In this section we discuss yet another source of impossible constructions. Recall that acircumtool produces a circle passing thru any given three points or a line if all three points lie on one line. Let us restate Exer- cise 10.9.

Exercise. Show that with a circumtool only, it is impossible to construct the center of a given circleΓ.

Remark. In geometric constructions, we allow to choose some free points, say any point on the plane, or a point on a constructed line, or a point which does not lie on a constructed line and so on.

In principle, when you make such a free choice it is possible to mark the center of Γ by accident. Nevertheless, we do not accept such a coincidence as true construction; we say that a construction produces the center if it produces it for any free choices.

Solution. Arguing by contradiction, assume we have a construction of the center.

Apply an inversion in a circle perpendicular to Γ to the whole con- struction. According to Corollary 10.16, the circle Γ maps to itself. Since the inversion sends a circline to a circline, we get that the whole construc- tion is mapped to an equivalent construction; that is, a constriction with a different choice of free points.

According to Exercise 10.8, the inversion sends the center of Γ to another point. That is, following the same construction, we can end up at a different point — a contradiction.

19.11. Exercise. Show that there is no circumtool-only construction which verifies if the given point is the center of a given circle. (We assume that we can only “verify” if two constructed points coincide.)

A similar example of impossible constructions for a ruler and a parallel tool is given in Exercise 14.6.

Let us discuss yet another example for a ruler-only construction. Note that ruler-only constructions are invariant with respect to the projective transformations. In particular, to solve the following exercise, it is suf- ficient to construct a projective transformation which fixes two pointsA

andB and moves its midpoint.

19.12. Exercise. Show that the midpoint of a given segment cannot be constructed with only a ruler.

19.13. Theorem. The center of a given circle cannot be constructed with only a ruler.

Sketch of the proof. It is sufficient to construct a projective transformation
which sends the given circle Γ to a circle Γ′ _{such that the center of Γ}′

is not the image of the center of Γ. (Compare the construction with Exercise 16.6.)

Let Γ be a circle which lies in the plane Π in the Euclidean space.
By Theorem 16.3, the inverse of a circle in a sphere is a circle or a
line. Fix a sphere Σ with the centerO so that the inversion Γ′ of Γ is a
circle and the plane Π′ _{containing Γ}′ _{is not parallel to Π; any sphere Σ in}

a general position will do.

LetZ and Z′ _{denote the centers of Γ and Γ}′_{. Note that}_{Z}′ _{∈}_{/} _{(}_{OZ}_{).}

It follows that the perspective projection Π→Π′ _{with center at}_{O} _{sends}

Γ to Γ′_{, but}_{Z}′ _{is not the image of}_{Z}_{.}

### Construction of a polar

In this section we describe a powerful trick that can be used in the con- structions only with ruler.

Assume Γ is a circle in the plane and P /∈Γ. Draw two lines x and

y thru P which intersect Γ at two pairs of points X, X′ and Y, Y′.
Let Z = (XY)∩(X′_{Y}′_{) and} _{Z}′ _{= (}_{XY}′_{)}_{∩}_{(}_{X}′_{Y}_{). Consider the line}

p= (ZZ′_{).}
Γ
P
p
x
y
X
X′
Y Y′
Z
Z′

The following claim will be used in the constructions without a proof.
19.14. Claim. The constructed line p= (ZZ′_{)} _{does not depend on the}
choice of the linesxandy. Moreover,P 7→pis a duality (see page 124).

The line pis called the polar ofP with respect to Γ. The same way the pointP is called the polar of the linepwith respect to Γ.

19.15. Exercise. Let p be the polar line of point P with respect to the circle Γ. Assume that p intersects Γ at points V andW. Show that the lines(P V)and(P W)are tangent toΓ.

Come up with a ruler-only construction of the tangent lines to the given circle Γ thru the given point P /∈Γ.

19.16. Exercise. Assume two concentric circles Γ and Γ′ _{are given.}
Construct the common center ofΓ andΓ′ _{with a ruler only.}

## Area

The area functional will be defined by Theorem 20.7. This theorem is given without proof, but it follows immediately from the properties of

Lebesgue measure on the plane. The construction of Lebesgue measure typically use the method of coordinates and it is included in any textbook in real analysis. Based on this theorem, we develop the concept of area with no cheating.

We choose this approach since any rigorous introduction to area is tedious. We do not want to cheat and at the same time we do not want to waste your time; soon or later you will have to learn Lebesgue measure if it is not done already.