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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>
  
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==Solution==
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==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>
Line 18: Line 18:
  
 
~Argonauts16 (Diagram by Brendanb4321)
 
~Argonauts16 (Diagram by Brendanb4321)
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 +
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==Solution 2==
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 +
Through the property of angles formed by intersecting chords, we find that
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<cmath>m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}</cmath>
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 +
Through the Outside Angles Theorem, we find that
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<cmath>m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}</cmath>
 +
 +
Adding the two equations gives us
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<cmath>m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB</cmath>
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 +
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
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<cmath>m\angle BFC=180-70=\boxed{\textbf{(D)}110}</cmath>
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~mn28407
  
 
==See Also==
 
==See Also==

Revision as of 21:32, 9 February 2019

Problem

Let $\Delta ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$. Contruct the circle with diameter $\overline{BC}$, and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$, respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$. What is the degree measure of $\angle BFC ?$

$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$

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]

Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find $\angle ABC=70^{\circ}$. We can find $\angle ECB=20^{\circ}$ and $\angle DBC=50^{\circ}$ by the triangle angle sum on $\triangle ECB$ and $\triangle DBC$.

\[\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}\]

\[\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}\]

Then, we take triangle $BFC$, and find $\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D)}110}.$

~Argonauts16 (Diagram by Brendanb4321)


Solution 2

Through the property of angles formed by intersecting chords, we find that \[m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}\]

Through the Outside Angles Theorem, we find that \[m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}\]

Adding the two equations gives us \[m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB\]

Since $\overarc{BC}$ is the diameter, $m\overarc{BC}=180$ and because $\triangle ABC$ is isosceles and $m\angle ACB=40$, $m\angle CAB=70$. Thus \[m\angle BFC=180-70=\boxed{\textbf{(D)}110}\]

~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

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