Difference between revisions of "2017 AMC 10B Problems/Problem 24"
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==Solution== | ==Solution== | ||
− | WLOG, let the centroid of <math>\triangle ABC</math> be <math>I = (-1,-1)</math>. The centroid of an equilateral triangle is the same as the | + | WLOG, let the centroid of <math>\triangle ABC</math> be <math>I = (-1,-1)</math>. The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must hit the graph exactly three times. Therefore, <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>. |
==Solution 2== | ==Solution 2== |
Revision as of 16:02, 24 February 2017
Contents
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 be
. The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must hit the graph exactly three times. Therefore,
, 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
.
Solution 2
WLOG, let the centroid of be
. Then, one of the vertices must be the other vertex of the hyperbola. WLOG, let
. Then, point
must be the reflection of
across the line
, so let
and
, where
. Because
is the centroid, the average of the
-coordinates of the vertices of the triangle is
. So we know that
. Multiplying by
and solving gives us
. So
and
. So
, and finding the square of the area gives us
.
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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