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 "2007 AMC 12B Problems/Problem 22"

(→‎Solution 2)
(→‎Solution 2)
Line 13: Line 13:
 
When one of the ants is on a vertex, the other will be on the midpoint of the opposite side. Marking these point gives us a rough view of what seems to be a smaller equilateral triangle that is concentric with the larger one. After observing or imagining the ants moving, we find that the area enclosed is indeed an equilateral circle.
 
When one of the ants is on a vertex, the other will be on the midpoint of the opposite side. Marking these point gives us a rough view of what seems to be a smaller equilateral triangle that is concentric with the larger one. After observing or imagining the ants moving, we find that the area enclosed is indeed an equilateral circle.
  
−
Let the line containing
+
Let the midpoint of <math>\overline{BC}</math> be <math>D</math>. Then we can find the coordinates of <math>D</math> using the equations of two lines: the extension of <math>\overline{AD}</math> and the extension of <math>\overline{BC}</math>. The equation of any line is simply the tangent of the angle between the line and the <math>x</math>-axis. The angle between <math>\overline{AD}</math> and the <math>x</math>-axis is <math>30^{\circ}</math>, and <math>tan(30) = \frac{\sqrt{3}}{3}</math>.
 +
 
 +
Therefore the equation of the line containing <math>\overline{AD}</math> is <math>y= \frac{\sqrt{3}}{3}x.</math>
 +
 
 +
The slope of the line containing <math>\overline{BC}</math> is <math>-\sqrt{3}</math>, and it goes through the point <math>(1, 0)</math>. After solving for the <math>y</math>-intercept, we get <math>y = -\sqrt{3}x+\sqrt{3}</math>. Setting these equal to each other yields
 +
 
 +
<math>D\left(\frac{3}{4}, \frac{\sqrt{3}}{4}\right).</math>
 +
 
 +
The midpoint of <math>D</math> and the <math>A</math> is the bottom-left vertex of the smaller equilateral triangle. Using the distance formula, we get the point
 +
 
 +
<math>\left(\frac{3}{8}, \frac{\sqrt{3}}{8}\right).</math>
 +
 
 +
Since the smaller triangle is simply a dilation of the larger one, they both have an axis of symmetry over <math>x=\frac{1}{2}</math>. This yields the bottom-right vertex of the smaller triangle as
 +
 
 +
<math>\left(\frac{5}{8}, \frac{\sqrt{3}}{8}\right).</math>
 +
 
 +
Therefore the side length of the smaller triangle is <math>\frac{1}{4}.</math> The problem asks us for the ratio of the areas, and since area is two-dimensional, we simply square the ratio to get <math>\frac{1}{16} \Rightarrow \boxed{\text{A}}.</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2007|ab=B|num-b=21|num-a=23}}
 
{{AMC12 box|year=2007|ab=B|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:22, 22 July 2020

Problem 22

Two particles move along the edges of equilateral $\triangle ABC$ in the direction \[A\Rightarrow B\Rightarrow C\Rightarrow A,\] starting simultaneously and moving at the same speed. One starts at $A$, and the other starts at the midpoint of $\overline{BC}$. The midpoint of the line segment joining the two particles traces out a path that encloses a region $R$. What is the ratio of the area of $R$ to the area of $\triangle ABC$?

$\mathrm {(A)} \frac{1}{16}\qquad \mathrm {(B)} \frac{1}{12}\qquad \mathrm {(C)} \frac{1}{9}\qquad \mathrm {(D)} \frac{1}{6}\qquad \mathrm {(E)} \frac{1}{4}$

Solution

First, notice that each of the midpoints of $AB$,$BC$, and $CA$ are on the locus. Suppose after some time the particles have each been displaced by a short distance $x$, to new positions $A'$ and $M'$ respectively. Consider $\triangle ABM$ and drop a perpendicular from $M'$ to hit $AB$ at $Y$. Then, $BA'=1-x$ and $BM'=1/2+x$. From here, we can use properties of a $30-60-90$ triangle to determine the lengths $YA'$ and $YM'$ as monomials in $x$. Thus, the locus of the midpoint will be linear between each of the three special points mentioned above. It follows that the locus consists of the only triangle with those three points as vertices. (A cheaper way to find the shape of the region is to look at the answer choices: if it were any sort of conic section then the ratio would not generally be rational.) Comparing inradii between this "midpoint" triangle and the original triangle, the area contained by $R$ must be $\textbf{(A)} \frac{1}{16}$ of the total area.


Solution 2

WLOG, let point $A$ be located on the origin. Let side $\overline{AB}$ be on the $x$-axis, and let point $B$ be $(0, 1)$. When one of the ants is on a vertex, the other will be on the midpoint of the opposite side. Marking these point gives us a rough view of what seems to be a smaller equilateral triangle that is concentric with the larger one. After observing or imagining the ants moving, we find that the area enclosed is indeed an equilateral circle.

Let the midpoint of $\overline{BC}$ be $D$. Then we can find the coordinates of $D$ using the equations of two lines: the extension of $\overline{AD}$ and the extension of $\overline{BC}$. The equation of any line is simply the tangent of the angle between the line and the $x$-axis. The angle between $\overline{AD}$ and the $x$-axis is $30^{\circ}$, and $tan(30) = \frac{\sqrt{3}}{3}$.

Therefore the equation of the line containing $\overline{AD}$ is $y= \frac{\sqrt{3}}{3}x.$

The slope of the line containing $\overline{BC}$ is $-\sqrt{3}$, and it goes through the point $(1, 0)$. After solving for the $y$-intercept, we get $y = -\sqrt{3}x+\sqrt{3}$. Setting these equal to each other yields

$D\left(\frac{3}{4}, \frac{\sqrt{3}}{4}\right).$

The midpoint of $D$ and the $A$ is the bottom-left vertex of the smaller equilateral triangle. Using the distance formula, we get the point

$\left(\frac{3}{8}, \frac{\sqrt{3}}{8}\right).$

Since the smaller triangle is simply a dilation of the larger one, they both have an axis of symmetry over $x=\frac{1}{2}$. This yields the bottom-right vertex of the smaller triangle as

$\left(\frac{5}{8}, \frac{\sqrt{3}}{8}\right).$

Therefore the side length of the smaller triangle is $\frac{1}{4}.$ The problem asks us for the ratio of the areas, and since area is two-dimensional, we simply square the ratio to get $\frac{1}{16} \Rightarrow \boxed{\text{A}}.$

See Also

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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=2007_AMC_12B_Problems/Problem_22&oldid=128899"