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

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<math>\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\textbf{(D)} \: 150 \qquad\textbf{(E)} \: 165</math>
 
<math>\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\textbf{(D)} \: 150 \qquad\textbf{(E)} \: 165</math>
  
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==Solution 1==
+
==Solution 1 (Area)==
  
 
Let the lengths of the two congruent sides of the triangle be <math>x</math>, then the product desired is <math>x^2</math>.
 
Let the lengths of the two congruent sides of the triangle be <math>x</math>, then the product desired is <math>x^2</math>.
Line 12: Line 12:
 
Set the vertex angle to be <math>a</math>, we derive the equation:
 
Set the vertex angle to be <math>a</math>, we derive the equation:
  
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x^2<math>=</math>4(\frac{1}{2}x^2\sin(a))
+
<math>x^2=4\left(\frac{1}{2}x^2\sin(a)\right)</math>
  
 
<math>\sin(a)=\frac{1}{2}</math>
 
<math>\sin(a)=\frac{1}{2}</math>
  
−
As the triangle is obtuse, <math>a=\frac{5\pi}{6}</math>. We get <math>\boxed{\textbf{(D)} \ 150}.</math>
+
As the triangle is obtuse, <math>a=150^\circ</math> only. We get <math>\boxed{\textbf{(D)} \ 150}.</math>
  
 
~Wilhelm Z
 
~Wilhelm Z
 +
 +
== Solution 2 ==
 +
Denote by <math>a</math> the length of each congruent side. Denote by <math>\theta</math> the degree measure of each acute angle.
 +
Denote by <math>\phi</math> the degree measure of the obtuse angle.
 +
 +
Hence, this problem tells us the following relationship:
 +
<cmath>
 +
\[
 +
a^2 = 2 a \cos \theta \cdot 2 a \sin \theta .
 +
\]
 +
</cmath>
 +
 +
Hence,
 +
<cmath>
 +
\begin{align*}
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1 & = 2 \cdot 2 \sin \theta \cos \theta \\
 +
& = 2 \sin 2 \theta \\
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& = 2 \sin \left( 180^\circ - 2 \theta \right) \\
 +
& = 2 \sin \phi .
 +
\end{align*}
 +
</cmath>
 +
 +
Hence, <math>\phi = 150^\circ</math>.
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(D) }150}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
 +
 +
 +
==Video Solution (Under 2 min!)==
 +
https://youtu.be/nXnS6pn8iJc
 +
 +
<i>~Education, the Study of Everything</i>
 +
 +
==Video Solution by TheBeautyofMath==
 +
https://www.youtube.com/watch?v=4qgYrCYG-qw&t=795
 +
 +
~IceMatrix
  
 
==See Also==
 
==See Also==
 +
{{AMC12 box|year=2021 Fall|ab=B|num-a=9|num-b=7}}
 +
 +
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 21:07, 10 July 2023

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:

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

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

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 (Under 2 min!)

https://youtu.be/nXnS6pn8iJc

~Education, the Study of Everything

Video Solution by TheBeautyofMath

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

~IceMatrix

See Also

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

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