Question

Two particles move along the edges of equilateral $\triangle ABC$ in the direction $A\to B\to C\to A,$ 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 $\triangle ABC?$

$\dfrac{1}{16}$

$\dfrac{1}{12}$

$\dfrac{1}{9}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

Answer: A

Concepts:similarity area ratio centroid

Difficulty rating: 2220

Solution:

Let $D,E,F$ be the midpoints of $BC,CA,AB,$ respectively, and let $X,Y,Z$ be the midpoints of $AD,BE,CF,$ respectively. Track a third point halfway between the two particles. It is at $X$ when the particles are at $A,D;$ at $Z$ when they are at $F,C;$ and at $Y$ when they are at $B,E.$

Between these instants both particle positions vary linearly, so their midpoint traces the segments $XZ,ZY,YX.$ Thus the enclosed path is the equilateral triangle $XYZ,$ which by symmetry shares the center $O$ of $\triangle ABC.$ Because $Z$ is the midpoint of the median $CF,$ $\begin{aligned} OZ&=OC-ZC \\ &=\dfrac23 CF-\dfrac12 CF \\ &=\dfrac16 CF, \end{aligned}$ while $OC=\dfrac23 CF.$

So the ratio of circumradii is $\dfrac{OZ}{OC}=\dfrac14,$ and the area ratio is $\left(\dfrac14\right)^2=\dfrac{1}{16}.$

Thus, the correct answer is A.

2007 AMC 12B Problem 22

Answer: A

Difficulty rating: 2220

Solution:

Problem 22 in Other Years