Difference between revisions of "2017 AMC 10B Problems/Problem 24"
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==Solution== | ==Solution== | ||
+ | WLOG, let the centroid of the triangle be <math>(-1,-1)</math>, and let it be <math>I</math>. By symmetry, <math>A = (1,1)</math>, so <math>AI = BI = CI = 2\sqrt{2}</math>, so since <math>\triangle AIB</math> is isosceles and <math>\angle AIB = 120^{\circ}</math>, then by Law of Cosines, <math>AB = 2\sqrt{6}</math>. Therefore, the area of the triangle is <math>\frac{(2\sqrt{6})^2\sqrt{3}}4 = 6\sqrt{3}</math>, so the square of the area of the triangle is <math>\boxed{\textbf{(C) } 108}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=23|num-a=25}} | {{AMC10 box|year=2017|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:37, 16 February 2017
Problem 24
The vertices of an equilateral triangle lie on the hyperbola , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
Solution
WLOG, let the centroid of the triangle be , and let it be
. By symmetry,
, so
, so since
is isosceles and
, then by Law of Cosines,
. Therefore, the area of the triangle is
, so the square of the area of the triangle is
.
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
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