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The question

Let $ABCDA'B'C'D'$ be a cube, and the points $O$ and $O'$ are the centres of the faces $ABCD$ and $A'B'C'D'$ respectively. We consider the points $P\in (AD')$ and $Q \in (B'C)$ and denote by $M$ the midpoint of the segment $PQ$. Prove that $PQ \parallel (ABC)$ if and only if $M \in (OO')$.

My drawing

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My idea

So we have only $2$ cases:

case $1$) We know that $PQ \parallel (ABC)$, we show that $M \in (OO')$.

case $2$) We know that $M \in (OO')$, we show that $PQ \parallel (ABC)$.

Even though I wasn't able to do any of these, I think that for the first case we have to show that any of $OM$ or $O'M$ are parallel to one of the heights of the cube, because $OO'$ is already parallel to them and this will mean by Euclid's theorem that $O, O', M$ are colinear and that $M\in OO'$. I think of this parallelism because $M$ and $O$ or $O'$ are already mid-points of $PQ$,$AC$ or $A'C'$.

A similar rationament would be for case $2$.

I tried showing this but it got me to nowhere.

I hope one of you can help me! Thank you!

IONELA BUCIU
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3 Answers3

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I think these kind of problems are better dealt with coordinates and vectors.

If you call $A = 0$ and $B = e_1, A' =e_2, D= e_3$, then you can complete the coordinates of the other points.

In particular, $D' = e_2 + e_3$, $B' = e_1 + e_2$ and $C = e_2 + e_3$. Thus you can write, for some $0\le t, s \le 1$ that \begin{equation*} P = t(e_2 + e_3),\; Q = e_1 + se_2 + (1 - s)e_3. \end{equation*} Thus we have \begin{align*} Q - P &= e_1 + (s - t)e_2 + (1 - s - t)e_3\\ M - O &= \dfrac{1}{2}((t + s)e_2 + (t - s)e_3) \end{align*}

With this you can rewrite both conditions: the first one, that is $Q-P$ has no $e_2$ component, becomes $s = t$. The second one, that is, $M\in OO'$, which is to say the vector $M - O$ only has $e_2$ component, also becomes $t = s$.

Thus they are equivalent to $t = s$. which mean they are equivalent.

MEEL
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  • Thank you so much for your idea. But is there any way to transform this answer into one that doesn't use coordinates and vectors? I didn't learn about them yet at school. Thank you again! – IONELA BUCIU Dec 24 '23 at 15:37
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We may suppose that the length of a side of the cube is $1$.

Let $E$ be a point on $AD$ such that $PE\perp AD$.

Let $F$ be a point on $BC$ such that $QF\perp BC$.

Let $G$ be the midpoint of $AB$.

Let $H$ be a point on the plane $ABCD$ such that $MH\perp (ABCD)$.

Then, we can see that $M$ is on the plane $OO'G$ which is parallel to the plane $AA'D'D$.

Let $s:=BF$ and $t:=AE$.

Then, we can write $QF=1-s$ and $PE=t$.

  • If $PQ\parallel (ABC)$, then $QF=PE$, so we have $1-s=t$. This can be written as $\frac{s+t}{2}=\frac 12$, so $\frac{BF+AE}{2}=\frac 12$, i.e. $HG=\frac 12$ (since we already know that $M$ is on the plane $OO'G$). So, we get $H=O$, and $M\in (OO')$ follows.

  • If $M\in (OO')$, then we have $H=O$ and $HG=\frac 12$, so $\frac{BF+AE}{2}=\frac 12$, i.e. $\frac{s+t}{2}=\frac 12$. This can be written as $1-s=t$, so $QF=PE$. Therefore, $PQ\parallel (ABC)$ follows.

mathlove
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  • Thank you so much for your answer! But may I ask why M is on the plane OO′G? – IONELA BUCIU Jan 03 '24 at 16:32
  • @IONELA BUCIU : Let $I$ be a point on the plane $BCC'B'$ such that $PI\perp (BCC'B')$. If $Q=I$, then it is obvious that $M$ is on the plane $OO'G$. If $Q\not=I$, then let $J$ be the intersection point of $PI$ with the plane $OO'G$, and let $K$ be the intersection point of $PQ$ with the plane $OO'G$. We have $\angle{KJP}=\angle{QIP}=90^\circ$, so $\triangle{PJK}\sim\triangle{PIQ}$ with $PI=2PJ$. So, we see that $K$ is the midpoint of $PQ$, which means that $K=M$. Therefore, $M$ is on the plane $OO'G$. – mathlove Jan 04 '24 at 03:46
  • @IONELA BUCIU : $I$ is a point on the plane $BCC′B′$ such that $PI\perp (BCC′B′)$. So, the line $PI$ is parallel to the line $AB$. Since $J$ is the intersection point of $PI$ with the plane $OO′G$, we can say that $J$ is the midpoint of $PI$. – mathlove Jan 04 '24 at 13:05
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Let us make some constructions, that are useful for both directions.

We relax the condition on $ABCDA'B'C'D'$ to be a cube, it is enough to work with a "box", a cuboid, i.e. a rectangular parallelipiped. (All faces are rectangles.) To have an ad-hoc quick manner to work with the faces, we name them up ($A'B'C'D'$) and down ($ABCD$), left ($ADD'A'$) and right ($BCC'B'$).

Let us draw the segments $PP_1$ and $QQ_1$ parallel and equal to the segments $AB\|DC\|D'C'\|A'B'$, and thus also to the up and down planes, so that $P_1\in BC'$ and $Q_1\in DA'$. They are perpendicular on the left and right planes. So $PP_1Q_1Q$ is a rectangle.

mse question 4832628

We may need also $R=BC'\cap CB'$ and $S=AD'\cap DA'$, the centers of symmetry of the right and left faces. Let $T$ also be the mid point of $RS$, and $OO'$, it is the center of symmetry of the "box".

It turns out that from the whole box we only need the (right) prism $$ SADRBC\ . $$ Its parallel faces $\Delta SAD$ and $\Delta RBC$ are isosceles triangles, $SA=SD$ and $RB=RC$. The point $M$ is the mid point of the diagonals of the rectangle $PP_1QQ_1$, and their intersection. An other way to introduce it is as the mid point of the segment delimited by the mid points $K, L$ of $PQ_1$, and respectively $P_1Q$.

It is time to extract the simpler equivalent problem for the prism, and give the proof.

Let $SADRBC$ be a right prism with parallel triangular faces $\Delta SAD$, $\Delta RBC$, which are isosceles in $R,S$. Let $P,P_1,Q,Q_1$ be points on the lines $SA,RB,RC,SD$ so that the segments $PP_1,QQ_1$ are parallel and equal to the segments (sides of the prism) $SR, AB,DC$. Let $T$ be the mid point of $RS$. Let $O=AC\cap BD$ be the center of the rectangle $ABCD$. Let $K,L$ be the mid points of $PQ_1,P_1Q$. Let $M$ be the mid point of $KL$, it is also the intersection of the diagonals of the rectangle $PP_1QQ_1$, and its center of symmetry.

Then the $PQ$ (and/or plane of the rectangle $PP_1QQ_1$) is parallel to $ABCD$ if and only if $M$ is on $TO$.

mse question 4832628 prism

Let us start the proof. (The curios, ambitious reader may want to search for a proof on her, his own. The used construction of supplementary points below is not shown in this picture.)

$(1)$ Assume that $PQ$ is parallel to $(ABCD)$, i.e. $PP_1QQ_1\|ABCD$. The $PQ_1\|AD$, $P_1Q\|BC$, so the triangles $\Delta SPQ_1$, $\Delta RP_1Q$ are isosceles (being similar to the corresponding faces of the prism). So $SK, RL$ are not only medians in these triangles, but also heights. SO $SK,RL$ are perpendicular also on $AD,BC$, intersect them in their mid points, $K',L'$ say, and the segment $K'L'$ joining them passes through $O$, its mid point. So $T,M,O$ are collinear, being the mid points of the segments (parallel and equal segments) $SR,KL,K'L'$.

We conclude $M\in TO$.

$\square$

$(2)$ Assume that $M$ is on $TO$. The plane $SRLK$ determined by the parallel and equal segments $SR\|KL$ thus passes through $O$, so it intersects $ABCD$ in the line through $O$ parallel to $SR$, which is $K'L'$ with $K',L'$ the mid points of $AD,BC$. The points $S,K,K'$ are thus in the intersection of the planes $SAD$ and $SRKLK'L'$, so they are collinear. So $K$ is not only on the median from $S$ in $\Delta SPQ_1$, but also on its angle bisector ($AK'$ of $\widehat {ASD}=\widehat{PSQ_1}$). So $\Delta SPQ_1$ is isosceles, giving $PQ_1\|AD$, so the plane $PP_1QQ_1$ is parallel to $ABCD$.

$\square$

dan_fulea
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