Mascheroni published in 1798 a book about the geometry of compass.
His aim was mainly a practical one :

To prove in general the superiority of using the compass instead of straightedge,
when we have to draw lines at the microscope accuracy [...]
drawing a line along the straightedge has in itself inaccuracy [...]
The compass is not subject to such [...] drawback, it requires just that the opening is fixed
and the tips are fine enough

He then discovered that

we can find with the compass alone all the points [found]
with the compass and straightedge together

This is the Mohr-Mascheroni theorem.

To prove this theorem, we have just to know how to construct with compass alone :

In practice, we don't "translate" a construction with compass and straightedge into a compass only using repeatedly these elementary steps.
The compass only constructions have their own elementary constructions.

Nearly following the Mascheroni's book, after a few lemmas as usefull metric relations, he begins with dividing the circle into 4, 8, 12, .. 240 equal parts.
This is fully useless to prove the theorem, but the book is mainly a collection of practical constructions. We'll come back to that later.
He goes on with the halving of any arc.

Then he proceeds with the division and multiplication of lengths, that is construction of points on a line.

Find the midpoint of a given segment AB

He gives no less than five different constructions for this problem !
One of them is already given there

Let's give another one, which in my opinion is more accurate, as the arcs intersect at a larger angle.

Construct the half circle ACDE, resulting into point E on the opposite of A, by copying three times radius AC = CD = DE = AB along the circle.
Circle centered in A with radius AE and circle centered in E with radius EC intersect in F and F'
Circles centered in F and F' with radius FE = F'E intersect in M, midpoint of AB.

Proof :
Of course (symmetry) M is on line AB.
EF² = EC² = 3AB² (AEC = 30°)
FH² = FE² - HE² = FA² - AH², that is (AH + HE)(AH - HE) = FA² - FE² = 4AB² - 3AB² = AB²
As AH + HE = 2AB, this gives AH - HE = AB/2.
and by substraction HE = 3AB/4, hence AM = AE - 2HE = AB/2.

Note : the cyan elements are neither constructed nor drawn, they are just used for the proof.

By the way, let's mention these two constructions can be done with a collapsing compass.
This is a (theoretical) compass that allows only to draw circles from a given center and going through a given point.
An "ordinary" compass allows also to draw circles from a given center and a given radius (= distance between two other given points).
The collapsing compass "collapses" as soon as it leaves the paper, hence doesn't allow to "memorize" radius.
We may get some idea of the difficulty to use a collapsing compass :
Construct with a collapsing compass alone (without the straightedge) a circle centered in given A and radius = given BC


In other words : Any allowed construction with an ordinary compass can be done with a collapsing compass
The use of a collapsing compass just reinforce the "legal" use of the compass : it is quite impossible to "cheat" with a collapsing compass (move along the compass until ... for instance)
The Mohr-Mascheroni theorem can then be reinforced as :

All points constructible with compass and straightedge can be constructed with the collapsing compass alone.

Mascheroni then proceeds with dividing a segment into 3 ... n. Then addition and substraction of distances, constructing parallel and perpendiculars.
Then about proportions, then square roots (construct √2, √3 etc...)
All this not beeing directly useful for prooving the theorem.
We have to wait until liber (chapter) seven to get at what we are interested in : prooving the theorem.

Intersection of a line and a circle

The line being defined by two of its points A and B
This construction being easy, search it by yourself before clicking on solution.


If line AB goes through the center of circle, that is to find intersection of OA with the circle :
From center A, draw any circle intersecting the given circle in B and C, symmetric with respect to line OA.
The construction proceeds by finding the midpoint of arc BC.
Circle centered in O with radius BC intersects the circles centered in B and C and radius OB = OC in D and E.
DBCO and OBCE are parallelograms hence DOE are in line.

Theorem : in any triangle with sides a,b,c the length m = AMa of the median through midpoint of side a is  4m² = 2b² + 2c² - a² 
Proof : develop the scalar product (AB> + AC>)² + (AB> - AC>)² = 2AB² + 2AC²

Here the diagonal CD, twice the median of triangle BOC, is CD² = 2BC² + 2OC² - OB² = 2BC² + OB² = 2OD² + OB²
Draw the circles centered in D and E with radius CD = BE, and intersecting in F.
OF² = DF² - OD² = CD² - OD² = OD² + OB²
The circles centered in D and E with radius OF intersect in M.
OM² = DM² - OD² = OF² - OD² = OB²
Hence M is on the given circle, and because of symmetry, at midpoint of arc BC.
This gives also the opposite point M'.
One of the two circles centered in D or E with radius OF is useless : M and M' are just intersections with the given circle.

Intersection of two lines

The construction by Mascheroni uses a previously given construction : Construct x with x/a = b/c and given a,b,c (proportions).
Let's see first of all this construction alone.

Draw two concentric circles with radii a and c.
Copy distance AB = b along the circle with radius c
Draw any two circles with the same radius R, centered in A and B,
and intersecting the circle with radius a in C and D.
then x = CD = a.b/c

Proof : triangles OAC and OBD are congruent (AC = BD = R, OA = OB, OC = OD).
Hence angle AOC = BOD , hence angle AOB = COD,
hence the isosceles triangles AOB and COD are similar, then CD/OC = AB/OA.

We may have to multiply the lengths a and c by a same number for the chord with length AB = b to be possible : b < 2c
we should even prefer b < 1.5c to get an accurate intersection point B.
We use the previously seen method of the half circle to double a distance.

We can now see the construction with compass alone of the intersection point of two lines defined by points AB and CD.
Circles centered in A and B and going through C intersect also in C', symmetric of C with respect to line AB. We similarily construct the symmetric D' of D.
Circle centered in C with radius DD', and circle centered in D' with radius CD intersect in E.
ECDD' is a parallelogram, and CM/ED' = C'C/C'E
The construction of CM is then as before, from that proportion :
Draw the circles centered in C' with radii C'C and C'E
Copy EF = ED'.
Copy any equal distance (here = EF) : EE' = FF'
E'F'/C'C = EF/C'E hence E'F' equals the searched CM = C'M.
Circles centered in C and C' with radius E'F' intersect in M, intersection of AB, CD and C'D'.

We are then finished with the elementary constructions which prove the Mohr-Mascheroni theorem.
Mascheroni's book is not just this theorem, and many usefull constructions are given.

Let's mention the construction of the center of a circle, given without its center.
The solution from Mascheroni is given there
We can see that it is a completely different solution from the usually given construction. (said to be from Napoleon)

Regular pentagon

Several constructions given by Mascheroni result into construction of a regular pentagon.
We already find one in the chapter about division of a given circle into up to 240. Dividing into 5 is :

Draw the half circle ABCD (OA = AB = BC = CD = R).
Circles centered in A and D with radius AC = DB intersect in E.
Recall that OE = R√2, this allows to construct F with AF = AE, on the perpendicular OE.
Circle centered in F with radius R = OF gives points I and J, dividing the semi-circle AIBFCJD into six 30° arcs.
Circles centered in I and J with radius OE = R√2 intersect in G.
AG is the side of the inscribed regular pentagon AMNPQ.
Proof : Line IJ intersects the radius OF in its midpoint H, and HI = R√3 / 2
Hence HG² = IG² - IH² = 2R² - 3R²/4 = 5R²/4
Let's leave here the proof from Mascheroni, who just calls to the Almageste by Ptolemy and to the "proof by Clavius in the scholie from proposition 10 in book 13 of Euclide" (sic), as these erudite books are not widely available.
We more directly deduce OG = R(√5 - 1)/2, hence AG² = OA² + OG² = R² + R²(√5 - 1)²/4 = R²(10 - 2√5)/4,
That is AG = R√(10 - 2√5) / 2 to be compared with the half side of the pentagon R.sin(36°) = R√(10 - 2√5) / 4
This value can be directly obtained from the De Moivre formula sin(5θ) = 16 sin5(θ) - 20sin3(θ) + 5sin(θ)
hence sin(36°) is the smallest > 0 solution of 5sin(θ) - 20sin3(θ) + 16 sin5(θ) = 0, that is 16sin4(θ) - 20sin2(θ) + 5 = 0.
The smallest solution of 16X² - 20X + 5 = 0 is X = (10 - √20)/16, hence sin(36°) = √X
Without using De Moivre, there is a pure elementary geometric proof (the above mentioned one from Clavius), given in some geometry class books.

Let's return to Mascheroni who gives as problem n░97 :

Divide a given segment OA in mean and extreme ratio

That is with today's words to find point M with MO/MA = OA/MO
We deduce at once that MO/OA is solution of x² + x - 1 = 0, in other words that x = 1/φ = (-1 + √5)/2
This results into a construction of the regular pentagon, because the ratio of diagonal over the side is the golden ratio φ.
Draw the 2/3 of circle ABCDE, from center O (OA = AB = BC = CD = DE)
Circles centered in A and D with radius AC = BD intersect in the now well known point F.
Circles centered in C and E with radius OF = R√2. intersect in the searched point M.
The proof that OM = OA(√5 - 1)/2 is as above, just the construction is turned by 90°.

We can then easily proceed with the regular pentagon of diagonal = OA, hence side = OM :
Circles centered in O and A with radius OM intersect in N.
Circle centered in A with radius OM intersects the ABCDE circle in Q.
Circles centered in Q and O with radius OM intersect in P.

Last he gives two constructions of a regular pentagon with given side.
One of them just proceeds from the previous construction, using the inverse ratio :
For DM = OA(√5 + 1)/2 = φ.OA, we get DM = diagonal of pentagon with side OA :
Draw circles centered in O and A with radius DM.
They intersect in P. The circle centered in A intersects the ABCDE circle in N, and the circle centered in O intersects the circle centered in A with radius OA (already drawn when constructing B) in Q.

Let's stop here with Mascheroni constructions.


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