Difference between revisions of "2019 AMC 12B Problems/Problem 25"
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==Problem== | ==Problem== | ||
− | ==Solution== | + | Let <math>ABCD</math> be a convex quadrilateral with <math>BC=2</math> and <math>CD=6.</math> Suppose that the centroids of <math>\triangle ABC,\triangle BCD,</math> and <math>\triangle ACD</math> form the vertices of an equilateral triangle. What is the maximum possible value of the area of <math>ABCD</math>? |
+ | |||
+ | <math>\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30</math> | ||
+ | |||
+ | ==Solution 1 (vectors)== | ||
+ | Place an origin at <math>A</math>, and assign position vectors of <math>B = \vec{p}</math> and <math>D = \vec{q}</math>. Since <math>AB</math> is not parallel to <math>AD</math>, vectors <math>\vec{p}</math> and <math>\vec{q}</math> are linearly independent, so we can write <math>C = m\vec{p} + n\vec{q}</math> for some constants <math>m</math> and <math>n</math>. Now, recall that the centroid of a triangle <math>\triangle XYZ</math> has position vector <math>\frac{1}{3}\left(\vec{x}+\vec{y}+\vec{z}\right)</math>. | ||
+ | |||
+ | Thus the centroid of <math>\triangle ABC</math> is <math>g_1 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}n\vec{q}</math>; the centroid of <math>\triangle BCD</math> is <math>g_2 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}(n+1)\vec{q}</math>; and the centroid of <math>\triangle ACD</math> is <math>g_3 = \frac{1}{3}m\vec{p} + \frac{1}{3}(n+1)\vec{q}</math>. | ||
+ | |||
+ | 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(\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== | ||
+ | |||
+ | Let <math>G_1</math>, <math>G_2</math>, <math>G_3</math> be the centroids of <math>ABC</math>, <math>BCD</math>, and <math>CDA</math> respectively, and let <math>M</math> be the midpoint of <math>BC</math>. <math>A</math>, <math>G_1</math>, and <math>M</math> are collinear due to well-known properties of the centroid. Likewise, <math>D</math>, <math>G_2</math>, and <math>M</math> are collinear as well. Because (as is also well-known) <math>3MG_1 = AM</math> and <math>3MG_2 = DM</math>, we have <math>\triangle MG_1G_2\sim\triangle MAD</math>. This implies that <math>AD</math> is parallel to <math>G_1G_2</math>, and in terms of lengths, <math>AD = 3G_1G_2</math>. (SAS Similarity) | ||
+ | |||
+ | We can apply the same argument to the pair of triangles <math>\triangle BCD</math> and <math>\triangle ACD</math>, concluding 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> (due to the triangle being equilateral), <math>AB = AD</math>, and the pair of parallel lines preserve the <math>60^{\circ}</math> angle, meaning <math>\angle BAD = 60^\circ</math>. Therefore <math>\triangle BAD</math> is equilateral. | ||
+ | |||
+ | At this point, we can finish as in Solution 1, or, to avoid using trigonometry, we can continue as follows: | ||
+ | |||
+ | Let <math>BD = 2x</math>, where <math>2 < x < 4</math> due to the Triangle Inequality in <math>\triangle BCD</math>. By breaking the quadrilateral into <math>\triangle ABD</math> and <math>\triangle BCD</math>, we can create an expression for the area of <math>ABCD</math>. We use the formula for the area of an equilateral triangle given its side length to find the area of <math>\triangle ABD</math> and Heron's formula to find the area of <math>\triangle BCD</math>. | ||
+ | |||
+ | After simplifying, | ||
+ | |||
+ | <cmath>[ABCD] = x^2\sqrt 3 + \sqrt{36 - (x^2-10)^2}</cmath> | ||
+ | |||
+ | Substituting <math>k = x^2 - 10</math>, the expression becomes | ||
+ | |||
+ | <cmath>[ABCD] = k\sqrt{3} + \sqrt{36 - k^2} + 10\sqrt{3}</cmath> | ||
+ | |||
+ | We can ignore the <math>10\sqrt{3}</math> for now and focus on <math>k\sqrt{3} + \sqrt{36 - k^2}</math>. | ||
+ | |||
+ | By the Cauchy-Schwarz inequality, | ||
+ | |||
+ | <cmath>\left(k\sqrt 3 + \sqrt{36-k^2}\right)^2 \leq \left(\left(\sqrt{3}\right)^2+1^2\right)\left(\left(k\right)^2 + \left(\sqrt{36-k^2}\right)^2\right)</cmath> | ||
+ | |||
+ | The RHS simplifies to <math>12^2</math>, meaning the maximum value of <math>k\sqrt{3} + \sqrt{36 - k^2}</math> is <math>12</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}} |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.