Napoleon & Fermat

by

Rhonda Iler

Napoleon’s Triangle:

Given any triangle ABC, construct equilateral triangles on each side. Construct the center of each equilateral triangle (the centroid). The triangle formed by the 3 centers is called Napoleon’s Triangle.

Napoleon’s Theorem states that the triangle formed by the 3 centers is an equilateral triangle.

Napolean's triangle.gsp

Proof: In an equilateral triangle, the centroid, the incenter, the circumcenter, and the orthocenter are the same point. Therefore, for triangle YAC point E is the incenter, thus lying on the angle bisector. Then angle EAC= 30 degrees. Similarly, angle DAB is 30 degrees. Then by the Law of Cosines:

DE² = AC² + AB² – 2(AC) (AB) cos (A + 60) where A is angle CAB.

Since E is also the centroid, E is 2/3 of the distance from A to YC. And D is 2/3 of the distance from A to XB. So we have

AE = (AC) = AC =

AD = (AB) = AB =

So we have 3(DE)² = AC² + AB² – 2(AC)(AB)cos(A + 60) since all the square roots are squared and we can multiply each side by 3.

Similarly: 3(EF)² = BC² + AC² – 2(BC)(AC)cos(C + 60)

Again, using the Law of Cosines:

BY² = AC² + AB² – 2(AC)(AB)cos (A + 60)

Or

BY² = BC² + AC² – 2(BC)(AC)cos (C + 60)

So we have: AC² + AB² – 2(AC)(AB)cos (A + 60) = BC² + AC² – 2(BC)(AC)cos (C + 60)

Substituting we get: 3(DE)² = 3(EF)²

So: DE = EF.

By similar argument for triangles BCX and BAX, EF = FD.

So we have DE = EF = FD. Therefore Triangle DEF is equilateral.

Constructing the circles with the vertices of Napoleon’s triangle being the centers and the radius being the sides of the triangle, we have 3 congruent circles. Then if we construct the points of intersection of the 3 circles and label them G, H, and I and then construct the triangle formed by those 3 points, we can measure the areas. The area of the new triangle is 4 times that of Napoleon’s triangle.

Napolean's triangle3.gsp

Interesting property of Napoleon’s Triangle (The Mathematical Intelligencer):

Two lines passing through the vertex of a triangle are called isogonal with respect to that vertex if they form equal angles with its internal angle bisector.

first nap point.gsp

AB’ and AC’, CB’ and BA’, BA’ and BC’ are all isogonal. The three lines AA’, BB’, and CC’ are concurrent. Napoleon’s theorem occurs when all three of the angles are equal to 30 degrees.

The common point is known as the First Napoleon Point, N.

Inner Napoleon Triangles:

We can also construct the equilateral triangles to the inside of the given triangle, triangle ABC. The same is true of the triangle formed by the centers of these triangles. Triangle GHI would also bean equilateral triangle.

The second Napoleon Point is obtained here. If you connect the points of the original triangle with the opposite vertices of the 3 equilateral triangle you get lines AZ, BY and CX. These three lines are concurrent at the second Napoleon point, N2.

N2.gsp

Further Investigation: Fermat’s Point

Construct the segments connecting the vertex of the equilateral triangle with the opposite vertex of the original triangle. This gives us segments AZ, XC, and YB. These segments are all congruent. The angles formed by the intersection of these segments are each 60 degrees. These 3 segments are all concurrent at point H. The circumcircles about each of the 3 equilateral triangles are also concurrent at point H. The point H is called the Fermat point of triangle ABC. This is the point in which the sum of the distances from the vertices of a triangle is minimized.

Fermat point.gsp

The Fermat point is also known as the Torricelli point and the first isogonic center. It is the solution to the problem of finding a point F inside a triangle ABC such that the total distance from the 3 vertices to a point F is the minimum possible. It is named such because the problem was first asked by Fermat in a private letter to Torricelli.

Fermat minimum.gsp

If we rotate triangle ABH 60 degrees about point B, we get triangle XBH’. By construction, BHH’ is equilateral. HB = H’B HH’, HA = H’X. Thus we have

HA + HB + HC = H’X + HH’ + HC

X does not depend on H since it is a rotation of A.

HA + HB+ HC ≥ CX since CHH’X is no shorter than CX. Thus HA + HB + HC reaches its minimum if and only if H lies on XC. Thus angle BHX = 60 degrees.

Similarly, if we had rotated the triangle about point A, angle AHX = 60 degrees.

The second Fermat point, F2, uses the three inner triangles to construct the three concurrent lines and circumcircles.

F2.gsp

References:

www.math.washington.edu

www.cut-the-knot.org

The Mathematical Intelligencer by David Gale