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

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~Wilhelm Z
~Wilhelm Z
== Solution 2 ==
Denote <math>A = \left( \cos 40^\circ , \sin 40^\circ \right)</math>,
<math>B = \left( \cos 60^\circ , \sin 60^\circ \right)</math>,
and <math>C = \left( \cos t^\circ , \sin t^\circ \right)</math>.
Case 1: <math>CA = CB</math>.
We have <math>t = 50</math> or <math>230</math>.
Case 2: <math>BA = BC</math>.
We have <math>t = 80</math>.
Case 3: <math>AB = AC</math>.
We have <math>t = 20</math>.
Therefore, the answer is <math>\boxed{\textbf{(E) }380}</math>.
~Steven Chen (www.professorchenedu.com)


{{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 23:10, 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

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

These problems are copyrighted © by the Mathematical Association of America.

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