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Difference between revisions of "2019 AMC 10A Problems/Problem 13"

m (Fixed formatting of Solution 4)
m (Fixed formatting)
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Finally, we deduce <cmath>\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110^{\circ}}.</cmath>
 
Finally, we deduce <cmath>\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110^{\circ}}.</cmath>
  
−
==Solution 3 (Outside Angles)==
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==Solution 3 (outside angles)==
  
 
Through the property of angles formed by intersecting chords, we find that
 
Through the property of angles formed by intersecting chords, we find that
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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>
 
<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
+
Since <math>\overarc{BC}</math> is the diameter, <math>m\overarc{BC}=180^\{circ}</math>, and because <math>\triangle ABC</math> is isosceles and <math>m\angle ACB=40^{\circ}</math>, <math>m\angle CAB=70^{\circ}</math>. Thus
−
<cmath>m\angle BFC=180-70=\boxed{\textbf{(D) } 110}</cmath>
+
<cmath>m\angle BFC=180^{\circ}-70^{\circ}=\boxed{\textbf{(D) } 110^{\circ}}</cmath>
  
 
==Solution 4==
 
==Solution 4==

Revision as of 23:49, 26 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 $\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110^{\circ}}.$

Solution 2

Alternatively, we could have used similar triangles. We start similarly to Solution 1.

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 deduce \[\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110^{\circ}}.\]

Solution 3 (outside angles)

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, $m\overarc{BC}=180^\{circ}$ (Error compiling LaTeX. Unknown error_msg), and because $\triangle ABC$ is isosceles and $m\angle ACB=40^{\circ}$, $m\angle CAB=70^{\circ}$. Thus \[m\angle BFC=180^{\circ}-70^{\circ}=\boxed{\textbf{(D) } 110^{\circ}}\]

Solution 4

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

See Also

Cheap Solution: Create an accurate diagram and measure the angle using a protractor. If you were accurate, the answer is $110^{\circ}$.

2019 AMC 10A (Problems • Answer Key • Resources)
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Problem 12 		Followed by
Problem 14
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All AMC 10 Problems and Solutions

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