We invoke barycentric coordinates on $ABC$. Let $P = (u:v:w)$, with $uv+vw+wu = 0$ (circumcircle equation with $a=b=c$). Then $D = (0:v:w)$, $E = (u:0:w)$, $F = (u:v:0)$. Hence \begin{align*} \frac{[DEF]}{[ABC]} &= \frac{1}{(u+v)(v+w)(w+u)} \det \begin{bmatrix} 0 & v & w \\ u & 0 & w \\ u & v & 0 \end{bmatrix} \\ &= \frac{2uvw}{(u+v)(v+w)(w+u)} \\ &= \frac{2uvw}{(u+v+w)(uv+vw+wu)-uvw} \\ &= \frac{2uvw}{-uvw} = -2 \end{align*}as desired (areas signed).

https://artofproblemsolving.com/community/c6h1296417_an_equilateral_triangle_and_two_areas cosmin our lord and savior he told everyone at amsp 2016 to look out for virgil nicula's posts

This was quite straightforward to complex bash as well.

Generic_Username wrote: https://artofproblemsolving.com/community/c6h1296417_an_equilateral_triangle_and_two_areas cosmin our lord and savior he told everyone at amsp 2016 to look out for virgil nicula's posts did he really i should not have slept during class

didn't sleep during class and still didn't solve it rip

Wait I trigbashed this one by using law of sines a dozen times, it worked out well and took only 1.5 hours

wu2481632 wrote: didn't sleep during class and still didn't solve it rip Did you try bashing?

Is there synthetic solution yet?, because some of us non-bary folk would like to see a non-bary solution.

I did similar triangles and then products and then converted the products to get the solution

i did the same as mathguy5041 (trig bash)

Interesting note: If you think the chord intersection formula in complex is $\frac{a+d-b-c}{ad-bc}$ it gives the correct area ratio!

JasperL wrote: Is there synthetic solution yet?, because some of us non-bary folk would like to see a non-bary solution. There's a synthetic solution in the link if you want to see one. I'm sure somebody will transfer it later.

i tried to pure complex bash, but putting stuff in shoelace was too ugly

redacted, bad at bashing

can't use area formula with unhomogenized

Just how detailed must a complex bash be? Say if one let $a=1,b=\omega,c=\omega^2$ and found $d,e,f$ and "showed" through expansion and stuff that $d\overline{e}-\overline{d}e+\overline{d}f-d\overline{f}+e\overline{f}-\overline{e}f = -6i\sqrt{3}$ so the complex shoelace thing gives the area as $\frac{3\sqrt{3}}{2}$?

agbdmrbirdyface wrote: @v_Enhance I just did this with bary and using the area formula I just got $2pqr$ as the ratio. That can't possibly be correct because the area ratio should not depend on the choice of homogenization factor.

Ok how was I supposed to be able to solve that then?

I remember I was in only in eighth grade when I took this test, only getting points on the first question (I excited to have solved anything back then). Looking back, the questions seem relatively straightforward now. Here is a relatively clean, low-powered solution I found: [asy][asy] import olympiad; import geometry; size(15cm); defaultpen(fontsize(10pt)); pair A, B, C, D, E, F, P, Q, E1, F1, E2; A = dir(210); B = dir(90); C = dir(330); P = dir(45); D = extension(B, C, A, P); E = extension(A, C, B, P); F = extension(A, B, C, P); fill(A--B--C--cycle, lightgrey); fill(D--E--F--cycle, grey); fill(B--D--F--cycle, mediumgrey); fill(C--D--E--cycle, mediumgrey); draw(A--B--C--cycle, linewidth(1.5)); draw(circumcircle(A, B, C), dotted); draw(B--F, linewidth(1)); draw(F--C, linewidth(1)); draw(B--E, linewidth(1)); draw(E--C, linewidth(1)); draw(F--E, linewidth(1)); draw(A--P, linewidth(1)); dot("$A$", A, dir(240)); dot("$B$", B, dir(120)); dot("$C$", C, dir(270)); dot("$D$", D, dir(170)); dot("$E$", E, dir(300)); dot("$F$", F, dir(90)); dot("$P$", P, dir(60)); [/asy][/asy] Diagram thanks to Awesome_guy Assume WLOG $\triangle ABC$ has side length $1$ and $P$ lies on minor arc of $BC$. This means $[ABC] = \frac{\sqrt{3}}{4}$, so we are tasked to prove that $[DEF] = \frac{\sqrt{3}}{2}$. Now set $BF = m, CE = n, BD = r, CD = s$ (note that $r + s = 1$) Then $$[DEF] = [FAE] - [BDF] - [CDE] - [ABC] = \frac{\sqrt{3}}{4}(1 + m)(1 + n) - \frac{\sqrt{3}}{4}mr - \frac{\sqrt{3}}{4}ns - \frac{\sqrt{3}}{4}$$where we used the $[ABC] = \frac{1}{2}ab\sin{C}$ to get the area of each expression. Taking out the factor of $\frac{\sqrt{3}}{4}$ from each expression and manipulating the variables, with the fact that $r + s = 1$, we get $$\frac{\sqrt{3}}{4}(ms + nr + mn)$$and we need to prove that this equals $\frac{\sqrt{3}}{2}$. Dividing out the constant, we just need to prove that $$ms + nr + mn = 2$$ To do so, we will reduce everything down to just one variable. Let $\angle PBC = (30 + k)^\circ$, thus $\angle PCB = (30 - k)^\circ$ for some $0 \leq k < 30$. Now let's find everything in terms of $k$ (everything will be measured in degrees). By Law of Sines on $\triangle DAB$, we have $\frac{r}{\sin(30 - k)} = \frac{1}{\sin(90 + k)} = \frac{1}{\cos(k)}$, so $r = \frac{\sin(30 - k)}{\cos(k)}$. It follows similarly from $\triangle DAC$ that $s = \frac{\sin(30 + k)}{\cos(k)}$. Now by Law of Sines on $\triangle FBC$, we have $\frac{m}{\sin(30 - k)} = \frac{1}{\sin(30 + k)}$, so $m = \frac{\sin(30 - k)}{\sin(30 + k)}$. It follows similarly from $\triangle ECB$ that $n = \frac{\sin(30 + k)}{\sin(30 - k)}$. Substituting these values in, our expression becomes $\frac{\sin(30 - k) + \sin(30 + k)}{\cos(k)} + 1$. Now notice that the fraction is exactly the same as $r + s$ from the expressions we obtained for $r$ and $s$ earlier (it turns out miraculously that $m = r/s$ and $n = s/r$), so it is just one, and this expression does indeed equal two and the proof in complete $\blacksquare$

More motivated synthetic solution ig: [asy][asy] import olympiad; import geometry; size(15cm); defaultpen(fontsize(10pt)); pair A, B, C, D, E, F, P, Q, E1, F1, E2; A = dir(210); B = dir(90); C = dir(330); P = dir(45); D = extension(B, C, A, P); E = extension(A, C, B, P); F = extension(A, B, C, P); Q = extension(A, foot(A, B, C), E, F); E1 = extension(A, E, D, Q); F1 = extension(A, F, D, Q); E2 = extension(A, P, C, Q); fill(B--C--Q--cycle, lightgrey); fill(D--E--F--cycle, lightgrey); fill(D--extension(D, E, C, Q)--Q--extension(D, F, B, Q)--cycle, mediumgrey); draw(A--B--C--cycle, linewidth(1.5)); draw(circumcircle(A, B, C), dotted); draw(B--Q, heavygrey + dashed); draw(C--E2, heavygrey + dashed); draw(E--E2, heavygrey + dashed); draw(B--F, linewidth(1)); draw(F--C, linewidth(1)); draw(B--E, linewidth(1)); draw(E--C, linewidth(1)); draw(F--E); draw(Q--E1, heavygrey); draw(A--E1, heavygrey); draw(A--E2, heavygrey); draw(circumcircle(C, E, E2), dotted); dot("$A$", A, dir(240)); dot("$B$", B, dir(120)); dot("$C$", C, dir(270)); dot("$D$", D, dir(170)); dot("$E$", E, dir(300)); dot("$F$", F, dir(90)); dot("$P$", P, dir(60)); dot("$Q$", Q, dir(0)); dot("$E_1$", E1, dir(240)); dot("$F_1$", F1, dir(150)); dot("$E_2$", E2, dir(90)); [/asy][/asy] Let the tangents to $(ABC)$ at $B$ and $C$ intersect at $Q$. Let $E_1 = AC\cap DQ$, $F_1 = AB\cap DQ$, and $E_2 = AP\cap CQ$. Since $\triangle ABC$ is equilateral, $\measuredangle CBQ = \measuredangle QCB = 60^{\circ}$, implying that $\triangle ABC \cong \triangle BCQ$. Thus, it suffices to show that $[DEF] = 2 [BCQ]$. By Pascal's theorem, $Q$ lies on $EF$. Thus, if $C$ and $B$ are the midpoints of $EE_1$ and $FF_1$, respectively, then $[DEQ] = 2[CDQ]$, $[DFQ] = 2[BDQ]$, and $$[DEF] = [DEQ] + [DFQ] = 2[CDQ] + 2[BDQ] = 2[BCQ].$$Thus, by symmetry, it suffices to show that $CE_1 = CE$. Note that $\measuredangle ECE_2 = 60^{\circ} = \measuredangle BCA = \measuredangle BPA = \measuredangle EPE_2$, implying that $ECPE_2$ is cyclic. Then since $\measuredangle PEA = \measuredangle PEC = \measuredangle PE_2C$ and $\measuredangle EAP = \measuredangle CAP = \measuredangle E_2CP$, $\triangle AEP \sim \triangle CE_2P$. Thus $\measuredangle CPE_2 = \measuredangle APE = 120^{\circ}$, $\measuredangle E_2EC = 60^{\circ}$, and $\triangle CEE_2$ is equilateral. Thus, by symmetry about $BC$, $CE_1 = CE_2 = CE$, as desired. $\blacksquare$

Complex bashing is fine if you do it correctly for this problem. I don't want to write the expressions out again (they're really not that bad), but if you just work with $A=1, B=z, C=z^2$ where $z=e^{2\pi i/3}$ instead of abstracting it to $a,b,c$ and spam $z^3=1$ as well as the crucial identity $1+z+z^2=0$, you should get that we want to prove $$\left|\begin{matrix}\frac{2p+1}{p-1}&\frac{p+2}{p-1}&1\\\frac{2pz+z^2}{p-z}&\frac{pz^2+2}{p-z}&1\\\frac{2pz^2+z}{p-z^2}&\frac{pz+2}{p-z^2}&1\end{matrix}\right|=-6\sqrt{3}i.$$The original matrix has the first column the negative of what it is here, but $DEF$ is oriented clockwise so we factor out the $-1$ and remove it because determinants give us the signed area (which would be negative). If you expand, say, the top left $2 \times 2$ submatrix, you get $$\frac{2(z-1)(p^2z-2p-(1+z))}{(p-1)(p-z)}.$$But since $1+z=-z^2$, the quadratic factor in $p$ actually equals $p^2z-2p+z^2$, which is then $z(p-z^2)^2$. The same is true (and maybe a bit more clear) for the other submatrices, so the problem actually reduces to proving that $$2z(z-1)((p-1)^3+(p-z)^3+(p-z^2)^3)=6\sqrt{3}i(p-1)(p-z)(p-z^2).$$Compute $z(z-1)=-\sqrt{3}i$, so the equation becomes $$(p-1)^3+(p-z)^3+(p-z^2)^3=3(p-1)(p-z)(p-z^2),$$but then both sides are equal to $3p^3-3$, so we're done.

Surprised this is a J3; thought I wasn't that good but I guess I got better, huh WLOG $P$ lies on arc $BC$ not containing $A$. Now recall the following theorem for equilateral triangles (the motivation behind this sol), which we will prove: \[ \frac{1}{PD} = \frac{1}{PB} + \frac{1}{PC}. \] Indeed by Ptolemy's theorem, we obtain that $PA = PB + PC$, whence $\frac{1}{PB} + \frac{1}{PC} = \frac{PA}{PB \cdot PC}$. Therefore if \[ PD \cdot PA = PB \cdot PC, \]we're done. Indeed, note that $\triangle PBA \sim \triangle PDC$ so we obtain that \[ \frac{PA}{PB} = \frac{PC}{PD}, \]which rearranges into the thing we want, as desired. This form of $PD$'s length is much nicer to work with. From here it isn't hard to see by more similar triangles ($\triangle PBC \sim \triangle EPA \sim \triangle APF$) that $PE = \frac{PA \cdot PC}{PB}$, and $PF = \frac{PA \cdot PB}{PC}$. Now, \[ [DEF] = \frac 12 \sin{120^{\circ}}(PD \cdot PE + PE \cdot PF + PF \cdot PD) = \frac 12 \sin{120^{\circ}}(PA^2 + PB^2 + PC^2). \]Hence it remains to show that $PA^2 + PB^2 + PC^2 = 2S^2$ where $S$ is the side length. However remember that $PA = PB + PC$, whence \[ PA^2 + PB^2 + PC^2 = 2(PB^2 + 2 PB \cdot PC + PC^2) = 2BC^2 \]where the last equality is just the law of cosines. Thus we're done.

We invoke barycentric coordinates with $ABC$ as the reference triangle For the point $P$ we know that if $P=(x_1, y_1, z_1)$ then $a^2y_1z_1 + b^2x_1z_1 + c^2x_1y_1 = 0$. Hoever since $a=b=c$ we know that $x_1y_1 +x_1z_1 + y_1z_1= 0$. We will use this later. We can see that $D = \left(0,\frac{y_1}{y_1+z_1},\frac{z_1}{y_1+z_1}\right)$, $E= \left(\frac{x_1}{x_1+z_1},0,\frac{z_1}{x_1+z_1}\right)$, and $F = \left(\frac{x_1}{x_1+y_1},\frac{y_1}{x_1+y_1},0\right)$ (normalized). This means that we need $\frac{1}{(x_1+y_1)(x_1+z_1)(y_1+z_1)} \cdot \det \begin{bmatrix} 0 & y_1 & z_1 \\ x_1 & 0 & z_1 \\ x_1 & y_1 & 0 \end{bmatrix} = 2$ (from the barycentric area formula). We can see that this determinant is equal to $2x_1y_1z_1$. Thus it remains to prove that $\frac{1}{(x_1+y_1)(x_1+z_1)(y_1+z_1)} = \left| \frac{1}{x_1y_1z_1} \right|$ (areas are signed). We have $$\frac{1}{(x_1+y_1)(x_1+z_1)(y_1+z_1)} = \frac{1}{(x_1y_1 +x_1z_1 + y_1z_1)(x_1+y_1+z_1) - x_1y_1z_1} = -\frac{1}{x_1y_1z_1}$$. $\blacksquare$

aleksam wrote: Not really bashing: Let's put the configuration in the complex plane, and let the circumcircle be the unit circle. Set $a=1$, $b, c$ easily follow. Note that by chord intersection formula, each of points/complex numbers $d, e, f$ is of the form $\frac{g(p)}{h(p)}$, where $g, h\in \mathbb{C}[ x ]$ and $\deg g=\deg h=1$($g$ and $h$ are different for different points), and so are $\bar{d}, \bar{e}, \bar{f}$. Now, if we apply the area formula, we see that the total area of $DEF$ is of the form $\frac{A(p)}{B(p)}$, where $A, B\in \mathbb{C}[ x ]$ and $\deg A=\deg B=1$. Our aim is to prove $A(p)=2[ABC]B(p)$, but as this is a polynomial equation in $p$, we know that it has either less than $3$ solutions or infinitely many of them. So we will just find three solutions, or geometrically, 3 point such that the problem statement holds. Those can be the midpoints of arcs $AB$, $BC$, $CA$ of the circumcircle. If $P$ is the midpoint of the arc $BC$, $P\not\equiv A$, then $D\in BC,DB=DC$, and $E\in AC,CE=CA$(because $\angle EBC=30, \angle BCE =120$, and $BC=CE$), $F\in AB,FB=AB$. So, by area formula, we get $[DEF]=\frac{d(D, EF)EF}{2}=\frac{2d(A, BC)AC}{2}=2[ABC]$., This solution is not correct. Firstly, the area formula gives you the sum of terms of the form $d\overline{e}$, which are $\frac{\text{quadratic}}{\text{quadratic}}$ (using $\overline{p}=\frac{1}{p}$). Second, when you add these terms you get $\frac{f(p)}{g(p)}$ where $f,g$ are of degree $6$ (using the fact that there are $6$ different linear factors appearing in the denominator). This solution can be made to work, however! Use complex numbers with $a=1, b=\omega, c=\omega^2$ for $\omega=e^{2\pi i/3}$. Notice that $d,e,f,$ are rational functions of $p$ that are $\frac{\text{linear}}{\text{linear}}$. Since $\overline{p}=\frac{1}{p}$, $\overline{d},\overline{e},\overline{f}$ are also $\frac{\text{linear}}{\text{linear}}$. There are $6$ different linear factors in the denominator among $\{d,e,f,\overline{d},\overline{e},\overline{f}\}$. We wish to show that the determinant $$\det \begin{vmatrix}d & \overline{d} & 1 \\ e & \overline{e} & 1 \\ f & \overline{f} & 1\end{vmatrix}$$is constant (and the correct constant). This determinant is hence of the form $\frac{f(p)}{g(p)}$, where $f,g$ are of degree $6$. Hence for the right constant $c$, we wish to verify $f(p)-cg(p)$ is identically $0$. This is a polynomial of degree $6$, so in order to verify it's identically zero we need to show that it is zero for $7$ points $p$. Now comes the weird part: For some $p$ we will argue algebraically and for other $p$ we will argue geometrically. Choose $p=1,\omega,\omega^2$. It can be seen that both $f$ and $g$ are identically zero here. Choose $P$ as the midpoint of the arcs $AB,BC,AC$. The argument given in post #66 applies. Choose $p=0$. For this $p$, the determinant is $$\det\begin{vmatrix}-1 & -1 & 1 \\ -\omega & -2/\omega & 1 \\ -1/\omega & -2/\omega^2 & 1\end{vmatrix}=-6\sqrt{3}i$$Which is the required value. We're done because $3+3+1=7$.

No bary, no complex, no trig, just lengths. Thanks Titu! Assume WLOG that $P$ is on minor arc $BC$. Let $AP=a$, $BP=b$, and $CP=c$. By Ptolemy on $ABPC$, we have $a=b+c$. Notice that $\tfrac{BD}{DC}=\tfrac{b}{c}$, $\tfrac{CE}{EA}=\tfrac{c}{a}$, and $\tfrac{AF}{FB}=\tfrac{a}{b}$ by the angle bisector theorem. Thus, we have \begin{align*} \frac{[DEF]}{[ABC]}&=\frac{[AEF]-[BFD]-[CDE]-[ABC]}{[ABC]} \\ &=\frac{AE}{AC} \cdot \frac{AF}{AB}-\frac{BF}{BA} \cdot \frac{BD}{BC}-\frac{CD}{CB} \cdot \frac{CE}{CA}-1 \\ &=\frac{a}{a-c} \cdot \frac{a}{a-b}-\frac{b}{a-b} \cdot \frac{b}{b+c}-\frac{c}{b+c} \cdot \frac{c}{a-c}-1 \\ &=\frac{a}{b} \cdot \frac{a}{c}-\frac{b}{c} \cdot \frac{b}{a}-\frac{c}{b} \cdot \frac{c}{a}-1 \\ &=\frac{a^3-b^3-c^3}{abc}-1. \end{align*} By Titu's favorite factoring trick on $a$, $-b$, and $-c$, we have $a^3-b^3-c^3=3abc$, so $\tfrac{[DEF]}{[ABC]}=2$, as desired.

Did someone ask for a bash with Complex Numbers Coordinates solution above? My favourite method https://infinityintegral.substack.com/p/usajmo-2017-contest-review

Observe DPF=FPE=EPD=120 Observe PE*PD=PC^2 Similarly for PA and PB Then find PA, PB, PC in terms of AB (using trig) and we are done

assume $AB=1$, the rest of the proof is the same since we can just scale the figure up draw $GH$ through $A$ and parallel to $BC$, where $GBE$ and $FCH$ are collinear we can easily prove that $AGFP$ and $AHEP$ are both concyclic, and by repeatedly plugging law of sines, we get $BF=\frac{\sin{PAF}}{\sin{CAP}}$, $CE=\frac{\sin{CAP}}{\sin{PAF}}$, $BD=\frac{\sin{PAF}}{\sin{CAP+60}}$, and $CD=\frac{\sin{CAP}}{\sin{PAF+60}}$, $S[ABC]=1*1*\sin{60}$,$S[BCEF]=\frac{\sin{PAF}^2+\sin{CAP}^2+\sin{PAF}\sin{CAP}}{\sin{PAF}\sin{CAP}}*\sin{60}$,$S[BDF]=\frac{\sin^2{PAF}}{\sin{CAP}\sin{CAP+60}}*\sin{60}$,$S[CDE]=\frac{\sin^2{CAP}}{\sin{PAF}\sin{PAF+60}}*\sin{60}$, and the rest comes from trig bashing

Would complex bash make more sense for this? Let, where $\omega = e^{\frac{2i\pi}{3}}$ $A = \omega^0$ $B = \omega^1$ $C = \omega^2$ $P = p$ Intersect it and use $Area = \frac{i}{4} \dots$ (I think this would work).

We use barycentric coordinates with $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$. Also WLOG the side length of $\triangle ABC$ is $1$. Let $P=:(x,y,z)$ so $xy+yz+zx=0$ since $P$ lies on $(ABC)$. Then \begin{align*} D&=(0:y:z)\\ E&=(x:0:z)\\ F&=(x:y:0) \end{align*}so \begin{align*} \frac{[DEF]}{[ABC]}&=\frac{1}{(y+z)(z+x)(x+y)} \begin{vmatrix} 0 & y & z\\ x & 0 & z\\ x & y & 0 \end{vmatrix}\\ &=\frac{2xyz}{(1-x)(1-y)(1-z)}\\ &=\frac{2xyz}{1-x-y-z+xy+yz+zx-xyz}\\ &=-2, \end{align*}so we are done. $\square$

Set $A=(1,0,0)$ and etc. Also, let $P=(x,y,z)$ for some $x,y,z\in \mathbb{R}$. Since $P$ lies on the circumcircle of $\triangle ABC$, we have that \[a^2(yz+zx+xy)=0 \implies yz+zx+xy=0\]since $x+y+z=1$, this implies that $\sum_{cyc}x^2y + 3xyz= (x+y+z)(xy+yz+zx)=0$. Now, since $D$ lies on $AP$, \[D=(0:y:z)=\left(0,\frac{y}{y+z},\frac{z}{y+z}\right)\]Similarly we also have that $E = \left(\frac{x}{x+z},0,\frac{z}{x+z}\right)$ and $F= \left(\frac{x}{x+y},\frac{y}{x+y},0\right)$. Now, we are left with a simple calculation. We know that \begin{align*} [DEF] &= [ABC] \begin{vmatrix} 0 & \frac{y}{y+z} & \frac{z}{y+z}\\ \frac{x}{x+z} & 0 & \frac{z}{x+z}\\ \frac{x}{x+y} & \frac{y}{x+z} &0 \end{vmatrix}\\ &= [ABC] \left(\frac{2xyz}{(x+y)(y+z)(x+z)}\right)\\ &= 2[ABC] \left(\frac{xyz}{\sum_{cyc}x^2y + 2xyz}\right)\\ &= 2[ABC] \left(\frac{xyz}{(\sum_{cyc}x^2y +3xyz) - xyz}\right)\\ &= 2[ABC] \left(\frac{xyz}{-xyz}\right)\\ &= -2[ABC] \end{align*}which proves the desired result.

WLOG $P$ is on minor arc $BC$ w.r.t. $(ABC)$, now since $\angle ADC = 60^\circ = \angle ADB$ we have \[ \angle APE = \angle APC + \angle CPE = \angle APB + \angle BPC = \angle FPA. \]Moreover, $\angle BAC = 60^\circ = \angle APB$ gives \[\angle PEA = 180^\circ - \angle BAC - \angle ABP = 180^\circ - \angle BPA - \angle ABP = \angle PAF. \]Therefore by AA Similarity we have $\triangle APE \sim \triangle FPA \implies (PE)(PF) = PA^2$. Analogously, $(PD)(PE) = PC^2, (PF)(PD) = PB^2$. It is easy to show $\angle DPE = \angle EPF = \angle FPD = 120^\circ$ by angle chasing, now by area decomposition and Law of Sines we have \[[DEF] = [DPE] + [EPF] + [FPD] = \dfrac{\sqrt{3}}{4} \bigg[ (PD)(PE) + (PE)(PF) + (PF)(PD) \bigg] = \dfrac{\sqrt{3}}{4} \bigg[ PA^2 + PB^2 + PC^2 \bigg] .\]WLOG $(ABC)$ is the unit circle, since $[ABC] = \dfrac{3\sqrt{3}}{4}$ we wish to show $PA^2 + PB^2 + PC^2 = 6$. We employ complex numbers, let $A, B, C$ be represented by $1, \omega, \omega^2$ where $\omega = e^{2 \pi i / 3}$ and let $P$ be denoted by $p$. Then \begin{align*} PA^2 + PB^2 + PC^2 &= |p - 1|^2 + |p - \omega|^2 + |p - \omega^2|^2 \\ &= (p - 1) (\overline{p} - 1) + (p - \omega)(\overline{p} - \omega^2) + (p - \omega^2)(\overline{p} - \omega) \\ &= (p - 1) \left( \dfrac{1}{p} - 1 \right) + (p - \omega) \left(\dfrac{1}{p} - \omega^2 \right) + (p - \omega^2) \left( \dfrac{1}{p} - \omega \right) \\ &= 6 - (1 + \omega + \omega^2) p - (1 + \omega + \omega^2) \dfrac{1}{p} \\ &= 6 \end{align*}as wanted. The end.