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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 10"

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<math>\textbf{(A)} \: 100 \qquad\textbf{(B)} \: 150 \qquad\textbf{(C)} \: 330 \qquad\textbf{(D)} \: 360 \qquad\textbf{(E)} \: 380</math>
 
<math>\textbf{(A)} \: 100 \qquad\textbf{(B)} \: 150 \qquad\textbf{(C)} \: 330 \qquad\textbf{(D)} \: 360 \qquad\textbf{(E)} \: 380</math>
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==Solution 1 (Quick Look for Symmetry)==
 
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By inspection, we may obtain the following choices for which symmetric isosceles triangles could be constructed within the unit circle described:
 
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<math>20^\circ</math>, <math>50^\circ</math>, <math>80^\circ</math>, and <math>230^\circ</math>.
 
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Thus we have <math>20+50+80+230=\boxed{(\textbf{E})\ 380}</math>.
 
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Note: You may check this with a diagram featuring a unit circle and the above angles for polar coordinates.
 
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~Wilhelm Z
 
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 01:17, 28 January 2022

Problem

What is the sum of all possible values of $t$ between $0$ and $360$ such that the triangle in the coordinate plane whose vertices are \[(\cos 40^\circ,\sin 40^\circ), (\cos 60^\circ,\sin 60^\circ), \text{ and } (\cos t^\circ,\sin t^\circ)\] is isosceles?

$\textbf{(A)} \: 100 \qquad\textbf{(B)} \: 150 \qquad\textbf{(C)} \: 330 \qquad\textbf{(D)} \: 360 \qquad\textbf{(E)} \: 380$

Solution 2

Denote $A = \left( \cos 40^\circ , \sin 40^\circ \right)$, $B = \left( \cos 60^\circ , \sin 60^\circ \right)$, and $C = \left( \cos t^\circ , \sin t^\circ \right)$.

Case 1: $CA = CB$.

We have $t = 50$ or $230$.

Case 2: $BA = BC$.

We have $t = 80$.

Case 3: $AB = AC$.

We have $t = 20$.

Therefore, the answer is $\boxed{\textbf{(E) }380}$.

~Steven Chen (www.professorchenedu.com)

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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All AMC 12 Problems and Solutions

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