Category: Geometry

Question Number 68664 by Maclaurin Stickker last updated on 14/Sep/19 $$\mathrm{In}\:\mathrm{a}\:\mathrm{equilateral}\:\mathrm{triangle}\:{ABC}\:\mathrm{whose} \\ $$$$\mathrm{side}\:\mathrm{is}\:\boldsymbol{{a}},\:\mathrm{the}\:\mathrm{points}\:{M}\:\mathrm{and}\:{N}\:\mathrm{are}\:\mathrm{taken} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{side}\:{BC},\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{triangles} \\ $$$${ABM},\:{AMN}\:\mathrm{and}\:{ANC}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\: \\ $$$$\mathrm{perimeter}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{distances}\:\mathrm{from} \\ $$$$\mathrm{vertex}\:{A}\:\mathrm{to}\:\mathrm{points}\:{M}\:\mathrm{and}\:{N}. \\ $$$$\left(\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{detail}}.\right) \\ $$…

Question Number 68611 by TawaTawa last updated on 14/Sep/19 Answered by mind is power last updated on 15/Sep/19 $$\frac{\mathrm{1}}{{r}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}\left({r}−\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{4}\left({r}+\mathrm{2}\right)} \\ $$$$\Rightarrow\sum_{{r}=\mathrm{3}} ^{\mathrm{100}} \frac{\mathrm{1}}{{r}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}\Sigma\left(\frac{\mathrm{1}}{{r}−\mathrm{2}}−\frac{\mathrm{1}}{{r}+\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{99}}−\frac{\mathrm{1}}{\mathrm{100}}−\frac{\mathrm{1}}{\mathrm{101}}−\frac{\mathrm{1}}{\mathrm{102}}\right)…

Question Number 68601 by Maclaurin Stickker last updated on 14/Sep/19 $${ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{side}\:\mathrm{square}\:\mathrm{1}.\: \\ $$$${B},\:{F}\:\mathrm{and}\:{E}\:\mathrm{are}\:\mathrm{collinear}. \\ $$$${FDE}\:\mathrm{is}\:\mathrm{a}\:\mathrm{right}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{hypotenuse}\:\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{the}\:{DE}\:\mathrm{cathetus}\:\mathrm{is}\:\mathrm{worth}\:\boldsymbol{{x}}.\: \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\boldsymbol{{x}}? \\ $$$$\left(\mathrm{Solve}\:\mathrm{with}\:\mathrm{algebra}\right) \\ $$ Commented by…

Question Number 134118 by ClarkeMelodyWenkeh last updated on 27/Feb/21 Answered by EDWIN88 last updated on 27/Feb/21 $$\left(\mathrm{i}\right)\:\angle\mathrm{POS}\:=\:\mathrm{2}×\angle\:\mathrm{PTS}\:=\:\mathrm{116}° \\ $$$$\left(\mathrm{ii}\right)\:\angle\:\mathrm{PRS}\:=\:\mathrm{180}°−\angle\mathrm{PTS}\:=\:\mathrm{122}°\: \\ $$ Commented by otchereabdullai@gmail.com last…

Question Number 68503 by TawaTawa last updated on 12/Sep/19 Commented by Prithwish sen last updated on 12/Sep/19 $$\mathrm{2x}\frac{\pi\mathrm{6}^{\mathrm{2}} }{\mathrm{4}}\:−\mathrm{2}\left[\mathrm{36}\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)\right]=\mathrm{18}\left[\sqrt{\mathrm{3}}−\frac{\pi}{\mathrm{3}}\right]\:\backsim\:\mathrm{12}.\mathrm{33} \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$ Commented by…