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2023 AMC 12B Problems/Problem 11: Difference between revisions

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==Solution 1==
==Solution 1==
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>
-PIDay
==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 (Calculus)==
==Solution 3 (Calculus)==


Derive the expression for area
Derive the expression for area
Line 44: Line 51:
<cmath>A = \frac{3}{4}\sqrt{4x^2-x^4}</cmath>
<cmath>A = \frac{3}{4}\sqrt{4x^2-x^4}</cmath>


Now to find the minimum, we can take the derivative of what is inside the square root.
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>
<cmath>f'(x)=8x-4x^3.</cmath>
This is equal to zero at <math>x=\pm\sqrt{2}</math> but the solution must be positive so <math>x=\sqrt{2}</math>. Putting that into the formula for area, we got <math>A = \boxed{(\text{D})\ \frac{3}{2}}</math>.
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 3 (Trigonometry)==
==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>.
Let the length of the shorter base of the trapezoid be <math>2x</math> and the height of the trapezoid be <math>y</math>.
Line 64: Line 71:
\end{align*}
\end{align*}
</cmath>
</cmath>
The maximum of <math>\sin(2t)</math> is <math>1</math>, thus the maximum of <math>3xy</math> is <math>\boxed{\frac{3}{2}}</math> which occurs at <math>t=\frac{\pi}{4}</math>, satisfying the inequality <math>0<t<\frac{\pi}{2}</math>.
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
~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==
==Video Solution 1 by OmegaLearn==
https://youtu.be/WlXBbaHl-z4
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
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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