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

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)
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All AMC 12 Problems and 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>
-==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:
-<math>20^\circ</math>, <math>50^\circ</math>, <math>80^\circ</math>, and <math>230^\circ</math>.
-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
 == Solution 2 ==

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

Revision as of 01:17, 28 January 2022

Problem

Solution 2