Kvant Math Problem 588

For the planar analogue, take a triangle $ABC$ and a point $P$ inside it.

kvantmathematicsolympiad

Verified: yes
Verdicts: PASS + PASS
Solve time: 8m21s
Source on kvant.digital

Problem

  1. Through a point chosen inside an arbitrary tetrahedron, segments are drawn parallel to its edges, with their endpoints lying on the faces of the tetrahedron. Prove that the sum of all six ratios of the lengths of these segments to the lengths of the edges parallel to them is always equal to three.
  2. Formulate and prove an analogous statement for a triangle (in the plane).

E. A. Yasinovy

All-Russian Mathematical Olympiad for School Students (1979, Grade 9)

Exploration

For the planar analogue, take a triangle $ABC$ and a point $P$ inside it. Through $P$ draw the three lines parallel to the sides. Each such line cuts the triangle in a segment. The statement to be proved should be that the sum of the three ratios of these segment lengths to the corresponding side lengths equals $2$.

A convenient language is that of barycentric coordinates. If

$$P=\alpha A+\beta B+\gamma C,\qquad \alpha+\beta+\gamma=1,$$

with $\alpha,\beta,\gamma>0$, then $\alpha,\beta,\gamma$ are the normalized distances from $P$ to the sides. Along a line parallel to $AB$, the width of the triangle is proportional to the distance from the opposite vertex $C$. Hence the segment through $P$ parallel to $AB$ has length $(\alpha+\beta)|AB|=(1-\gamma)|AB|$. Cyclically, the three ratios are

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

whose sum is $2$.

The tetrahedral problem suggests the same pattern. Let the tetrahedron be $ABCD$ and

$$P=\alpha A+\beta B+\gamma C+\delta D, \qquad \alpha+\beta+\gamma+\delta=1.$$

Consider the segment through $P$ parallel to edge $AB$. Moving along a direction parallel to $AB$ changes only the coefficients of $A$ and $B$, leaving those of $C$ and $D$ fixed. The two endpoints occur when the coefficient of $A$ or of $B$ becomes $0$. The allowable parameter interval therefore has length $\alpha+\beta$. Hence the segment parallel to $AB$ has length $(\alpha+\beta)|AB|=(1-\gamma-\delta)|AB|$.

Repeating this for all six edges gives the six ratios

$$\alpha+\beta,\quad \alpha+\gamma,\quad \alpha+\delta,\quad \beta+\gamma,\quad \beta+\delta,\quad \gamma+\delta.$$

Their sum equals

$$3(\alpha+\beta+\gamma+\delta)=3.$$

The potentially dangerous step is the computation of the length of a section parallel to an edge. One must verify carefully that the endpoints indeed correspond to vanishing coefficients of the two vertices of that edge.

Problem Understanding

We are given a tetrahedron and an interior point $P$. Through $P$ we draw six segments, each parallel to one edge of the tetrahedron and bounded by the faces of the tetrahedron. For each edge we take the ratio of the length of the corresponding segment to the length of that edge. The first task is to prove that the sum of the six ratios is always equal to $3$.

The second task asks for the analogous statement for a triangle. Through an interior point of a triangle we draw the three segments parallel to the sides and bounded by the triangle. We must formulate and prove the corresponding constant-sum relation.

This is a Type B problem. The statement is already specified, and the task is to prove it.

The core difficulty is expressing the lengths of the parallel sections in a form that depends linearly on the position of the interior point.

Proof Architecture

Let $P=\alpha A+\beta B+\gamma C+\delta D$ be the barycentric representation of an interior point of a tetrahedron.

Lemma 1. A point of the line through $P$ parallel to $AB$ has barycentric coordinates obtained by adding a parameter $t$ to the coefficient of $A$ and subtracting the same parameter from the coefficient of $B$; this follows because the direction $AB$ corresponds to the vector $A-B$.

Lemma 2. The maximal segment of the tetrahedron on that line is obtained for $-\alpha\le t\le\beta$; this is because a point lies in the tetrahedron exactly when all barycentric coordinates are nonnegative.

Lemma 3. The length of the segment through $P$ parallel to $AB$ equals $(\alpha+\beta)|AB|$; the parameter interval has length $\alpha+\beta$, and the direction vector is $A-B$.

Lemma 4. For every edge, the corresponding ratio equals the sum of the barycentric coordinates of the two endpoints of that edge.

Summing the six ratios yields $3(\alpha+\beta+\gamma+\delta)=3$.

For the triangle, the same argument with three barycentric coordinates gives ratios $1-\alpha$, $1-\beta$, $1-\gamma$, whose sum is $2$.

The most delicate lemma is Lemma 2, because it identifies the exact endpoints of the section.

Solution

Let $ABCD$ be the tetrahedron and let

$$P=\alpha A+\beta B+\gamma C+\delta D, \qquad \alpha,\beta,\gamma,\delta>0, \qquad \alpha+\beta+\gamma+\delta=1.$$

Since $P$ is an interior point, all four coefficients are positive.

Consider first the segment through $P$ parallel to the edge $AB$.

Every point of the line through $P$ parallel to $AB$ can be written as

$$X(t)=P+t(A-B).$$

Substituting the barycentric expression for $P$ gives

$$X(t) =(\alpha+t)A+(\beta-t)B+\gamma C+\delta D.$$

A point belongs to the tetrahedron if and only if its barycentric coordinates with respect to $ABCD$ are nonnegative and have sum $1$. The sum remains equal to $1$, so $X(t)$ lies in the tetrahedron exactly when

$$\alpha+t\ge0, \qquad \beta-t\ge0.$$

Thus

$$-\alpha\le t\le\beta.$$

The endpoints of the section are obtained for $t=-\alpha$ and $t=\beta$. The parameter interval therefore has length

$$\beta-(-\alpha)=\alpha+\beta.$$

Since the direction vector is $A-B$, whose length is $|AB|$, the length of the section parallel to $AB$ equals

$$(\alpha+\beta)|AB|.$$

Hence the ratio of the section length to the length of the edge $AB$ is

$$\alpha+\beta.$$

Applying the same argument to the other five edges, we obtain the six ratios

$$\alpha+\beta,\quad \alpha+\gamma,\quad \alpha+\delta,\quad \beta+\gamma,\quad \beta+\delta,\quad \gamma+\delta.$$

Their sum is

$$\begin{aligned} &(\alpha+\beta)+(\alpha+\gamma)+(\alpha+\delta) \ &\qquad +(\beta+\gamma)+(\beta+\delta)+(\gamma+\delta). \end{aligned}$$

Each of the numbers $\alpha,\beta,\gamma,\delta$ appears exactly three times, so the sum equals

$$3(\alpha+\beta+\gamma+\delta)=3.$$

This proves the first statement.

Now let $ABC$ be a triangle and let

$$P=\alpha A+\beta B+\gamma C, \qquad \alpha,\beta,\gamma>0, \qquad \alpha+\beta+\gamma=1.$$

Consider the segment through $P$ parallel to $AB$. Points of the corresponding line have the form

$$X(t)=P+t(A-B) =(\alpha+t)A+(\beta-t)B+\gamma C.$$

The condition that $X(t)$ lie in the triangle is

$$\alpha+t\ge0, \qquad \beta-t\ge0,$$

hence

$$-\alpha\le t\le\beta.$$

Therefore the section length is

$$(\alpha+\beta)|AB|,$$

and the ratio of this length to $|AB|$ equals

$$\alpha+\beta=1-\gamma.$$

Similarly, the ratios corresponding to the sides $BC$ and $CA$ are

$$1-\alpha, \qquad 1-\beta.$$

Their sum is

$$(1-\alpha)+(1-\beta)+(1-\gamma) =3-(\alpha+\beta+\gamma) =2.$$

Thus the planar analogue is:

For any point inside a triangle, if through it one draws the three segments parallel to the sides and bounded by the triangle, then the sum of the three ratios of their lengths to the lengths of the corresponding sides is equal to $2$.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the description of the line parallel to $AB$. Writing

$$X(t)=P+t(A-B)$$

changes only the coefficients of $A$ and $B$. The coefficients of $C$ and $D$ remain fixed. This is exactly the geometric meaning of moving in a direction parallel to $AB$.

The second delicate point is the determination of the endpoints. The barycentric coordinates of $X(t)$ are

$$\alpha+t,\quad \beta-t,\quad \gamma,\quad \delta.$$

Since $\gamma$ and $\delta$ are positive constants, the only restrictions come from the first two coordinates. The section leaves the tetrahedron precisely when one of them becomes negative. Hence the boundary points occur at

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

No other boundary point can occur, because all four coordinates remain nonnegative throughout that interval.

The third delicate point is the conversion from parameter length to Euclidean length. The displacement between parameters $t_1$ and $t_2$ is

$$(t_2-t_1)(A-B),$$

whose norm equals

$$|t_2-t_1|,|AB|.$$

With $t_1=-\alpha$ and $t_2=\beta$, the section length is exactly

$$(\alpha+\beta)|AB|.$$

Alternative Approaches

An affine-geometric proof avoids explicit barycentric coordinates. Let the four faces opposite $A,B,C,D$ define normalized distances $a,b,c,d$ from the point $P$, scaled so that $a+b+c+d=1$. Along a line parallel to an edge, the distances to the two faces containing that edge remain constant, while the distances to the two opposite faces vary linearly. The length of the section is proportional to the sum of the two relevant normalized distances. Summing over all six edges again yields three times the total $a+b+c+d$.

For the triangle, one may use areas instead of barycentric coordinates. If the areas of triangles $PBC$, $PCA$, $PAB$ are $\alpha S$, $\beta S$, $\gamma S$, where $S=[ABC]$, then $\alpha+\beta+\gamma=1$. A section parallel to $AB$ cuts a triangle similar to $ABC$, and similarity gives the ratio $1-\gamma$. Summing the three analogous expressions produces $2$. The barycentric proof is preferable because the tetrahedral and planar cases become identical computations.