Question Number 224592 by Nicholas666 last updated on 20/Sep/25 Commented by Nicholas666 last updated on 20/Sep/25 $$\:\:\:\:\:\mathrm{The}\:\mathrm{medians}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:{ABC}\:\:\mathrm{cut}\:\mathrm{it}\:\mathrm{into}\:\mathrm{6}\:\mathrm{triangles}.\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{centers}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circum}\:\mathrm{scribed}\:\mathrm{circles}\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{of}\:\mathrm{these}\:\mathrm{triangles}\:\mathrm{lie}\:\mathrm{ln}\:\mathrm{the}\:\mathrm{same}\:\mathrm{circle}.\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{How}\:\mathrm{is}\:\mathrm{this}\:\mathrm{theorem}\:\mathrm{called}? \\ $$$$…
Question Number 224594 by Nicholas666 last updated on 20/Sep/25 $$ \\ $$$$\:\:\:\:\mathrm{let}\:{I}\:\mathrm{be}\:\mathrm{the}\:\mathrm{incenter}\:\mathrm{of}\:\mathrm{a}\:\mathrm{non}−\mathrm{isosceles}\:\:\Delta{ABC}\:\:\:\:\:\: \\ $$$$\:\:\:\:\mathrm{and}\:\mathrm{let}\:\mathrm{the}\:\mathrm{incircle}\:\mathrm{be}\:\mathrm{tanget}\:\mathrm{to}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{point}\:{D},{E},{F}.\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{line}\:{AI}\:\mathrm{intersects}\: \left({ABC}\right)\:\mathrm{at}\:{A}\:\mathrm{and}\:{S}. \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{line}\:{SD}\:\mathrm{intersects}\: \left({ABC}\right)\:\mathrm{at}\:{S}\:\mathrm{and}\:{T}.\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{let}\:{IT}\:\cap\:{EF}\:={M},\: \left({BIC}\right)\:\cap\: \left({DEF}\right)={K},{L}. \\…
Question Number 224558 by ajfour last updated on 19/Sep/25 Commented by ajfour last updated on 19/Sep/25 Commented by ajfour last updated on 19/Sep/25 Commented by…
Question Number 224568 by fkwow344 last updated on 19/Sep/25 $$\sqrt{{x}}={a\begin{cases}{{a}\in\mathbb{Z}=\mathrm{0}}\\{{a}\notin\mathbb{Z}=\mathrm{1}}\end{cases}} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{3}} \:\sqrt{{x}}{dx}=? \\ $$ Commented by Ghisom_ last updated on 19/Sep/25 $$\mathrm{what}\:\mathrm{does}\:\mathrm{this}\:\mathrm{mean}? \\…
Question Number 224570 by Abdulazim last updated on 19/Sep/25 Answered by fantastic last updated on 19/Sep/25 $$\mathrm{30}^{\mathrm{0}} \\ $$$${AB}=\mathrm{2}{R}\mathrm{sin}\:\alpha \\ $$$${CH}=\mathrm{2}{R}\mathrm{cos}\:\alpha\left[{R}={circumradius}\right] \\ $$$${CH}=\sqrt{\mathrm{3}}{AB} \\ $$$$\mathrm{2}{R}\mathrm{cos}\:\alpha=\sqrt{\mathrm{3}}×\mathrm{2}{R}\mathrm{sin}\:\alpha…
Question Number 224562 by gregori last updated on 19/Sep/25 $$\:\:\: \\ $$ Commented by Frix last updated on 19/Sep/25 $${r}=\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }=\frac{{a}_{\mathrm{3}} }{{a}_{\mathrm{2}} }\:\Leftrightarrow\:\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}\:\Leftrightarrow\:\frac{\mathrm{cos}^{\mathrm{3}}…
Question Number 224556 by mr W last updated on 18/Sep/25 Commented by mr W last updated on 18/Sep/25 $${if}\:{the}\:{friction}\:{coefficient}\:{between} \\ $$$${a}\:{small}\:{ball}\:{with}\:{mass}\:\boldsymbol{{m}}\:{and}\:{a} \\ $$$${thin}\:{tube}\:{ring}\:{with}\:{radius}\:\boldsymbol{{R}}\:{and} \\ $$$${mass}\:\boldsymbol{{M}}\:{is}\:\boldsymbol{\mu},\:{find}\:{the}\:{angle}\:\boldsymbol{\theta}\:{which}…
Question Number 224538 by MirHasibulHossain last updated on 17/Sep/25 $$\mathrm{32}^{\mathrm{4r}^{\mathrm{2}} −\mathrm{8}} =\mathrm{1}\:\:\:\:\mathrm{then}\:\mathrm{find}\:\mathrm{r}=? \\ $$ Answered by fantastic last updated on 17/Sep/25 $$\left(\mathrm{4}{r}^{\mathrm{2}} −\mathrm{8}\right)=\mathrm{log}\:_{\mathrm{32}} \mathrm{1}=\mathrm{0} \\…
Question Number 224539 by fantastic last updated on 17/Sep/25 $${a}^{{x}} ={m},\:{a}^{{y}} ={n}\:,{a}^{\mathrm{2}} =\left({m}^{{y}} {n}^{{x}} \right)^{{z}} \\ $$$${prove}\:{xyz}=\mathrm{1} \\ $$ Answered by som(math1967) last updated on…
Question Number 224529 by ajfour last updated on 17/Sep/25 Commented by ajfour last updated on 17/Sep/25 $${Find}\:{r}/{R}\:{if}\:{r}_{\mathrm{1}} ={r}_{\mathrm{2}} ={r}_{\mathrm{3}} \:{and}\:{R}_{\mathrm{2}} ={R}_{\mathrm{1}} . \\ $$ Answered…