Cube question without coordinates, cross sections and planes

Question

We have a cube with corners $ABCDA_1B_1C_1D_1$. The points $A_1B_1C_1D_1$ lie above the square $ABCD$. We also have a plane that goes through the middle of line $BC$, through the middle of square $ABA_1B_1$, through the middle of $A_1B_1C_1D_1$. At what ratio does the plane cut line $AB$.

What I did, I've put the cube into the coordinate system $B(0,0,1), A_1 (0,1,0) \text{ and } A(0,0,0)$, $D(1,0,0)$ and then it was pretty easy to solve. I just got the equation of the plane and then solved for intersection of the line that goes through AB, to get the point. Then I knew the ratio.

My question is, how to solve this problem without putting the cube into the coordinate system?

SOLUTION: solution should be that the plane cuts AB in ratio 3:1

Presumably you also don't want to use vectors (despite saying vector BC) — user10354138, Feb 08 '21 at 13:43
If the middle of square $ABA_1B_1$ means its center then you have only two points a point on $BC$ and the center of square $ABA_1B_1$. That may not lead to a unique solution. — Math Lover, Feb 08 '21 at 13:45
@MathLover There are three points. The middle of $ABA_1B_1$ and middle of $A_1B_1C_1D_1$ besides middle of BC. — VLC, Feb 08 '21 at 13:47
@DavidK The solution should be that the plane cuts AB in ratio 3:1 — VLC, Feb 08 '21 at 14:22
Please show your full solution -- coordinates of points, equation of plane. I thought I had a clear picture in my head but it does not match your result. — David K, Feb 08 '21 at 14:25
I found my point of confusion: "through the middle of $A_1B_1C_1D_1$" means through the middle of a face. I misread the name. — David K, Feb 08 '21 at 19:28

score 3 · Accepted Answer · answered Feb 08 '21 at 14:31

3

Fairly quickly:

(1) the plane (call it $\Pi$) also passes through the centre $M_{ABC_2D_2}$ of square $ABC_2D_2$ ($C_2,D_2$ are the reflection of $C,D$ in line $AB$). This is just extending $M_{A_1B_1C_1D_1}M_{ABB_1A_1}$ to meet the bottom plane $ABCD$.

(2) So the line $M_{BC}M_{ABC_2D_2}$ lies in $\Pi$. This line intersects AD at $X$ where $AX=\frac32 AD$ and $X$ is opposite to $D$.

Now look at the configuration $A,B,X,M_{BC}$ and you have similar (right-angled) triangles $\triangle AXY$ and $\triangle YBM_{BC}$ where $Y$ is the intersection of diagonals, i.e., intersection of $AB$ with $\Pi$. So $AY:YB=AX:BM_{BC}=3:1$.

answered Feb 08 '21 at 14:31

user10354138

33,239

You could also look at the quadrilateral $M_{BC}BM_{ABC_2D_2}M_{AB}$ where $M_{AB}$ is the midpoint of $AB.$ This is a parallelogram, one diagonal lies on $AB$ and the other lies in the plane $\Pi.$ Their intersection $X$ is therefore the midpoint of each diagonal, and $BX$ is half the distance $BM_{AB}$ which is half of $AB$. – David K Feb 08 '21 at 19:39
@DavidK I thought about writing that way but decided against it. It is either do the extending to my $X$ and read off the ratio, or don't extend and do the mid-point of half segment. – user10354138 Feb 09 '21 at 04:19
Your way is fine too. And your way gets the $3:1$ ratio directly rather than via $2:1:1$. – David K Feb 09 '21 at 04:20

score -1 · Answer 2 · answered Feb 08 '21 at 17:37

In response to the question how I solved with coordinate system:

Let's say:

$$AA_1 = (0,0,1) = c$$ $$AB = (0,1,0) = a$$ $$AD = (1,0,0) = b$$

Now we get all three points as mentioned above:

$$P_1 = c + (1/2)a + (1/2)b = (1/2,1/2,1/2)$$ $$P_2 = (1/2)c + (1/2)b = (0,(1/2),(1/2))$$ $$P_3 = a + (1/2)b = ((1/2),1,0)$$ Now we get two vectors $P_2P_3$ and $P_2P_1$. When we do cross product we get: $$n = (1,-2,-1)$$

The equation of the plane is thus: $$x - 2y - z = \frac{-3}{2}$$ Because: $\text{We use the second point and get:}0*1 + (1/2)*(-2)+(1/2)(-1) = \frac{-3}{2}$

Now we know that $AB$ is on $y$ axis , or the second coordinate as we may call it is a parameter $t$. So the equation of the line on which $AB$ is on is $y = t, x =0, z=0$, where $t$ is the parameter.

Now we input $y =t$ into the plane equation and get $$(-2t = \frac{-3}{2}\iff t =\frac{3}{4} )\implies \text{ratio is: } 1:3$$

$P_1$ should be $(1/2,1/2,1)$ rather than $(1/2,1/2,1/2).$ This didn't help when I tried to reproduce the rest of the answer. — David K, Feb 08 '21 at 19:31

Cube question without coordinates, cross sections and planes

2 Answers2