Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 12B Problems/Problem 25"
Line 10: Line 10:
  
 
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\leq x\leq 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,
 +
 +
<math>[ABCD] = x^2\sqrt 3 + \sqrt{36 - (x^2-10)^2}</math>
 +
 +
Substitute <math>k = x^2 - 10</math> and then the expression becomes
 +
 +
<math>[ABCD] = k\sqrt{3} + \sqrt{36 - k^2} + 10\sqrt{3}</math>
 +
 +
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,
 +
 +
<math>(k\sqrt 3 + \sqrt{36-k^2})^2 \leq ((\sqrt{3})^2+1^2)((k)^2 + (\sqrt{36-k^2})^2).</math>
 +
 +
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>.
 +
 +
Finally, the maximum value of the area of <math>[ABCD]</math> is <math>\boxed{\textbf{(C) }12 + 10\sqrt{3}}</math>.
  
 
==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}}

Revision as of 18:16, 14 February 2019

Problem

Let $ABCD$ be a convex quadrilateral with $BC=2$ and $CD=6.$ Suppose that the centroids of $\triangle ABC,\triangle BCD,$ and $\triangle ACD$ form the vertices of an equilateral triangle. What is the maximum possible value of $ABCD$?

$\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30$

Solution

Set $G_1$, $G_2$, $G_3$ as the centroids of $ABC$, $BCD$, and $CDA$ respectively, while $M$ is the midpoint of line $BC$. $A$, $G_1$, and $M$ are collinear due to the centroid. Likewise, $D$, $G_2$, and $M$ are collinear as well. Because $AG_1 = 3AM$ and $DG_2 = 3DM$, $\triangle MG_1G_2\sim\triangle MAD$. From the similar triangle ratios, we can deduce that $AD = 3G_1G_2$. The similar triangles implies parallel lines, namely $AD$ is parallel to $G_1G_2$.

We can apply the same strategy to the pair of triangles $\triangle BCD$ and $\triangle ACD$. We can conclude that $AB$ is parallel to $G_2G_3$ and $AB = 3G_2G_3$. Because $3G_1G_2 = 3G_2G_3$, $AB = AD$ and the pair of parallel lines preserve the 60 degree angle, meaning $\angle BAD = 60^\circ$. Therefore, $\triangle BAD$ is equilateral.

Set $BD = 2x$ where $2\leq x\leq 4$ due to the triangle inequality. By breaking the quadrilateral into $[ABD]$ and $[BCD]$, we can create an expression for the area of $[ABCD]$. We will use the formula for the area of an equilateral triangle given its side length to find the area of $[ABD]$ and Heron's formula to find the area of $[BCD]$.

After simplifying,

$[ABCD] = x^2\sqrt 3 + \sqrt{36 - (x^2-10)^2}$

Substitute $k = x^2 - 10$ and then the expression becomes

$[ABCD] = k\sqrt{3} + \sqrt{36 - k^2} + 10\sqrt{3}$

We can ignore the $10\sqrt{3}$ for now and focus on $k\sqrt{3} + \sqrt{36 - k^2}$.

By the Cauchy-Schwarz Inequality,

$(k\sqrt 3 + \sqrt{36-k^2})^2 \leq ((\sqrt{3})^2+1^2)((k)^2 + (\sqrt{36-k^2})^2).$

The RHS simplifies to $12^2$, meaning the maximum value of $k\sqrt{3} + \sqrt{36 - k^2}$ is $12$.

Finally, the maximum value of the area of $[ABCD]$ is $\boxed{\textbf{(C) }12 + 10\sqrt{3}}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_25&oldid=102611"