Difference between revisions of "2019 AMC 10A Problems/Problem 13"
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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>. | 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 DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}</cmath> | + | <cmath>\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC=180^{\circ}\implies\angle DBC=50^{\circ}</cmath> |
− | <cmath>\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}</cmath> | + | <cmath>\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB=180^{\circ}\implies\angle ECB=20^{\circ}</cmath> |
− | Then, we take triangle <math>BFC</math>, and find <math>\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.</math> | + | Then, we take triangle <math>BFC</math>, and find <math>\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110^{\circ}}.</math> |
− | + | ==Solution 2== | |
+ | |||
+ | Alternatively, we could have used similar triangles. We start similarly to Solution 1. | ||
+ | |||
+ | Drawing it out, we see <math>\angle BDC</math> and <math>\angle BEC</math> are right angles, as they are inscribed in a semicircle. Therefore, <cmath>\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.</cmath> | ||
+ | |||
+ | So, <math>\triangle BEF \sim BDA</math> by AA Similarity, since <math>\angle EBF = \angle DBA</math> and <math>\angle BEC = 90^{\circ} = \angle BDA</math>. | ||
+ | Thus, we know <cmath>\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.</cmath> | ||
− | ==Solution | + | Finally, we deduce <cmath>\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110^{\circ}}.</cmath> |
+ | |||
+ | ==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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Adding the two equations gives us | Adding the two equations gives us | ||
− | <cmath>m\angle BFC | + | <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^{\circ}</math>, and because <math>\triangle ABC</math> is isosceles and <math>m\angle ACB=40^{\circ}</math>, we have <math>m\angle CAB=70^{\circ}</math>. Thus | ||
+ | <cmath>m\angle BFC=180^{\circ}-70^{\circ}=\boxed{\textbf{(D) } 110^{\circ}}</cmath> | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Notice that if <math>\angle BEC = 90^{\circ}</math>, then <math>\angle BCE</math> and <math>\angle ACE</math> must be <math>20^{\circ}</math>. Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that <math>\angle EBD \cong \angle ECD = 20^{\circ}</math>. Thus <math>\angle CBF = 70 - 20 = 50^{\circ}</math>, and so <math>\angle BFC = 180 - 20 - 50 = 110^{\circ}</math>, which is <math>\boxed{\textbf{(D)}}</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | <math>\triangle{ABC}</math> is isosceles so <math>\angle{CAB}=70^{\circ}</math>. Since <math>CB</math> is a diameter, <math>\angle{CDB}=\angle{CEB}=90^{\circ}</math>. Quadrilateral <math>ADFE</math> is cyclic since <math>\angle{ADF}+\angle{AEF}=180^{\circ}</math>. Therefore <math>\angle{BFC}=\angle{DFE}=180^{\circ}-\angle{CAB}=\boxed{110^{\circ}}</math> | ||
+ | |||
+ | ==Video Solution== | ||
− | + | https://youtu.be/GmQIEX4Izt4 | |
− | |||
− | + | Education, the Study of Everything | |
==See Also== | ==See Also== |
Latest revision as of 02:57, 20 September 2021
Contents
Problem
Let be an isosceles triangle with and . Construct 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
Solution 2
Alternatively, we could have used similar triangles. We start similarly to Solution 1.
Drawing it out, we see and are right angles, as they are inscribed in a semicircle. Therefore,
So, by AA Similarity, since and . Thus, we know
Finally, we deduce
Solution 3 (outside angles)
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 , we have . Thus
Solution 4
Notice that if , then and must be . Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that . Thus , and so , which is .
Solution 5
is isosceles so . Since is a diameter, . Quadrilateral is cyclic since . Therefore
Video Solution
Education, the Study of Everything
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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