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Difference between revisions of "2019 AMC 10A Problems/Problem 13"
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~mn28407
 
~mn28407
  
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==Solution 4==
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==Solution 4=
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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>
 
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Notice that if <BEC is 90 degrees, then <BCE and <ACE must be 20 degrees. Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that <EBD ~ <ECD = 20 degrees. Thus <CBF is 70 - 20 = 50 degrees, and so <BFC is 180 - 20 - 50 = 110 degrees, which is D.
 
  
 +
Notice that if <math>\angle BEC</math> is <math>\text{90}</math> degrees, then <math>\angle BEC</math> and <math>\angle ACE</math> must be <math>\text{20}</math> degrees. Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that <math>\angle EBD \cong \angle ECD = 20\text{degrees}</math>. Thus <math>\angle CBF</math> is<math> 70 - 20 = 50 \text{degrees}</math>, and so <math>\angle BFC</math> is <math>180 - 20 - 50 = 110\text{degrees}</math>, which is <math>\boxed{\textbf{(D)}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 15:58, 12 February 2019

Problem

Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$. Construct 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 $\triangle DBC$.

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

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

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

Alternatively, we could have used similar triangles. We start similarly to 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. Therefore, \[\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.\]


So, $\triangle BEF \sim BDA$ by AA Similarity, since $\angle EBF = \angle DBA$ and $\angle BEC = 90^{\circ} = \angle BDA$. Thus, we know: \[\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.\]


Finally, we know: \[\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.\]

~ alleycat (Diagram by Brendanb4321)

Solution 3

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

=Solution 4

Notice that if $\angle BEC$ is $\text{90}$ degrees, then $\angle BEC$ and $\angle ACE$ must be $\text{20}$ degrees. Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that $\angle EBD \cong \angle ECD = 20\text{degrees}$. Thus $\angle CBF$ is$70 - 20 = 50 \text{degrees}$, and so $\angle BFC$ is $180 - 20 - 50 = 110\text{degrees}$, which is $\boxed{\textbf{(D)}}$.

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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