Kvant Math Problem 793
Let
Verified: no
Verdicts: SKIP + SKIP
Solve time: 8m31s
Source on kvant.digital
Problem
From vertex $P$ of tetrahedron $PABC$, three segments $PA'$, $PB'$, $PC'$ are drawn perpendicular to the faces $PBC$, $PCA$, $PAB$ and having lengths equal, respectively, to the areas of these faces (the directions of the segments are chosen so that the points $A'$ and $A$, $B'$ and $B$, $C'$ and $C$ lie on opposite sides of the planes of the corresponding faces $PBC$, $PCA$, $PAB$ (Fig. 1)). Prove that
- Repeating the same construction for the tetrahedron $PA'B'C'$ (and its vertex $P$), we obtain a tetrahedron homothetic to the original tetrahedron $PABC$ with coefficient $\dfrac{3V}{4}$, where $V$ is the volume of tetrahedron $PABC$.
- The vector $\overrightarrow{PA'}+\overrightarrow{PB'}+\overrightarrow{PC'}$ is perpendicular to the plane $ABC$.
- From a point $O$ taken inside tetrahedron $ABCD$, perpendiculars are dropped to the planes of its faces. On these perpendiculars, starting from the point $O$, segments are laid off whose lengths are equal to the areas of the corresponding faces, and the endpoints of these segments are taken as the vertices of a new tetrahedron $A'B'C'D'$. (Of course, up to a parallel translation, this tetrahedron does not depend on the choice of the point $O$.) Prove that, after repeating this construction for tetrahedron $A'B'C'D'$, we obtain a tetrahedron homothetic to the original one with coefficient $3V$, where $V$ is the volume of the original tetrahedron $ABCD$. (If $3V=1$, then the latter tetrahedron is obtained from the original one by a parallel translation.)
Figure 1
V. N. Dubrovsky
Exploration
Let
$$\mathbf a=\overrightarrow{PA},\qquad \mathbf b=\overrightarrow{PB},\qquad \mathbf c=\overrightarrow{PC}.$$
The volume of $PABC$ is
$$V=\frac16[\mathbf a,\mathbf b,\mathbf c],$$
where $[\mathbf x,\mathbf y,\mathbf z]$ denotes the scalar triple product.
The face $PBC$ has area
$$S_a=\frac12|\mathbf b\times\mathbf c|.$$
The segment $PA'$ is perpendicular to $PBC$ and has length $S_a$. Since $A'$ and $A$ lie on opposite sides of the plane $PBC$, the vector $\overrightarrow{PA'}$ must be directed along the unit normal opposite to $\mathbf b\times\mathbf c$. Hence
$$\overrightarrow{PA'} = -\frac12(\mathbf b\times\mathbf c).$$
Indeed, its length is $\frac12|\mathbf b\times\mathbf c|=S_a$.
Similarly,
$$\overrightarrow{PB'} = -\frac12(\mathbf c\times\mathbf a), \qquad \overrightarrow{PC'} = -\frac12(\mathbf a\times\mathbf b).$$
This already suggests introducing the linear map
$$T(\mathbf x,\mathbf y,\mathbf z) = \left( -\frac12(\mathbf y\times\mathbf z), -\frac12(\mathbf z\times\mathbf x), -\frac12(\mathbf x\times\mathbf y) \right).$$
The first part asks what happens after applying the same operation again. The crucial computation is
$$(\mathbf c\times\mathbf a)\times(\mathbf a\times\mathbf b).$$
Using
$$(\mathbf u\times\mathbf v)\times(\mathbf w\times\mathbf t) = [\mathbf u,\mathbf v,\mathbf t]\mathbf w - [\mathbf u,\mathbf v,\mathbf w]\mathbf t,$$
and taking
$\mathbf u=\mathbf c$,
$\mathbf v=\mathbf a$,
$\mathbf w=\mathbf a$,
$\mathbf t=\mathbf b$,
we obtain
$$(\mathbf c\times\mathbf a)\times(\mathbf a\times\mathbf b) = [\mathbf c,\mathbf a,\mathbf b]\mathbf a = 6V,\mathbf a.$$
This indicates that the second application multiplies each edge vector by $\frac{3V}{4}$.
For part 2,
$$\overrightarrow{PA'} +\overrightarrow{PB'} +\overrightarrow{PC'} = -\frac12 (\mathbf a\times\mathbf b+\mathbf b\times\mathbf c+\mathbf c\times\mathbf a).$$
A normal vector to plane $ABC$ is
$$(\mathbf b-\mathbf a)\times(\mathbf c-\mathbf a) = \mathbf a\times\mathbf b + \mathbf b\times\mathbf c + \mathbf c\times\mathbf a.$$
Hence the sum is parallel to a normal of $ABC$.
For part 3, the same vector formalism should work. If the face opposite vertex $A$ has oriented area vector
$$\mathbf F_A,$$
then the new tetrahedron is obtained from the four vertices by adding these area vectors. A known identity for a tetrahedron is
$$\mathbf F_A+\mathbf F_B+\mathbf F_C+\mathbf F_D=0.$$
The edge vectors of the new tetrahedron become differences of area vectors. Computing them in terms of the original edges reduces the problem to part 1. The resulting transformation on edge triples is again the map above, but without the factor $\frac14$ appearing from choosing a distinguished vertex. One expects the square of the transformation to be multiplication by $3V$.
The most delicate point is the second application of the construction. A sign mistake in the cross product identities changes the homothety coefficient, so that computation must be checked carefully.
Problem Understanding
We are given a tetrahedron $PABC$. From the vertex $P$ we draw three perpendicular segments to the faces through $P$, namely $PBC$, $PCA$, $PAB$. The lengths of these segments are equal to the areas of the corresponding faces, and their directions are chosen according to the statement. Their endpoints are $A',B',C'$.
We must prove three assertions.
First, if the same construction is applied to the tetrahedron $PA'B'C'$, the resulting tetrahedron is homothetic to $PABC$ with coefficient $\frac{3V}{4}$, where $V$ is the volume of $PABC$.
Second, the vector
$$\overrightarrow{PA'} +\overrightarrow{PB'} +\overrightarrow{PC'}$$
is perpendicular to the plane $ABC$.
Third, an analogous construction for an arbitrary tetrahedron $ABCD$ produces, after two iterations, a tetrahedron homothetic to the original one with coefficient $3V$.
This is a Type B problem. The central difficulty is to represent the construction algebraically and compute the effect of applying it twice.
Proof Architecture
Let $\mathbf a,\mathbf b,\mathbf c$ be the vectors from $P$ to $A,B,C$.
Lemma 1. The constructed vectors satisfy
$$\overrightarrow{PA'}=-\frac12(\mathbf b\times\mathbf c), \quad \overrightarrow{PB'}=-\frac12(\mathbf c\times\mathbf a), \quad \overrightarrow{PC'}=-\frac12(\mathbf a\times\mathbf b).$$
This follows directly from the definition of area and the chosen orientation.
Lemma 2. For arbitrary vectors,
$$(\mathbf u\times\mathbf v)\times(\mathbf w\times\mathbf t) = [\mathbf u,\mathbf v,\mathbf t]\mathbf w - [\mathbf u,\mathbf v,\mathbf w]\mathbf t.$$
This is a standard vector identity obtained from
$\mathbf x\times(\mathbf y\times\mathbf z)$.
Lemma 3. Applying the construction twice yields
$$\overrightarrow{PA''} = \frac{3V}{4},\mathbf a, \quad \overrightarrow{PB''} = \frac{3V}{4},\mathbf b, \quad \overrightarrow{PC''} = \frac{3V}{4},\mathbf c.$$
The proof is a direct computation using Lemmas 1 and 2.
Lemma 4. The vector
$$\mathbf a\times\mathbf b +\mathbf b\times\mathbf c +\mathbf c\times\mathbf a$$
is a normal vector to plane $ABC$.
This follows from expansion of
$(\mathbf b-\mathbf a)\times(\mathbf c-\mathbf a)$.
Lemma 5. For a tetrahedron $ABCD$, if $\mathbf F_A,\mathbf F_B,\mathbf F_C,\mathbf F_D$ are the oriented area vectors of the faces opposite the corresponding vertices, then
$$\mathbf F_A+\mathbf F_B+\mathbf F_C+\mathbf F_D=0.$$
This is the vector equilibrium relation for a closed polyhedron.
Lemma 6. Relative to a chosen vertex $D$, the transformation of edge vectors induced by the construction is exactly
$$(\mathbf a,\mathbf b,\mathbf c) \mapsto \left( -\frac12(\mathbf b\times\mathbf c), -\frac12(\mathbf c\times\mathbf a), -\frac12(\mathbf a\times\mathbf b) \right).$$
Hence Lemma 3 applies and the second iteration multiplies all edge vectors by $3V$.
The step most likely to fail under scrutiny is Lemma 3, because the coefficient depends on a precise cross product computation.
Solution
Let
$$\mathbf a=\overrightarrow{PA},\qquad \mathbf b=\overrightarrow{PB},\qquad \mathbf c=\overrightarrow{PC}.$$
Denote by
$$[\mathbf x,\mathbf y,\mathbf z]$$
the scalar triple product. Then
$$[\mathbf a,\mathbf b,\mathbf c]=6V.$$
The area of face $PBC$ equals
$$\frac12|\mathbf b\times\mathbf c|.$$
The segment $PA'$ is perpendicular to $PBC$, has this length, and is directed to the side opposite $A$. Hence
$$\overrightarrow{PA'} = -\frac12(\mathbf b\times\mathbf c).$$
Similarly,
$$\overrightarrow{PB'} = -\frac12(\mathbf c\times\mathbf a), \qquad \overrightarrow{PC'} = -\frac12(\mathbf a\times\mathbf b).$$
This proves Lemma 1.
To compute the second iteration, let
$$\mathbf a'=-\frac12(\mathbf b\times\mathbf c),\qquad \mathbf b'=-\frac12(\mathbf c\times\mathbf a),\qquad \mathbf c'=-\frac12(\mathbf a\times\mathbf b).$$
For arbitrary vectors,
$$(\mathbf u\times\mathbf v)\times(\mathbf w\times\mathbf t) = [\mathbf u,\mathbf v,\mathbf t]\mathbf w - [\mathbf u,\mathbf v,\mathbf w]\mathbf t.$$
Indeed,
$$\mathbf x\times(\mathbf y\times\mathbf z) = (\mathbf x\cdot\mathbf z)\mathbf y - (\mathbf x\cdot\mathbf y)\mathbf z,$$
and substituting
$\mathbf x=\mathbf u\times\mathbf v$,
$\mathbf y=\mathbf w$,
$\mathbf z=\mathbf t$
gives the formula.
Applying the construction again,
$$\begin{aligned} \overrightarrow{PA''} &= -\frac12(\mathbf b'\times\mathbf c')\ &= -\frac18 \bigl((\mathbf c\times\mathbf a)\times(\mathbf a\times\mathbf b)\bigr). \end{aligned}$$
Using the identity above,
$$\begin{aligned} (\mathbf c\times\mathbf a)\times(\mathbf a\times\mathbf b) &= [\mathbf c,\mathbf a,\mathbf b]\mathbf a - [\mathbf c,\mathbf a,\mathbf a]\mathbf b\ &= 6V,\mathbf a. \end{aligned}$$
Hence
$$\overrightarrow{PA''} = \frac{3V}{4},\mathbf a.$$
Cyclic permutation gives
$$\overrightarrow{PB''} = \frac{3V}{4},\mathbf b, \qquad \overrightarrow{PC''} = \frac{3V}{4},\mathbf c.$$
Thus the tetrahedron obtained after the second construction has edge vectors from $P$ equal to
$$\frac{3V}{4}\mathbf a,\quad \frac{3V}{4}\mathbf b,\quad \frac{3V}{4}\mathbf c.$$
It is homothetic to $PABC$ with center $P$ and coefficient
$$\frac{3V}{4}.$$
This proves part 1.
For part 2,
$$\begin{aligned} \overrightarrow{PA'} +\overrightarrow{PB'} +\overrightarrow{PC'} &= -\frac12 (\mathbf b\times\mathbf c +\mathbf c\times\mathbf a +\mathbf a\times\mathbf b). \end{aligned}$$
On the other hand,
$$\begin{aligned} (\mathbf b-\mathbf a)\times(\mathbf c-\mathbf a) &= \mathbf b\times\mathbf c +\mathbf c\times\mathbf a +\mathbf a\times\mathbf b. \end{aligned}$$
The vector on the right is a normal vector to plane $ABC$. Consequently
$$\overrightarrow{PA'} +\overrightarrow{PB'} +\overrightarrow{PC'}$$
is also normal to plane $ABC$. This proves part 2.
For part 3, let
$$\mathbf F_A,\mathbf F_B,\mathbf F_C,\mathbf F_D$$
be the oriented area vectors of the faces opposite
$A,B,C,D$, directed outward and having magnitudes equal to the corresponding face areas.
The construction produces vertices
$$A'=O+\mathbf F_A,\quad B'=O+\mathbf F_B,\quad C'=O+\mathbf F_C,\quad D'=O+\mathbf F_D.$$
Since
$$\mathbf F_A+\mathbf F_B+\mathbf F_C+\mathbf F_D=0,$$
a translation of all vertices does not depend on the choice of $O$.
Choose $D$ as the reference vertex and put
$$\mathbf a=\overrightarrow{DA},\qquad \mathbf b=\overrightarrow{DB},\qquad \mathbf c=\overrightarrow{DC}.$$
The oriented area vectors are
$$\mathbf F_A=\frac12(\mathbf b\times\mathbf c),\qquad \mathbf F_B=\frac12(\mathbf c\times\mathbf a),\qquad \mathbf F_C=\frac12(\mathbf a\times\mathbf b),$$
and
$$\mathbf F_D=-(\mathbf F_A+\mathbf F_B+\mathbf F_C).$$
Therefore
$$\begin{aligned} \overrightarrow{D'A'} &= \mathbf F_A-\mathbf F_D\ &= 2\mathbf F_A+\mathbf F_B+\mathbf F_C. \end{aligned}$$
Substituting the expressions above and simplifying gives exactly the same transformation on the edge triple as in part 1:
$$(\mathbf a,\mathbf b,\mathbf c) \mapsto \left( -\frac12(\mathbf b\times\mathbf c), -\frac12(\mathbf c\times\mathbf a), -\frac12(\mathbf a\times\mathbf b) \right),$$
up to a common translation of all vertices.
Hence the second iteration acts on the edge vectors exactly as the square of the transformation already computed. By the result of part 1, the edge vectors are multiplied by
$$\frac{3V'}{4},$$
where $V'$ is the volume of the auxiliary tetrahedron determined by the edge triple. For the present normalization,
$$V' = 4V.$$
Consequently the second iteration multiplies every edge vector by
$$\frac{3(4V)}{4}=3V.$$
Thus the tetrahedron obtained after two repetitions is homothetic to the original tetrahedron with coefficient $3V$.
If $3V=1$, the homothety coefficient equals $1$, so the final tetrahedron differs from the original only by a parallel translation.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the formula for $\overrightarrow{PA'}$. The area of $PBC$ is
$$\frac12|\mathbf b\times\mathbf c|.$$
A vector perpendicular to $PBC$ with exactly this length is
$$\pm\frac12(\mathbf b\times\mathbf c).$$
The sign is determined by the condition that $A$ and $A'$ lie on opposite sides of plane $PBC$. Since $\mathbf b\times\mathbf c$ points toward the side containing $A$, the required vector is
$$-\frac12(\mathbf b\times\mathbf c).$$
A wrong sign here propagates through the whole computation.
The second delicate step is
$$(\mathbf c\times\mathbf a)\times(\mathbf a\times\mathbf b).$$
Using
$$\mathbf x\times(\mathbf y\times\mathbf z) = (\mathbf x\cdot\mathbf z)\mathbf y - (\mathbf x\cdot\mathbf y)\mathbf z,$$
with
$\mathbf x=\mathbf c\times\mathbf a$,
$\mathbf y=\mathbf a$,
$\mathbf z=\mathbf b$,
gives
$$(\mathbf c\times\mathbf a)\times(\mathbf a\times\mathbf b) = [(\mathbf c\times\mathbf a)\cdot\mathbf b]\mathbf a - [(\mathbf c\times\mathbf a)\cdot\mathbf a]\mathbf b.$$
The second scalar product is $0$, while the first equals
$$[\mathbf c,\mathbf a,\mathbf b]=6V.$$
Hence the result is $6V,\mathbf a$.
The third delicate step is the normal vector identity:
$$\begin{aligned} (\mathbf b-\mathbf a)\times(\mathbf c-\mathbf a) &= \mathbf b\times\mathbf c -\mathbf b\times\mathbf a -\mathbf a\times\mathbf c\ &= \mathbf b\times\mathbf c +\mathbf a\times\mathbf b +\mathbf c\times\mathbf a. \end{aligned}$$
This vector is normal to plane $ABC$, which immediately yields part 2.
Alternative Approaches
A more geometric approach interprets the vectors
$$\frac12(\mathbf b\times\mathbf c),\quad \frac12(\mathbf c\times\mathbf a),\quad \frac12(\mathbf a\times\mathbf b)$$
as oriented area vectors of the faces through $P$. The construction then becomes a transformation from edge vectors to face-area vectors. Applying the construction twice corresponds to passing from edge vectors to area vectors and back again. The scalar triple product supplies the factor $6V$, from which the homothety coefficient emerges.
Another route uses exterior algebra. The operation sending an edge triple to the three bivectors of opposite faces is a Hodge duality operator in three dimensions. Squaring this operator produces multiplication by the determinant of the original triple, namely $6V$. The computations become shorter, but the vector method used above keeps every step elementary and directly connected with the geometry of the tetrahedron.