Difference between revisions of "2010 AMC 12A Problems/Problem 8"
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\angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}</cmath> | \angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}</cmath> | ||
− | Since <math>\frac{AC}{AB} = \frac{1}{2}</math>, triangle <math>ABC</math> is a <math>30-60-90</math> triangle, so <math>\angle BCA = \boxed{90^\circ\,\textbf{(C)}}</math>. | + | Since <math>\frac{AC}{AB} = \frac{1}{2}</math> and the angle between the hypotenuse and the shorter side is <math>60^\circ</math>, triangle <math>ABC</math> is a <math>30-60-90</math> triangle, so <math>\angle BCA = \boxed{90^\circ\,\textbf{(C)}}</math>. |
== Solution 2(Trig and Angle Chasing) == | == Solution 2(Trig and Angle Chasing) == |
Revision as of 19:07, 19 October 2020
Problem
Triangle has
. Let
and
be on
and
, respectively, such that
. Let
be the intersection of segments
and
, and suppose that
is equilateral. What is
?
Solution
![AMC 2010 12A Problem 8.png](https://wiki-images.artofproblemsolving.com//9/90/AMC_2010_12A_Problem_8.png)
Let .
Since and the angle between the hypotenuse and the shorter side is
, triangle
is a
triangle, so
.
Solution 2(Trig and Angle Chasing)
Let . Let
. Because
is equilateral, we get
, so
. Because
is equilateral, we get
. Angles
and
are vertical, so
. By triangle
, we have
, and because of line
, we have
. Because Of line
, we have
, and by line
, we have
. By quadrilateral
, we have
.
By the Law of Sines, we have . By the sine addition formula(which states
by the way), we have
. Because cosine is an even function, and sine is an odd function, we have
. We know that
, and
, hence
. The only value of
that satisfies
(because
is an angle of the triangle) is
. We seek to find
, which as we found before is
, which is
. The answer is
-vsamc
Video Solution
https://youtu.be/kU70k1-ONgM?t=785
~IceMatrix
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.