Napoleon Problem: Magic of the circles

Napoleon Bonaparte (1769 - 1821) well known as a genius general, a political leader, an Emperor of the French (1804 - 1814). Napoleon also was known to be an excellent amateur mathematician! The following problem to find a centre of a given circle widely known as "Napoleon Problem":

pic 1: Find a center of a given circle, using the pair of compasses alone (no straight edge)

The problem looks simple to be understood, but the solution is surprised magic of 7 circles! The following figure describes how to detect a centre of a given circle:

pic 2: How to detect a centre of a given circle

Constructing steps:

1) Select an arbitrary point A on the circle and draw a circle C2 of centre A, which intersects with a given circle (C1) at points B and C.

2) Draw circles C3, C4 of centres B, C and radius r = BA, which intersect each other at point D (D ≢ A), and draw a circle C5 of centre D and radius r = DA.

3) Circle C 5 intersects with circle C2 at points E, F. Draw circles C6, C7 of centres E, F and radius r = EA, the two circles intersect each other at A and an other point - this point is a centre of given circle C1!

It is a magic, isn't it? The problem is solved by drawing of 6 new cicles by compass alone, and there is not any lines or straight segments on the figue!

Proof:

To prove that the above solution works, we will prove the following lemma:

Lemma: Lets (C) be a circle of centre O radius a, AE be a diameter of circle (C), B, C be points on (C) and BC is perpendicular to OA. Point D is on AE and BD = AD, b is length of BC. So we will have AD = b*b/r (see picture)

pic 3: Lemma

Lemma's proof: Triangle EBA has a right angle B, and EA ⊥ BC => ∆EBA ∾ ∆BHA => EA/BA = BA/AH => AH = BA * BA / EA = (b x b) / (2 x a) => AD = 2AH = b^2 / a.

In the above construction of centre (pic 2), the configuration of lemma appears twice:

1) Let O be a centre of a circle C1, radius OA = r, see a circle C1, points A, B, C, D and O. Application of lemma give us: R = BA = CA = BD =  CD => AD = R x R / r

2) See a circle C5 of centre D, radius R, points E, F, A lie on C5, and point O' = C6 ∩ C7 => O' ∈ DA and EO' = EA = FO' = FA. Application of lemma gives us: AO' = R * R / AD = R * R / (R * R / r) = r => O' ≣ O (or O' is a centre of a circle C1).

Reference: wikipedia.org

6.2.2014

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