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 $(\cos 40^\circ,\sin 40^\circ), (\cos 60^\circ,\sin 60^\circ), \text{ and } (\cos t^\circ,\sin t^\circ)$ is isosceles?

$\textbf{(A)} \: 100 \qquad\textbf{(B)} \: 150 \qquad\textbf{(C)} \: 330 \qquad\textbf{(D)} \: 360 \qquad\textbf{(E)} \: 380$

Solution

Let $A = (\cos 40^{\circ}, \sin 40^{\circ}), B = (\cos 60^{\circ}, \sin 60^{\circ}),$ and $C = (\cos t^{\circ}, \sin t^{\circ}).$ We apply casework to the legs of isosceles $\triangle ABC:$

$AB=AC$
Note that $A$ must be the midpoint of arc $\widehat{BC}.$ We conclude that $C = (\cos 20^{\circ}, \sin 20^{\circ}),$ so $t=20.$

$BA=BC$
Note that $B$ must be the midpoint of arc $\widehat{AC}.$ We conclude that $C = (\cos 80^{\circ}, \sin 80^{\circ}),$ so $t=80.$

$CA=CB$
Note that $C$ must be the midpoint of arc $\widehat{AB}.$ We conclude that $C = (\cos 50^{\circ}, \sin 50^{\circ})$ or $C = (\cos 230^{\circ}, \sin 230^{\circ}),$ so $t=50$ or $t=230.$

Together, the sum of all such possible values of $t$ is $20+80+50+230=\boxed{\textbf{(E)} \: 380}.$

~Steven Chen (www.professorchenedu.com) ~Wilhelm Z ~MRENTHUSIASM

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, as part of the American Mathematics Competitions.

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2021 Fall AMC 12B Problems/Problem 10

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