Kvant Math Problem 653

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Problem

Fig. 1

There is a ruler with two markings (Fig. 1). Using this ruler, one can draw arbitrary straight lines and lay off segments of a given length. Using it, construct:

any right angle;
a line perpendicular to a given line.

V. L. Gutenmacher

Correspondence Mathematical Olympiads

Exploration

The ruler has two fixed marks. Let the distance between them be $a$. Since the ruler can be placed through any two points, we can transfer a segment of length $a$ anywhere in the plane.

The problem is to obtain a right angle using only a straightedge together with the ability to mark off the fixed length $a$.

A natural idea is to construct an equilateral triangle. Since a segment of length $a$ can be laid off repeatedly, we can construct a triangle whose three sides are all $a$. In an equilateral triangle, the median from a vertex is also an altitude. Thus, if we can construct the midpoint of a segment, a perpendicular follows immediately.

The crucial point is therefore the construction of the midpoint of a segment using only the given ruler. Since the ruler permits laying off the same length repeatedly, we can construct rhombi and parallelograms.

Take a segment $AB$ of length $2a$. Construct equilateral triangles on $AB$ on opposite sides, with third vertices $C$ and $D$. Then $AC=BC=AD=BD=2a$. Hence $ABCD$ is a rhombus. The diagonals of a rhombus bisect each other. Therefore the intersection of $AB$ and $CD$ is the midpoint of $AB$.

Once a midpoint of a segment of length $2a$ is available, we obtain a segment of length $a$ with a known midpoint. In an equilateral triangle built on that segment, the line from the opposite vertex to the midpoint is perpendicular to the base. This yields a right angle.

For a perpendicular to a given line, construct a segment of length $a$ on the given line, build an equilateral triangle on it, find the midpoint of the base by the previous construction, and draw the altitude.

The only potentially hidden step is the midpoint construction. It must be checked carefully that the quadrilateral obtained is indeed a rhombus and that the relevant diagonal is the original segment whose midpoint is sought.

Problem Understanding

We are given a ruler carrying two marks whose distance apart is fixed. Denote this distance by $a$. The ruler may be used to draw arbitrary straight lines and to lay off segments of length $a$.

We must construct, first, a right angle, and second, a line perpendicular to an arbitrary given line.

This is a Type D problem. We must give explicit constructions and verify that they produce the required objects.

The core difficulty is that ordinary perpendicular constructions usually require a compass. Here the only metric information available is the fixed distance $a$ between the two marks. The essential idea is to use repeated copies of the length $a$ to build equilateral triangles and rhombi; a rhombus provides a midpoint, and the altitude of an equilateral triangle provides a perpendicular.

Proof Architecture

The first lemma is that a segment of length $2a$ can be constructed. This is done by laying off the length $a$ twice on the same line.

The second lemma is that the midpoint of a segment of length $2a$ can be constructed. Construct equilateral triangles on that segment on opposite sides; the four vertices form a rhombus, whose diagonals bisect each other.

The third lemma is that a perpendicular to a segment of length $a$ can be constructed. Build an equilateral triangle on the segment and join the opposite vertex to the midpoint of the base.

The final step applies the third lemma to a segment of length $a$ chosen on any given line.

The most delicate lemma is the second one, because the midpoint is obtained indirectly through a rhombus, and the identification of the diagonals must be justified carefully.

Solution

Let the distance between the two marks on the ruler be $a$.

First construct a segment $AB$ of length $2a$. Choose an arbitrary point $A$ and an arbitrary line through it. Lay off a segment $AC$ of length $a$ on that line, and then lay off a second segment $CB$ of length $a$ in the same direction. Then

$$AB=AC+CB=2a.$$

We now construct the midpoint of $AB$.

Construct an equilateral triangle $ABC_1$ on one side of the line $AB$ and an equilateral triangle $ABC_2$ on the other side. Since $AB=2a$,

$$AC_1=BC_1=AB=2a,$$

and similarly

$$AC_2=BC_2=AB=2a.$$

Consider the quadrilateral $AC_1BC_2$. Its four sides are

$$AC_1,\quad C_1B,\quad BC_2,\quad C_2A,$$

and each of them has length $2a$. Hence $AC_1BC_2$ is a rhombus.

The diagonals of a rhombus bisect each other. The diagonals are $AB$ and $C_1C_2$. Let

$$M=AB\cap C_1C_2.$$

Then $M$ is the midpoint of $AB$.

Since $AB=2a$, it follows that

$$AM=MB=a.$$

We now construct a right angle.

Take the segment $AM$, whose length is $a$. Construct an equilateral triangle $AMP$.

Since $M$ is the midpoint of the side $AM$, the point $M$ is also the midpoint of the base of triangle $AMP$. In an equilateral triangle, the median from a vertex is simultaneously an altitude. Therefore the line through $P$ and the midpoint of the opposite side is perpendicular to that side.

Let $N$ be the midpoint of $AM$. Draw the line $PN$. Then

$$PN\perp AM.$$

Thus a right angle has been constructed.

To construct a perpendicular to a given line $\ell$, choose on $\ell$ a segment $RS$ of length $a$. Construct an equilateral triangle $RST$ on that segment. Construct the midpoint $K$ of $RS$ by the midpoint construction described above. Draw the line $TK$.

Since $RST$ is equilateral, the median from $T$ to the midpoint $K$ of the side $RS$ is an altitude. Hence

$$TK\perp RS.$$

Because $RS$ lies on the given line $\ell$,

$$TK\perp \ell.$$

This gives the required perpendicular.

The required constructions are therefore obtained by constructing midpoints via a rhombus and then using the altitude of an equilateral triangle. The constructed objects are

$$\boxed{\text{a right angle and a perpendicular to any given line}.}$$

Verification of Key Steps

The midpoint construction relies on the quadrilateral $AC_1BC_2$ being a rhombus. The equalities

$$AC_1=C_1B=BC_2=C_2A=2a$$

come directly from the fact that $ABC_1$ and $ABC_2$ are equilateral with side $AB=2a$. Thus all four sides of the quadrilateral are equal, which is exactly the definition of a rhombus.

The identification of the diagonals requires care. The vertices of the rhombus occur in the order

$$A,\ C_1,\ B,\ C_2.$$

Hence the opposite vertex pairs are $(A,B)$ and $(C_1,C_2)$. Therefore the diagonals are precisely $AB$ and $C_1C_2$. Their intersection bisects $AB$, giving its midpoint.

For the perpendicular construction, the relevant theorem is that in an equilateral triangle the median from a vertex is an altitude. This follows because the two triangles formed by the median have equal corresponding sides and are congruent; the adjacent angles at the midpoint are equal and supplementary, so each is $90^\circ$. Thus the median is perpendicular to the base.

Alternative Approaches

Instead of constructing a rhombus from two equilateral triangles on a segment of length $2a$, one may first construct any rhombus with side length $a$. The intersection of its diagonals gives the midpoint of each diagonal. After obtaining a midpoint of a segment of known length, the rest of the argument proceeds exactly as above through the altitude of an equilateral triangle.

The chosen construction is preferable because every step uses only the fixed length supplied by the marked ruler and requires no auxiliary parallelogram constructions. The midpoint emerges directly from a single rhombus whose sides are forced by two equilateral triangles.