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

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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>
<math>\textbf{(A)} \: 100 \qquad\textbf{(B)} \: 150 \qquad\textbf{(C)} \: 330 \qquad\textbf{(D)} \: 360 \qquad\textbf{(E)} \: 380</math>


== Solution 2 ==
== Solution ==
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>.
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>
<ol style="margin-left: 1.5em;">
  <li><math>AB=AC</math><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>
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>
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>
</ol>
Together, the sum of all such possible values of <math>t</math> is <math>20+80+50+230=\boxed{\textbf{(E)} \: 380}.</math>


We have <math>t = 50</math> or <math>230</math>.
<u><b>Remark</b></u>


Case 2: <math>BA = BC</math>.
The following diagram shows all possible locations of <math>C:</math>


We have <math>t = 80</math>.
<asy>
/* Made by MRENTHUSIASM */
size(200);


Case 3: <math>AB = AC</math>.
int xMin = -1;
int xMax = 1;
int yMin = -1;
int yMax = 1;
int numRays = 36;


We have <math>t = 20</math>.
//Draws a polar grid that goes out to a number of circles
//equal to big, with numRays specifying the number of rays:
void polarGrid(int big, int numRays)
{
  for (int i = 1; i < big+1; ++i)
  {
    draw(Circle((0,0),i), gray+linewidth(0.4));
  }
  for (int i=0;i<numRays;++i)
  draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4));
}


Therefore, the answer is <math>\boxed{\textbf{(E) }380}</math>.
polarGrid(xMax,numRays);
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$x$",(xMax,0),2*E);
label("$y$",(0,yMax),2*N);


~Steven Chen (www.professorchenedu.com)
pair A, B, C1, C2, C3, C4;
A = dir(40);
B = dir(60);
C1 = dir(20);
C2 = dir(80);
C3 = dir(50);
C4 = dir(230);


dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C_1$",C1,1.5*dir(C1),red+linewidth(4));
dot("$C_2$",C2,1.5*dir(C2),red+linewidth(4));
dot("$C_3$",C3,1.5*dir(C3),red+linewidth(4));
dot("$C_3$",C4,1.5*dir(C4),red+linewidth(4));
</asy>
~Steven Chen (www.professorchenedu.com) ~Wilhelm Z ~MRENTHUSIASM
==Video Solution (Just 1 min!)==
https://youtu.be/F_Hy5OWBC54
<i>~Education, the Study of Everything</i>
== Video Solution by TheBeautyofMath ==
https://www.youtube.com/watch?v=4qgYrCYG-qw&t=1304
~IceMatrix
== See Also ==
{{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}}

Latest revision as of 01:01, 14 December 2024

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 $20+80+50+230=\boxed{\textbf{(E)} \: 380}.$

Remark

The following diagram shows all possible locations of $C:$

[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -1; int xMax = 1; int yMin = -1; int yMax = 1; int numRays = 36; //Draws a polar grid that goes out to a number of circles //equal to big, with numRays specifying the number of rays: void polarGrid(int big, int numRays) { for (int i = 1; i < big+1; ++i) { draw(Circle((0,0),i), gray+linewidth(0.4)); } for (int i=0;i<numRays;++i) draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4)); } polarGrid(xMax,numRays); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),2*E); label("$y$",(0,yMax),2*N); pair A, B, C1, C2, C3, C4; A = dir(40); B = dir(60); C1 = dir(20); C2 = dir(80); C3 = dir(50); C4 = dir(230); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C_1$",C1,1.5*dir(C1),red+linewidth(4)); dot("$C_2$",C2,1.5*dir(C2),red+linewidth(4)); dot("$C_3$",C3,1.5*dir(C3),red+linewidth(4)); dot("$C_3$",C4,1.5*dir(C4),red+linewidth(4)); [/asy]

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

Video Solution (Just 1 min!)

https://youtu.be/F_Hy5OWBC54

~Education, the Study of Everything

Video Solution by TheBeautyofMath

https://www.youtube.com/watch?v=4qgYrCYG-qw&t=1304

~IceMatrix

See Also

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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