Area related question for an equilateral triangle

Question

The question is

Let $ABC$ be an equilateral triangle with side $a$. $M$ is a point such that $MS=d$ where $S$ is the centre of $\Delta ABC$. Prove that the area of the triangle whose sides are of length $MA,MB,MC$  is  $[\sqrt 3/12] |a^2−3d^2|.$

I have been trying this question since long with many approaches but none makes me solve this question

My approach was confined to find $MA ,MB ,MC$ for that I dropped perpendicular from $M$ to the base $BC$ and assumed the unknown length to be $x$ and $\angle {MBC}$ as $\theta$ , then I tried to applied trigonometry and properties of an equilateral triangle but I finally had one variable always left $\theta$

Secondly i have tried certain constructions also but i will not mention since they may not take us to any result

This is a pure geometry question (includes construction,trigonometry, theorems related to triangle and circles) question so please do not use co-ordinate geometry to solve the question and please be generalise in your solution and do not take $M$ as any special point

This question although have an answer on a site called doubtnut but again they have used coordinate geometry and have assumed $M$ to lie on $AS$ which reduces our work.

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Edit : I put the bars of restriction away you can now provide a solution using any sort of mathematical tool but the solution must be short elegant and simple.

Answer 1

If you want to prove this using only basic geometry and trigonometry, you'll need the Heron's formula (which can be proven without any usage of coordinates, say with the law of cosines)

Given a triangle with sides of length $\alpha, \beta, \gamma$ and $p=(\alpha+\beta+\gamma)/2$, the area of the triangle is $$ \mathrm{Area} = \sqrt{p(p-\alpha)(p-\beta)(p-\gamma)}=\frac{1}{4}\sqrt{2E-F} $$ with $E=\alpha^2\beta^2+\alpha^2\gamma^2+\beta^2\gamma^2$ and $F=\alpha^4+\beta^4+\gamma^4$.

Now let $\theta =\angle ASM$. Then $\angle BSM$ and $\angle CSM$ are $\theta \pm 2\pi/3$ (which one is which doesn't matter, we'll assume $\angle BSM=\theta+2\pi/3$). Let $s=|AS|=|BS|=|CS|$ which one computes to be $s=a/\sqrt{3}$. Then, by law of cosines $$ \begin{aligned} \alpha^2&:=|MA|^2 = d^2+s^2-2ds\cos \theta\\ \beta^2&:=|MB|^2 = d^2+s^2-2ds\cos (\theta+2\pi/3)\\ \gamma^2&:=|MC|^2 = d^2+s^2-2ds\cos (\theta-2\pi/3) \end{aligned} $$ So it remains to combine the above formulas with Heron and simplify using trigonometry and algebra. I'll leave most of this as an exercise. First show that following (at least for $n=3$) $$ \begin{aligned} &\sum_{k=0}^{n-1}\cos(t+2\pi k/n)=0\\ &\sum_{0\leq k<l\leq n-1}\cos(t+2\pi k/n)\cos(t+2\pi l/n)=-\frac{n}{4}\qquad (n>2) \end{aligned} $$ Using that show $$ E=3(d^4+d^2s^2+s^4), \qquad F = 3(d^4+4d^2s^2+s^4) $$ So $2E-F = 3(d^2-s^2)^2$ $$ \mathrm{Area} = \frac{\sqrt{3}}{4}|d^2-s^2| = \frac{\sqrt{3}}{12}|a^2-3d^2| $$

Answer 2

Let $S(0,0)$ and $M(x,y)$.

Thus, $$\sum_{cyc}MA^2=\sum_{cyc}((x-x_A)^2+(y-y_A)^2)=$$ $$=3(x^2+y^2)-2x\sum_{cyc}x_A-2y\sum_{cyc}y_A+\sum_{cyc}(x_A^2+y_A^2)=$$ $$=3d^2+3\cdot\frac{a^2}{3}=3d^2+a^2.$$ Now, let $M$ is placed inside the triangle and $M_A,$ $M_B$ and $M_C$ are placed outside the triangle such that

$B$ and $M_A$ are placed in the different half planes in respect to $AC$,

$C$ and $M_B$ are placed in the different half planes in respect to $AB$,

$A$ and $M_C$ are placed in the different half planes in respect to $BC$ and triangles

$AMM_A,$ $BMM_B$ and $CMM_C$ are equilaterals.

Now, we see that $\Delta AMB\cong AM_AC,$ $\Delta BMC\cong\Delta BM_BA$ and $\Delta CMA\cong CM_CB,$ which gives $$S_{AM_BBM_CCM_A}=2S_{\Delta ABC}=\frac{a^2\sqrt3}{2}.$$ In another hand, $$\Delta AM_BM\cong\Delta M_CMB\cong\Delta MCM_A$$ and they are congruent to the triangle, which area we need to get.

Let this area is equal to $S$.

Thus, $$S_{AM_BBM_CCM_A}=3S+\sum_{cyc}S_{\Delta AMM_A}=3S+\frac{\sqrt3}{4}\sum_{cyc}MA^2=3S+\frac{\sqrt3}{4}(3d^2+a^2).$$ Id est, $$\frac{a^2\sqrt3}{2}=3S+\frac{\sqrt3}{4}(3d^2+a^2),$$ which gives $$S=\frac{\sqrt3}{12}(a^2-3d^2).$$ If $M$ is placed outside the triangle, so we can use the same way.

Answer 3

General framework, notations, etc.

First, please let me use an other letter for the side of $ABC$, say $e$, instead of $a$, sorry, later i am using formulas (Heron) for the area of a triangle with sides $a,b,c$, and really need a free variable $a$.

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We can rescale and show the formula for a specific value of $e=AB=BC=CA$, my choice of a "normed value" is such that $SA=SB=SC=1$, so $$ e=\sqrt 3\ . $$

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1.st solution, we use complex numbers.

Let us use complex numbers, then see if there is a pattern to follow.

We arrange now that $S$ has the afix $0\in \Bbb C$, and the vertices $A,B,C$ are placed in the points $1,u,u^2$, here $u$ being the primitive root of order three in the upper half plane, $$ u=\frac 12(-1+i\sqrt 3)=\exp\frac{2\pi}3\ . $$ We will try to isolate only algebraic numbers in $u$, so using $1+u+u^2=0$ we can rephrase powers of $u$, or $\bar u$, in terms of affine expressions in $u$.

Let $M$ have affix $z$. The relation $MS=d$ translates as $$ d^2 = MS^2=|z-0|^2=z\;\bar z\ . $$ Then the distances $MA^2$, $MB^2$, $MC^2$ are, using now the letters $a,b,c$: $$ \begin{aligned} a^2=|z-1|^2 &= (z-1)\overline{(z-1)} = d^2-z-\bar z+1\ ,\\ b^2=|z-u|^2 &= (z-u)\overline{(z-u)} = d^2-uz-u^2\bar z+1\ ,\\ c^2=|z-u^2|^2 &= (z-u^2)\overline{(z-u^2)} = d^2-u^2z-u\bar z+1\ ,\\ &\qquad\text{ so using $1+u+u^2=0$, we get a first useful geometrical relation,}\\ a^2+b^2+c^2 &= 3(d^2+1)\ , \end{aligned} $$ and the area of the triangle with sides $a,b,c$ and half perimeter $p=\frac 12(a+b+c)$ is (Heron) $$ \sqrt{p(p-a)(p-b)(p-c)} = \frac 14\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)} $$ and the expression under the square root can be easily expanded in terms of $a^2$, $b^2$, $c^2$. This is the computational idea, we only need to fill in the details to have a dry, analytical proof. $$ \begin{aligned} &(a+b+c)(a+b-c)(a-b+c)(-a+b+c) \\ &\qquad =2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4 \\ & \qquad =4(a^2b^2+b^2c^2+c^2a^2)-(a^2+b^2+c^2)^2 \\ & \qquad =(a^2+b^2+c^2)^2-2(a^4+b^4+c^4) \ . \end{aligned} $$ We can compute explicitly one of the last two expression, i will do it for the last one.

Let $v$ run in the set with elements $1,u,u^2$ below, $$ \begin{aligned} a^4+b^4+c^4 &= \sum_v((d^2+1)-(uz+\bar u\bar z))^2 \\ &= 3(d^2+1)^2 - 2(d^2+1)z\underbrace{\sum v}_{=0} - 2(d^2+1)\bar z\underbrace{\sum \bar v}_{=0} + \sum_v(vz+\bar v\bar z)^2 \\ &= 3(d^2+1)^2 + \sum_v 2\; vz\;\bar v\bar z \\ &=3(d^2+1)^2+6d^2 \ . \end{aligned} $$ Putting all together: $$ \begin{aligned} &(a+b+c)(a+b-c)(a-b+c)(-a+b+c) \\ & \qquad =(a^2+b^2+c^2)^2-2(a^4+b^4+c^4) \\ & \qquad =3^2(d^2+1)^2-2\cdot 3(d^2+1)^2-2\cdot 6d^2 \\ & \qquad =3(d^2+1)^2-2\cdot 6d^2 \\ & \qquad =3(d^2-1)^2 \\ & \qquad \text{and we pass to the homogenized version} \\ &\qquad =3\left(d^2-\frac 13 e^2\right)^2 =\frac 13(3d^2-e^2)^2 \ . \end{aligned} $$ We apply Heron, extract the square root from the above, divide by four, get the needed result in terms of $|e^2-3d^2|$.

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Note: This was of course not satisfactory, i am trying to convert the above into a synthetical or at least more structural proof.

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2.nd solution, algebraic geometry and Galois theory. This is exactly the first solution, but without (straightforward) calculus.

Since it rather needs some more advanced knowledge, this solution is written for the audience which feels home here.

As above, we can write the area of the triangle $\Delta$ with sides $a=MA$, $b=MB$, $c=MC$, using a certain symmetric polynomial expression in $a^2,b^2,c^2$, fixed by the Galois morphism $u\to u^2$ on the coefficients, and by $z\to\bar z$, and depending only on $$ \begin{aligned} a^2+b^2+c^2&=\sum_{v=1,u,u^2}(d^2-vz-\bar v\bar z+1) \ ,\\ a^4+b^4+c^4&=\sum_{v=1,u,u^2}((d^2+1)-(uz+\bar u\bar z))^2 \ , \end{aligned} $$ in the first one there is no $z$ surviving, in the second one only expressions in $z\bar z=d^2$ are remaining. We do not need more.

So far we have the following algebraic information:

Using the Heron formula, the expression for the squared area of $\Delta$, is a polynomial of degree four in $d$, even more, a polynomial of degree two in $d^2$.

Now, this polynomial vanishes iff $d^2=1$ as a consequence of the degenerated cases $M\in\{A,B,C\}$, and/or by the Theorem of Ptolemy, applied in our special case for a point on the circle $(ABC)$. (We need Ptolemey to see there is no other vanishing.) So far we expect a polynomial formula with the factor $(d^2-1)$, but since the squared area is positive we need at least the factor $(d^2-1)^2$. We reach the maximal degree, so the formula for the area is of the shape $$ C(d^2-1)^2\ , $$ with a suitable constant $C$ which can be obtained from the special case $M=S$, $d=0$.

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Note: From the second we know that the formula for the squared area of $\Delta$ depends only on $d^2=z\bar z$, we may go to the special case from the OP where $M$ is taken on one of the lines $AS$, $BS$, $CS$. I suspected the proof went in some earlier versions in this way, but lost some of the difficult arguments in time.

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3.rd solution, (same) computations are done with basic geometry knowledge, Heron, generalized Pythagoras.

It is finally time for a picture:

Here, $M$ was chosen without loss of generality in the ("smaller") angle $\widehat{ASC}$. Perpendiculars were from $M$ drawn to $AS$, $CS$, and the prolongation of $BS$. The foot points of the perpendiculars are $A_1\in[SA$, $C_1\in [SC$, and $B_1$ on the halfline $[BS$ ("after $S$). Note the slight asymmetry. Then one has by the above colored picture proof: $$ SB_1=SA_1+SC_1\ . $$
And now we come back to the same computational idea, in the simpler still most general situation $SA=SB=SC=1$, $MS=d$, we express everything in terms of $$ \begin{aligned} a^2 &=AM^2=SA^2+SM^2-2\; SA\cdot SM\;\cos\widehat{ASM} \\ &=1^2 +d^2-2SA_1\ ,\\ b^2 &=BM^2=SB^2+SM^2-2\; SB\cdot SM\;\cos\widehat{BSM} \\ &=1^2 +d^2\color{red}{+}2SB_1\ ,\\ c^2 &=CM^2=SC^2+SM^2-2\; SC\cdot SM\;\cos\widehat{CSM} \\ &=1^2 +d^2-2SC_1\ . \\[3mm] 16\operatorname{Area}(\Delta)^2 &=(a+b+c)(a+b-c)(a-b+c)(-a+b+c) \\ & =2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4 \\ & =(a^2+b^2+c^2)^2-2(a^4+b^4+c^4) \\ &= \Big(\ 3(1^2 +d^2)-2\underbrace{(SA_1-SB_1+SC_1)}_{=0}\ \Big)^2 \\ &\qquad-2(1^2 +d^2-2SA_1)^2\\ &\qquad-2(1^2 +d^2+2SB_1)^2\\ &\qquad-2(1^2 +d^2-2SC_1)^2 \ . \end{aligned} $$ We further only need $SA_1^2+SB_1^2+SC_1^2$ to conclude. Using the generalized Pythagoras and $\cos \widehat {C_1SA_1}=\cos 120^\circ=-\frac 12$, $$ SA_1^2+SB_1^2+SC_1^2\\ = SA_1^2+(SA_1+SC_1)^2+SC_1^2\\ = 2(SA_1^2+SA_1\cdot SC1 +SC_1^2)\\ =2A_1C_1^2=2SV^2= 2\cdot \frac 34 SM^2=\frac 32 d^2\ , $$ using $\widehat{VSM}=30^\circ$. (Since $SM$ diameter, and we know a $60^\circ$ angle.)

The (same) computation finishes now in mind.

$\square$