6Et9KBkRy8

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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