Difference between revisions of "2019 AMC 12B Problems/Problem 25"
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We can apply the same strategy to the pair of triangles <math>\triangle BCD</math> and <math>\triangle ACD</math>. We can conclude that <math>AB</math> is parallel to <math>G_2G_3</math> and <math>AB = 3G_2G_3</math>. Because <math>3G_1G_2 = 3G_2G_3</math>, <math>AB = AD</math> and the pair of parallel lines preserve the 60 degree angle, meaning <math>\angle BAD = 60^\circ</math>. Therefore, <math>\triangle BAD</math> is equilateral. | We can apply the same strategy to the pair of triangles <math>\triangle BCD</math> and <math>\triangle ACD</math>. We can conclude that <math>AB</math> is parallel to <math>G_2G_3</math> and <math>AB = 3G_2G_3</math>. Because <math>3G_1G_2 = 3G_2G_3</math>, <math>AB = AD</math> and the pair of parallel lines preserve the 60 degree angle, meaning <math>\angle BAD = 60^\circ</math>. Therefore, <math>\triangle BAD</math> is equilateral. | ||
− | Set <math>BD = 2x</math> where <math>2 | + | Set <math>BD = 2x</math> where <math>2 < x < 4</math> due to the triangle inequality. By breaking the quadrilateral into <math>[ABD]</math> and <math>[BCD]</math>, we can create an expression for the area of <math>[ABCD]</math>. We will use the formula for the area of an equilateral triangle given its side length to find the area of <math>[ABD]</math> and Heron's formula to find the area of <math>[BCD]</math>. |
After simplifying, | After simplifying, |
Revision as of 19:22, 14 February 2019
Problem
Let be a convex quadrilateral with
and
Suppose that the centroids of
and
form the vertices of an equilateral triangle. What is the maximum possible value of
?
Solution
Set ,
,
as the centroids of
,
, and
respectively, while
is the midpoint of line
.
,
, and
are collinear due to the centroid. Likewise,
,
, and
are collinear as well. Because
and
,
. From the similar triangle ratios, we can deduce that
. The similar triangles implies parallel lines, namely
is parallel to
.
We can apply the same strategy to the pair of triangles and
. We can conclude that
is parallel to
and
. Because
,
and the pair of parallel lines preserve the 60 degree angle, meaning
. Therefore,
is equilateral.
Set where
due to the triangle inequality. By breaking the quadrilateral into
and
, we can create an expression for the area of
. We will use the formula for the area of an equilateral triangle given its side length to find the area of
and Heron's formula to find the area of
.
After simplifying,
Substitute and then the expression becomes
We can ignore the for now and focus on
.
By the Cauchy-Schwarz Inequality,
The RHS simplifies to , meaning the maximum value of
is
.
Finally, the maximum value of the area of is
.
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.