What Is Napoleon's Theorem?

Napoleon's Theorem

If we construct equilateral triangles on the sides of any triangle (all outward or all inward), the centers of those equilateral triangles themselves form an equilateral triangle.

We will prove this with the equilateral triangle condition for complex numbers. Let $j$ be a $120^\circ$ rotation in the complex plane. Then, by the equilateral triangle condition, we have: $$A + jB + j^2C_1 = 0$$ $$C + jA + j^2B_1 = 0$$ $$B + jC + j^2A_1 = 0$$ Then, it suffices to prove: $$X + jY + j^2Z = 0$$ We note that the center of triangle $ABC$ in the complex plane is $(A+B+C)/3$. Then, it suffices to show that the following equals $0$: $$A + B + C_1 + j(C + A + B_1) + j^2(B + C + A_1)$$ $$= (B + jC + j^2A_1) + j(A + jB + j^2C_1) + j^2(C + jA + j^2B_1)$$ $$= 0$$ So we are done. $\Box$

Remark: Note that this proof works when the triangles are constructed internally as well, because $j$ is just then the opposite rotation, but it is still a $120^\circ$ rotation, and hence a third root of unity.

Napoleon's theorem says that in any triangle $ABC$, if equilateral triangles are constructed externally, the centroids of those triangles form an equilateral triangle.

One proof is using the law of cosines. Note the angles $ZAB$ and $YAC$ are both $60$ degrees, so $ZAY$ is $a+60$ degrees. It suffices to show $3YZ^2$ is symmetric in $a$, $b$, $c$.

Now, $3YZ^2=3AZ^2+3AY^2-6(AY)(AZ)(\cos(a+60^\circ))$.

This is $c^2+b^2-(2bc)\cos(a+60^\circ)$. So it suffices to show now that $-a^2-(2bc)(\cos(a+60^\circ))$ is symmetric in $a$, $b$, $c$, or that $a^2+(bc)(2\cos(a+60^\circ))$ is symmetric in $a$, $b$, $c$. This is because subtracting $a^2+b^2+c^2$ and multiplying by $-1$ preserve symmetry.

But $2\cos(a+60^\circ)=\cos(a)-\sin(a)\sqrt{3}$.

So it suffices to show that $a^2+(bc)(\cos(a))-bc\sin(a)\sqrt{3}$ is symmetric in the three sides.

Note $bc\sin(a)\sqrt{3}$ is symmetric because it is a fixed constant multiple of areas. So it suffices to show $a^2+(bc)(\cos(a))$ is symmetric.

But $b^2+c^2-(2bc)\cos(a)=a^2$, so $(bc)(\cos(a))=\frac {b^2+c^2-a^2} {2}$. So this expression is $\frac {a^2+b^2+c^2} {2}$, certainly symmetric.

Therefore, $YZ$ is some symmetric function of the sides, and $XY$, $XZ$ are this same function except with the inputs rearranged. But because the function is symmetric, they are equal so $YZ=XZ=XY$ as desired. We are done.

Napoleon's theorem states that

Given a triangle $ABC$, construct equilateral triangles on the sides of $ABC$, either all outward or all inward. Then the centers of these three triangles form an equilateral triangle. Now we will look at a synthetic proof of this. Let $A_1BC$, $AB_1C$, and $ABC_1$ be the three equilateral triangles.

Lemma 1: The circumcircles of the three triangle all intersect at a common point. We will call this point $K$.

Hint: Easy Angle Chase

Proof: Suppose we define $K \not= C$ as the intersection of the circumcircles of $A_1BC$ and $AB_1C$. Then $\angle BKC = 180^\circ - \angle BA_1C = 120^\circ$ and $\angle AKC = 180^\circ - \angle AB_1C = 120^\circ$. Thus $\angle BKA = 120^\circ$, so $K$ lies on the circumcircle of $ABC_1$. $\Box$

Lemma 2: $AA_1$, $BB_1$, and $CC_1$ concur at $K$.

Proof: $\angle BKC + \angle BKC_1 = 180^\circ$ so $K$ lies on $CC_1$. The $K$ also lies on the two other lines, so they concur. $\Box$

Lemma 3: $AA_1$, $BB_1$ and $CC_1$ have the same lengths.

Hint: What theorem helps use use lengths with cyclic quadrilaterals?

Proof: By Ptolemy's theorem, we have that $$(A_1K)(BC) = (A_1B)(KC) + (A_1C)(KB)$$ so $$A_1K = KC + KB \implies AA_1 = KC + KB + KA.$$ Similarly, $BB_1$ and $CC_1$ equal the same value. $\Box$

Now try to prove the theorem using these facts.

Hint: Look for a homothety.

Proof of the Theorem: Let $D$ be the midpoint of $BC$, the centroid of $BCA_1$ be $A_2$, and the centroid of $ABC$ be $G$. Then since $A_2$ cuts $DA_1$ and $G$ cuts $DA$ into $2:1$ ratio, There exists a homothety centered at $D$ that maps $GA_2$ to $AA_1$ so $A_2G = \frac 13 A_1A_2$ and $A_2G || A_1A$. Similarly $B_2G = \frac 13 B_1B$, $C_2G = \frac 13 C_1C$, $B_2G || B_1B$, and $C_2G || C_1C$. Thus $GA_2 = GB_2 = GC_2$. Also, $\angle B_2GC_2 = B_1KC_1 = 120^\circ$, and similarly, $\angle A_2GB_2 = \angle A_2GC_2 = 120^\circ$. Thus, we can conclude that the sides of $A_2B_2C_2$ are congruent. $\Box$

Napoleon's Theorem is an interesting geometry theorem allegedly proven by Napoleon Bonaparte. It states the following:

Let $ABC$ be a triangle. Let $D$, $E$, and $F$ be points such that $BCD$, $AEC$ and $ABF$ are equilateral triangles and $D$ is on the opposite side of $\overline{BC}$ as $A$, $E$ is on the opposite side of $\overline{AC}$ as $B$, and $F$ is on the opposite side of $\overline{AB}$ as $C$. Let $X$, $Y$, and $Z$ be the centres of equilateral triangles $BCD$, $AEC$, and $ABF$, respectively.

Napoleon's Theorem states that $XYZ$ is an equilateral triangle.

A concise proof of Napoleon's theorem:

This proof makes use of the several similar and congruent triangles in this configuration.

First we will show that $\triangle FAC \sim \triangle ZAY$, and $FC/ZY=\sqrt{3}$. To prove this, note that $FA/AZ=AY/AC=\sqrt{3}$ and both $\angle FAC$ and $\angle ZAY$ are 60 degrees more than $\angle BAC$, so the result is true by side-angle-side triangle similarity.

With the same argument, $FC/ZX=\sqrt{3}$, so $ZY=ZX$.

With the same argument, $ZX=YX=ZY$, so $\triangle XYZ$ is clearly equilateral.