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2007 AMC 12B Problem 22

Problem 22 of 25HarderGeometry

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

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Solution

Let D,E,FD,E,F be the midpoints of BC,CA,AB,BC,CA,AB, respectively, and let X,Y,ZX,Y,Z be the midpoints of AD,BE,CF,AD,BE,CF, respectively. Track a third point halfway between the two particles. It is at XX when the particles are at A,D;A,D; at ZZ when they are at F,C;F,C; and at YY when they are at B,E.B,E. Between these instants both particle positions vary linearly, so their midpoint traces the segments XZ,ZY,YX.XZ,ZY,YX. Thus the enclosed path is the equilateral triangle XYZ,XYZ, which by symmetry shares the center OO of △ABC.\triangle ABC. Because ZZ is the midpoint of the median CF,CF, OZ=OC−ZC=23CF−12CF=16CF, \begin{aligned} OZ&=OC-ZC \\ &=\dfrac23 CF-\dfrac12 CF \\ &=\dfrac16 CF, \end{aligned} while OC=23CF.OC=\dfrac23 CF. So the ratio of circumradii is OZOC=14,\dfrac{OZ}{OC}=\dfrac14, and the area ratio is (14)2=116.\left(\dfrac14\right)^2=\dfrac{1}{16}. Thus, the correct answer is A.
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Tagged: similarity · area ratio · centroid

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