2017 AMC 10B Problems/Problem 24
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 curve 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
.
Solution 3
WLOG, let a vertex of equilateral triangle
be at
on hyperbola
.
We are given that the vertex of the hyperbola is the centroid of the triangle. Hence, WLOG, the centroid of the triangle is at . Mark the centroid to be point
.
The length of .
Now, using the information that is
the height of equilateral triangle
(centroid), we find that the height of equilateral triangle
is
Hence, since the height of triangle , its base is
Using the formula for the area of an equilateral triangle...
Hence, the area squared $={6\sqrt{3}}^2 \implies 108\implies \boxed{\textbf{(C) } 108$ (Error compiling LaTeX. Unknown error_msg)
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 | ||
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