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

Revision as of 18:45, 15 November 2023

Solution

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)

@@ Line 1: / Line 1: @@
 ==Solution==
-Denote by <math>x</math> the length of the shorten base.
+Denote by <math>x</math> the length of the shorter base.
 Thus, the height of the trapezoid is
 <cmath>

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

Revision as of 18:45, 15 November 2023

Solution