Let $ABC$ be a triangle. Find all the triangles $XYZ$ with vertices inside triangle $ABC$ such that $XY,YZ,ZX$ and six non-intersecting segments from the following $AX, AY, AZ, BX, BY, BZ, CX, CY, CZ$ divide the triangle $ABC$ into seven regions with equal areas.
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Tags: geometry, geometry proposed
Arslan
27.02.2022 15:59
Could someone solve this problem?
MathMotion
18.07.2022 20:10
Arslan wrote: Could someone solve this problem? I have solved this one. Took me some time but I got it. The only thing I wasn't trying so hard to find detailed explanation for last part of the problem exc with analytic geometry and coordinates. So would be nice if someone can try to do that maybe I was just tired to see. First shift all 3 sides toward inside of the triangle by $\frac{1}{7}$ of corresponding altitude. That will give us triangle $MNP$ and our points lie one on each side of this triangle. Now it took me some time but eventually I calculated that these points has to split those sides such that $BX=3AX, CY=3BY and AZ=3CZ$ in order for triangle $XYZ$ to have required area. Now it wasn't obvious to me why triangles $AXZ, BXY\ and\ CYZ$ have the same area, I just haven't done those problems in a while I guess so I calculated that shortly with using coordinates. This is very interesting and wild problem