<prompt> Acha A na B ziwe mwisho wa safu ya nusu duara ya radius 2. Safu imegawanywa katika safu saba zinazolingana kwa alama sita zilizo na nafasi sawa C_1, C_2, dots, C_6. Nyimbo zote za fomu overline AC_i au overline BC_i zimechorwa. Tafuta bidhaa ya urefu wa chodi hizi kumi na mbili.  <response>: Hebu omega e2 pi i14. Tunaweza kutambua A na 2, B na -2, na C_k na nambari changamano 2 omegak. asyukubwa wa kitengo cm 3;int i;jozi A, B;jozi C;A 1,0;B -1,0;C1 dir11807;C2 dir21807;C3 dir31807;C4 dir41807;C5 dir51807;C6 dir61807;choraA--B;choraarc0,0,1,0,180;kwa i 1; i 6; i kuchoraA--Ci--B; nuktaC_ stringi , Ci, Ci;nukta A, A, E;nuktaB, B, W;asyKisha AC_k 2 - 2 omegak 2 1 - omegak naBC_k -2 - 2 omega_k 2 1 omegak.Kwa kuwa omega7 -1, tunaweza pia kuandika hili kamaBC_k 2 1 - omegak 7.Kwa hiyo,AC_1 cdot AC_2 dotsm AC_6 26 1 - omega1 - omega2 dotsm 1 - omega6naBC_1 cdot BC_2 dotsm BC_6 26 1 - omega81 - omega9 dotsm 1 - omega13.Kumbuka kwamba 1, omega, omega2, dots, omega13 zote ni mizizi ya z14 - 1 0. Hivyoz14 - 1 z - 1z - omegaz - omega2 dotsm z - omega13.Kipengele kimoja upande wa kulia ni z - 1, na kipengele kingine upande wa kulia ni z - omega7 z 1. Hivyo,z14 - 1 z - 1z 1 cdot z - omegaz - omega2 dotsm z - omega6z - omega8z - omega9 dotsm z - omega13.Kwa kuwa z14 - 1 z2 - 1z12 z10 z8 dots 1, tunaweza kuandikaz12 z10 z8 vitone 1 z - omegaz - omega2 dotsm z - omega6z - omega8z - omega9 dotsm z - omega13.Kuweka z 1, tunapata7 1 - omega1 - omega2 dotsm 1 - omega61 - omega81 - omega9 dotsm 1 - omega13.Kwa hiyo,anzaalignAC_1 cdot AC_2 dotsm AC_6 cdot BC_1 cdot BC_2 dotsm BC_6 26 1 - omega1 - omega2 dotsm 1 - omega6 cdoti 26 1 - omega81 - omega9 dotsm 1 - omega13 212 1 - omega1 - omega2 dotsm 1 - omega61 - omega81 - omega9 dotsm 1 - omega13 7 cdot 212 28672.mwishopanga 