2007 AMC 12B Problem 22

Problem 22 of 25HarderGeometry

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?

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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;

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

\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.