Could someone solve this problem?

WLOG let $X$ be on $AB$. There are two ways $\bigtriangleup ABC$ 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 $[\bigtriangleup AXY]=[\bigtriangleup BXY]$, $|AX|=|BX|$ and $X=(1/2, 1/2, 0)$. Next, $[\bigtriangleup ACY]=[\bigtriangleup BCY]=\frac{1}{4}[\bigtriangleup ABC]$ and $[\bigtriangleup ABY]=\frac{1}{2}[\bigtriangleup ABC]$, 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 $[\bigtriangleup BCX]=\frac{1}{4}[\bigtriangleup ABC]$, so $|AX|=3|BX|$ and $X=(1/4,3/4,0)$. Clearly $Y$ must be the centroid of $\bigtriangleup ACX$. Let the intersection of $CY$ and $AB$ meet at $P$. Then $|AP|=|PX|$ and $|AP|=\frac{3}{8}|AB|$, 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.

What does $Y(1/4,1/4,1/2)$ mean? Is it the coordinates of the point Y in the space?!

I believe those are the barycentric coordinates of Y with respect to reference triangle ABC.