Difference between revisions of "2019 AMC 10A Problems/Problem 13"
Sevenoptimus (talk | contribs) m (Fixed formatting of Solution 4) |
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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 ( | + | ==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 | ||
Line 39: | Line 39: | ||
<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 00:49, 27 February 2019
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, $m\overarc{BC}=180^\{circ}$ (Error compiling LaTeX. ! Extra }, or forgotten $.), and because is isosceles and , . 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 .
See Also
Cheap Solution: Create an accurate diagram and measure the angle using a protractor. If you were accurate, the answer is .
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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