# Four interesting geometrical problems

• Duplicating the cube
• Trisecting the angle
• Finding the mid-point of two given ones
• Theorem of G.Pick
(Both delivered and written by M.R. Most of the text is adapted from [3].)

### 1 Trisecting the angle and duplicating the cube

Two of the three classical Greek problems, which cannot be solved by using ruler and compasses alone, are to trisect a general angle and to duplicate a cube. Truly speaking, I myself do not know by heart the proofs that these constructions are impossible by the above mentioned classical method, but let us believe the authorities [3], at least for a moment.

1.1 Duplicating the cube

When the Athenians were suffering from a plague in 430 BC, they consulted the oracle of the god Apollo at Deos, and were instructed to double the size of their altar, which was a cube. Their first idea was to double every edge, but the ravages of the plague increased. (Surprised?) Finally they came to the observation that the problem consists of constructing a length 2^(1/3) (the cube root of 2) times the length given. (I don't know if this knowledge decreased the ravages of the plague, but an indirect proof of this could be the fact that there are some Greeks who survived.) The problem to duplicate the cube by using the ruler and compasses only became known as the Delian problem.

One can never find a solution requiring the ruler and compasses only, as was shown later [1]. However, Greeks obtained many solutions using either a "human judgement" or special curves invented for the purpose, which themselves couldn't be constructed by ruler and compasses only, as are the conchoid of Nicomedes or the cissoid of Diocles, see [2]. A natural question arises what would be the simplest set of instruments (as close to "ruler + compasses" as possible) which would allow this construction.

The aim of this section is to show that, in fact, a ruler and compasses is enough if one strange pre-step is allowed, namely the following: given ruler, compasses and a segment of unit length, mark (using compasses) two points on the ruler 1 unit apart, as shown on the picture.

And this is how to duplicate the cube (cf. the figure):

• Given a distance "1 unit", construct an equalsided triangle ABC, with side length 1 unit.
• Prolong the line through A,B and construct the point D such that |DB|=1.
• Draw the lines through B,C and through D,C making both of them "long enough".
• Finally, and this is the only step requiring the "marked ruler", adjust the ruler by hand so that it passes through the vertex A, and the distance between the points where it intercepts the lines BC and DC is just 1 unit.
It is not difficult to show (which still could mean that you would spent the whole afternoon practicing your high-school math) that the distance between point A and E is just 2^(1/3).

1.2 Trisecting the angle

The problem of trisecting the angle (finding an angle the size of which is exactly one third of the size of a given one) can be solved e.g. by using the so-called conchoid curve. In fact, Nicomedes invented the curve to solve this problem. We show that the same set of instruments as in the previous section (namely a compasses and marked ruler) is enough to allow the construction. However, the conchoid curve is behind the construction, so let us spend some time discussing it.

To construct a conchoid, or more precisely, a conchoid of the straight line with respect to a fixed point, take straight line and a point, A, not on this line, and a constant distance, k. Draw a straight line through A to meet the given line at Q. If P and P' are the points on the line AQ such that P'Q = QP=k, then P and P' trace out both branches of the conchoid.

Using the conchoid, the problem of trisecting the angle can be easily solved: in the figure, let AQ=QP/2=k/2, and let QR be perpendicular to the line. Then the angle RAB equals one third of the angle PAB. (Prove it.)

A practical method can be the following:

• Given an angle PAB, find a point Q within a segment PA such that AQ = 1/3 AP; this can be done by ruler and compasses only.
• Draw a line through Q perpendicular to the segment AB and call it a Q-line.
• Draw a line through Q perpendicular to the Q-line and call it R-line. The point R which is to be constructed lies on R-line.
• It remains to construct a part of conchoid intersecting the R-line. To do this, mark using the compasses two points on the ruler, having a distance k. Then adjust the ruler by hand so that it passes through the vertex A, and the distance between the points where it intercepts the Q- and the R-lines is just k. In such a way, the point R is obtained.
This shows that a general angle can be trisected by using compasses and a "marked ruler", the same set of instruments as in the case of duplicating the cube.

### 2 Finding the mid-point of two given ones

This in fact is a part of a problem given to me by one of my students during the discussion. After I finished my lecture, he came to the blackboard and said: "I understand you like puzzles of all kind. So try this one: Given an arc (a part of a circle), find a point on it having the same distance from both end-points of the arc, but using a compasses only."

Well, I tried to solve it myself, I discussed with friends. Finally I gave up, came to him and said: "Well, you won, tell me". He said: "I don't know myself, I just thought it would be a good idea to ask you." Ok, so I said to myself: "it could well be the problem is unsolvable..." and quit trying.

After some time, I was browsing through an interesting book [3] of David Wells on geometrical problems. And it was in this book that I happened to discover that "Mohr and Mascheroni showed the surprising fact that any construction that can be performed with a ruler and compasses can also be done with compasses only." This proves the original problem given to me was solvable, since the solution using both ruler and compasses can trivially be constructed.

Truth is, I still don't know how to solve the original problem, but at least first step can be done by showing how to find the mid-point of two given ones, say A, B, using a compasses only.

Given two points A, B (with no segment drawn connecting them!), first construct circles centered at A and B with radii equal to AB. Them draw a circle centered at C with radius AC and a circle centered at D with radius DB. With centre A and radius AE, draw the next circle, followed by the circle centered at E with radius EC. Finally, circles centered at F and G, passing through E, cross at the centre of A,B.

All readers are encouraged to solve the problem completely.

### 3 Theorem of G.Pick

Finally, I want to present (again without any proof leaving all of them to the kind reader) an interesting theorem. Not surprisingly, I have picked this theorem out because of the family name of the author, coinciding with beerer Lubos' one.

In 1899 G.Pick discovered a simple method of finding the area of a (not necessarily convex) polygon whose vertices coincide with the points of a plane square grid: If i is the number of points of the lattice inside the polygon, and b is the number of lattice points on the boundary, including the vertices, then

A = i+b/2-1

is the area of the polygon measured in square units defined by the grid. In the picture, i=4 and b=6, so the area is 4+6/2-1=6 square units.

### 4 References

1. Dorrie, H.: One Hundred Great Problems of Elementary Mathematics. Dover, N.Y., 1963.
2. Lockwood, E.H.: A Book of Curves. Cambridge University Press, Cambridge, 1961.
3. Wells, D.: The Penguin Dictionary of Curious and Interesting Geometry. Penguin Books, 1991.