<prompt> Katika pembetatuABC na AB 12, BC 13, na AC 15, acha M iwe kigezo kwenye overlineAC kiasi kwamba miduara ya pembetatuABM na pembetatuBCM zina radii sawa. Kisha fracAMCM fracpq, ambapo p na q ni nambari kamili chanya. Tafuta p q.  <response>: asy kuagiza grafu; defaultpenlinewidth0.7fontsize10; ukubwa 200; sehemu na takwimu chora0,0--15,0; kuchora15,0--6.66667,9.97775; kuchora 6.66667,9.97775--0,0; kuchora 7.33333,0--6.66667,9.97775; kuchora mduara 4.66667,2.49444, 2.49444; kuchora mduara 9.66667,2.49444, 2.49444; kuchora 4.66667,0--4.66667,2.49444; kuchora 9.66667,2.49444--9.66667,0; pointi na lebo lebor, 10.19662,1.92704,SE; lebor,5.02391,1.8773,SE; nukta 0,0; leboA,-1.04408,-0.60958,NE; nukta 15,0; leboC,15.41907,-0.46037,NE; nukta 6.66667,9.97775; leboB,6.66525,10.23322,NE; lebo15,6.01866,-1.15669,NE; lebo13,11.44006,5.50815,NE; lebo12,2.28834,5.75684,NE; nukta 7.33333,0; leboM,7.56053,-1.000,NE; leboH_1,3.97942,-1.200,NE; leboH_2,9.54741,-1.200,NE; nukta 4.66667,2.49444; leboI_1,3.97942,2.92179,NE; nukta 9.66667,2.49444; leboI_2,9.54741,2.92179,NE; klipu-3.72991,-6.47862---3.72991,17.44518--32.23039,17.44518--32.23039,-6.47862--mzunguko; asyAcha AM x, kisha CM 15 - x. Pia acha BM d Ni wazi, frac ABMCBM frac x15 - x. Tunaweza pia kueleza kila eneo kwa fomula ya rs. Kisha frac ABMCBM frac pABMpCBM frac 12 d x28 d - x. Kusawazisha na kuzidisha mazao 25x 2dx 15d 180 au d frac 25x - 18015 - 2x. Kumbuka kuwa ili d iwe chanya, lazima tuwe na 7.2 x 7.5.Kulingana na Nadharia ya Stewart, tuna 12215 - x 132x d215 15x15 - x au 432 3d2 40x - 3x2. Kulazimisha kwa nguvu kwa kuchomeka matokeo ya awali ya d, tuna 432 frac 325x - 180215 - 2x2 40x - 3x2. Kufuta sehemu na kukusanya kama sheria na masharti, tunapata 0 12x4 - 340x3 2928x2 - 7920x.Kando: Kwa kuwa x lazima ziwe na mantiki ili jibu letu liwe katika umbo tunalotaka, tunaweza kutumia Nadharia ya Rational Root kufichua kwamba 12x ni nambari kamili. x kama hiyo pekee katika safu iliyotajwa hapo juu ni frac 223.Ukisuluhisha quartic hiyo kihalali, kumbuka kuwa x 0 na x 15 inapaswa kuwa suluhisho wazi, zinazolingana na pande za pembetatu na kwa hivyo kuharibika kwa cevians. Kuweka zile nje, tunapata 0 4xx - 153x2 - 40x 132 xx - 15x - 63x - 22. Suluhisho pekee katika safu inayotakikana ni hivyo frac 223. Kisha CM frac 233, na uwiano wetu tunaoutaka frac AMCM frac 2223, ikitupa jibu la 45. 