Difference between revisions of "2019 AMC 10A Problems/Problem 13"
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<asy> unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label("$A$",(1,0),SE);label("$C$",(0,2.75),N);label("$B$",(-1,0),SW);label("$E$",(0,0),S);label("$D$",(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));</asy> | <asy> unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label("$A$",(1,0),SE);label("$C$",(0,2.75),N);label("$B$",(-1,0),SW);label("$E$",(0,0),S);label("$D$",(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));</asy> | ||
| − | Drawing it out, we see <math>\angle BDC</math> and <math>\angle BEC</math> are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find <math>\angle ABC=70^{\circ}</math>. We can find <math>\angle ECB=20</math> and <math>\angle DBC=50</math> by the triangle angle sum on <math>\triangle ECB</math> and <math>\triangle DBC</math>. | + | Drawing it out, we see <math>\angle BDC</math> and <math>\angle BEC</math> are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find <math>\angle ABC=70^{\circ}</math>. We can find <math>\angle ECB=20^{\circ}</math> and <math>\angle DBC=50^{\circ}</math> by the triangle angle sum on <math>\triangle ECB</math> and <math>\triangle DBC</math>. |
| − | <cmath>\angle{BDC}+\angle{DBC}+\angle{ | + | <cmath>\angle{BDC}+\angle{DCB}+\angle{DBC}=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle{DBC}\implies\angleDBC=50^{\circ}</cmath> |
| + | <cmath>\angle{BEC}+\angle{EBC}+\angle{ECB}=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle{DBC}\implies\angle{DBC}=20^{\circ}</cmath> | ||
Then, we take triangle <math>BFC</math>, and find <math>\angle BFC=180-50-20=\boxed{(\textbf{\text{D}})110}.</math> | Then, we take triangle <math>BFC</math>, and find <math>\angle BFC=180-50-20=\boxed{(\textbf{\text{D}})110}.</math> | ||
Revision as of 20:06, 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
![[asy] unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label("$A$",(1,0),SE);label("$C$",(0,2.75),N);label("$B$",(-1,0),SW);label("$E$",(0,0),S);label("$D$",(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));[/asy]](http://latex.artofproblemsolving.com/5/e/1/5e1c2532c45ec05a88a74ea36b442cbd49851637.png)
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 
.
\[\angle{BDC}+\angle{DCB}+\angle{DBC}=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle{DBC}\implies\angleDBC=50^{\circ}\] (Error compiling LaTeX. Unknown error_msg)
![\[\angle{BEC}+\angle{EBC}+\angle{ECB}=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle{DBC}\implies\angle{DBC}=20^{\circ}\]](http://latex.artofproblemsolving.com/c/8/8/c88d7cc9a296b2eb99e110b70cd6b9d29cb7c928.png)
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. 
