How did I solve this (triply logarithmic) equation?

In an optimiation problem I came across the following daunting equation: $$ \log \left(\frac{1-t_2}{1-y}\right) \log \left(\frac{(1-x) (t_1-t_2)}{(1-t_1) (x-y)}\right) \log \left(\frac{t_2 x}{t_1 y}\right)\\=\log \left(\frac{t_2}{y}\right) \log \left(\frac{x (t_1-t_2)}{t_1 (x-y)}\right) \log \left(\frac{(1-t_2) (1-x)}{(1-t_1) (1-y)}\right) $$ in which $x,y\in[0,1]$ are given constants and $t_1,t_2\in[0,1]$ are the variables.

I ended up plotting it numerically and noting that at the solutions (over $t_1,t_2$) the logarithmic factors would be pairwise equal, which in particular implied the hyperbola $$t_1 (1-x) y=t_1 t_2 (y-x)+t_2 x (1-y)$$ along with a few trivial solutions ($t_2=y$, $t_1=x$, $x t_2 = t_1 y$ and $t_2=t_1$). Checking that these are indeed solutions is easy, but coming up with them was not.

I wonder if there was a more structural approach I could have taken to come up with this result?

In fact I have some even more formidable equations which I now hope have simple polynomial solutions, but those I haven't been lucky enough to guess.

Solution 1:

HINT

$\color{brown}{\textbf{Standing of the task.}}$

Denote $$u=t_2,\quad v=t_1,$$ $$p = \log\dfrac{u}{x},\quad q = \log\dfrac{v}{y},\quad r = \log\dfrac{1-u}{1-x},\quad s = \log\dfrac{1-v}{1-y},\quad w = \log\dfrac{v-u}{y-x},\quad $$ $$f(u,v) = s(w-r)(q-p) - q(w-p)(s-r) = sr(p-q) - swp + pq(s-r) +qwr,$$ $$f(u,v) = \begin{vmatrix}w-p-s & w-q-r \\ ps & rq\end{vmatrix},$$ then \begin{align} &f_{mn}(u,v) = \dfrac{\partial^{m+n}f_{mn}(u,v)}{\partial u^m\partial v^n}\\[4pt] &= \sum\limits_{i=0}^m\sum\limits_{j=0}^n \binom mi \binom nj \begin{vmatrix} w_{ij}-\delta_{0j}\,p_i-\delta_{i0}\,s_{j} & w_{ij}-\delta_{0j}\,q_i-\delta_{i0}\,r_j \\ p_{m-i}\,s_{n-j} & r_{m-i}\,q_{n-j} \end{vmatrix},\quad\text{where}\\ &p_i = p^{(i)}(u) = \delta_{i0}\log\dfrac{u}{x} + (\delta_{i0}-1)^{i-1}\,\dfrac{(i-1)!}{u^i},\\ &q_j = q^{(j)}(v) = \delta_{0j}\log\dfrac{v}{y} + (\delta_{0j}-1)^{j-1}\,\dfrac{(j-1)!}{v^j},\\ &r_i= r^{(i)}(u) = \delta_{i0}\log\dfrac{1-u}{1-x} + (\delta_{i0}-1)\,\dfrac{(i-1)!}{(1-u)^i},\\ &s_{ij} = s^{(j)}(v) =\delta_{0j}\log\dfrac{1-v}{1-y} + (\delta_{0j}-1)\,\dfrac{(j-1)!}{(1-v)^j},\\ &w_{ij}=\dfrac{\partial^{i+j}w}{\partial u^i\partial v^j} = \delta_{i0}\delta_{0j}\log\dfrac{v-u}{y-x} + (\delta_{i0}\delta_{0j}-1)^{\delta_{i0}+j-1}\,\dfrac{(i-1)!(j-1)!}{(v-u)^{i+j}},\\[4pt] \end{align}

$\color{brown}{\textbf{Derivatives in the given point.}}$

Note that \begin{align} &p(x) = 0,\quad r(x) = 0,\quad q(y) = 0,\quad s(y) = 0,\quad w(x,y) = 0, \quad f(x,v) = f(u,y) = 0,\\[4pt] &p_i(x) = (\delta_{i0}-1)^{i-1}\,\dfrac{(i-1)!}{x^i},\\ &q_j(y) = (\delta_{0j}-1)^{j-1}\,\dfrac{(j-1)!}{y^j},\\ &r_i(x) = (\delta_{i0}-1)\,\dfrac{(i-1)!}{(1-x)^i},\\ &s_j(y) = (\delta_{0j}-1)\,\dfrac{(j-1)!}{(1-y)^j},\\ &w_{ij}(x,y) = (\delta_{i0}\delta_{0j}-1)^{\delta_{i0}+j-1}\,\dfrac{(i-1)!(j-1)!}{(y-x)^{i+j}}\\[4pt] &=\begin{cases} 0,\quad\text{if}\quad i=0\wedge j=0\\ (-1)^j\,\dfrac{(j-1)!}{(y-x)^j},\quad\text{if}\quad i=0 \wedge j>0\\ -\dfrac{(i-1)!}{(y-x)^i},\quad\text{if}\quad i>0 \wedge j=0\\ (-1)^{j-1}\,\dfrac{(i-1)!(j-1)!}{(y-x)^{i+j}},\quad\text{if}\quad i>0 \wedge j>0,\\ \end{cases}\\[4pt] &f_{mn}(x,y) = \sum\limits_{i=1}^m\sum\limits_{j=1}^n \binom mi \binom nj \begin{vmatrix} a_{ij}(x,y) & b_{ij}(x,y) \\ p_{m-i}(x)s_{n-j}(y) & r_{m-i}(x)q_{n-j}(y) \end{vmatrix},\\ &a_{ij}(x,y) = w_{ij}(x,y)-\delta_{0j}p_i(x)-\delta_{i0}s_j(y),\\ & b_{ij}(x,y) = w_{ij}(x,y)-\delta_{0j}r_i(x)-\delta_{i0}q_j(y),\\ \end{align} Taking in account zero values of $p,q,r,s,w$ in the given point $\dbinom uv = \dbinom xy,$ unzero derivatives can start from $i+j=3.$

$\color{brown}{\textbf{2D Taylor series.}}$

The previous results allow to present the issue equality in the form of $f_l(u,v)=0,$ where $$f_l(u,v) = \sum\limits_{k=3}^l \dfrac1{k!}\sum\limits_{m=1}^{k-1}\binom km f_{m,k-m}(x,y)(u-x)^m(v-y)^{k-m}.$$ is $2D$ Taylor series of $l-$th order.

The calculated values of the derivatives are in the table below.

\begin{vmatrix} & i=0 & i = 1 & i=2 & i = 3 \\ p_i(x) & 0 & \dfrac1x & -\dfrac1{x^2} & \dfrac2{x^3} \\ q_i(y) & 0 & \dfrac1y & -\dfrac1{y^2} & \dfrac2{y^3} \\ r_i(x) & 0 & -\dfrac1{1-x} & -\dfrac1{(1-x)^2} & -\dfrac2{(1-x)^3}\\ s_i(y) & 0 & -\dfrac1{1-y} & -\dfrac1{(1-y)^2} & -\dfrac2{(1-y)^3} \\ w_{i0}(x,y) & 0 & -\dfrac1{y-x} & -\dfrac1{(y-x)^2} & -\dfrac2{(y-x)^3} \\ w_{i1}(x,y) & \dfrac1{y-x} & \dfrac1{(y-x)^2} & \dfrac2{(y-x)^3} & \dfrac2{(y-x)^4} \\ w_{i2}(x,y) & -\dfrac1{(y-x)^2} & -\dfrac2{(y-x)^3} & -\dfrac6{(y-x)^4} & -\dfrac{24}{(y-x)^5} \\ \end{vmatrix}

$\color{brown}{\textbf{The third order model.}}$

One can get \begin{align} &a_{01} = w_{01}(x,y)-p_0(x)-s_1(y) = \dfrac1{y-x} + \dfrac1{1-y}=\dfrac{1-x}{(y-x)(1-y)},\\ &b_{01} = w_{01}(x,y)-r_0(x)-q_1(y) = \dfrac1{y-x}-\dfrac1y = -\dfrac x{y(y-x)},\\ &a_{10} = w_{01}(x,y)-p_1(x)-s_0(y) = -\dfrac1{y-x} - \dfrac1x = -\dfrac{y}{x(y-x)},\\ &b_{10} = w_{01}(x,y)-r_1(x)-q_0(y) = \dfrac1{y-x}+\dfrac1{1-x} = \dfrac{1+y-2x}{(1-x)(y-x)},\\ &f_{12} = a_{01}r_1(x)q_1(y) - b_{01}p_1(x)s_1(y) = -\dfrac{a_{01}}{(1-x)y} + \dfrac{b_{01}}{x(1-y)} = -\dfrac2{(y-x)y(1-y)},\\ &f_{21} = a_{10}r_1(x)q_1(y) - b_{10}p_1(x)s_1(y) = -\dfrac{a_{10}}{(1-x)y} + \dfrac{b_{10}}{x(1-y)}\\ &= \dfrac1{x(y-x)(1-x)} + \dfrac{1+y-2x}{x(1-x)(y-x))(1-y)} =\dfrac2{x(y-x)(1-y)},\\ &f_3(u,v) = \dfrac12(f_{12}(v-y)+f_{21}(u-x))(u-x)(v-y) =\left(\dfrac{u-x}{x}-\dfrac{v-y}{y}\right)\dfrac{(u-x)(v-y)}{(y-x)(1-y)},\\ &f_3(u,v)=0\rightarrow (u=x)\vee(v=y)\vee(uy=vx). \end{align}

The order of model can be increased.