Additional solutions and generalizations of an 8th grade Olympiad problem about a cube

Question

This year, I was the author of the following problem which came at the county level of the Math Olympiad at 8th grade:

Consider cube $ABCDA^\prime B^\prime C^\prime D^\prime$ with faces $\square ABCD$ and $\square A^\prime B^\prime C^\prime D^\prime$ connected by edges $[AA^\prime]$, $[BB^\prime]$, $[CC^\prime]$, $[DD^\prime]$. Let $M$ be the middle of the side $[AB]$, and $E$ be the symmetric of $D^\prime$ with respect to $M$. Prove that $DB^\prime\perp(A^\prime C^\prime E)$.

The interesting thing was that this was regarded as difficult, even by teachers I talked to about this problem. I will leave to you to find the solutions to this (there is a very simple one:

There is actually only one plane perpendicular on $DB^\prime$ which contains the line $A^\prime C^\prime$, i.e. $A^\prime C^\prime B$, so we only have to prove that this plane contains $E$, which can be done by looking at the triangle $D^\prime C^\prime E$.

another one would be:

This time, let's prove by seeingthe $DB^\prime\perp A^\prime C^\prime$. Also, we will prove that $DB^\prime\perp A^\prime E$. Let $F$ the symmetric of $C$ with respect to $D$. The idea is to calculate each of the sides of the triangle $A^\prime FE$ and prove that it is right angled, thus we will get the conclusion.

I'm really interested in how many new ones we can find).

Also, I am really curious on this particular configuration which can lead to many questions (as an example: if we denote by $O$ the center of the face $A^\prime B^\prime C^\prime D^\prime$, we can ask many questions about the trapezoid determined by the intersection of the plane $(OAE)$. Can you figure out some?). Even further, I think it would be very interesting to generalize the original problem in $4$ dimensions.

I really hope that this can lead to many problems for us to solve.

Answer 1

Perhaps I'm being pessimistic, but this problem seems to simply give free points to those who know basic vector operations.

Let $A (0,0,0), B (1,0,0), C (1,1,0), D (0,1,0)$ and $X'=X+(0,0,1).$

It is easy to verify that $DB'= (1,-1,1), M = (0.5,0,0), E (1,-1,-1)$

Let $V = A'E \times A'C' = (1,-1,-2) \times (1,1,0) = (2,-2,2)$ be a vector normal to $(A'C'E).$ We have $V = 2 \cdot DB',$ from which the result follows.

Answer 2

It is enough to prove $DB'\bot AC'$ and $DB'\bot AE$. Say $AB =a$

We will use this property:

A line $XY$ is perpendicular to a line $AB$ iff $$AX^2-AY^2 =BX^2-BY^2$$

$\bullet DB'\bot AC':$ Since $A'D = DC'$ and $A'B' = C'B'$ we have $$DC'^2-B'C'^2 = DA'^2-B'A'^2 $$ $\bullet DB'\bot AE:$ Since $AB' = a$, $A'D =a\sqrt{2}$, $EB' = a\sqrt{5}$ and $DE = a\sqrt{6}$ we have $$DE^2-B'E^2 = 6a^2- 5a^2 = 2a^2-a^2 =DA'^2-B'A'^2 $$

and we are done.