where p and q are known intervals. In this case we don't have to know exact values of p and q i.e.
lengths of intervals. We don't need to know the unit interval in this case either.
If a solution of this equation exists, we can construct with ruler and compass intervals x1 and x2,
which are roots of this
equations. We'll apply the formula of Vieta:
and we'll use the following property of right triangles:
Suppose AB is a diameter of the circumscribed circle on ABC. Hence, angle C is the right angle. Let CD be
the height of ABC of the vertex C. Then
In other words: height of the right triangle is a geometrical mean value of intervals it divides hypotenuse.
If AB=p, h=q, then roots of our equation are AD and DB, which follows from the previously mentioned
equalities of Vieta.
Exactly, we proceed the following way:
1. Take 2 intervals: one of length p, second of length q.
2. Take an interval AB of length p and circle S such that AB is a diameter of S.
3. Draw a line L parallel to the line AB at the distance q from line AB.
4. Take point D - one of points of intersection of L with S. If there are not such points - it means
that equation has no roots.
5. Find point D on the line AB such that CD is perpendicular to AB.
6. AD and DB are roots of our equation.
For a second coefficient positive we have an equation
If roots of this equation exist, they are negative. Of course we cannot construct intervals with negative
lengths. That is why we construct intervals of lengths of absolute value of roots and we add few words, that
roots are opposite numbers to these lengths.
To construct intervals of lengths of absolute value of roots we change the sign corresponding to coefficient
p receiving an equation of Case 1. Then we proceed according to procedures of Case 1.
For third coefficient negative and second negative, we have an equation:
In this case roots have the different signs.
We proceed similar way as before, now regarding the length of AB as unknown and DB equal p.
1. Build a right triangle CDE such that CD has length q, DE has
length p/2 and an angle D is the right angle (90 degrees).
2. Draw line DB.
3. Draw a circle S with the center at E and radius EC.
4. Take points A and B, where S intersects line DB.
5. Length of AD is an absolute value of one root and
DB is an absolute value of the second root.
6. Sign of a root corresponding to longer interval is plus, sign of a root
corresponding to shorter interval is minus.
Knowing properties of right triangle, it is easy to check that equalities of Vieta are satisfied with
such defined roots.
For third coefficient negative and second positive, we have an equation:
We construct roots as in the case III and then reverse signs of these roots.
Let us consider an equation
where r and s are positive numbers. We want to construct intervals with ruler and compass,
which have lengths
equal absolute value of roots. It has only a sense if we can construct intervals of length equal
r and s. Of course then we need to know a unit interval i.e. interval of length 1.
Of course if we have a unit interval and for example r and s are rational, we can construct the roots.
We just reduce this case to one of previous cases plugging p=r and finding q such that
It is easy to construct such an interval q knowing mentioned above property of the right triangle.
It seems to Literka that the following picture explains everything, since q is a geometrical mean value
of s and 1
The hypotenuse of above right triangle consists of 2 intervals: a unit interval and interval of length s.
Some mathematicians are using for construction of roots so called Carlyle circles. Literka does not know
how much easier theirs constructions are. Literka hesitated to do the same, but decided not, because
either way constructions are straightforward.