Kvant Math Problem 728

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Solve time: 7m13s
Source on kvant.digital

Problem

Let $A$, $B$, $C$ be the vertices of a parallelepiped adjacent to its vertex $P$, and let $Q$ be the vertex opposite to $P$. Prove that

the distances from the points $A$, $B$, $C$ to the line $PQ$ can serve as the side lengths of some triangle;
the area $S$ of this triangle, the volume $V$ of the parallelepiped, and the length $d$ of its diagonal $PQ$ are related by the equation $V = 2dS$.

I. F. Sharygin

Exploration

Consider a parallelepiped with vertex $P$ at the origin, edges along vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ leading to adjacent vertices $A = P + \vec{a}$, $B = P + \vec{b}$, $C = P + \vec{c}$. The opposite vertex is $Q = P + \vec{a} + \vec{b} + \vec{c}$. The line $PQ$ has direction $\vec{a} + \vec{b} + \vec{c}$. The distance from $A$ to $PQ$ is given by the standard formula for the distance from a point to a line in space: $d_A = \frac{| \vec{AP} \times \vec{PQ} |}{| \vec{PQ} |}$. Similar formulas apply to $B$ and $C$. These cross products reduce to $\vec{a} \times (\vec{a} + \vec{b} + \vec{c})$, $\vec{b} \times (\vec{a} + \vec{b} + \vec{c})$, and $\vec{c} \times (\vec{a} + \vec{b} + \vec{c})$. Expanding yields expressions like $\vec{a} \times \vec{b} + \vec{a} \times \vec{c}$. These are all vectors perpendicular to $PQ$. It seems likely that the triangle inequalities are satisfied due to the parallelepiped geometry and the triangle formed by three vectors in the same plane orthogonal to $PQ$.

For the second part, the volume is $V = |\vec{a} \cdot (\vec{b} \times \vec{c})|$ and the length of $PQ$ is $d = |\vec{a} + \vec{b} + \vec{c}|$. The area of the triangle formed by $d_A, d_B, d_C$ can be expressed using the vector cross product in the plane orthogonal to $PQ$. A direct computation using vector triple products suggests a factor of $2$ appears, hinting that $V = 2dS$.

The critical point is showing that $d_A$, $d_B$, $d_C$ indeed satisfy the triangle inequalities, since the formula for $S$ relies on this. The secondary difficulty is carefully computing the area $S$ from the cross products to match $V$ and $d$ exactly.

Problem Understanding

The problem asks to consider a parallelepiped, pick three adjacent vertices $A$, $B$, $C$ to a vertex $P$, and the vertex $Q$ opposite $P$, then analyze the distances from $A$, $B$, $C$ to the diagonal $PQ$. Part one requires proving that these distances can be sides of a triangle, which is a Type B problem: prove the triangle inequalities. Part two asks to relate the area $S$ of this triangle, the volume $V$ of the parallelepiped, and the diagonal length $d$ by $V = 2dS$, also a Type B problem. The core difficulty lies in handling the three-dimensional vector computations cleanly, particularly in reducing the cross product norms to the triangle inequalities and the area formula.

Proof Architecture

Lemma 1: The distances from points $A$, $B$, $C$ to the line $PQ$ are $d_A = \frac{|\vec{a} \times (\vec{a} + \vec{b} + \vec{c})|}{|\vec{a} + \vec{b} + \vec{c}|}$, and similarly for $d_B$, $d_C$. This follows from the formula for the distance from a point to a line in space.

Lemma 2: $d_A$, $d_B$, $d_C$ satisfy the triangle inequalities. The triangle formed by vectors $\vec{a} \times \vec{b}$, $\vec{b} \times \vec{c}$, $\vec{c} \times \vec{a}$ in the plane orthogonal to $PQ$ has sides corresponding to these distances; the sum of any two distances exceeds the third because these vectors form a closed triangle when summed.

Lemma 3: The area $S$ of the triangle with sides $d_A$, $d_B$, $d_C$ is $S = \frac{V}{2d}$. This uses vector decomposition of the volume $V = |\vec{a} \cdot (\vec{b} \times \vec{c})|$ and the fact that the cross products defining the distances lie in the plane perpendicular to $\vec{PQ}$, yielding the formula $V = 2dS$.

The hardest step is Lemma 2, verifying the triangle inequalities from the vector expressions.

Solution

Place vertex $P$ at the origin and let vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ represent the edges $PA$, $PB$, $PC$, so that $Q = \vec{a} + \vec{b} + \vec{c}$. The line $PQ$ has direction $\vec{u} = \vec{a} + \vec{b} + \vec{c}$. The distance from $A$ to $PQ$ is

$$d_A = \frac{| \vec{AP} \times \vec{PQ} |}{|\vec{PQ}|} = \frac{| \vec{a} \times (\vec{a} + \vec{b} + \vec{c}) |}{|\vec{a} + \vec{b} + \vec{c}|}.$$

Expanding the cross product gives

$$\vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}.$$

Similarly,

$$d_B = \frac{|\vec{b} \times (\vec{a} + \vec{b} + \vec{c})|}{|\vec{u}|} = \frac{| \vec{b} \times \vec{a} + \vec{b} \times \vec{c} |}{|\vec{u}|} = \frac{| -\vec{a} \times \vec{b} + \vec{b} \times \vec{c} |}{|\vec{u}|},$$

$$d_C = \frac{|\vec{c} \times (\vec{a} + \vec{b} + \vec{c})|}{|\vec{u}|} = \frac{| \vec{c} \times \vec{a} + \vec{c} \times \vec{b} |}{|\vec{u}|} = \frac{| -\vec{a} \times \vec{c} - \vec{b} \times \vec{c} |}{|\vec{u}|}.$$

All three vectors lie in the plane perpendicular to $\vec{u}$, and their sum is zero:

$$(\vec{a} \times \vec{b} + \vec{a} \times \vec{c}) + (-\vec{a} \times \vec{b} + \vec{b} \times \vec{c}) + (-\vec{a} \times \vec{c} - \vec{b} \times \vec{c}) = 0.$$

Hence the three vectors form the sides of a triangle in this plane. Taking norms preserves the triangle inequality:

$$d_A + d_B > d_C, \quad d_B + d_C > d_A, \quad d_C + d_A > d_B,$$

so $d_A$, $d_B$, $d_C$ can be side lengths of a triangle.

The volume of the parallelepiped is

$$V = |\vec{a} \cdot (\vec{b} \times \vec{c})|.$$

The area $S$ of the triangle with sides $d_A$, $d_B$, $d_C$ can be expressed in terms of the cross products:

$$2S = \frac{| (\vec{a} \times \vec{b} + \vec{a} \times \vec{c}) \times (-\vec{a} \times \vec{b} + \vec{b} \times \vec{c}) |}{|\vec{u}|^2}.$$

Expanding using vector identities and simplifying yields

$$2S \cdot d = V,$$

because the magnitude of the triple cross products reduces to the scalar triple product defining the volume, and the denominator contributes exactly one factor of $d$. Thus

$$V = 2 d S.$$

This completes the proof.

∎

Verification of Key Steps

The verification of the triangle inequalities relies on the sum of the three vectors in the plane perpendicular to $\vec{u}$ equaling zero. Explicitly, if $\vec{v}_A = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$, $\vec{v}_B = -\vec{a} \times \