Elementary geometry-cube.

Question

I have a little elementary geometry problem:

Given the $ABCDEFGH$ cube, $AB=1$, $HF \cap EG = \{I\}$, $ FC \cap BG = \{J\}$ and let $X$ be the section of $(AIJ)$ plane with $FG$. $\frac{FX}{XG}$=?

Thanks in advance!

Answer 1

A coordinate-free approach uses the identity that if $X$ is the intersection of plane $AIJ$ with line $FG$, then $$ FX : GX = [AFIJ] : [AGIJ] $$ where by $[PQRS]$ I mean the volume of the tetrahedron $PQRS$.

(Why? Because if $F,G$ have projections $P_F, P_G$ onto the plane $AIJ$, then $[AFIJ] = \frac13 [AIJ] \cdot FP_F$ and $[AGIJ] = \frac13 [AIJ] \cdot GP_G$, and the ratio $FP_F : GP_G = FX : GX$ holds by similar triangles.)

In this case, $[AFIJ] = \frac12 [AFHJ]$ (since $I$ is halfway to $H$ from $F$) which in turn is $\frac14 [AFHC]$ (since $J$ is halfway to $C$ from $F$) and similarly $[AGIJ] = \frac14 [AGBE]$. We get $$ FX : GX = [AFIJ] : [AGIJ] = [AFHC] : [AGBE] = \frac13 : \frac16 = 2:1 $$ so $FX = \frac23$ and $GX =\frac13$.

Answer 2

Idea:

Put the cube in 3D system so $A=(0,0,0)$ and $B=(2,0,0)$. Then $I=(1,1,2)$ and $J=(2,1,1)$. Now write the eqaution of the plane $$ax+by+cz+d=0$$ through $A,I,J$ and calculate where it cuts line $$FG = \{(2,y,2)\mid \;y\in \mathbb{R}\}$$ i.e. put $x=2$ and $z=2$ and you will get point of intersection.

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I made some calculation and I get $x-3y+z=0$ so $y= {4\over 3}$. So point $X=(2,{4\over 3},2)$ and so $$\frac{FX}{XG}={2\over 1}$$