2017 AMC 10B Problems/Problem 24
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 intersect the graph exactly three times. Therefore,
, so
, so since
is isosceles and
, then by Law of Cosines,
. Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to
. 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
.
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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