# $M$ is a point in an equalateral $ABC$ of area $S$. $S'$ is the area of the triangle with sides $MA,MB,MC$. Prove that $S'\leq \frac{1}{3}S$.

$$M$$ is a point in an equilateral triangle $$ABC$$ with the area $$S$$. Prove that $$MA, MB, MC$$ are the lengths of three sides of a triangle which has area $$S'\leq \frac{1}{3}S$$

We can assume that side of a triangle $$ABC$$ is $$1$$. Further let $$CM =x$$ and $$\angle KCM =\gamma$$. Rotate $$M$$ around $$C$$ for $$-60^{\circ}$$ in to $$F$$. Then the area of triangle $$AMF$$ is the one we are looking for and it's area is area of $$AMCF$$ minus area of equilateral triangle $$CFM$$, so $$4S' = -x^2\sqrt{3}+2x\sin (60^{\circ}+\gamma)$$ and this should be easy to calculate that is less than $${\sqrt{3}\over 3}$$.

If we see $$S'$$ quadratic function on $$x$$ we get: $$4S'\leq {1\over \sqrt{3}}\sin (60^{\circ}+\gamma)\leq {1\over \sqrt{3}}$$ From here we can see that equality is achieved iff $$\gamma = 30^{\circ}$$ and $$x= {\sqrt{3}\over 3} = {2\over 3}v$$ where $$v$$ is altitude of triangle $$ABC$$. That is, equality is achieved iff $$M$$ is gravity centre of $$ABC$$.

I'm interested in different solutions (for example without trigonometry).

• This problem has been posted here before. Unfortunately, it was closed and deleted. But high-reputation users like you can access it: math.stackexchange.com/questions/2993965. (But all answers there use trigonometry or complex numbers to solve the problem.) Nov 18 '18 at 20:55
• Then, I will do this. @achillehui, could you please copy and paste your solution from that link here? I could do it, but I want the upvotes to go to you. Nov 18 '18 at 20:57
• @QuangHoang Could you also please copy and paste your solution from that link here. Nov 18 '18 at 20:58
• I have no problem with that, I will also upvote it. It is nonsense to delete such a solution with artificial reasons.
– Aqua
Nov 18 '18 at 20:59
• @Batominovski answer copied. Nov 18 '18 at 21:34

Reflect $$M$$ with respect to the sides of $$ABC$$. You get an hexagon whose area is $$2S$$: The hexagon can be decomposed as the union between $$A'B'C'$$ (whose side lengths are $$\sqrt{3}MA,\sqrt{3}MB,\sqrt{3}MC$$) and the isosceles triangles $$CA'B',BC'A',AB'C'$$. It follows that $$2S = \frac{\sqrt{3}}{4}(AM^2+BM^2+CM^2)+ 3 S'$$ where $$S'$$ is the area of a triangle with side lengths $$MA,MB,MC$$. By Weitzenbock's inequality $$AM^2+BM^2+CM^2 \geq 4\sqrt{3}S'$$, hence $$S'\leq \frac{S}{3}$$ as wanted.

Solution using complex numbers. Copied from a deleted question per request.

( update - I have added another solution using circle inversion at end)

Solution 1 - using complex numbers.

Choose a coordinate system so that triangle $$ABC$$ is lying on the unit circle centered at origin and $$A$$ on the $$x$$-axis. Let $$a = AM$$, $$b = BM$$, $$c = CM$$ and $$S'$$ be the area of a triangle with sides $$a,b,c$$. In this coordinate system, $$S = \frac{3\sqrt{3}}{4}$$, we want to show $$S' \le \frac{\sqrt{3}}{4}$$. Using Heron's formula, this is equivalent to

$$16S'^2 = (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) \stackrel{?}{\le} 3$$

Identify euclidean plane with the complex plane. The vertices $$A,B,C$$ corresponds to $$1, \omega, \omega^2 \in \mathbb{C}$$ where $$\omega = e^{\frac{2\pi}{3}i}$$ is the cubic root of unity. Let $$z$$ be the complex number corresponds to $$M$$ and $$\rho = |z|$$, we have

$$\begin{cases} a^2 = |z-1|^2 = \rho^2 + 1 - (z + \bar{z})\\ b^2 = |z-\omega|^2 = \rho^2 + 1 - (z\omega + \bar{z}\omega^2)\\ c^2 = |z-\omega^2|^2 = \rho^2 + 1 - (z\omega^2 + \bar{z}\omega) \end{cases} \implies a^2 + b^2 + c^2 = 3(\rho^2 + 1)$$ Thanks to the identity $$\omega^2 + \omega + 1 = 0$$, all cross terms involving $$\omega$$ explicitly get canceled out.

Doing the same thing to $$a^4 + b^4 + c^4$$, we get \begin{align}a^4 + b^4 + c^4 &= \sum_{k=0}^2 (\rho^2 + 1 + (z\omega^k + \bar{z}\omega^{-k}))^2\\ &= \sum_{k=0}^2\left[ (\rho^2 + 1)^2 + (z\omega^k + \bar{z}\omega^{-k})^2\right]\\ &= 3(\rho^2 + 1)^2 + 6\rho^2\end{align} Combine these, we obtain

$$16S'^2 = 3(\rho^2+1)^2 - 12\rho^2 = 3(1 - \rho^2)^2$$ Since $$M$$ is inside triangle $$ABC$$, we have $$\rho^2 \le 1$$. As a result,

$$S' = \frac{\sqrt{3}}{4}(1-\rho^2) \le \frac{\sqrt{3}}{4} = \frac13 S$$

Solution 2 - using circle inversion.

Let $$a = AM, b = BM, c = CM$$ again. Let $$\Delta(u,v,w)$$ be the area of a triangle with sides $$u,v,w$$. In particular, $$S = \Delta(1,1,1)$$ and $$S' = \Delta(a,b,c)$$. We will use the fact $$\Delta(u,v,w)$$ is homogeneous in $$u,v,w$$ with degree $$2$$.

Consider the circle inversion with respect to a unit circle centered at $$A$$. Under such an inversion, $$B,C$$ get mapped to itself while $$M$$ mapped to a point $$M'$$ with $$AM' = \frac{1}{a}, BM' = \frac{b}{a}, CM' = \frac{c}{a}$$

We can decompose the quadrilateral $$ABM'C$$ in two manners. $$\triangle ABC + \triangle BM'C$$ and $$\triangle ABM' + \triangle AM'C$$. This leads to

\begin{align} &\verb/Area/(ABC) + \verb/Area/(BM'C) = \verb/Area/(ABM') + \verb/Area/(AM'C)\\ \iff & S + \Delta(1,\frac{b}{a},\frac{c}{a}) = \Delta(1,\frac{b}{a},\frac{1}{a}) + \Delta(1,\frac{c}{a},\frac{1}{a})\\ \iff & Sa^2 + S' = \Delta(1,a,b) + \Delta(1,a,c) \end{align} By a similar argument, we have $$Sb^2 + S' = \Delta(1,b,a) + \Delta(1,b,c)\quad\text{ and }\quad Sc^2 + S' = \Delta(1,c,a) + \Delta(1,c,b)$$

Summing these three equalities together and notice

$$\Delta(1,a,b) + \Delta(1,b,c) + \Delta(1,c,a) = \verb/Area/(ABM) + \verb/Area/(BCM) + \verb/Area/(CAM) = S$$

We obtain

$$3S' = 2S - S(a^2+b^2+c^2)$$

For any $$\triangle XYZ$$ and point $$P$$ in the plane, we know the expression $$XP^2 + YP^2 + ZP^2$$ is minimized when $$P$$ is the centroid of $$\triangle XYZ$$. For an equilateral triangle of side $$1$$, the centroid is at a distance $$\frac{1}{\sqrt{3}}$$ from the vertices. This implies $$a^2 + b^2 + c^2 \ge 1$$.

As a result, $$3S' \le S$$ and we are done.

• As said...........+1
– Aqua
Nov 18 '18 at 21:42
• @greedoid Thanks, I haved added a new approach which use circle inversion instead of complex numbers. Nov 18 '18 at 22:24