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 $\triangle ABC?$

$\dfrac{1}{16}$

$\dfrac{1}{12}$

$\dfrac{1}{9}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

Solution:

Track a third point always halfway between the two particles. Between the moments when the particles are at vertices/midpoints, both particles move linearly, so the midpoint moves linearly too, tracing straight segments that form a small triangle $XYZ.$

By symmetry $XYZ$ shares its center with $\triangle ABC.$ If $O$ is that center and $F$ is the midpoint of a side, then $OZ=OC-ZC=\dfrac23 CF-\dfrac12 CF=\dfrac16 CF,$ 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

Solution:

Problem 22 in Other Years