"i have found a problem of all time" "do p3" - DottedCaculator By Angle Bisector Theorem on $\triangle ADC$ it follows $AC=\tfrac{5}{3}AD,$ and on $\triangle ABE$ it follows $AB=\tfrac{1}{3}AE.$ By Stewart's Theorem we obtain \begin{align*} 3\left(\frac{1}{3}AE\right)^2+(AE)^2=4((AD)^2+3), \\ 8\left(\frac{1}{3}AE\right)^2+\left(\frac{5}{3}AE\right)^2=9((AD)^2+8, \end{align*}which produce imaginary lengths.

genius $ $

bryanguo wrote: "i have found a problem of all time" "do p3" - DottedCaculator By Angle Bisector Theorem on $\triangle ADC$ it follows $AC=\tfrac{5}{3}AD,$ and on $\triangle ABE$ it follows $AB=\tfrac{1}{3}AE.$ By Stewart's Theorem we obtain \begin{align*} 3\left(\frac{1}{3}AE\right)^2+(AE)^2=4((AD)^2+3), \\ 8\left(\frac{1}{3}AE\right)^2+\left(\frac{5}{3}AE\right)^2=9((AD)^2+8, \end{align*}which produce imaginary lengths. Yes, as the title suggests, the lengths are imaginary. That is an error from the problem-writers' end.