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

Revision as of 04:02, 25 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 $20+50+80+230=\boxed{(\textbf{E})\ 380}$ .

Note: You may check this with a diagram featuring a unit circle and the above angles for polar coordinates.

~Wilhelm Z

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

Revision as of 04:02, 25 November 2021

Problem

Solution 1 (Quick Look for Symmetry)

@@ Line 13: / Line 13: @@
 Thus we have <math>20+50+80+230=\boxed{(\textbf{E})\ 380}</math>.
+Note: You may check this with a diagram featuring a unit circle and the above angles for polar coordinates.
 ~Wilhelm Z