Doctor A wrote: (B.Frenkin, A.Zaslavsky) A convex quadrilateral was drawn on the blackboard. Boris marked the centers of four excircles each touching one side of the quadrilateral and the extensions of two adjacent sides. After this, Alexey erased the quadrilateral. Can Boris define its perimeter? Let $ ABCD$ is convex quadrilateral. $ 0_1$ is a center of excircle touching $ AB$ at $ K_1$ , extension $ BC$ at $ N_1$ and extension $ DA$ at $ M_1$ $ 0_2$ is a center of excircle touching $ BC$ at $ K_2$ , extension $ CD$ at $ N_2$ and extension $ AB$ at $ M_2$ $ 0_3$ is a center of excircle touching $ CD$ at $ K_3$ , extension $ DA$ at $ N_3$ and extension $ BC$ at $ M_3$ $ 0_4$ is a center of excircle touching $ DA$ at $ K_4$ , extension $ AB$ at $ N_4$ and extension $ CD$ at $ M_4$ $ R_1,R_2,R_3,R_4$ - radii of circles with centers $ O_1,O_2,O_3,O_4$ $ s$ - semiperimeter of $ ABCD$ $ N_1M_3 = M_1N_3 = N_4M_2 = M_4N_2 = s$; $ K_1N_4 = M_1K_4 = K_2M_3 = N_2K_3 = a$; $ a^2 = O_1O_4^2 - (R_1 + R_4)^2 = O_2O_3^2 - (R_2 + R_3)^2$; $ K_2N_1 = M_2K_1 = K_3M_4 = N_3K_4 = b$; $ b^2 = O_2O_1^2 - (R_2 + R_1)^2 = O_3O_4^2 - (R_3 + R_4)^2$; So we have : $ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -$ $ O_1O_4^2 - O_2O_3^2 = (R_1 + R_2 + R_3 + R_4)((R_1 - R_3) - (R_2 - R_4))$ $ O_2O_1^2 - O_3O_4^2 = (R_1 + R_2 + R_3 + R_4)((R_1 - R_3) + (R_2 - R_4))$ $ O_1O_3^2 = s^2 + (R_1 - R_3)^2$ $ O_2O_4^2 = s^2 + (R_2 - R_4)^2$ $ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -$ from this system we can easy find : $ s^2 = \frac {1}{2}(O_1O_3^2 + O_2O_4^2) - \frac {1}{4}(O_1O_3^2 - O_2O_4^2)(\frac {O_2O_3^2 - O_1O_4^2}{O_3O_4^2 - O_2O_1^2} + \frac {O_3O_4^2 - O_2O_1^2}{O_2O_3^2 - O_1O_4^2} )$;

Let $ I_1,I_2,I_3,I_4$ be the excenters of the quadrilateral. Let $ I_1'I_2I_3I_4'$ be the reflection of $ I_1I_2I_3I_4$ in $ I_2I_3$, and $ I_1''I_2'I_3I_4'$ the reflection of $ I_1'I_2I_3I_4'$ in $I_3I_4'$, and $ I_1''I_2'I_3I_4'$ the reflection of $I_1''I_2'I_3I_4'$ in $I_1''I_4'$. Easily we see that the perimeter of the quadrilateral equals to $I_1I_1'$.