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 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
.
Solution 3
WLOG, let the centroid of be
and let point
be
. It is known that the centroid is equidistant from the three vertices of
. Because we have the coordinates of both
and
, we know that the distance from
to any vertice of
is
. Therefore,
. It follows that from
, where
and
,
using the formula for the area of a triangle with sine
. Because
and
are congruent to
, they also have an area of
. Therefore,
. Squaring that gives us the answer of
.
Solution 4 (5-second sol)
WLOG let the centroid of the triangle be . By symmetry, one of the vertices is
. The distance between these two points is
, so the height of the triangle is
, the side length is
, and the area squared 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 | ||
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