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

(→‎Solution 1 (Quick Look for Symmetry): About to combine solutions.)
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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 2 ==
+
== Solution ==
 
Denote <math>A = \left( \cos 40^\circ , \sin 40^\circ \right)</math>,
 
Denote <math>A = \left( \cos 40^\circ , \sin 40^\circ \right)</math>,
 
<math>B = \left( \cos 60^\circ , \sin 60^\circ \right)</math>,
 
<math>B = \left( \cos 60^\circ , \sin 60^\circ \right)</math>,
Line 25: Line 25:
 
Therefore, the answer is <math>\boxed{\textbf{(E) }380}</math>.
 
Therefore, the answer is <math>\boxed{\textbf{(E) }380}</math>.
  
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~Steven Chen (www.professorchenedu.com)
+
~Steven Chen (www.professorchenedu.com) ~Wilhelm Z ~MRENTHUSIASM
  
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=11|num-b=9}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=11|num-b=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:18, 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

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) ~Wilhelm Z ~MRENTHUSIASM

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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