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. Therefore, <cmath>\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.</cmath> | 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> | ||
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Finally, we know: <cmath>\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.</cmath> | Finally, we know: <cmath>\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.</cmath> | ||
− | ~ alleycat | + | ~ alleycat |
==Solution 3== | ==Solution 3== |
Revision as of 15:59, 12 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
~Argonauts16 (Diagram by Brendanb4321)
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 know:
~ alleycat
Solution 3
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
,
. Thus
~mn28407
=Solution 4
Notice that if is
degrees, then
and
must be
degrees. Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that
. Thus
is
, and so
is
, which is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.