Difference between revisions of "2021 Fall AMC 12B Problems/Problem 10"
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== Solution == | == Solution == | ||
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− | + | 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;"> | |
− | We | + | <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> | |
− | + | <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> | |
− | We | + | <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> | |
− | + | </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> | |
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~Steven Chen (www.professorchenedu.com) ~Wilhelm Z ~MRENTHUSIASM | ~Steven Chen (www.professorchenedu.com) ~Wilhelm Z ~MRENTHUSIASM |
Revision as of 03:07, 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 arc
We conclude that
so
Note that
must be the midpoint of arc
We conclude that
so
Note that
must be the midpoint of arc
We conclude that
or
so
or
Together, the sum of all such possible values of is
~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.