Difference between revisions of "2021 Fall AMC 12B Problems/Problem 8"

Revision as of 00:48, 30 December 2022

Problem

The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?

$\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\textbf{(D)} \: 150 \qquad\textbf{(E)} \: 165$

Solution 1 (Area)

Let the lengths of the two congruent sides of the triangle be $x$ , then the product desired is $x^2$ .

Notice that the product of the base and twice the height is $4$ times the area of the triangle.

Set the vertex angle to be $a$ , we derive the equation:

$x^2=4(\frac{1}{2}x^2\sin(a))$

$\sin(a)=\frac{1}{2}$

As the triangle is obtuse, $a=150^\circ$ only. We get $\boxed{\textbf{(D)} \ 150}.$

~Wilhelm Z

Solution 2

Denote by $a$ the length of each congruent side. Denote by $\theta$ the degree measure of each acute angle. Denote by $\phi$ the degree measure of the obtuse angle.

Hence, this problem tells us the following relationship: $a^2 = 2 a \cos \theta \cdot 2 a \sin \theta .$

Hence, $\begin{align*} 1 & = 2 \cdot 2 \sin \theta \cos \theta \\ & = 2 \sin 2 \theta \\ & = 2 \sin \left( 180^\circ - 2 \theta \right) \\ & = 2 \sin \phi . \end{align*}$

Hence, $\phi = 150^\circ$ .

Therefore, the answer is $\boxed{\textbf{(D) }150}$ .

~Steven Chen (www.professorchenedu.com)

Video Solution by TheBeautyofMath

https://www.youtube.com/watch?v=4qgYrCYG-qw&t=795

~IceMatrix

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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