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

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Thus we have <math>20+50+80+230=\boxed{(\textbf{E})\ 380}</math>.
 
Thus we have <math>20+50+80+230=\boxed{(\textbf{E})\ 380}</math>.
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~Wilhelm Z
 
~Wilhelm Z

Revision as of 03:40, 24 November 2021

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 $(\cos(40^\circ),\sin(40^\circ))$, $(\cos(60^\circ),\sin(60^\circ))$, 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 1 (Quick Look for Symmetry)

By inspection, we may obtain the following choices for which symmetric isosceles triangles could be constructed within the unit circle described:

$20^\circ$, $50^\circ$, $80^\circ$, and $230^\circ$.

Thus we have $20+50+80+230=\boxed{(\textbf{E})\ 380}$.

~Wilhelm Z

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