Real numbers $a,b,c,d$ satisfy the equality $$\frac{1-ab}{a+b}=\frac{bc-1}{b+c}=\frac{1-cd}{c+d}=\sqrt{3}$$Find all possible values of $ad$.
Problem
Source: Belarusian National Olympiad 2023
Tags: algebra
aidan0626
31.12.2024 21:19
We can split the equality up into three different equations,
\begin{align*}
\frac{1-ab}{a+b}=\sqrt3\\
\frac{bc-1}{b+c}=\sqrt3\\
\frac{1-cd}{c+d}=\sqrt3.
\end{align*}Solving them gives $b=\frac{1-a\sqrt3}{a+\sqrt3}$, $c=\frac{1+b\sqrt3}{b-\sqrt3}$, and $d=\frac{1-c\sqrt3}{c+\sqrt3}.$ Plugging in the value for $b$ into the second one gives $c=\frac{a-\sqrt3}{1+a\sqrt3}.$ Plugging that into the last one gives $d=\frac{1}{a}.$ Therefore, the only possible value for $ad$ is $1.$
Double07
31.12.2024 21:24
I been waiting for too long for a trigonometric substitution in an olympiad problem. Take $a=\tan x, b=\tan y, c=\tan z, d=\tan t$, with $x, y, z, t\in[0,\pi)$. The relation rewirtes as $\cot (x+y)=-\cot (y+z)=\cot (z+x)=\sqrt{3}$, so $\begin{cases} x+y\in\left\{\dfrac{\pi}{6}, \dfrac{7\pi}{6}\right\}\\ y+z\in\left\{\dfrac{5\pi}{6}, \dfrac{11\pi}{6}\right\}\\ z+t\in\left\{\dfrac{\pi}{6}, \dfrac{7\pi}{6}\right\} \end{cases}$ Case 1: $x+y=\dfrac{\pi}{6}\implies y\leq\dfrac{\pi}{6}\implies y+z=\dfrac{5\pi}{6}\implies z\geq \dfrac{4\pi}{6}\implies z+t=\dfrac{7\pi}{6}\implies t=\dfrac{\pi}{2}-x\implies$ $\implies d=\tan t=\tan \left(\dfrac{\pi}{2}-x\right)=\cot x\implies ad=1$ Case 2: $x+y=\dfrac{7\pi}{6}$ is split into $2$ different cases for $y+z$ and is treated similarly.