Difference between revisions of "2023 AMC 12B Problems/Problem 11"
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+ | ==Solution== | ||
+ | Denote by <math>x</math> the length of the shorten base. | ||
+ | Thus, the height of the trapezoid is | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \sqrt{1^2 - \left( \frac{x}{2} \right)^2} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Thus, the area of the trapezoid is | ||
+ | <math></math> | ||
+ | \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) <math>\frac{3}{2}</math>}} , | ||
+ | \end{align*} | ||
+ | <math></math> | ||
+ | |||
+ | where the inequality follows from the AM-GM inequality and it is binding if and only if <math>x^2 = 4 - x^2</math>. | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Revision as of 18:29, 15 November 2023
Solution
Denote by the length of the shorten base.
Thus, the height of the trapezoid is
Thus, the area of the trapezoid is
$$ (Error compiling LaTeX. Unknown error_msg)
\begin{align*}
\frac{1}{2} \left( x + 2 x \right) \sqrt{1^2 - \left( \frac{x}{2} \right)^2}
& = \frac{3}{4} \sqrt{x^2 \left( 4 - x^2 \right)} \\
& \leq \frac{3}{4} \frac{x^2 + \left( 4 - x^2 \right)}{2} \\
& = \boxed{\textbf{(D) }} ,
\end{align*}
$$ (Error compiling LaTeX. Unknown error_msg)
where the inequality follows from the AM-GM inequality and it is binding if and only if .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)