Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Estonia

The midpoints of the sides $BC$ , $CA$ , and $AB$ of triangle $ABC$ are $D$ , $E$ , and $F$ , respectively. The reflections of centroid $M$ of $ABC$ around points $D$ , $E$ , and $F$ are $X$ , $Y$ , and $Z$ , respectively. Segments $XZ$ and $YZ$ intersect the side $AB$ in points $K$ and $L$ , respectively. Prove that $AL = BK$ .

Solutions — 2

Solution 1

As $\frac{MY}{MB} = \frac{2ME}{2ME} = 1$ and analogously $\frac{MZ}{MC} = 1$ we have $BC \parallel YZ$ . Let $G$ be the intersection of lines $YZ$ and $AD$ (Fig. 14).

Then $\frac{MG}{MD} = \frac{MY}{MB} = 1$ from which $MG = MD = \frac{1}{3}AD$ and $AG = AD - MD - MG = \frac{1}{3}AD$ . Hence $\frac{AL}{AB} = \frac{AG}{AD} = \frac{1}{3}$ .

By swapping the roles of $A$ and $B$ , the roles of $D$ and $E$ , roles of $X$ and $Y$ , and finally the roles of $K$ and $L$ , we get analogously $\frac{BK}{AB} = \frac{1}{3}$ . Therefore $AL = BK$ .

Fig. 14

Solution 2

Notice that triangle $XYZ$ is a homothetic transformation of triangle $ABC$ with centre $M$ and ratio $-1$ . Homothety preserves the directions of the lines, therefore $KL \parallel XY$ and $F$ lies on the median drawn from vertex $Z$ of triangle $XYZ$ . Thus $FK = FL$ and $AL = AF - FL = BF - FK = BK$ .

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