Difference between revisions of "2017 AMC 10B Problems/Problem 19"
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<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math> | <math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math> | ||
==Solution== | ==Solution== | ||
| − | + | Note that by symmetry, <math>\triangle A'B'C'</math> is also equilateral. Therefore, we only need to find one of the sides of <math>A'B'C'</math> to determine the area ratio. WLOG, let <math>AB = BC = CA = 1</math>. Therefore, <math>BB' = 3</math> and <math>BC' = 4</math>. Also, <math>\angle B'BC' = 120^{\circ}</math>, so by the Law of Cosines, <math>B'C' = \sqrt{37}</math>. Therefore, the answer is <math>(\sqrt{37})^2 : 1^2 = \boxed{\textbf{(E) } 37 : 1}</math> | |
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=18|num-a=20}} | {{AMC10 box|year=2017|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:25, 16 February 2017
Problem
Let 
be an equilateral triangle. Extend side 
beyond 
to a point 
so that 
. Similarly, extend side 
beyond 
to a point 
so that 
, and extend side 
beyond 
to a point 
so that 
. What is the ratio of the area of 
to the area of 
?
Solution
Note that by symmetry, is also equilateral. Therefore, we only need to find one of the sides of 
to determine the area ratio. WLOG, let 
. Therefore, 
and 
. Also, 
, so by the Law of Cosines, 
. Therefore, the answer is 
See Also
| 2017 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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. 
