Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

← Older editNewer edit →
Professorchenedu (talk | contribs)
editor
175 edits
No edit summary
Professorchenedu (talk | contribs)
editor
175 edits
No edit summary
Line 10: Line 10:


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


where the inequality follows from the AM-GM inequality and it is binding if and only if <math>x^2 = 4 - x^2</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)
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 17:30, 15 November 2023

Solution

Denote by $x$ the length of the shorten 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)

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_11&oldid=203363"