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Napoleon Triangle

Problem

One of the greatest military leaders in history, Napoleon Bonaparte, was also an amateur mathematician and is credited for the following result.

If equilateral triangles are constructed on the sides of any triangle then the centres joining the constructed triangles will always form an equilateral triangle.

Prove "Napoleon's Theorem".


Solution

We shall begin by drawing circumcircles of each constructed equilateral triangle with centres $X$, $Y$, and $Z$ respectively.

First we shall show that all three circles pass through the common point, $D$. Consider circle $Z$.

Because quadrilateral $ADCE$ is cyclic, $\angle AEC + \angle ADC = 180$ degrees, and as $\angle AEC = 60$ degrees by construction, it follows that $\angle ADC = 120$ degrees.

Similarly we can show that each of the angles $ADB$ and $BDC$ are 120 degrees.

Therefore $\angle ADC + \angle ADB + \angle BDC = 360$ degrees, and we demonstrate that the three circles are concurrent at $D$.

Now as $AD$ is a common chord to circles $X$ and $Z$ the segment joining their centres, $XZ$, is a perpendicular bisector of $AD$ (see Common Chord).

Therefore $\angle DFZ = 90$ degrees; and in the same way $\angle DGZ = 90$ degrees.

Hence in quadrilateral $FDGZ$, we can see that $\angle FZG = 60$ degrees.

In the same way we can show that $\angle ZXY = \angle XYZ = 60$ degrees and we prove that the constructed triangle is indeed equilateral.

Problem ID: 313 (15 Feb 2007)     Difficulty: 4 Star

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