Difference between revisions of "2021 Fall AMC 12B Problems/Problem 10"
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==Problem== | ==Problem== | ||
− | What is the sum of all possible values of <math>t</math> between <math>0</math> and <math>360</math> such that the triangle in the coordinate plane whose vertices are < | + | What is the sum of all possible values of <math>t</math> between <math>0</math> and <math>360</math> such that the triangle in the coordinate plane whose vertices are <cmath>(\cos 40^\circ,\sin 40^\circ), (\cos 60^\circ,\sin 60^\circ), \text{ and } (\cos t^\circ,\sin t^\circ)</cmath> |
is isosceles? | is isosceles? | ||
<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 | + | == Solution == |
− | + | 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> | ||
− | < | + | <u><b>Remark</b></u> |
− | + | The following diagram shows all possible locations of <math>C:</math> | |
− | + | <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 | ||
+ | |||
+ | <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 21:50, 10 July 2023
Contents
Problem
What is the sum of all possible values of between and such that the triangle in the coordinate plane whose vertices are is isosceles?
Solution
Let and We apply casework to the legs of isosceles
Note that must be the midpoint of It follows that so
Note that must be the midpoint of It follows that so
Note that must be the midpoint of It follows that or so or
Together, the sum of all such possible values of is
Remark
The following diagram shows all possible locations of
~Steven Chen (www.professorchenedu.com) ~Wilhelm Z ~MRENTHUSIASM
Video Solution (Just 1 min!)
~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 |
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