Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.2 National Olympiad, round 2 Prove it Romania

Let $ABC$ be a triangle and let $X, Y, Z$ be interior points on the sides $BC, CA, AB$ , respectively. Show that the magnified image of the triangle $XYZ$ under a homothety of factor $4$ from its centroid covers at least one of the vertices $A, B, C$ .

Solutions — 2

Solution 1

Since the problem is of an affine nature, we may (and will) assume that the triangle $XYZ$ is equilateral. The triangle $ABC$ has at least one vertex angle, say at $A$ , greater than or equal to $60^\circ$ , so $A$ is covered by the closed circumdisc $OYZ$ , where $O$ is the center of the triangle $XYZ$ . Since the latter is covered by the $4$ -fold blow-up of the triangle $XYZ$ from $O$ , the conclusion follows.

Solution 2

Suppose, if possible, that none of the vertices $A, B, C$ is covered by the $4$ -fold blow-up of the triangle $XYZ$ from its centroid. Then the distance of the point $A$ to the line $YZ$ is greater than the distance of the point $X$ to this line, so the area of the triangle $AYZ$ is greater than the area of the triangle $XYZ$ . Similarly, the triangles $BZX$ and $CXY$ both have an area greater than that of the triangle $XYZ$ , in contradiction with the well known fact that of the four triangles $AYZ, BZX, CXY, XYZ$ , the latter has not the smallest area.

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