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I recently solved a question of complex numbers which was this:

$A\left( \frac{2}{\sqrt{3}} e^\frac{i\pi}{2}\right)$, $B\left( \frac{2}{\sqrt{3}} e^\frac{-i\pi}{6}\right)$, $C\left( \frac{2}{\sqrt{3}} e^\frac{-5i\pi}{6}\right)$ are the vertices of an equilateral triangle. If $P$ be a point on the incircle of the triangle, prove that $AP^2 + BP^2 + CP^2 = 5$.

My approach:

The point $P$ is given by $z = \frac{1}{\sqrt{3}}e^{i\theta}$, since the radius of the incircle of an equilateral triangle is half its circumradius. Then,

$$ AP² = |A - z|^2 \\ = (A - z)(A^* - z^*) \\ = AA^* - Az^* - A^*z + zz^* \\ = |A|² + |z|² - Az^* - A^*z \\ = \frac43 + \frac13 - Az^* - A^*z \\ = \frac53 - Az^* - A^*z. $$

Now,

$$ AP^2 + BP^2 + CP^2 \\ = 3\times\frac53 - z^*(A+B+C) - z(A^*+B^*+C^*) \\ = 5 $$

since $A+B+C = A^*+B^*+C^* = 0$ because the position vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$ are coplanar and are mutually separated by $120^\circ$.

This, I guess, is a pretty neat solution. But what I'm looking for is a more intuitive solution, rather a geometric interpretation. Does anyone know such an approach?

Nilabro Saha
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  • Your question is about an intuitive understanding that the sum of the squares is constant? – anderstood Jan 30 '17 at 18:49
  • Absolutely. And, if possible, also that it equals 5. – Nilabro Saha Jan 30 '17 at 18:50
  • I would be funny to check if this still holds for $P$ on the sphere (centered in $O$ and of radius $2/\sqrt{3}$ of course). I don't think there is a "intuitive" interpretation of the value 5 (which is an integer only because you adapted the radius to get an integer). – anderstood Jan 30 '17 at 18:51
  • Does this hold on the sphere too? I mean, in 3D I guess point P will be on the sphere inside a perfect tetrahedron. – Nilabro Saha Jan 30 '17 at 18:54

3 Answers3

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Hint: for any triangle $\triangle ABC$ and any point $P$ the following holds, where $G$ is the centroid of the triangle (see here for example):

$$ PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3 PG^2 $$

For an equilateral triangle all the centers coincide, so $G$ is also the center of the incircle, thus $PG$ is constant for any point $P$ on the incircle. Therefore the RHS is constant, and you can choose for example $P$ to be the midpoint of $BC$ in order to calculate the value of the constant.

dxiv
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The intuition lies in the fact that the three points in question $A$, $B$, $C$ has a bary-center which is at the center $O$ of the inscribed circle.

If $P$ is on the inscribed circle then using scalar product and letting $a=\vec{OA}$, $b=\vec{OB}$, $c=\vec{OC}$ and $m=\vec{OP}$: $$ |AP|^2+|BP|^2+|CP|^2= \sum_{p\in \{a,b,c\}} (p-m,p-m) = 3 |p|^2 + 3 |m|^2 - 2(m,\sum_{p\in \{a,b,c\}} p) = 3(4/3+1/3) -0=5$$

H. H. Rugh
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The sides of your triangle are $2$. E.g. the distance of $A$ and $B$ is the length of $$\frac{2}{\sqrt{3}}(e^{-\frac{i\pi}{6}}-e^{\frac{i\pi}{2}})=\frac{2}{\sqrt{3}}(\frac{\sqrt{3}}{2}-\frac{3}{2}i)=1-\sqrt{3}i.$$

We can calculate $PA^2+PB^2+PC^2$ using elementary methods, if $P$ is a point on the incircle of an equilateral triangle $ABC$ of side $a$.

In this case the radius of the incircle is $r=\frac{a\cdot\sqrt{3}}{6}$. Let $O$ be the incenter. Then $OA=OB=OC=\frac{a\sqrt{3}}{3}$. Let $\angle POC$ be $\varphi$. Then using the cosine theorem for the triangle $POC$ $$PC^2=r^2+OC^2-2r\cdot OC\cdot\cos\varphi = \frac{a^2}{12}+\frac{a^2}{3}-\frac{a^2}{3}\cos\varphi.$$ Since $\angle POA=120^{\circ}-\varphi$ and $\angle POB=120^{\circ}+\varphi$, similarly we get $$PA^2 = \frac{a^2}{12}+\frac{a^2}{3}-\frac{a^2}{3}\cos(120^{\circ}-\varphi)$$ and $$PB^2 = \frac{a^2}{12}+\frac{a^2}{3}-\frac{a^2}{3}\cos(120^{\circ}+\varphi).$$ Summarizing and using the addition theorem

$$PA^2+PB^2+PC^2=a^2(\frac{1}{4}+1-\frac{1}{3}\cos\varphi-\frac{1}{3}(-\frac{1}{2}\cos\varphi+\frac{\sqrt{3}}{2}\sin\varphi)-\frac{1}{3}(-\frac{1}{2}\cos\varphi-\frac{\sqrt{3}}{2}\sin\varphi))=\frac{5}{4}a^2.$$

Since in our case $a=2$, this gives $PA^2+PB^2+PC^2=5.$

Sz_Z
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