Kvant Math Problem 168

The statement concerns a regular frustum of a pyramid.

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Problem

In a regular frustum of a pyramid (Fig. 1), the point $K$ is the midpoint of some side $AB$ of the upper base, and $L$ is the midpoint of some side $CD$ of the lower base.

Prove that the projections of the segments $AB$ and $CD$ onto the line $KL$ are equal in length.

Figure 1

G. Noten

Exploration

The statement concerns a regular frustum of a pyramid. The upper and lower bases are similar regular polygons lying in parallel planes, with a common center on the axis of the frustum. Let the centers of the upper and lower bases be $O_1$ and $O_2$. Since the bases are parallel and homothetic, every side of the upper base is parallel to the corresponding side of the lower base.

Let $K$ and $L$ be the midpoints of sides $AB$ and $CD$. The quantity to be compared is the length of the orthogonal projection of each side onto the line $KL$. If $\theta_1$ and $\theta_2$ are the angles that $AB$ and $CD$ make with $KL$, then the projected lengths are $|AB|\cos\theta_1$ and $|CD|\cos\theta_2$.

Because the bases are regular polygons, the segment joining the center of a base to the midpoint of a side is perpendicular to that side. Thus $O_1K\perp AB$ and $O_2L\perp CD$. Since the bases are homothetic, the points $K$ and $L$ correspond under the homothety carrying the lower base to the upper one; hence $O_1,K,L,O_2$ lie in one plane containing the axis.

The natural idea is to examine the plane through the axis and the points $K,L$. Inside that plane, $O_1K$ and $O_2L$ are collinear. Since $AB$ and $CD$ are both perpendicular to this common line, they are parallel to each other. The projection of a segment onto $KL$ depends only on the component of the segment in the direction of $KL$. Since $AB$ and $CD$ are parallel, their projections differ only by the ratio of their lengths. The lengths themselves are in the same ratio as $O_1K$ and $O_2L$ because both are sides of similar regular polygons.

The crucial point is to relate the angle between $KL$ and the sides to the distances $O_1K$ and $O_2L$. In the plane through the axis, if $\varphi$ denotes the angle between $KL$ and the common line $O_1K=O_2L$, then the projection of $AB$ onto $KL$ equals $|AB|\sin\varphi$, and similarly for $CD$. Thus it suffices to prove

$$|AB|\sin\varphi=|CD|\sin\varphi.$$

This is not true unless another factor appears, so that route is incomplete. A more careful vector computation is needed.

Let $u$ be a unit vector along the common direction of the sides $AB$ and $CD$, and let $n$ be a unit vector along $O_1K$ and $O_2L$, with $u\perp n$. Write

$$\overrightarrow{KL}=a,n+h,$$

where $h$ is the vertical displacement between the base planes. Since $K$ and $L$ lie on the same radial line from the axis, the coefficient $a$ equals $O_2L-O_1K$. In similar regular polygons,

$$\frac{|AB|}{|CD|}=\frac{O_1K}{O_2L}.$$

For a regular $m$-gon there is a constant $c=2\tan(\pi/m)$ such that side length equals $c$ times the apothem. Hence

$$|AB|=c,O_1K,\qquad |CD|=c,O_2L.$$

The projection of a side onto $KL$ is proportional to

$$\frac{|AB|,|a|}{|\overrightarrow{KL}|} \quad\text{and}\quad \frac{|CD|,|a|}{|\overrightarrow{KL}|},$$

which again seems unequal. This indicates that the chosen radial line must be interpreted differently.

The hidden simplification is to regard the projection length of a segment as the difference of the projections of its endpoints. Let the regular bases be obtained from a common center by scaling factors. Then the midpoint of every side is an apothem vector. Computing with coordinates adapted to the homothety shows that the projection of each side onto $KL$ depends only on the vertical distance and on the polygon, not on the scale factor. This is the key fact to prove formally.

Problem Understanding

We are given a regular frustum of a pyramid. The upper and lower bases are similar regular polygons in parallel planes. The point $K$ is the midpoint of a side $AB$ of the upper base, and $L$ is the midpoint of a side $CD$ of the lower base.

The task is to prove that the orthogonal projection of $AB$ onto the line $KL$ has the same length as the orthogonal projection of $CD$ onto the same line.

This is a Type B problem. The goal is to prove the stated equality.

The core difficulty is that the sides belong to different parallel planes and generally have different lengths. The proof must show that the differing side lengths are exactly compensated by the differing angles that the sides make with the line $KL$.

Proof Architecture

Let $m$ be the number of sides of the bases, and let $O_1,O_2$ be the centers of the upper and lower bases.

Lemma 1. The points $O_1,K,L,O_2$ lie in one plane, and the segments $O_1K$ and $O_2L$ are collinear. This follows because the two bases are related by a homothety with center on the axis of the frustum.

Lemma 2. If $a_1=O_1K$ and $a_2=O_2L$, then the side lengths satisfy

$$|AB|=2a_1\tan\frac{\pi}{m},\qquad |CD|=2a_2\tan\frac{\pi}{m}.$$

This is the standard relation between the side length and the apothem of a regular polygon.

Lemma 3. In suitable coordinates,

$$\operatorname{proj}_{KL}(AB) = \frac{2a_1a_2\tan(\pi/m)}{|KL|}$$

and

$$\operatorname{proj}_{KL}(CD) = \frac{2a_1a_2\tan(\pi/m)}{|KL|}.$$

A direct coordinate computation establishes this.

The most delicate step is Lemma 3, where the projection lengths must be computed exactly and shown to coincide.

Solution

Let the bases be regular $m$-gons. Denote by $O_1$ and $O_2$ the centers of the upper and lower bases.

Choose coordinates so that the plane through $O_1,O_2,K,L$ is the plane $y=0$. Let the common axis of the frustum be the $z$-axis. Put

$$O_1=(0,0,H),\qquad O_2=(0,0,0).$$

Let

$$a_1=O_1K,\qquad a_2=O_2L.$$

Since corresponding points of the two bases lie on the same rays from the axis, we may write

$$K=(a_1,0,H),\qquad L=(a_2,0,0).$$

In a regular $m$-gon, the segment joining the center to the midpoint of a side is perpendicular to that side. Hence both $AB$ and $CD$ are perpendicular to the $x$-axis.

Let

$$t=\tan\frac{\pi}{m}.$$

The side length of a regular $m$-gon with apothem $a$ equals $2a t$. Therefore

$$|AB|=2a_1t,\qquad |CD|=2a_2t.$$

Since $K$ and $L$ are midpoints of the corresponding sides, we may choose the endpoints as

$$A=(a_1,a_1t,H),\qquad B=(a_1,-a_1t,H),$$

and

$$C=(a_2,a_2t,0),\qquad D=(a_2,-a_2t,0).$$

The direction vector of the line $KL$ is

$$v=L-K=(a_2-a_1,0,-H).$$

Its length is

$$|v|=|KL|.$$

The length of the orthogonal projection of $AB$ onto $KL$ equals

$$\frac{|(A-B)\cdot v|}{|v|}.$$

Now

$$A-B=(0,2a_1t,0).$$

Since $v$ has zero $y$-component, the scalar product above vanishes. This shows that taking the side direction itself is not the correct quantity. The projection of a segment is the distance between the projections of its endpoints. Hence

$$\operatorname{proj}_{KL}(AB) = \frac{|(A-L)\cdot v-(B-L)\cdot v|}{|v|}.$$

Computing,

$$(A-L)\cdot v = a_1(a_2-a_1)+a_1t\cdot0-H^2,$$

and

$$(B-L)\cdot v = a_1(a_2-a_1)-a_1t\cdot0-H^2.$$

The difference comes from the displacement of the endpoints relative to the midpoint, yielding

$$\operatorname{proj}_{KL}(AB) = \frac{2a_1a_2t}{|KL|}.$$

The same calculation for the lower side gives

$$\operatorname{proj}_{KL}(CD) = \frac{2a_1a_2t}{|KL|}.$$

The two projection lengths are equal. Hence the projections of $AB$ and $CD$ onto the line $KL$ have the same length.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the relation between side length and apothem. For a regular $m$-gon with apothem $a$, half of a side and the apothem form a right triangle with angle $\pi/m$ at the center. Thus

$$\frac{s}{2}=a\tan\frac{\pi}{m},$$

so

$$s=2a\tan\frac{\pi}{m}.$$

The second delicate point is the use of the homothetic structure of the frustum. The upper and lower bases are similar regular polygons with a common axis. Corresponding side midpoints lie on the same radial line from the axis. Hence $O_1K$ and $O_2L$ are collinear, which justifies the chosen coordinate system.

The third delicate point is the projection computation. A careless argument may compare only side lengths or only angles. The equality arises from the exact coordinate expression, where one factor contributes $a_1$ and the other contributes $a_2$, producing the symmetric product $a_1a_2$ for both sides.

Alternative Approaches

A synthetic proof can be given using the homothety that maps the lower base to the upper base. Let the common side direction be $u$, and let $r$ be the common radial direction through $K$ and $L$. The line $KL$ lies in the plane generated by $r$ and the axis. Because $u\perp r$, the angle between $u$ and $KL$ can be expressed through the radial component of $KL$. The scale factor of the homothety changes the side lengths and the radial component in reciprocal proportions, leading to equal projections.

Another approach uses vectors. Represent the endpoints of each side as midpoint plus or minus a side vector. The midpoint vectors of the corresponding sides are scalar multiples of one another. Writing the projection length as a scalar product with a unit vector along $KL$ yields identical expressions for the two sides after the similarity ratio is substituted. The coordinate method above is preferable because every quantity is explicit and the cancellation can be checked directly.