# Napoleon problem

Napoleon (Buonaparte) is known as military stratege and imperator, but he was also a chess player and amateur mathematician.
Several problems are called "Napoleon problem"

### Construct using compass alone the center of a given circle

Given a circle without its center, construct the center using only a compass (and no straightedge).

Solution
There are other methods, among them the one from Napoleon himself (really from Mascheroni and carried back to France by Napoleon).

### Construct with compass alone a square inscribed in a given circle

If the circle is given without the center, first construct the center A as above.

Choose any point = 1st vertex B on this circle.

Remains to construct E opposite to B on a diameter, then M and P with BM=BP=AB√2

Solution
### Similar triangles on sides of given triangle ABC

Given any triangle ABC, draw the three triangles PBA, BMC and ACN directly similar
to ABC (that is without flip), outside of ABC.

Prove that centroids R, S and T of these three triangles build a triangle similar to ABC.

(That is also true for any corresponding points in the three triangles, as the orthocenters etc...)

### Equilateral triangles on a given ABC triangle

Draw the three equilateral triangles ABP, BCM and CAN on the sides of any given triangle ABC.

The three centers I, J and K of these triangles build an equilateral triangle (Napoleon theorem).

Animated

JavaSketchpad (patience when loading and initializing Java engine).

Equilateral triangles on a segment - "Napoléon's hat"

It is a degenerate case of the previous problem, with C on AB.

Find the locus of the centroid of IJK when C moves along AB ?
Hint

Signed sum of altitudes of the three triangles is 0.

Proof of Napoleon's theorem :

Let X the intersection point of circumcircles to ACN and BCM.

The inscribed angles AXC and ANC sum up to 180° : AXC = 120°.
Similarily BXC = 120°, hence AXB = 120°,
and X is then on the circumcircle of ABP.
The three circumcircles then concur in one point X (Fermat/Torricelli point), also radical center of the three circles.

Even more AM, BN and CP are concurent in X, as the 6 angles in X all equal π/3 (inscribed).

Because CX is perpendicular to IJ, and similarily BX_|_IK and AX_|_JK, the angles of triangle IJK all equal to 60°.

This proof is valid only when X is inside ABC, otherwise adapt the values of inscribed angles, with the same final conclusion :

AXC = 120° or 60° etc... all angles of IJK are 60° or 120°, hence 60°.