For each $A_i$ Notice that : $\overline{PA_{i-1}}\cdot \overline{PA_{i+1}}=\overline{PA_{i+2}}\cdot \overline{PB_i}\Leftrightarrow \overline{A_iB_i}=\frac{\overline{PA_{i-1}}\cdot \overline{PA_{i+1}}-\overline{PA_{i+2}}\cdot \overline{PA_i}}{\overline{PA_{i+2}}}$ Now : $\frac{1}{OA_i^2-r_i^2}=\frac{1}{\overline{A_iA_{i+2}}\cdot \overline{A_iB_i}}=\frac{\overline{PA_{i+2}}}{\overline{A_iA_{i+2}}\cdot ( \overline{PA_{i-1}}\cdot \overline{PA_{i+1}}-\overline{PA_{i+2}}\cdot \overline{PA_i}} )$ $\small \frac{1}{OA_i^2-r_i^2}+\frac{1}{OA_{i+2}^2-r_{i+2}^2}=\frac{1}{\overline{PA_{i-1}}\cdot \overline{PA_{i+1}}-\overline{PA_{i+2}}\cdot \overline{PA_i}}$ Our result immediately follows

This basically dies immediately if you realize that the standard barycentric equation for a circle (namely, $-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)$) actually gives the power of the point $(x,y,z)$ when $x+y+z=1$. (The reason is the same as the corresponding statement in Cartesian coordinates.) Set $A_2A_3A_4$ as the reference triangle, and let $A_1 = (x_1, y_1, z_1)$ with sum one. Do a few computations and the quantity in the problem becomes \[ \frac{1}{-a^2y_1z_1 -b^2z_1x_1 - c^2x_1y_1} \left( -1 + x_1 + y_1 + z_1 \right) = 0 \]

Another one of the official solutions: suppose the circumcircle of $A_{i}A_{i+1}A_{i+2}$ has equation $C_{i}(x,y) = x^{2}+y^{2}+\ell_{i}(x,y)=0$, where $\ell_{i}(x,y)$ is a linear polynomial in $x,y$. Consider the quadratic polynomial $\sum_{i=1}^{4}\dfrac{C_{i}(x,y)}{O_{i}A_{i}^{2}-r_{i}^{2}}-1$, whose zero locus must be a circle (it is the sum of four circles), and observe that each of the $A_{i}$ is in the zero locus because three terms will evaluate to zero and the fourth will evaluate to 1. However, quadrilateral $A_{1}A_{2}A_{3}A_{4}$ is not cyclic, meaning that in fact $P(x,y) = 0$ (or is a line, but this is also impossible). Computing the $x^{2}+y^{2}$ coefficient of $P$ immediately gives the desired result.

v_Enhance wrote: This basically dies immediately if you realize that the standard barycentric equation for a circle (namely, $-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)$) actually gives the power of the point $(x,y,z)$ when $x+y+z=1$. (The reason is the same as the corresponding statement in Cartesian coordinates.) Set $A_2A_3A_4$ as the reference triangle, and let $A_1 = (x_1, y_1, z_1)$ with sum one. Do a few computations and the quantity in the problem becomes \[ \frac{1}{-a^2y_1z_1 -b^2z_1x_1 - c^2x_1y_1} \left( -1 + x_1 + y_1 + z_1 \right) = 0 \] can you tell me how to get a power of the pointA4 ? There is no information about the center of the circle in your circle's barycentric equation.

sami9000 wrote: v_Enhance wrote: This basically dies immediately if you realize that the standard barycentric equation for a circle (namely, $-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)$) actually gives the power of the point $(x,y,z)$ when $x+y+z=1$. (The reason is the same as the corresponding statement in Cartesian coordinates.) Set $A_2A_3A_4$ as the reference triangle, and let $A_1 = (x_1, y_1, z_1)$ with sum one. Do a few computations and the quantity in the problem becomes \[ \frac{1}{-a^2y_1z_1 -b^2z_1x_1 - c^2x_1y_1} \left( -1 + x_1 + y_1 + z_1 \right) = 0 \] can you tell me how to get a power of the pointA4 ? There is no information about the center of the circle in your circle's barycentric equation. Ah I mean the power of A1

Well, if you look closely at the proof of the circle formula, it's actually just $d^2-R^2=0$, so instead of setting the quantity equal to zero we just evaluate it at the (normalized) point we care about to get the power of the point. In particular, since the circumcircle has equation $\omega: -a^2yz-b^2zx-c^2xy=0$, the power of $A_1=(x_1, y_1 z_1)$ to $(A_2A_3A_4)$ in \[ \text{Pow}_{(A_2A_3A_4)} (A_1) = -a^2y_1z_1 - b^2z_1x_1 - c^2x_1y_1\] I guess I might as well write out the other details now: the other circles are \begin{align*} (A_1A_3A_4): \qquad & 0 = -a^2yz-b^2zx-c^2xy + (x+y+z)\left( \frac{N}{x_1} x \right) \\ (A_1A_2A_4): \qquad & 0 = -a^2yz-b^2zx-c^2xy + (x+y+z)\left( \frac{N}{y_1} y \right) \\ (A_1A_2A_3): \qquad & 0 = -a^2yz-b^2zx-c^2xy + (x+y+z)\left( \frac{N}{z_1} z \right) \end{align*} where $N=a^2y_1z_1 + b^2z_1x_1 + c^2x_1y_1$. You can get all of these just by plugging in the three points and solving for $u,v,w$ (you get $0$'s for two of them by plugging in the vertices, and then solve for the last value.) So now, since $A_2 = (1,0,0)$, \begin{align*} \text{Pow}_{(A_1A_3A_4)} (A_2) &= -a^2 \cdot 0 \cdot 0 - b^2 \cdot 0 \cdot 1 - c^2 \cdot 1 \cdot 0 + (1+0+0) \left( \frac{N}{x_1} \cdot 1 \right) \\ &= \frac{N}{x_1} \end{align*} So doing the same thing with the others, we find that the sum we seek is just \[ \frac{1}{-N} + \frac{x_1}{N} + \frac{y_1}{N} + \frac{z_1}{N} =0 \]

My solution uses complex numbers: It is clear that $|o_1-a_3|=r_1$ and $|o_3-a_1|=r_3$. So $\frac{1}{{O_1A_1}^2-{r_1}^2}+\frac{1}{{O_3A_3}^2-{r_3}^2}=\frac{|o_1-a_1|^2+|o_3-a_3|^2-|o_1-a_3|^2-|o_3-a_1|^2}{(|o_1-a_1|^2-|o_1-a_3|^2)(|o_3-a_3|^2-|o_3-a_1|^2}=\frac{2(a_3-a_1)\cdot(o_1-o_3)}{(a_1 \cdot a_1+2o_1 \cdot a_3-2o_1 \cdot a_1-a_3 \cdot a_3)(a_3 \cdot a_3-a_1 \cdot a_1-2o_3 \cdot a_3+2o_3 \cdot a_1)}$ Obtain a similar expression for $\frac{1}{{O_2A_2}^2-{r_2}^2}+\frac{1}{{O_4A_4}^2-{r_4}^2}$. Clearing all the denominators and using the fact that $\frac{a_3-a_1}{o_2-o_4}=c_1i,\frac{a_2-a_4}{o_1-o_3}=c_2i$ for some $c_1,c_2 \in \mathbb{R}$ we get the result. BTW I think there is a more smart way of bashing.

Metric bashing works! Let the circles be $\omega_1, \omega_2, \omega_3, \omega_4$ respectively, let $A_1A_3 \cap A_2A_4 = I$, and let $IA_i = y_i$ for $1 \le i \le 4$. The desired quantity is just $\sum_{i=1}^{4} \frac{1}{p(A_i, \omega_i)}$. Consider $p(A_1, \omega_1) = A_1B_1 \cdot A_1A_3$. By Power of a Point, $IB_1 = \frac{y_2 y_4}{y_3}$. Thus $p{(A_1, \omega_1}) = A_1 B_1 \cdot A_1A_3 = (y_1 - \frac{y_2 y_4}{y_3}) (y_1+y_3)$. Analogous computations and summing yields the desired. To avoid configuration issues and what not in the above solution, one could use directed segments, as a side note.

A funny solution. Note that in the pursuit of avoiding a full-bash, we discover some interesting general ideas, at the cost of inflating the solution. WakeUp wrote: Let $A_1A_2A_3A_4$ be a non-cyclic quadrilateral. Let $O_1$ and $r_1$ be the circumcentre and the circumradius of the triangle $A_2A_3A_4$. Define $O_2,O_3,O_4$ and $r_2,r_3,r_4$ in a similar way. Prove that \[\frac{1}{O_1A_1^2-r_1^2}+\frac{1}{O_2A_2^2-r_2^2}+\frac{1}{O_3A_3^2-r_3^2}+\frac{1}{O_4A_4^2-r_4^2}=0.\] Proposed by Alexey Gladkich, Israel Consider the following result. Claim: Let $ABC$ be a triangle and $\ell$ be a line. Point $P$ varies uniformly over line $\ell$, say, at distance $p \in \mathbb{R}$ from a fixed point; then the power of point $A$ in circle $(BPC)$ is given by a rational function $\frac{f(x)}{g(x)}$ in $p$ with $\text{deg} \, f \le 2$ and $\text{deg} \, g \le 1$. (Proof) Notice that the equation of $(BPC)$ is $$ux(x+y+z)-a^2yz-b^2zx-c^2xy=0.$$Plug $A=(1,0,0)$ to see that it has power equal to $u$. Plug $P=(x', y', z')$ to see that $u$ has the desired form since $x', y', z'$ are linear functions of $p$ with $x'+y'+z'=1$. $\blacksquare$ Fix $\triangle A_1A_2A_3$ and vary $A_4$ uniformly along a fixed line $\ell$ intersecting $(A_1A_2A_3)$ twice, with parameter $z$. Consider $$\frac{f_i(z)}{g_i(z)} \overset{\text{def}}{:=} O_iA_i^2-R_i^2,$$for all $i$; then we wish to show that $$\sum_{1 \le i \le 4} \frac{g_i(z)}{f_i(z)}=0.$$Here, $g$'s are linear (or constant) and $f$'s are quadratic (or lower degree). Notice that when $A_4$ is on $(A_1A_2A_3)$, all these rational functions vanish. Consequently, there exists a quadratic polynomial $h(z)$ which is a factor of $f_i(z)$ for all $1 \le i \le 4$. Hence, the desired equation is a linear equation in $z$. Notice that $\ell$ intersects at least two of the sides of $\triangle A_1A_2A_3$. For $A_4$ on line $A_2A_3$, we see that $$\sum_{i} \frac{1}{O_iA_i^2-R_i^2}=\frac{1}{\overline{A_2A_4} \cdot \overline{A_2A_3}}+\frac{1}{\overline{A_3A_4} \cdot \overline{A_3A_2}}+\frac{1}{\overline{A_4A_2} \cdot \overline{A_4A_3}},$$which is clearly zero. Hence, for all points on a line $\ell$ meeting $(A_1A_2A_3)$ twice, the result is true. It is clear now that it is true for every point $A_4$ on the plane. $\blacksquare$

Purely synthetic: [asy][asy] unitsize(0.3inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -17.78447466400611, xmax = 10.716608257349337, ymin = -24.204266321400723, ymax = 10.590805745087307; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-2.431816913201228,2.1907507675132574), 4.675230780284219), linewidth(2) + wrwrwr); draw(circle((-4.962982129945255,2.630953413903523), 6.661451793359984), linewidth(2) + wrwrwr); draw(circle((-4.717642515394023,0.8078262781866162), 5.403143342102441), linewidth(2) + wrwrwr); draw(circle((-2.82584980263733,0.02715199279138752), 6.496479391938767), linewidth(2) + wrwrwr); draw((-6.56636972167397,9.09656152342805)--(1.7253148666398948,-4.6087022589089175), linewidth(2) + wrwrwr); draw((1.22,5.11)--(-8.6,-2.95), linewidth(2) + wrwrwr); /* dots and labels */ dot((-4.82,6.21),dotstyle); label("$A$", (-4.721478325051531,6.458148721490775), NE * labelscalefactor); dot((1.22,5.11),dotstyle); label("$B$", (1.5012581127777422,5.436859916808873), NE * labelscalefactor); dot((0.02,-1.79),dotstyle); label("$C$", (0.36121479592352435,-2.0921761549158404), NE * labelscalefactor); dot((-8.6,-2.95),dotstyle); label("$D$", (-9.305402494902864,-3.374724886376833), NE * labelscalefactor); dot((-6.56636972167397,9.09656152342805),linewidth(4pt) + dotstyle); label("$E$", (-6.4790451052017834,9.28450611119185), NE * labelscalefactor); dot((1.7253148666398948,-4.6087022589089175),linewidth(4pt) + dotstyle); label("$F$", (2.332539697983943,-4.823529934879065), NE * labelscalefactor); dot((-0.2749610454290614,3.8829749464197305),linewidth(4pt) + dotstyle); label("$G$", (-1.0163375452753223,3.798047648830939), NE * labelscalefactor); dot((-6.007065238946761,-0.8217867439827797),linewidth(4pt) + dotstyle); label("$H$", (-6.764055934415338,-0.9521328380616251), NE * labelscalefactor); dot((-2.371216156447541,2.1624233990868444),linewidth(4pt) + dotstyle); label("$I$", (-2.9876624473357407,2.1354844784185416), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We will assume the above configuration, the others can be handled simalarly (I could do the whole thing out with directed lengths but this is easier). Let $\omega_i=(A_{i+1}A_{i+2}A_{i+3})$. Let $\mathcal{P}$ be the statement that \[\sum_i\frac{1}{OA_i^2-r_i^2}=0.\]Note that $OA_i^2-r_i^2$ is the power of $A_i$ with respect to $\omega_i$, so in our configuration, we want to show that \[\frac{1}{AE\cdot AC}+\frac{1}{CA\cdot CF} = \frac{1}{BG\cdot BD}+\frac{1}{DH\cdot DB}.\]Thus, \[\mathcal{P}\iff \frac{1}{AC}\left(\frac{1}{AC}+\frac{1}{CF}\right) = \frac{1}{BD}\left(\frac{1}{BG}+\frac{1}{DH}\right).\]Let $I=AC\cap BD$. Then power of a point on $I$ tells us that \[IA\cdot IC = IG\cdot ID = IH\cdot IB =: \alpha\]and \[IB\cdot ID = IE\cdot IC = IF\cdot IA =: \beta.\]Therefore, \begin{align*} \frac{1}{AC}\left(\frac{1}{AE}+\frac{1}{CF}\right) &= \frac{1}{IA+IC}\left(\frac{1}{IE-IA}+\frac{1}{IF-IC}\right) \\ &= \frac{1}{IA+IC}\left(\frac{1}{\beta/IC-IA}+\frac{1}{\beta/IA-IC}\right) \\ &= \frac{1}{\beta-\alpha} \end{align*}and \begin{align*} \frac{1}{BD}\left(\frac{1}{BG}+\frac{1}{DH}\right) &= \frac{1}{IB+ID}\left(\frac{1}{IB-IG}+\frac{1}{ID-IH}\right) \\ &= \frac{1}{IB+ID}\left(\frac{1}{IB-\alpha/ID}+\frac{1}{ID-\alpha/IB}\right) \\ &= \frac{1}{\beta-\alpha}, \end{align*}so the truth of $\mathcal{P}$ is verified, as desired.

Congratulations!

s t o r a g e We use barycentric coordinates with $A_1A_2A_3$ as the reference triangle. Let the coordinates of $A_4$ be $(l,m,n)$ (so $l+m+n=1$). We can easily compute the equation of each circle to be \[(A_1A_2A_3): \sum{-a^2yz}=0\]\[(A_2A_3A_4): \sum{-a^2yz}+(x+y+z)\left(\frac{\sum{a^2mn}}{l}\right)x=0\]\[(A_3A_4A_1): \sum{-a^2yz}+(x+y+z)\left(\frac{\sum{a^2mn}}{m}\right)y=0\]\[(A_4A_1A_2): \sum{-a^2yz}+(x+y+z)\left(\frac{\sum{a^2mn}}{n}\right)z=0\]where all sums are cyclic. Noting that we are simply summing the reciprocal of the power of each point wrt the circumcircle of the other three points, the sum evaluates to \[\frac{1}{\sum{-a^2yz}}\left(\frac{1}{1}+\frac{1}{-\frac{1}{l}}+\frac{1}{-\frac{1}{m}}+\frac{1}{-\frac{1}{n}}\right)=\frac{1}{\sum{-a^2yz}}\left(1-l-m-n\right)=0\]as desired.

This one can be simply solved using directed lengths and power of point

Use barycentric coordinates, and let $\triangle A_2A_3A_4$ be the reference triangle, with $A_2=(1:0:0)$, $A_3=(0:1:0)$, and $A_4=(0:0:1)$. Let $A_1=(a_{11},a_{12},a_{13})$, so $a_{11}+a_{12}+a_{13}=0$. The equation of $(A_2A_3A_4)$ is $-a^2yz-b^2xz-c^2xy=0$, which means \[ \frac{1}{O_1A_1^2-R_1^2} = \frac{1}{\text{Pow}(A_1,(A_2A_3A_4))} = \frac{1}{-a^2a_{12}a_{13} - b^2a_{11}a_{13}- c^2a_{11}a_{12}}. \]The equation of $(A_1A_3A_4)$ can easily be computed to be \[ -a^2yz-b^2xz-c^2xy + \left( \frac{a^2a_{12}a_{13} + b^2a_{11}a_{13} + c^2a_{11}a_{12}}{a_{11}}\right)x(x+y+z) = 0. \]Therefore, since $A_2=(1,0,0)$, we have \[ \text{Pow}(A_2,(A_1A_3A_4)) = -0+ \frac{a^2a_{12}a_{13} + b^2a_{11}a_{13} + c^2a_{11}a_{12}}{a_{11}} \cdot 1. \]Therefore, \[ \frac{1}{O_2A_2^2-R_2^2} = \frac{1}{\text{Pow}(A_2,(A_1A_3A_4))}=\frac{a_{11}}{a^2a_{12}a_{13} + b^2a_{11}a_{13} + c^2a_{11}a_{12}}. \]Cyclicaly summing the above over $2,3,4$ gives \[ \sum_{i=2}^4 \frac{1}{O_iA_i^2-R_i^2} = \frac{1}{a^2a_{12}a_{13} + b^2a_{11}a_{13} + c^2a_{11}a_{12}},\]so summing with the value for $i=1$ finishes.

pad wrote: The main idea is breaking symmetry by letting one triangle be the reference triangle And Grant Sanderson says that "Math has a tendency to reward you, if you respect it's symmetries".

Note that the denominators are just the powers of $A_1$ wrt $(A_2A_3A_4)$, $A_2$ wrt $(A_1A_3A_4)$, $A_3$ wrt $(A_1A_2A_4)$, and $A_4$ wrt $(A_1A_2A_3)$, respectively. We will employ barycentric coordinates with $A_2=(1,0,0), A_3=(0,1,0), A_4=(0,0,1)$. Let $A_1=(m,n,p)$ with $m+n+p=1$. Let $C=-a^2np-b^2mp-c^2mn$. Note that $(A_2A_3A_4)$ has equation $-a^2yz-b^2xz-c^2xy=0$. Now, the power of $A_1$ wrt the circle is simply $C$. Letting the equation of $(A_1A_3A_4)$ be $-a^2yz-b^2xz-c^2xy+(x+y+z)(ux+vy+wz)=0$, plugging in $A_3,A_4$ gives $v=0, w=0$. Plugging in $A_1$ gives $u=-\frac{C}{m}$.Then, the power of $A_2$ wrt $(A_1A_3A_4)$ is simply $u=-\frac{C}{m}$ Similarly, the equation of $(A_1A_2A_4)$ is $C+(x+y+z)(-\frac{C}{n}x)=0$, so the power of $A_3$ wrt that circle is $-\frac{C}{n}$. The equation of $(A_1A_2A_3)$ is $C+(x+y+z)(-\frac{C}{p}x)=0$, so the power of $A_4$ wrt that circle is $-\frac{C}{p}$. The sum reduces to: $$\frac{1}{C}-\frac{m}{C}-\frac{n}{C}-\frac{p}{C}=\frac{1-(m+n+p)}{C}=0$$ as required.

Let $A$ be the intersection of the diagonals of $A_1A_2A_3A_4$, and let $a_i$ be the distance $AA_i$ for $1\le i\le 4$. We will only compute $\frac{1}{A_1O_1^2-r_1^2}$, as the other three fractions will simply be cyclic expressions. Notice that the expression $O_1A_1^2-r^2$ simply denotes the power of $A_1$ with respect to circle $O_1$. Therefore, if we let $B$ be the second intersection of circle $O_1$ with line $A_1A_3$, then we have $O_1A_1^2-r_1^2=A_1B\cdot A_1A_3$. Notice that $AB\cdot a_3=a_2\cdot a_4 \Rightarrow AB=\frac{a_2a_4}{a_3}$ by power of a point, so $A_1B=A_1A-AB=a_1-\frac{a_2a_4}{a_3}$. Since $A_1A_3=a_1+a_3$, we get that $$O_1A_1^2-r_1^2=(a_1+a_3)\left(a_1-\frac{a_2a_4}{a_3}\right) \Rightarrow \frac{1}{A_1O_1^2-r_1^2}=\frac{1}{(a_1+a_3)\left(a_1-\frac{a_2a_4}{a_3}\right)}=\frac{a_3}{(a_1+a_3)(a_1a_3-a_2a_4)}.$$ Similarly, $$\frac{1}{A_3O_3^2-r_3^2}=\frac{a_1}{(a_1+a_3)(a_1a_3-a_2a_4)}.$$Therefore, adding these two together cancels out the $a_1+a_3$ term in the denominator and gives $\frac{1}{a_1a_3-a_2a_4}$. Similarly, $$\frac{1}{A_2O_2^2-r_2^2}+\frac{1}{A_4O_4^2-r_4^2}=\frac{1}{a_2a_4-a_1a_3}.$$Since these expressions are negations of each other, they sum to zero, and we are done.

Let $B=\overline{A_1A_3}\cap\overline{A_2A_4},$ $C=\overline{A_1A_3}\cap(A_2A_3A_4),$ and $x_i=BA_i.$ By PoP, $A_2B\cdot BA_4=CB\cdot BA_3$ so $$A_1C=A_1C-CB=A_1B-\frac{A_2B\cdot BA_4}{BA_3}=\frac{x_1x_3-x_2x_4}{x_3}.$$Thus, \begin{align*}O_1A_1^2-r_1^2&=A_1C\cdot A_1A_3=A_1C\cdot(x_1+x_3)=\frac{(x_1+x_3)(x_1x_3-x_2x_4)}{x_3}\\&\implies \frac{1}{O_1A_1^2-r_1^2}=\frac{x_3}{(x_1+x_3)(x_1x_3-x_2x_4)}.\end{align*}Hence, $$\sum_{i=1}^{4}{\frac{1}{O_iA_i^2-r_i^2}}=\sum_{\text{cyc}}{\frac{x_3}{(x_1+x_3)(x_1x_3-x_2x_4)}}=\sum_{\text{cyc}}{\frac{x_1+x_3}{(x_1+x_3)(x_1x_3-x_2x_4)}}=0.$$$\square$

$A_1A_3$ and $A_2A_4$ intersection $X$ $A_1X = a,A_2X=b,A_3X=c,A_4X=d$ $$\frac{1}{O_1A_1^2-r_1^2} = \frac{c}{(a+c)(ac-bd)}$$the expression is 0 if all these sums are calculated by connecting them to $a, b, c, d$. I think it’s easier than the G1.

Note the equation is equivalent to $$\sum_\text{cyc} \frac1{\text{Pow}_{(A_2A_3A_4)}(A_1)}=0.$$Let $X$ be the intersection of $A_1A_3$ and $A_2A_4$. Let $P_i$ be the intersection of $A_iA_{i+2}$ and $(A_{i+1}A_{i+2}A_{i+3})$ besides $A_{i+2}$ with indices taken mod 4. Let $x_i=A_iX$. We have $\text{Pow}_{(A_2A_3A_4)}(A_1)=A_1P_1\cdot A_1A_2$ using signed lengths. By power of a point on circle $(A_2A_3A_4)$, $P_1X=\frac{x_2x_4}{x_3}$. So, using signed lengths, $$\text{Pow}_{(A_2A_3A_4)}(A_1)=A_1P_1\cdot A_1A_2=(x_1-\frac{x_2x_4}{x_3})(x_1+x_2)=\frac1{x_3}(x_1x_3-x_2x_4)(x_1+x_3)$$So, using similar logic on the other points, the whole expression is $$\frac{x_3}{(x_1x_3-x_2x_4)(x_1+x_3)}+\frac{x_4}{(x_2x_4-x_1x_3)(x_2+x_4)}+\frac{x_1}{(x_1x_3-x_2x_4)(x_1+x_3)}+\frac{x_2}{(x_2x_4-x_1x_3)(x_2+x_4)}$$$$=\frac{1}{x_1x_3-x_2x_4}+\frac{1}{x_2x_4-x_1x_3}=0.\qquad\square$$

Note that the equation is equivalent to $$\sum_{cyc}\frac{1}{\text{pow}_{(A_2A_3A_4)}(A_1)} =0.$$We use barycentric coordinates with respect to $A_1A_2A_3$. Let $A_4=(d,e,f)$. We have that $$\frac{1}{\text{pow}_{(A_2A_3A_4)}(A_1)} = \frac{1}{-a^2ef-b^2df-c^2de}.$$Note that the circumcircle of $A_1A_2A_4$ is of the form $-a^2yz-b^2xz-c^2xy+(x+y+z)(wz)=0$. Plugging in $a_4$, we have $w=\frac{a^2ef+b^2df+c^2de}{f}$. We thus get that $\frac{1}{\text{pow}_{(A_1A_2A_4)}(A_3)} = \frac{f}{a^2ef+b^2df+c^2de}$. Symmetrically applying this argument, we have $$\sum_{cyc}\frac{1}{\text{pow}_{(A_2A_3A_4)}(A_1)} = \frac{1}{a^2ef+b^2df+c^2de}(-1+d+e+f)=0$$as desired so we are done.

From here every length mentioned is directed. Denote by $\omega_1,\omega_2,\omega_3,\omega_4$ the circumcircles of $\triangle A_2A_3A_4,\triangle A_3A_4,A_1,\triangle A_4A_1A_2,\triangle A_1A_2A_3$ respectively. Let $A_1A_3$ and $A_2A_4$ intersect at a point $P$, and let $A_iO_i$ intersect $\omega_i$ at $B_i$ for $i=1,2,3,4.$ Let $x_1$ be the directed length $PA_i.$ Note that \[\text{pow}_{w_1}(A_1)=A_1B_1\cdot A_1A_3=(PB_1-PA_1)(PA_3+PA_1)=(x_3+x_1)(\frac{x_2x_4}{x_3}-x_1).\]Now use this advanced method called "adding" to get $0.$

very fun problem

(In this solution, consider all lengths in the directed sense, and take all indices modulo $4$.) First of all, notice that the desired result is equivalent to \[ \sum_{i=1}^4 \frac{1}{\text{Pow}(A_i, (A_{i+1}A_{i+2}A_{i+3}))} = 0. \]Let $P=\overline{A_1A_3} \cap \overline{A_2A_4}$ and $X=\overline{A_1A_3} \cap (A_2A_3A_4)$. Furthermore, let $PA_i=x_i$ for $1 \le i \le 4$. By PoP on $X$ with respect to $(A_2A_3A_4)$, $PX=\frac{x_2x_4}{x_3}$. Thus, \[ \frac{1}{\text{Pow}(A_1, (A_2A_3A_4))} = \frac{1}{(A_1P+PA_3)(PX-PA_1)} = \frac{1}{(x_3-x_1)(\frac{x_2x_4}{x_3}-x_1)} = \frac{x_3}{(x_3-x_1)(x_2x_4-x_1x_3)}. \]Replacing $A_1$ with $A_i$ and summing over $1 \le i \le 4$, we have \[ \frac{1}{O_1A_1^2-r_1^2}+\frac{1}{O_2A_2^2-r_2^2}+\frac{1}{O_3A_3^2-r_3^2}+\frac{1}{O_4A_4^2-r_4^2} = \sum_{i=1}^4 \frac{x_{i+2}}{(x_{i+2}-x_i)(x_{i+1}x_{i+3}-x_ix_{i+2})} = 0, \]which completes the proof.

"can i get geometry" "we have geometry at home" geometry at home: The expression is just $$\sum \frac{1}{Pow_{\omega_i}(A_i)}$$if we let $\omega_i$ be the circumcircle of the three vertices other than $A_i$. Let $A_1A_3$ and $A_2A_4$ intersect at $P$. Let $d_i=A_iP.$ Claim: $$Pow_{\omega_4}(A_4)=(d_2+d_4)\frac{d_2d_4-d_1d_3}{d_2}.$$ Let $A_2A_4$ intersect $\omega_4$ again at $B_4.$ Then, by Power of a Point, $B_4P=\frac{d_1d_3}{d_2}$. Thus, $$Pow_{\omega_4}(A_4)=(d_2+d_4)(d_4-\frac{d_1d_3}{d_2})=(d_2+d_4)\frac{d_2d_4-d_1d_3}{d_2}.$$ Let $P_i=Pow_{\omega_i}(A_i)$ For the sake of simplicity, let $$S_O=d_1+d_3,S_E=d_2+d_4,Orz=d_1d_3-d_2d_4.$$We can then rewrite $$P_4=\frac{S_E(-Orz)}{d_3}.$$Similarly, $$P_1=\frac{S_O(Orz)}{d_3},P_2=\frac{S_E(-Orz)}{d_4},P_3=\frac{S_O(Orz)}{d_1},P_4=\frac{S_E(-Orz)}{d_2}$$Then, we have $$d_1d_2d_3d_4P_1P_2P_3=-d_2S_O^2S_E(Orz)^3,$$$$d_1d_2d_3d_4P_2P_3P_4=d_3S_OS_E^2(Orz)^3,$$$$d_1d_2d_3d_4P_3P_4P_1=-d_4S_O^2S_E(Orz)^3,$$$$d_1d_2d_3d_4P_4P_1P_2=d_1S_OS_E^2(Orz)^3,$$so $$d_1d_2d_3d_4(\sum_{cyc}P_1P_2P_3)=S_OS_E(Orz)^3(-d_2S_O-d_4S_O+d_3S_E+d_1S_E)$$$$=S_OS_E(Orz)^3(S_O(-d_2-d_4)+S_E(d_1+d_3)$$$$=S_OS_E(Orz)^3(-S_OS_E+S_ES_O)=0.$$Hence, $\sum_{cyc}P_1P_2P_3=0$, and thus $\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}+\frac{1}{P_4}=0,$ as desired.

nice app of signed distances plus pop edit: oops wrong order but i cant be bothered to rearrange from liog (I don't want to write up all of my latex, so this is just for my own storage)

Oops apparently I missed the non-trig sol. However the bash surprisingly clean, with areas appearing. To avoid config issues, just replace lengths with directed lengths and same thing for angles and areas. Let $\alpha_i=\angle A_{i-1}A_iA_{i+1}$, with indices taken modulo $4$. We also let $\gamma_1$ be the circumcircle of $A_2A_3A_4$ and define $\gamma_2,\gamma_3,\gamma_4$ in a similar way. We need to prove that \[\sum_i \frac{1}{\mathcal P_{\gamma_i}(A_i)}=0\]We now examine each term individually : Claim : We have $\mathcal P_{\gamma_1}(A_1)=-\sin (\alpha_4+\alpha_2)\frac{A_1A_2\cdot A_1A_4}{\sin \alpha_3}$, and similarly for the other $A_i$'s. Proof : Suppose circle $\gamma_1$ intersects line $(A_1A_2)$ again at a point $X$. An angle chase gives $\angle A_1A_4X=\alpha_4+\alpha_2-180$, as well as $\angle A_1XA_4=\alpha_3$. Sine law in triangle $A_1A_2A_4$ then gives \[A_1X=-\sin (\alpha_4+\alpha_2)\frac{A_1A_4}{\sin \alpha_3},\]and the conclusion follows since $\mathcal P_{\gamma_1}(A_1)=A_1X\cdot A_1A_2$. $\square$ We're actually done now. Indeed, pair up the $A_1$ and $A_3$ terms to get \begin{eqnarray*} \frac{1}{\mathcal P_{\gamma_1}(A_1)}+\frac{1}{\mathcal P_{\gamma_3}(A_3)}&=&-\frac{1}{\sin(\alpha_4+\alpha_2)}\cdot \left(\frac{\sin \alpha_3}{A_1A_2\cdot A_1A_4}+\frac{\sin \alpha_1}{A_3A_2\cdot A_3A_4}\right)\\ &=& -\frac{1}{\sin(\alpha_4+\alpha_2)}\cdot \left(\frac{2S_1+2S_3}{A_1A_2\cdot A_2A_3\cdot A_3A_4\cdot A_4A_1}\right)\\ &=& -\frac{1}{\sin(\alpha_4+\alpha_2)}\cdot \underbrace{\left(\frac{2S}{A_1A_2\cdot A_2A_3\cdot A_3A_4\cdot A_4A_1}\right)}_{:\overset{def}{=} T}, \end{eqnarray*}where we let $S$ be the area of $A_1A_2A_3A_4$ and $S_i$ be the area of the triangle $A_{i+1}A_{i+2}A_{i+3}$. By symmetry, we also have \[\frac{1}{\mathcal P_{\gamma_2}(A_2)}+\frac{1}{\mathcal P_{\gamma_4}(A_4)}=-\frac{1}{\sin(\alpha_3+\alpha_1)}\cdot T\]Thus summing up we'll get the desired result, since $\sin(\alpha_3+\alpha_1)=-\sin(\alpha_4+\alpha_2)$. $\blacksquare$