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Difference between revisions of "2023 AMC 12B Problems/Problem 11"

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==Solution 1==
 
==Solution 1==
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Let the trapezoid be <math>ABCD</math> with <math>AD = BC = 1, \; AB = x, CD = 2x</math>. Extend <math>AD</math> and <math>BC</math> to meet at point <math>E</math>. Then, notice <math>\triangle ABE \sim \triangle DCE</math> with side length ratio <math>1:2</math> and <math>AE = BE = 1</math>. Thus, <math>[DCE] = 4 \cdot [ABE]</math> and <math>[ABCD] = [DCE] - [ABE] = \frac{3}{4} \cdot [DCE]</math>.
 +
 +
The problem reduces to maximizing the area of <math>[DCE]</math>, an isosceles triangle with legs of length <math>2</math>. Analyzing the sine area formula, this is clearly maximized when <math>\angle DEC = 90^{\circ}</math>, so <math>[DCE] = 2</math> and <math>[ABCD] = \frac{3}{4} \cdot 2 = \boxed{\frac{3}{2}}.</math>
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 +
-PIDay
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 +
==Solution 2==
  
 
Denote by <math>x</math> the length of the shorter base.
 
Denote by <math>x</math> the length of the shorter base.
Line 28: Line 35:
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
−
==Solution 2==
+
==Solution 3 (Calculus)==
 +
 
 +
Derive the expression for area
 +
<cmath>A = \frac{3}{4}x\sqrt{4-x^2}</cmath>
 +
as in the solution above. To find the minimum, we can take the derivative with respect to <math>x</math>:
 +
<cmath>\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}}.</cmath>
 +
This expression is equal to zero when <math>x=\pm\sqrt{2}</math>, so <math>A</math> has two critical points at <math>\pm\sqrt{2}</math>. But given the bounds of the problem, we can conclude <math>x = \sqrt{2}</math> maximizes <math>A</math> (alternatively you can do first derivative test). Plugging that value back in, we get <math>A_{\text{max}} = \boxed{(\text{D})\ \frac{3}{2}}</math>.
  
 
~cantalon
 
~cantalon
 +
 +
(Slightly Simpler)
 +
 +
Or rewrite the expression for area to be
 +
 +
<cmath>A = \frac{3}{4}\sqrt{4x^2-x^4}</cmath>
 +
 +
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 <math>4x^2-x^4</math>,
 +
<cmath>f'(x)=8x-4x^3.</cmath>
 +
This is equal to zero at <math>x=0,\pm\sqrt{2}</math> but the solution must be positive so <math>x=\sqrt{2}</math>.
 +
 +
==Solution 4 (Trigonometry)==
 +
 +
Let the length of the shorter base of the trapezoid be <math>2x</math> and the height of the trapezoid be <math>y</math>.
 +
 +
<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 <math>1</math> if and only if <math>x^2+y^2=1</math>, where <math>x</math> and <math>y</math> are positive real numbers. The general solution to this equation is <cmath>(x,y)=(\cos t,\sin t)</cmath> for any number <math>0<t<\frac{\pi}{2}</math> so that <math>x</math> and <math>y</math> are positive. The area to maximize is <cmath>\frac{1}{2}(2x+4x)y=3xy</cmath> Hence, we maximize <math>3\sin t\cos t</math> for <math>0<t<\frac{\pi}{2}</math>.
 +
<cmath>
 +
\begin{align*}
 +
    3xy &= 3\sin t\cos t \\
 +
    &= \frac{3}{2}(2\sin t\cos t) \\
 +
    &= \frac{3}{2}\sin(2t)
 +
\end{align*}
 +
</cmath>
 +
The maximum of <math>\sin(2t)</math> is <math>1</math>, thus the maximum of <math>3xy</math> is <math>\boxed{\text{(D) }\frac{3}{2}}</math> which occurs at <math>t=\frac{\pi}{4}</math>, satisfying the inequality <math>0<t<\frac{\pi}{2}</math>.
 +
 +
~Robabob1
 +
 +
 +
==Solution 5==
 +
 +
Denote <math>x</math> and <math>2x</math> as the two bases. We can straightforwardly find that the height of the trapezoid is <math>\sqrt{1-\frac{x^2}{4}}</math> by the Pythagorean theorem.
 +
 +
The area of this trapezoid is then given by the expression <math>\frac{x+2x}{2}\cdot\sqrt{1-\frac{x^2}{4}}.</math>
 +
 +
Let <math>m</math> be the maximum value that this expression achieves. Since <math>x</math> is positive and assuming <math>m\geq 1</math>, we can perform the following operations to maximize <math>16m^2</math>, thus <math>m^2</math>, and thus <math>m</math>. We have
 +
 +
<cmath>\frac{x+2x}{2}\cdot\sqrt{1-\frac{x^2}{4}} = m</cmath>
 +
<cmath>\frac{9x^2}{4}\cdot\frac{4-x^2}{4} = m^2</cmath>
 +
<cmath>-9x^4 + 36x^2 = 16m^2.</cmath>
 +
 +
The maximum vaue of this quartic occurs at <math>x^2 = \frac{-36}{2(-9)} = 2</math>, or when <math>x = \sqrt{2}.</math>
 +
 +
It follows that the area of the trapezoid is equal to
 +
 +
<cmath>\frac{3\sqrt{2}}{2}\cdot\frac{1}{\sqrt{2}} = \boxed{\text{(D) }\frac{3}{2}}.</cmath>
 +
 +
-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)
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2023|ab=B|num-b=10|num-a=12}}
 
{{AMC12 box|year=2023|ab=B|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:55, 29 November 2023

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 $x$ is positive and assuming $m\geq 1$, 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)

See Also

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AMC 12 Problems and Solutions

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