Menu Close

Category: Geometry

Given-an-acute-triangle-with-AB-lt-AC-is-inscribed-in-the-circle-O-Let-D-and-E-be-the-midpoints-of-the-minor-arc-and-major-arc-BC-respectively-Let-I-and-J-be-the-incenters-of-trianges-ABD-and-AC

Question Number 209329 by Huy250 last updated on 07/Jul/24 $${Given}\:{an}\:{acute}\:{triangle}\:{with}\:{AB}\:<{AC} \\ $$$${is}\:{inscribed}\:{in}\:{the}\:{circle}\:\left({O}\right).\:{Let}\:{D} \\ $$$${and}\:{E}\:{be}\:{the}\:{midpoints}\:{of}\:{the}\:{minor}\:{arc} \\ $$$${and}\:{major}\:{arc}\:{BC},\:{respectively}.\:{Let}\:{I}\:{and}\:{J} \\ $$$${be}\:{the}\:{incenters}\:{of}\:{trianges}\:{ABD}\:{and}\:{ACD}, \\ $$$${respectively}.\:{Prove}\:{that}\:{EI}={EJ}. \\ $$ Commented by Huy250…

Question-209288

Question Number 209288 by Jubr last updated on 06/Jul/24 Answered by A5T last updated on 06/Jul/24 $$\frac{{x}}{{r}}=\frac{\mathrm{8}}{{r}}\Rightarrow{x}=\mathrm{8} \\ $$$${r}^{\mathrm{2}} =\left(\mathrm{2}{x}\right)^{\mathrm{2}} +\left({r}−\mathrm{8}\right)^{\mathrm{2}} \Rightarrow{r}=\mathrm{20} \\ $$$$\left[{Yellow}\right]=\frac{\mathrm{90}}{\mathrm{360}}×\mathrm{400}\pi−\frac{\mathrm{20}×\mathrm{20}}{\mathrm{2}}=\mathrm{100}\pi−\mathrm{200} \\…

Question-209289

Question Number 209289 by Jubr last updated on 06/Jul/24 Answered by A5T last updated on 06/Jul/24 $$\frac{{s}}{{x}}=\frac{\mathrm{7}}{\mathrm{21}}\Rightarrow{x}=\mathrm{3}{s} \\ $$$$\frac{{s}}{{y}}=\frac{\mathrm{14}}{\mathrm{21}}\Rightarrow{y}=\frac{\mathrm{3}{s}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{21}^{\mathrm{2}} \Rightarrow\mathrm{9}{s}^{\mathrm{2}} +\frac{\mathrm{9}{s}^{\mathrm{2}}…

Question-208940

Question Number 208940 by efronzo1 last updated on 28/Jun/24 Answered by kapoorshah last updated on 28/Jun/24 $${r}\:=\:{OD}\:=\:\sqrt{\mathrm{10}} \\ $$$${BE}\:=\:\sqrt{\mathrm{15}} \\ $$$${BOE}\:\sim\:{BCA} \\ $$$$\frac{{BO}}{{BE}}\:=\:\frac{{BC}}{{AB}} \\ $$$$\frac{\sqrt{\mathrm{10}}}{\:\sqrt{\mathrm{15}}}\:=\:\frac{{BC}}{\mathrm{2}\sqrt{\mathrm{10}}}\:\:\Rightarrow\:{BC}\:=\:\frac{\mathrm{20}}{\:\sqrt{\mathrm{15}}}\:=\:\mathrm{5}.\mathrm{164}…