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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 10"
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== Solution ==
 
== Solution ==
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Denote <math>A = \left( \cos 40^\circ , \sin 40^\circ \right)</math>,
 
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<math>B = \left( \cos 60^\circ , \sin 60^\circ \right)</math>,
 
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and <math>C = \left( \cos t^\circ , \sin t^\circ \right)</math>.
 
  
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Case 1: <math>CA = CB</math>.
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Let <math>A = (\cos 40^{\circ}, \sin 40^{\circ}), B = (\cos 60^{\circ}, \sin 60^{\circ}),</math> and <math>C = (\cos t^{\circ}, \sin t^{\circ}).</math> We apply casework to the legs of isosceles <math>\triangle ABC:</math>
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+
<ol style="margin-left: 1.5em;">
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We have <math>t = 50</math> or <math>230</math>.
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  <li><math>AB=AC</math><p>
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+
Note that <math>A</math> must be the midpoint of arc <math>\widehat{BC}.</math> We conclude that <math>C = (\cos 20^{\circ}, \sin 20^{\circ}),</math> so <math>t=20.</math></li><p>
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Case 2: <math>BA = BC</math>.
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  <li><math>BA=BC</math><p>
−
 
+
Note that <math>B</math> must be the midpoint of arc <math>\widehat{AC}.</math> We conclude that <math>C = (\cos 80^{\circ}, \sin 80^{\circ}),</math> so <math>t=80.</math></li><p>
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We have <math>t = 80</math>.
+
  <li><math>CA=CB</math><p>
−
 
+
Note that <math>C</math> must be the midpoint of arc <math>\widehat{AB}.</math> We conclude that <math>C = (\cos 50^{\circ}, \sin 50^{\circ})</math> or <math>C = (\cos 230^{\circ}, \sin 230^{\circ}),</math> so <math>t=50</math> or <math>t=230.</math>
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Case 3: <math>AB = AC</math>.
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</li><p>
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</ol>
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We have <math>t = 20</math>.
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Together, the sum of all such possible values of <math>t</math> is <math>20+80+50+230=\boxed{\textbf{(E)} \: 380}.</math>
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Therefore, the answer is <math>\boxed{\textbf{(E) }380}</math>.
 
  
 
~Steven Chen (www.professorchenedu.com) ~Wilhelm Z ~MRENTHUSIASM
 
~Steven Chen (www.professorchenedu.com) ~Wilhelm Z ~MRENTHUSIASM

Revision as of 02:07, 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

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 $\triangle ABC:$

  1. $AB=AC$

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

  2. $BA=BC$

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

  3. $CA=CB$

    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

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