Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 Fall AMC 12B Problems/Problem 10"
m
Line 11: Line 11:
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
 
   <li><math>AB=AC</math><p>
 
   <li><math>AB=AC</math><p>
−
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>
+
Note that <math>A</math> must be the midpoint of <math>\widehat{BC}.</math> It follows that <math>C = (\cos 20^{\circ}, \sin 20^{\circ}),</math> so <math>t=20.</math></li><p>
 
   <li><math>BA=BC</math><p>
 
   <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>
+
Note that <math>B</math> must be the midpoint of <math>\widehat{AC}.</math> It follows that <math>C = (\cos 80^{\circ}, \sin 80^{\circ}),</math> so <math>t=80.</math></li><p>
 
   <li><math>CA=CB</math><p>
 
   <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>
+
Note that <math>C</math> must be the midpoint of <math>\widehat{AB}.</math> It follows 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>
 
</li><p>
 
</li><p>
 
</ol>
 
</ol>

Revision as of 03:11, 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 $\widehat{BC}.$ It follows that $C = (\cos 20^{\circ}, \sin 20^{\circ}),$ so $t=20.$

  2. $BA=BC$

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

  3. $CA=CB$

    Note that $C$ must be the midpoint of $\widehat{AB}.$ It follows 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. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_10&oldid=170384"