Kvant Math Problem 502

The three segments $AA_1$, $BB_1$, $CC_1$ are parallel and not coplanar.

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Verified: no
Verdicts: SKIP + SKIP
Solve time: 8m45s
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Problem

Given three parallel segments $AA_1$, $BB_1$, $CC_1$ that do not lie in a single plane. Let $M$ be the point of intersection of the planes $ABC_1$, $BCA_1$, and $CAB_1$, and let $M_1$ be the point of intersection of the planes $A_1B_1C$, $B_1C_1A$, and $C_1A_1B$. Prove that the segment $MM_1$ is parallel to the three original segments.

Z. Skopets

Exploration

The three segments $AA_1$, $BB_1$, $CC_1$ are parallel and not coplanar. This suggests regarding them as translates of a common vector. Let that vector be $v$. Then there exist points $A,B,C$ and a nonzero vector $v$ such that

$$A_1=A+v,\qquad B_1=B+v,\qquad C_1=C+v.$$

The planes appearing in the definition of $M$ are

$$ABC_1,\qquad BCA_1,\qquad CAB_1.$$

A point $X$ belongs to the plane $ABC_1$ exactly when its oriented distance from the plane $ABC$ is represented by the barycentric coefficient of $C$ multiplied by $v$. This suggests introducing affine coordinates with respect to the tetrahedron generated by $A,B,C$ and the direction $v$.

Choose coordinates so that

$$A=(0,0,0),\quad B=(1,0,0),\quad C=(0,1,0),\quad v=(0,0,1).$$

Then

$$A_1=(0,0,1),\quad B_1=(1,0,1),\quad C_1=(0,1,1).$$

Computing the three planes gives

$$ABC_1:\ z=y,$$

$$BCA_1:\ z=1-x-y,$$

$$CAB_1:\ z=x.$$

Their intersection is determined by

$$x=z,\qquad y=z,\qquad z=1-x-y,$$

hence

$$x=y=z=\frac13.$$

Thus

$$M=\left(\frac13,\frac13,\frac13\right).$$

Repeating the computation for the planes defining $M_1$ yields

$$A_1B_1C:\ z=1-y,$$

$$B_1C_1A:\ z=x+y,$$

$$C_1A_1B:\ z=1-x.$$

Their intersection satisfies

$$1-y=x+y,\qquad 1-x=x+y,$$

which gives

$$x=y=\frac13,\qquad z=\frac23.$$

Therefore

$$M_1=\left(\frac13,\frac13,\frac23\right),$$

and

$$M_1-M=(0,0,\tfrac13),$$

which is parallel to $v$.

The only potentially dangerous step is proving that the chosen coordinate model does not lose generality. Since affine transformations preserve parallelism, incidence, and intersections of planes, any configuration with three parallel segments can be sent to this model.

Problem Understanding

We are given three parallel segments $AA_1$, $BB_1$, $CC_1$ that are not contained in one plane. The point $M$ is the common intersection of the planes $ABC_1$, $BCA_1$, and $CAB_1$. The point $M_1$ is the common intersection of the planes $A_1B_1C$, $B_1C_1A$, and $C_1A_1B$.

The task is to prove that the line joining $M$ and $M_1$ is parallel to the common direction of the segments $AA_1$, $BB_1$, and $CC_1$.

This is a Type B problem, a pure proof problem.

The core difficulty is to describe the two triple intersections in a convenient coordinate system and then compare their positions.

Proof Architecture

Lemma 1. By an affine change of coordinates, the configuration may be placed in the form

$$A=(0,0,0),\quad B=(1,0,0),\quad C=(0,1,0),$$

and

$$A_1=(0,0,1),\quad B_1=(1,0,1),\quad C_1=(0,1,1).$$

This is true because the vectors $AB$, $AC$, and the common direction of the parallel segments are linearly independent.

Lemma 2. In these coordinates, the three planes defining $M$ have equations

$$z=y,\qquad z=1-x-y,\qquad z=x.$$

This follows from direct computation of plane equations through the corresponding triples of points.

Lemma 3. Their common intersection is

$$M=\left(\frac13,\frac13,\frac13\right).$$

This is obtained by solving the three linear equations.

Lemma 4. The three planes defining $M_1$ have equations

$$z=1-y,\qquad z=x+y,\qquad z=1-x.$$

Again this follows from direct computation.

Lemma 5. Their common intersection is

$$M_1=\left(\frac13,\frac13,\frac23\right).$$

This is obtained by solving the corresponding linear system.

The final step is to compute $M_1-M$ and show that it is parallel to the common direction of the original segments.

The most delicate point is Lemma 1, namely the justification that the chosen coordinates represent the general situation.

Solution

Since the segments $AA_1$, $BB_1$, $CC_1$ are parallel and do not lie in a single plane, the vectors

$$AB,\qquad AC,\qquad AA_1$$

are linearly independent.

Hence there exists an affine coordinate system in which

$$A=(0,0,0),\qquad B=(1,0,0),\qquad C=(0,1,0),$$

and the common direction of the three segments is the $z$-direction. After scaling that direction we may assume

$$AA_1=BB_1=CC_1=(0,0,1).$$

Therefore

$$A_1=(0,0,1),\qquad B_1=(1,0,1),\qquad C_1=(0,1,1).$$

Affine transformations preserve planes, intersections, and parallelism. Thus it suffices to prove the statement for this coordinate model.

We first determine $M$.

The plane $ABC_1$ passes through

$$A=(0,0,0),\quad B=(1,0,0),\quad C_1=(0,1,1).$$

Its equation is

$$z=y.$$

The plane $BCA_1$ passes through

$$B=(1,0,0),\quad C=(0,1,0),\quad A_1=(0,0,1).$$

Its equation is

$$x+y+z=1,$$

or

$$z=1-x-y.$$

The plane $CAB_1$ passes through

$$C=(0,1,0),\quad A=(0,0,0),\quad B_1=(1,0,1).$$

Its equation is

$$z=x.$$

The point $M$ satisfies

$$z=y,\qquad z=x,\qquad z=1-x-y.$$

Substituting $x=z$ and $y=z$ into the third equation gives

$$z=1-2z,$$

hence

$$3z=1,\qquad z=\frac13.$$

Therefore

$$x=y=\frac13,$$

and

$$M=\left(\frac13,\frac13,\frac13\right).$$

Now determine $M_1$.

The plane $A_1B_1C$ passes through

$$A_1=(0,0,1),\quad B_1=(1,0,1),\quad C=(0,1,0),$$

hence

$$z=1-y.$$

The plane $B_1C_1A$ passes through

$$B_1=(1,0,1),\quad C_1=(0,1,1),\quad A=(0,0,0),$$

hence

$$z=x+y.$$

The plane $C_1A_1B$ passes through

$$C_1=(0,1,1),\quad A_1=(0,0,1),\quad B=(1,0,0),$$

hence

$$z=1-x.$$

The point $M_1$ satisfies

$$1-y=x+y,$$

and

$$1-x=x+y.$$

Subtracting these equations yields

$$x=y.$$

Substituting into either equation gives

$$1=3x,$$

so

$$x=y=\frac13.$$

Then

$$z=x+y=\frac23.$$

Consequently

$$M_1=\left(\frac13,\frac13,\frac23\right).$$

Finally,

$$M_1-M = \left(0,0,\frac13\right).$$

This vector is parallel to

$$AA_1=(0,0,1),$$

and therefore also parallel to $BB_1$ and $CC_1$.

Hence the segment $MM_1$ is parallel to the three original segments.

This completes the proof.

∎

Verification of Key Steps

The reduction to coordinates uses the fact that $AB$, $AC$, and $AA_1$ are linearly independent. If they were dependent, the points $A,B,C,A_1$ would lie in one plane. Since $BB_1$ and $CC_1$ are parallel to $AA_1$, all six endpoints would then belong to that plane, contradicting the hypothesis. Thus the affine coordinate model is legitimate.

For the computation of $M$, the equation of the plane $BCA_1$ deserves checking independently. Writing

$$\alpha(1,0,0)+\beta(0,1,0)+\gamma(0,0,1),$$

with

$$\alpha+\beta+\gamma=1,$$

gives all points of the plane. Hence

$$x=\alpha,\quad y=\beta,\quad z=\gamma,$$

and therefore

$$x+y+z=1.$$

This confirms the equation used in the solution.

For the computation of $M_1$, substituting $x=y=\frac13$ into all three plane equations gives

$$z=1-y=\frac23,$$

$$z=x+y=\frac23,$$

$$z=1-x=\frac23.$$

Thus the obtained point indeed belongs to all three planes.

Alternative Approaches

A coordinate-free affine proof can be given with barycentric coordinates relative to the triangle $ABC$. If a point of space is written as

$$X=\alpha A+\beta B+\gamma C+t,v, \qquad \alpha+\beta+\gamma=1,$$

where $v=AA_1=BB_1=CC_1$, then the equations of the planes defining $M$ become

$$t=\alpha,\qquad t=\beta,\qquad t=\gamma.$$

Together with

$$\alpha+\beta+\gamma=1,$$

this yields

$$\alpha=\beta=\gamma=t=\frac13.$$

Similarly, the planes defining $M_1$ give

$$t=1-\alpha,\qquad t=1-\beta,\qquad t=1-\gamma,$$

hence

$$\alpha=\beta=\gamma=\frac13,\qquad t=\frac23.$$

The two points differ only in the parameter $t$, so their difference is a multiple of $v$. Thus $MM_1\parallel v$.

The coordinate solution is preferable because it reduces the entire problem to two small linear systems and makes the geometry completely transparent.