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Category: Geometry

Question-224592

Question Number 224592 by Nicholas666 last updated on 20/Sep/25 Commented by Nicholas666 last updated on 20/Sep/25 $$\:\:\:\:\:\mathrm{The}\:\mathrm{medians}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:{ABC}\:\:\mathrm{cut}\:\mathrm{it}\:\mathrm{into}\:\mathrm{6}\:\mathrm{triangles}.\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{centers}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circum}\:\mathrm{scribed}\:\mathrm{circles}\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{of}\:\mathrm{these}\:\mathrm{triangles}\:\mathrm{lie}\:\mathrm{ln}\:\mathrm{the}\:\mathrm{same}\:\mathrm{circle}.\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{How}\:\mathrm{is}\:\mathrm{this}\:\mathrm{theorem}\:\mathrm{called}? \\ $$$$…

let-I-be-the-incenter-of-a-non-isosceles-ABC-and-let-the-incircle-be-tanget-to-the-sides-point-D-E-F-the-line-AI-intersects-ABC-at-A-and-S-the-line-SD-intersect

Question Number 224594 by Nicholas666 last updated on 20/Sep/25 $$ \\ $$$$\:\:\:\:\mathrm{let}\:{I}\:\mathrm{be}\:\mathrm{the}\:\mathrm{incenter}\:\mathrm{of}\:\mathrm{a}\:\mathrm{non}−\mathrm{isosceles}\:\:\Delta{ABC}\:\:\:\:\:\: \\ $$$$\:\:\:\:\mathrm{and}\:\mathrm{let}\:\mathrm{the}\:\mathrm{incircle}\:\mathrm{be}\:\mathrm{tanget}\:\mathrm{to}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{point}\:{D},{E},{F}.\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{line}\:{AI}\:\mathrm{intersects}\: \left({ABC}\right)\:\mathrm{at}\:{A}\:\mathrm{and}\:{S}. \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{line}\:{SD}\:\mathrm{intersects}\: \left({ABC}\right)\:\mathrm{at}\:{S}\:\mathrm{and}\:{T}.\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{let}\:{IT}\:\cap\:{EF}\:={M},\: \left({BIC}\right)\:\cap\: \left({DEF}\right)={K},{L}. \\…

Question-224531

Question Number 224531 by ajfour last updated on 17/Sep/25 Commented by ajfour last updated on 17/Sep/25 A plank inclined at angle phi to the level ground has coefficient of friction such that it just permits enough friction for a ring of mass M and radius R to purely roll down the fixed incline. Now our plan is to weld a bead at the periphery of the ring so that this new ring plus bead combination when olaced upright on the same incline only slides down not rotating at all. we hsve to find the minimum bead mass. assume coefficient of static and kinetic friction to be same. Commented by mahdipoor last updated on 17/Sep/25 $${i}\:{get}\:…

Question-224337

Question Number 224337 by wongb1506 last updated on 04/Sep/25 Answered by BaliramKumar last updated on 04/Sep/25 $${cos}\left({A}=\mathrm{60}°\right)\:=\:\frac{\mathrm{5}^{\mathrm{2}} +\left({x}−\mathrm{6}\right)^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}×\left({x}−\mathrm{6}\right)} \\ $$$${x}=\mathrm{14}\:\checkmark\:\:\:\:\:{or}\:\:\:\:\:\:\cancel{\mathrm{3}}\: \\ $$$$ \\…

Question-224336

Question Number 224336 by wongb1506 last updated on 04/Sep/25 Answered by som(math1967) last updated on 04/Sep/25 $${let}\:{EC}={x},\:\angle{BEC}=\alpha \\ $$$$\therefore\angle{AED}=\mathrm{180}−\mathrm{60}−\alpha=\mathrm{120}−\alpha \\ $$$$\:\frac{\mathrm{4}}{{x}}={sin}\alpha\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{4}} \\ $$$$\:\:\frac{\mathrm{3}}{{x}}={sin}\left(\mathrm{120}−\alpha\right)\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{{sin}\left(\mathrm{120}−\alpha\right)}{\mathrm{3}} \\ $$$$\therefore\:\frac{{sin}\alpha}{\mathrm{4}}=\frac{{sin}\left(\mathrm{120}−\alpha\right)}{\mathrm{3}}…