Difference between revisions of "2010 AMC 12A Problems/Problem 8"
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+ | == Solution 3 (Similar Triangles) == | ||
+ | Notice that <math>\angle AEB=\angle AFC = 120^{\circ}</math> and <math>\angle ACF=\angle AEB</math>. Hence, triangle AEB is similar to triangle CFA. Since <math>AB=2BC</math>, <math>AE=2CF=2FE</math>, since triangle CFE is equilateral. Therefore, <math>AF=FE=FC</math>, and since <math>\angle AFC=120^{\circ}</math>,<math>x=30</math>. Thus, the measure of <math>\angle ACE=\angle FCE+\angle ACF=90^{\circ}, \text{or} \textbf{(C)}</math> | ||
+ | -HarryW | ||
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==Video Solution by the Beauty of Math== | ==Video Solution by the Beauty of Math== |
Revision as of 10:05, 13 December 2020
Contents
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
Solution 3 (Similar Triangles)
Notice that and
. Hence, triangle AEB is similar to triangle CFA. Since
,
, since triangle CFE is equilateral. Therefore,
, and since
,
. Thus, the measure of
-HarryW
Video Solution by the Beauty of Math
https://youtu.be/kU70k1-ONgM?t=785
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.