Difference between revisions of "2010 AMC 12A Problems/Problem 8"
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Notice that <math>\angle AEB=\angle AFC = 120^{\circ}</math> and <math>\angle ACF=\angle BAE</math>. Hence, triangle AEB is similar to triangle CFA. Since <math>AB=2AC</math>, <math>AE=2CF=2FE</math>, as 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</math> equals to <math>\angle FCE+\angle ACF=90^{\circ}, \text{or} \textbf{(C)}</math> | Notice that <math>\angle AEB=\angle AFC = 120^{\circ}</math> and <math>\angle ACF=\angle BAE</math>. Hence, triangle AEB is similar to triangle CFA. Since <math>AB=2AC</math>, <math>AE=2CF=2FE</math>, as 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</math> equals to <math>\angle FCE+\angle ACF=90^{\circ}, \text{or} \textbf{(C)}</math> | ||
-HarryW | -HarryW | ||
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+ | == Solution 4 == | ||
+ | Notice that <math>\triangle ADF \sim \triangle CDA</math> (by AA Similarity.) Since the corresponding angles of a pair of similar triangles are congruent, we have <math>\angle DAC = \angle DFA = 60^\circ.</math> Since <math>AB = 2 \cdot AC</math> and <math>\angle DFA = 60^\circ,</math> we have that <math>\triangle ABC</math> is congruent by SAS to a <math>30-60-90</math> right triangle, which gives the answer <math>\boxed{\textbf{(C)}}</math>. | ||
+ | ~clever14710owl | ||
==Video Solution by the Beauty of Math== | ==Video Solution by the Beauty of Math== |
Latest revision as of 20:33, 19 August 2024
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 1
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:
By the sine addition formula():
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 , , as triangle CFE is equilateral. Therefore, , and since , . Thus, the measure of equals to -HarryW
Solution 4
Notice that (by AA Similarity.) Since the corresponding angles of a pair of similar triangles are congruent, we have Since and we have that is congruent by SAS to a right triangle, which gives the answer . ~clever14710owl
Video Solution by the Beauty of Math
https://youtu.be/kU70k1-ONgM?t=785
Video Solution by OmegaLearn
https://youtu.be/O_o_-yjGrOU?t=58
~ pi_is_3.14
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 |
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.