2021 Fall AMC 12B Problems/Problem 10

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

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
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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2021 Fall AMC 12B Problems/Problem 10

Problem

Solution 1 (Quick Look for Symmetry)

Solution 2