2021 Fall AMC 12B Problems/Problem 10
Problem
What is the sum of all possible values of 
between 
and 
such that the triangle in the coordinate plane whose vertices are ![\[(\cos 40^\circ,\sin 40^\circ), (\cos 60^\circ,\sin 60^\circ), \text{ and } (\cos t^\circ,\sin t^\circ)\]](http://latex.artofproblemsolving.com/f/5/f/f5f86750dcbb985e9fd6602ccf4fdfe67a1b94c0.png)
is isosceles?
Solution 1
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 
must be the midpoint of 
It follows that 
so 
must be the midpoint of 
It follows that 
so 
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 
![[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]](http://latex.artofproblemsolving.com/5/8/6/5869cb1e15961ad7094e23093a30f622bad26ae3.png)
Solution 2 Cheese 🧀
It is fairly unlikely that the sum of all angles is less than 360, therefore we choose 380 as our answer.
vids
~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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