Difference between revisions of "2019 AMC 10A Problems/Problem 13"
Argonauts16 (talk | contribs) (→Solution) |
Argonauts16 (talk | contribs) (→Solution) |
||
Line 15: | Line 15: | ||
<cmath>\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}</cmath> | <cmath>\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}</cmath> | ||
− | Then, we take triangle <math>BFC</math>, and find <math>\angle BFC=180-50-20=\boxed{ | + | Then, we take triangle <math>BFC</math>, and find <math>\angle BFC=180^{\circ}-50^{circ}-20^{\circ}=\boxed{\textbf{(D)}110}.</math> |
~Argonauts16 (Diagram by Brendanb4321) | ~Argonauts16 (Diagram by Brendanb4321) |
Revision as of 20:12, 9 February 2019
Problem
Let be an isosceles triangle with and . Contruct the circle with diameter , and let and be the other intersection points of the circle with the sides and , respectively. Let be the intersection of the diagonals of the quadrilateral . What is the degree measure of
Solution
Drawing it out, we see and are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find . We can find and by the triangle angle sum on and .
Then, we take triangle , and find
~Argonauts16 (Diagram by Brendanb4321)
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.