Kvant Math Problem 1104
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Problem
In the tetrahedron $ABCD$, the faces $ABC$ and $BCD$ are perpendicular, $\angle BAC=90^\circ$. Prove that from segments whose lengths are equal to the products of the lengths of opposite edges of the tetrahedron, one can form a right triangle.
V. N. Dubrovsky
Exploration
Let
$$x=AB\cdot CD,\qquad y=AC\cdot BD,\qquad z=AD\cdot BC.$$
The statement says that the three numbers $x,y,z$ are the side lengths of a right triangle. Thus one should prove that one of them squared equals the sum of the squares of the other two.
The geometric conditions concern the faces $ABC$ and $BCD$. Since these faces share the edge $BC$ and are perpendicular, a coordinate system adapted to these planes is natural.
Take $B=(0,0,0)$ and $C=(c,0,0)$. Put the face $ABC$ into the plane $z=0$. Since $\angle BAC=90^\circ$, the point $A$ lies on the sphere with diameter $BC$. Hence we may write
$$A=(u,v,0),$$
with
$$u(c-u)=v^2.$$
The face $BCD$ can be placed in the plane $y=0$, which is perpendicular to $z=0$ along the common line $BC$. Thus
$$D=(p,0,q).$$
The desired relation should emerge from direct computation of the six edge lengths.
Compute first
$$AB^2=u^2+v^2=uc, \qquad AC^2=(c-u)^2+v^2=c(c-u).$$
This is a pleasant simplification coming from the right angle at $A$.
Also,
$$BD^2=p^2+q^2, \qquad CD^2=(c-p)^2+q^2.$$
The crucial point is to determine which Pythagorean relation holds among $x,y,z$. Expanding,
$$z^2=(AD\cdot BC)^2=c^2AD^2.$$
Since
$$AD^2=(u-p)^2+v^2+q^2,$$
and $v^2=u(c-u)$,
$$\begin{aligned} z^2 &=c^2\bigl(u^2-2up+p^2+u(c-u)+q^2\bigr)\ &=c^2\bigl(uc-2up+p^2+q^2\bigr). \end{aligned}$$
On the other hand,
$$\begin{aligned} x^2+y^2 &=(AB\cdot CD)^2+(AC\cdot BD)^2\ &=uc\bigl((c-p)^2+q^2\bigr) +c(c-u)\bigl(p^2+q^2\bigr). \end{aligned}$$
After expansion all quadratic terms in $p$ cancel except $cp^2$, and the result becomes
$$c^2\bigl(uc-2up+p^2+q^2\bigr),$$
exactly $z^2$.
Thus the relation is
$$x^2+y^2=z^2.$$
The computation appears complete.
Problem Understanding
We are given a tetrahedron $ABCD$ such that the faces $ABC$ and $BCD$ are perpendicular and $\angle BAC=90^\circ$. We must prove that the three numbers
$$AB\cdot CD,\qquad AC\cdot BD,\qquad AD\cdot BC$$
can be taken as the side lengths of a right triangle.
This is a Type B problem. The task is to prove a stated property.
The core difficulty is to translate the two geometric orthogonality conditions into algebraic relations among the edge lengths and then derive a Pythagorean identity for the three products of opposite edges.
Proof Architecture
First place the tetrahedron in coordinates so that $BC$ is the $x$-axis, the face $ABC$ lies in the plane $z=0$, and the face $BCD$ lies in the plane $y=0$; this encodes the perpendicularity of the two faces.
Next express the condition $\angle BAC=90^\circ$ as the equation $u(c-u)=v^2$ for suitable coordinates $A=(u,v,0)$.
Then compute the squared lengths of all six edges and derive the simplified formulas
$$AB^2=uc,\qquad AC^2=c(c-u).$$
After that expand
$$(AB\cdot CD)^2+(AC\cdot BD)^2$$
and simplify.
Finally compute
$$(AD\cdot BC)^2$$
and show that the two expressions are equal.
The most delicate step is the algebraic simplification proving
$$(AB\cdot CD)^2+(AC\cdot BD)^2=(AD\cdot BC)^2.$$
Solution
Choose a coordinate system such that
$$B=(0,0,0),\qquad C=(c,0,0).$$
Since the faces $ABC$ and $BCD$ are perpendicular and intersect along the edge $BC$, we may place the face $ABC$ in the plane $z=0$ and the face $BCD$ in the plane $y=0$.
Write
$$A=(u,v,0),\qquad D=(p,0,q).$$
The condition $\angle BAC=90^\circ$ means that the vectors
$$B-A=(-u,-v,0),\qquad C-A=(c-u,-v,0)$$
are perpendicular. Hence
$$(-u)(c-u)+v^2=0,$$
or
$$v^2=u(c-u). \tag{1}$$
Now compute the relevant edge lengths.
From (1),
$$\begin{aligned} AB^2 &=u^2+v^2 \ &=u^2+u(c-u) \ &=uc, \end{aligned} \tag{2}$$
and similarly
$$\begin{aligned} AC^2 &=(c-u)^2+v^2 \ &=(c-u)^2+u(c-u) \ &=c(c-u). \end{aligned} \tag{3}$$
Also,
$$BD^2=p^2+q^2, \tag{4}$$
and
$$CD^2=(c-p)^2+q^2. \tag{5}$$
Using (2), (3), (4), and (5),
$$\begin{aligned} &(AB\cdot CD)^2+(AC\cdot BD)^2 \ &=AB^2,CD^2+AC^2,BD^2 \ &=uc\bigl((c-p)^2+q^2\bigr) +c(c-u)\bigl(p^2+q^2\bigr). \end{aligned}$$
Expanding,
$$\begin{aligned} &uc(c^2-2cp+p^2+q^2) +c(c-u)(p^2+q^2) \ &=uc^3-2uc^2p+ucp^2+ucq^2 +cp^2(c-u)+cq^2(c-u). \end{aligned}$$
The terms containing $p^2$ and $q^2$ combine to give
$$cp^2\bigl(u+c-u\bigr)=c^2p^2,$$
and
$$cq^2\bigl(u+c-u\bigr)=c^2q^2.$$
Therefore
$$\begin{aligned} &(AB\cdot CD)^2+(AC\cdot BD)^2 \ &=uc^3-2uc^2p+c^2p^2+c^2q^2 \ &=c^2\bigl(uc-2up+p^2+q^2\bigr). \tag{6} \end{aligned}$$
Next,
$$\begin{aligned} AD^2 &=(u-p)^2+v^2+q^2 \ &=u^2-2up+p^2+u(c-u)+q^2 \ &=uc-2up+p^2+q^2, \end{aligned}$$
again by (1). Since $BC=c$,
$$\begin{aligned} (AD\cdot BC)^2 &=c^2AD^2 \ &=c^2\bigl(uc-2up+p^2+q^2\bigr). \end{aligned} \tag{7}$$
Comparing (6) and (7), we obtain
$$(AB\cdot CD)^2+(AC\cdot BD)^2=(AD\cdot BC)^2.$$
Hence the three numbers
$$AB\cdot CD,\qquad AC\cdot BD,\qquad AD\cdot BC$$
satisfy the Pythagorean theorem, so they are the side lengths of a right triangle.
This completes the proof.
∎
Verification of Key Steps
The first delicate point is the derivation of (2) and (3). From $\angle BAC=90^\circ$ we have
$$(B-A)\cdot(C-A)=0,$$
which yields $v^2=u(c-u)$. Substituting this into
$$AB^2=u^2+v^2$$
gives
$$AB^2=u^2+u(c-u)=uc.$$
The same substitution gives
$$AC^2=(c-u)^2+u(c-u)=c(c-u).$$
No other geometric assumption is used here.
The second delicate point is the simplification of
$$uc\bigl((c-p)^2+q^2\bigr) +c(c-u)(p^2+q^2).$$
After expansion, the coefficient of $p^2$ is
$$uc+c(c-u)=c^2,$$
and the coefficient of $q^2$ is also
$$uc+c(c-u)=c^2.$$
Missing either of these cancellations would prevent the expression from factoring as $c^2(\cdots)$.
The third delicate point is the computation of $AD^2$. Starting from
$$AD^2=(u-p)^2+v^2+q^2,$$
one must substitute $v^2=u(c-u)$ before expanding. This gives
$$AD^2=uc-2up+p^2+q^2,$$
which matches exactly the factor obtained from the previous calculation. The proof hinges on this coincidence.
Alternative Approaches
A more invariant approach uses vectors. Let $BC$ be represented by a vector $c$. Because the planes $ABC$ and $BCD$ are perpendicular, one may decompose vectors relative to the orthogonal directions determined by these planes. The condition $\angle BAC=90^\circ$ implies that $A$ lies on the sphere with diameter $BC$, yielding identities equivalent to
$$AB^2+AC^2=BC^2.$$
Expressing $AD$ through orthogonal projections onto the two face planes and expanding scalar products leads directly to
$$(AD\cdot BC)^2=(AB\cdot CD)^2+(AC\cdot BD)^2.$$
The coordinate method is preferable because the perpendicularity of the faces becomes the simple statement that one face lies in $z=0$ and the other in $y=0$. The required identity then emerges from straightforward computations with no auxiliary geometric constructions.