A triangle ABC is given. Find all the pairs of points X,Y so that X is on the sides of the triangle, Y is inside the triangle, and four non-intersecting segments from the set {XY,AX,AY,BX,BY,CX,CY} divide the triangle ABC into four triangles with equal areas.
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Tags: geometry, geometry proposed, geometry unsolved
20.02.2021 20:52
Could someone solve this problem?
30.04.2021 22:14
WLOG let X be on AB. There are two ways △ABC can be divided into four triangles. Case 1: XY,AY,BY,CY [asy][asy] draw((0,0)--(4,0)--(3,3)--(0,0)); draw((0,0)--(2.5,1.5)--(4,0)); draw((2,0)--(2.5,1.5)--(3,3)); label("A",(0.1,0.1),SW); label("B",(3.97,0.1),SE); label("C",(3,2.98),N); label("X",(2,0.02),S); label("Y",(2.6,1.5),NW); [/asy][/asy] Since [△AXY]=[△BXY], |AX|=|BX| and X=(1/2,1/2,0). Next, [△ACY]=[△BCY]=14[△ABC] and [△ABY]=12[△ABC], so Y=(1/4,1/4,1/2). Considering the permutations, there are 3 total pairs of points in this case. Case 2: XY,AY,CX,CY (WLOG |AX|>|BX|) [asy][asy] draw((0,0)--(4,0)--(3,3)--(0,0)); draw((0,0)--(2,1)); draw((2.7,0)--(2,1)--(3,3)--(2.7,0)); label("A",(0.1,0.1),SW); label("B",(3.97,0.1),SE); label("C",(3,2.98),N); label("X",(2.7,0.02),S); label("Y",(2.1,1),NW); [/asy][/asy] We have that [△BCX]=14[△ABC], so |AX|=3|BX| and X=(1/4,3/4,0). Clearly Y must be the centroid of △ACX. Let the intersection of CY and AB meet at P. Then |AP|=|PX| and |AP|=38|AB|, so Y=((5/8)(2/3),(3/8)(2/3),1/3)=(5/12,1/4,1/3). There are 6 pairs of points in this case.
01.03.2022 16:23
What does Y(1/4,1/4,1/2) mean? Is it the coordinates of the point Y in the space?!
08.05.2023 07:20
I believe those are the barycentric coordinates of Y with respect to reference triangle ABC.