Difference between revisions of "2021 Fall AMC 12B Problems/Problem 10"
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The following diagram shows Cases 1, 2, and 3 in red, blue, and green, respectively: | The following diagram shows Cases 1, 2, and 3 in red, blue, and green, respectively: | ||
+ | <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),blue+linewidth(4)); | ||
+ | dot("$C_3$",C3,1.5*dir(C3),heavygreen+linewidth(4)); | ||
+ | dot("$C_3$",C4,1.5*dir(C4),heavygreen+linewidth(4)); | ||
+ | </asy> | ||
~Steven Chen (www.professorchenedu.com) ~Wilhelm Z ~MRENTHUSIASM | ~Steven Chen (www.professorchenedu.com) ~Wilhelm Z ~MRENTHUSIASM |
Revision as of 04:02, 28 January 2022
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 Cases 1, 2, and 3 in red, blue, and green, respectively:
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.