Difference between revisions of "2019 AMC 12B Problems/Problem 25"
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Hence <math>\overrightarrow{G_{1}G_{2}} = \frac{1}{3}\vec{q}</math>, <math>\overrightarrow{G_{2}G_{3}} = -\frac{1}{3}\vec{p}</math>, and <math>\overrightarrow{G_{3}G_{1}} = \frac{1}{3}\vec{p} - \frac{1}{3}\vec{q}</math>. For <math>\triangle G_{1}G_{2}G_{3}</math> to be equilateral, we need <math>\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{2}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{q}\right| \Rightarrow AB = AD</math>. Further, <math>\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{1}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{p} - \vec{q}\right| = BD</math>. Hence we have <math>AB = AD = BD</math>, so <math>\triangle ABD</math> is equilateral. | Hence <math>\overrightarrow{G_{1}G_{2}} = \frac{1}{3}\vec{q}</math>, <math>\overrightarrow{G_{2}G_{3}} = -\frac{1}{3}\vec{p}</math>, and <math>\overrightarrow{G_{3}G_{1}} = \frac{1}{3}\vec{p} - \frac{1}{3}\vec{q}</math>. For <math>\triangle G_{1}G_{2}G_{3}</math> to be equilateral, we need <math>\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{2}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{q}\right| \Rightarrow AB = AD</math>. Further, <math>\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{1}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{p} - \vec{q}\right| = BD</math>. Hence we have <math>AB = AD = BD</math>, so <math>\triangle ABD</math> is equilateral. | ||
− | Now let the side length of <math>\triangle ABD</math> be <math>k</math>, and let <math>\angle BCD = \theta</math>. By the Law of Cosines in <math>\triangle BCD</math>, we have <math>k^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \cos{\theta} = 40 - 24\cos{\theta}</math>. Since <math>\triangle ABD</math> is equilateral, its area is <math>\frac{\sqrt{3}}{4}k^2 = 10\sqrt{3} - 6\sqrt{3}\cos{\theta}</math>, while the area of <math>\triangle BCD</math> is <math>\frac{1}{2} \cdot 2 \cdot 6 \cdot \sin{\theta} = 6 \sin{\theta}</math>. Thus the total area of <math>ABCD</math> is <math>10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3} + 12\left( | + | Now let the side length of <math>\triangle ABD</math> be <math>k</math>, and let <math>\angle BCD = \theta</math>. By the Law of Cosines in <math>\triangle BCD</math>, we have <math>k^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \cos{\theta} = 40 - 24\cos{\theta}</math>. Since <math>\triangle ABD</math> is equilateral, its area is <math>\frac{\sqrt{3}}{4}k^2 = 10\sqrt{3} - 6\sqrt{3}\cos{\theta}</math>, while the area of <math>\triangle BCD</math> is <math>\frac{1}{2} \cdot 2 \cdot 6 \cdot \sin{\theta} = 6 \sin{\theta}</math>. Thus the total area of <math>ABCD</math> is <math>10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3} + 12\left(\frac{1}{2} \sin{\theta} - \frac{\sqrt{3}}{2}\cos{\theta}\right) = 10\sqrt{3}+12\sin{\left(\theta-60^{\circ}\right)}</math>, where in the last step we used the subtraction formula for <math>\sin</math>. Alternatively, we can use calculus to find the local maximum. Observe that <math>\sin{\left(\theta-60^{\circ}\right)}</math> has maximum value <math>1</math> when e.g. <math>\theta = 150^{\circ}</math>, which is a valid configuration, so the maximum area is <math>10\sqrt{3} + 12(1) = \boxed{\textbf{(C) } 12+10\sqrt3}</math>. |
==Solution 2== | ==Solution 2== | ||
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Thus the maximum possible area of <math>ABCD</math> is <math>\boxed{\textbf{(C) }12 + 10\sqrt{3}}</math>. | Thus the maximum possible area of <math>ABCD</math> is <math>\boxed{\textbf{(C) }12 + 10\sqrt{3}}</math>. | ||
+ | |||
+ | ==Solution 3 (Complex Numbers)== | ||
+ | Let <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> correspond to the complex numbers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>, respectively. Then, the complex representations of the centroids are <math>(a+b+c)/3</math>, <math>(b+c+d)/3</math>, and <math>(a+c+d)/3</math>. The pairwise distances between the centroids are <math>\lvert (d-a)/3 \rvert</math>, <math>\lvert (b-a)/3 \rvert</math>, and <math>\lvert (b-d)/3 \rvert</math>, all equal. Thus, <math>\lvert (b-a)/3 \rvert=\lvert (d-a)/3 \rvert=\lvert (b-d)/3 \rvert</math>, so <math>\lvert (b-a) \rvert=\lvert (d-a) \rvert=\lvert (b-d) \rvert</math>. Hence, <math>\triangle DBA</math> is equilateral. | ||
+ | |||
+ | By the Law of Cosines, | ||
+ | <math>[ABCD]=[ABD]+[BCD]=\frac{(\sqrt{2^2+6^2-2 \cdot 2 \cdot 6 \cos(\angle BCD)})^2 \cdot \sqrt{3}}{4}+1/2 \cdot 2 \cdot 6 \sin(\angle BCD)</math>. | ||
+ | |||
+ | <math>[ABCD]=10\sqrt{3}+6(\sin{\angle BCD}-\sqrt{3}\cos(\angle BCD))= 10\sqrt{3}+12\sin(\angle BCD-60^{\circ}) \le 12 + 10\sqrt{3}</math>. Thus, the maximum possible area of <math>ABCD</math> is <math>\boxed{\textbf{(C) }12 + 10\sqrt{3}}</math>. | ||
+ | |||
+ | ~ Leo.Euler | ||
+ | |||
+ | ==Solution 4 (Homothety)== | ||
+ | Let <math>G_1, G_2</math>, and <math>G_3</math> be the centroids of <math>\triangle ABC, \triangle BCD</math>, and <math>\triangle ACD</math>, respectively, and let <math>X, Y,</math> and <math>Z</math> be the midpoints of <math>\overline{AB}, \overline{BD},</math> and <math>\overline{AD}</math>, respectively. Note that <math>G_1, G_2,</math> and <math>G_3</math> are <math>\frac{2}{3}</math> of the way from <math>C</math> to <math>X, Y,</math> and <math>Z</math>, respectively, by a well-known property of centroids. Then a homothety centered at <math>C</math> with ratio <math>\frac{3}{2}</math> maps <math>G_1, G_2,</math> and <math>G_3</math> to <math>X, Y,</math> and <math>Z</math>, respectively, implying that <math>\triangle XYZ</math> is equilateral too. But <math>\triangle XYZ</math> is the medial triangle of <math>\triangle ABD</math>, so <math>\triangle ABD</math> is also equilateral. We may finish with the methods in the solutions above. | ||
+ | |||
+ | ~ numberwhiz | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtu.be/c26N2w2MMQE | ||
+ | |||
+ | ~r00tsOfUnity | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=24|after=Last Problem}} | {{AMC12 box|year=2019|ab=B|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:03, 15 September 2023
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
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 |
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