Difference between revisions of "2019 AMC 12B Problems/Problem 25"
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Let <math>X, Y, Z</math> be the centroids of <math>\triangle ABC, \triangle BCD, \triangle ACD</math> respectively, then | Let <math>X, Y, Z</math> be the centroids of <math>\triangle ABC, \triangle BCD, \triangle ACD</math> respectively, then | ||
− | <math>XZ//BD</math>, since <math>EX=\frac13 | + | <math>XZ//BD</math>, since <math>EX=\frac13 EB, EZ=\frac13 ED</math>, |
<math>XY//GE</math>, since <math>BX=\frac23BE, BY=\frac23BG, EG//AD</math> by midsegment theorem, so <math>XY//AD</math> | <math>XY//GE</math>, since <math>BX=\frac23BE, BY=\frac23BG, EG//AD</math> by midsegment theorem, so <math>XY//AD</math> | ||
Line 77: | Line 77: | ||
Assume <math>\alpha=\angle BCD</math>, then <math>BD^2=BC^2+CD^2-2\cdot BC\cdot CD\cos \alpha=40-24\cos\alpha</math>, the area | Assume <math>\alpha=\angle BCD</math>, then <math>BD^2=BC^2+CD^2-2\cdot BC\cdot CD\cos \alpha=40-24\cos\alpha</math>, the area | ||
− | <math>[ABCD]=[ABD]+[BCD]=\frac{\sqrt3}4 BD^2+\frac12\cdot BC\cdot CD\sin\alpha= | + | <math>[ABCD]=[ABD]+[BCD]=\frac{\sqrt3}4 BD^2+\frac12\cdot BC\cdot CD\sin\alpha=</math> |
+ | <math>10\sqrt3+6(\sin\alpha-\sqrt3\cos \alpha)=10\sqrt3+12\sin(\alpha-60^\circ)</math> | ||
− | The maximal value happens when <math>\sin(\alpha-60^\circ)=1</math>, and the value is <math> | + | |
+ | The maximal value happens when <math>\sin(\alpha-60^\circ)=1</math>, and the value is <math>10\sqrt3+12</math>, and the answer is <math>\boxed{\textbf{(C)} 12+10\sqrt3}</math>. | ||
<asy> | <asy> |
Latest revision as of 13:37, 29 August 2024
Contents
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 the area of
?
Solution 1 (vectors)
Place an origin at , and assign position vectors of
and
. Since
is not parallel to
, vectors
and
are linearly independent, so we can write
for some constants
and
. Now, recall that the centroid of a triangle
has position vector
.
Thus the centroid of is
; the centroid of
is
; and the centroid of
is
.
Hence ,
, and
. For
to be equilateral, we need
. Further,
. Hence we have
, so
is equilateral.
Now let the side length of be
, and let
. By the Law of Cosines in
, we have
. Since
is equilateral, its area is
, while the area of
is
. Thus the total area of
is
, where in the last step we used the subtraction formula for
. Alternatively, we can use calculus to find the local maximum. Observe that
has maximum value
when e.g.
, which is a valid configuration, so the maximum area is
.
Solution 2
Let ,
,
be the centroids of
,
, and
respectively, and let
be the midpoint of
.
,
, and
are collinear due to well-known properties of the centroid. Likewise,
,
, and
are collinear as well. Because (as is also well-known)
and
, we have
. This implies that
is parallel to
, and in terms of lengths,
. (SAS Similarity)
We can apply the same argument to the pair of triangles and
, concluding that
is parallel to
and
. Because
(due to the triangle being equilateral),
, and the pair of parallel lines preserve the
angle, meaning
. Therefore
is equilateral.
At this point, we can finish as in Solution 1, or, to avoid using trigonometry, we can continue as follows:
Let , where
due to the Triangle Inequality in
. By breaking the quadrilateral into
and
, we can create an expression for the area of
. We 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,
Substituting , 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
.
Thus the maximum possible area of is
.
Solution 3 (Complex Numbers)
Let ,
,
, and
correspond to the complex numbers
,
,
, and
, respectively. Then, the complex representations of the centroids are
,
, and
. The pairwise distances between the centroids are
,
, and
, all equal. Thus,
, so
. Hence,
is equilateral.
By the Law of Cosines,
.
. Thus, the maximum possible area of
is
.
~ Leo.Euler
Solution 4 (Homothety)
Let , and
be the centroids of
, and
, respectively, and let
and
be the midpoints of
and
, respectively. Note that
and
are
of the way from
to
and
, respectively, by a well-known property of centroids. Then a homothety centered at
with ratio
maps
and
to
and
, respectively, implying that
is equilateral too. But
is the medial triangle of
, so
is also equilateral. We may finish with the methods in the solutions above.
~ numberwhiz
Video Solution by MOP 2024
~r00tsOfUnity
Solution 5
Let be the centroids of
respectively, then
, since
,
, since
by midsegment theorem, so
![]()
Similarly,,
Sois an equilateral triangle
Assume, then
, the area
The maximal value happens when, and the value is
, and the answer is
.
~szhangmath
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.