Difference between revisions of "2019 AMC 10A Problems/Problem 13"
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<math>\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120</math> | <math>\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120</math> | ||
− | ==Solution== | + | ==Solution 1== |
<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> | ||
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~Argonauts16 (Diagram by Brendanb4321) | ~Argonauts16 (Diagram by Brendanb4321) | ||
+ | |||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Through the property of angles formed by intersecting chords, we find that | ||
+ | <cmath>m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}</cmath> | ||
+ | |||
+ | Through the Outside Angles Theorem, we find that | ||
+ | <cmath>m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}</cmath> | ||
+ | |||
+ | Adding the two equations gives us | ||
+ | <cmath>m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB</cmath> | ||
+ | |||
+ | Since <math>\overarc{BC}</math> is the diameter, <math>m\overarc{BC}=180</math> and because <math>\triangle ABC</math> is isosceles and <math>m\angle ACB=40</math>, <math>m\angle CAB=70</math>. Thus | ||
+ | <cmath>m\angle BFC=180-70=\boxed{\textbf{(D)}110}</cmath> | ||
+ | |||
+ | ~mn28407 | ||
==See Also== | ==See Also== |
Revision as of 21:32, 9 February 2019
Contents
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 1
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)
Solution 2
Through the property of angles formed by intersecting chords, we find that
Through the Outside Angles Theorem, we find that
Adding the two equations gives us
Since is the diameter, and because is isosceles and , . Thus
~mn28407
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.