Difference between revisions of "2017 AMC 10B Problems/Problem 24"
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==Solution 2== | ==Solution 2== | ||
− | WLOG, let the centroid of <math>\triangle ABC</math> be <math>G = (-1,-1)</math>. Then, one of the vertices must be the other curve of the hyperbola. WLOG, let <math>A = (1,1)</math>. Then, point <math>B</math> must be the reflection of <math>C</math> across the line <math>y=x</math>, so let <math>B = (a,\frac{1}{a})</math> and <math>C=(\frac{1}{a},a)</math>, where <math>a <-1</math>. Because <math>G</math> is the centroid, the average of the <math>x</math>-coordinates of the vertices of the triangle is <math>-1</math>. So we know that <math>a + 1/a+ 1 = -3</math>. Multiplying by <math>a</math> and solving gives us <math>a=-2-\sqrt{3}</math>. So <math>B=(-2-\sqrt{3},-2+\sqrt{3})</math> and <math>C=(-2+\sqrt{3},-2-\sqrt{3})</math>. So <math>BC=2\sqrt{6}</math>, and finding the square of the area gives us <math>\boxed{\textbf{(C) } 108}</math>. | + | WLOG, let the centroid of <math>\triangle ABC</math> be <math>G = (-1,-1)</math>. Then, one of the vertices must be the other curve of the hyperbola. WLOG, let <math>A = (1,1)</math>. Then, point <math>B</math> must be the reflection of <math>C</math> across the line <math>y=x</math>, so let <math>B = \left(a,\frac{1}{a}\right)</math> and <math>C=\left(\frac{1}{a},a\right)</math>, where <math>a <-1</math>. Because <math>G</math> is the centroid, the average of the <math>x</math>-coordinates of the vertices of the triangle is <math>-1</math>. So we know that <math>a + 1/a+ 1 = -3</math>. Multiplying by <math>a</math> and solving gives us <math>a=-2-\sqrt{3}</math>. So <math>B=(-2-\sqrt{3},-2+\sqrt{3})</math> and <math>C=(-2+\sqrt{3},-2-\sqrt{3})</math>. So <math>BC=2\sqrt{6}</math>, and finding the square of the area gives us <math>\boxed{\textbf{(C) } 108}</math>. |
+ | ~minor LaTeX edit by dolphin7 | ||
==Solution 3== | ==Solution 3== |
Revision as of 15:43, 12 January 2020
Contents
[hide]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 1
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
.
~minor LaTeX edit by dolphin7
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 solution)
WLOG, let the centroid of the triangle be . By symmetry, the other vertex is
. The distance between these two points is
, so the height of the triangle is
, the side length is
, and the area is
, yielding an answer of
.
-Stormersyle
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.