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 →
Eevee9406 (talk | contribs)
editor
662 edits
insert question
Cantalon (talk | contribs)
editor
40 edits
Line 4: Line 4:
<math>\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</math>
<math>\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</math>


==Solution==
==Solution 1==


Denote by <math>x</math> the length of the shorter base.
Denote by <math>x</math> the length of the shorter base.
Line 27: Line 27:


~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
==Solution 2==
~cantalon


==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}}

Revision as of 20:09, 15 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

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 2

~cantalon

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_11&oldid=203622"