Question Number 120215 by benjo_mathlover last updated on 30/Oct/20 $${Prove}\:{that}\:\sqrt[{\mathrm{3}}]{\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{\mathrm{8}}\:}\:\geqslant\:\sqrt{\frac{{ab}+{bc}+{ca}}{\mathrm{3}}} \\ $$$${for}\:{a},{b},{c}\:>\:\mathrm{0} \\ $$ Answered by TANMAY PANACEA last updated on 30/Oct/20 $$\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}}\: \\ $$$$\frac{{b}+{c}}{\mathrm{2}}\geqslant\sqrt{{bc}}…
Question Number 54661 by ajfour last updated on 08/Feb/19 Commented by ajfour last updated on 08/Feb/19 $${Given}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}},\:{find}\:\boldsymbol{{c}}. \\ $$ Commented by behi83417@gmail.com last updated on…
Question Number 185729 by Mingma last updated on 26/Jan/23 Commented by HeferH last updated on 26/Jan/23 $${not}\:{clear}\:{what}\:{segments}\:{are}\:{a},{b},{c} \\ $$ Commented by Mingma last updated on…
Question Number 185685 by cherokeesay last updated on 25/Jan/23 Answered by mr W last updated on 25/Jan/23 $$\frac{{r}}{\mathrm{2}{R}−{r}}=\mathrm{sin}\:\mathrm{30}°=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{{green}}{{yellow}}=\frac{\pi{r}^{\mathrm{2}} }{\pi{R}^{\mathrm{2}} −\pi{r}^{\mathrm{2}} }=\frac{\mathrm{1}}{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 185674 by ajfour last updated on 25/Jan/23 Commented by ajfour last updated on 25/Jan/23 $${Find}\:{the}\:{radii}\:\left({equal}\right)\:{of}\:{the}\: \\ $$$${circles},\:{given}\:{outer}\:{figure}\:{is} \\ $$$${a}\:{square}\:{of}\:{side}\:{a}. \\ $$ Answered by…
Question Number 54602 by behi83417@gmail.com last updated on 07/Feb/19 $${in}\:{a}\:{given}\:{triangle}: \\ $$$$\:\:\:\boldsymbol{\mathrm{tg}}\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{a}}.\boldsymbol{\mathrm{tgA}}+\boldsymbol{\mathrm{b}}.\boldsymbol{\mathrm{tgB}}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}}\:. \\ $$$$\boldsymbol{\mathrm{define}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{kind}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{triangle}}. \\ $$ Commented by MJS last updated on 08/Feb/19 $$\mathrm{using}\:\mathrm{the}\:\mathrm{laws}\:\mathrm{of}\:\mathrm{sines}\:\mathrm{and}\:\mathrm{cosines}\:\mathrm{I}\:\mathrm{get} \\…
Question Number 120133 by prakash jain last updated on 29/Oct/20 Commented by nimnim last updated on 29/Oct/20 $${Let}\:{the}\:{smallest}\:{diameter}\:{be}\:{x}\:{and}\:{the}\:{second}\:{be}\:{y} \\ $$$${then}\:{largest}\:{diameter}=\left({x}+{y}\right) \\ $$$${xy}=\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$${Area}_{{shaded}}…
Question Number 185637 by Mingma last updated on 24/Jan/23 Answered by HeferH last updated on 24/Jan/23 $$\:\frac{{a}}{{b}}\:=\:\frac{\mathrm{1}\:+\:\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:\frac{{a}\:+\:{b}}{\mathrm{sin}\:\left(\mathrm{18}°+{x}\right)}\:=\:\frac{{a}}{\mathrm{sin}\:\mathrm{18}°} \\ $$$$\:\mathrm{1}\:+\:\frac{{b}}{{a}}\:=\:\frac{\mathrm{sin}\:\left(\mathrm{18}°\:+\:{x}\right)}{\mathrm{sin}\:\mathrm{18}°}\: \\ $$$$\:\mathrm{1}\:+\:\frac{\mathrm{2}}{\mathrm{1}\:+\:\sqrt{\mathrm{5}}}\:=\:\frac{\mathrm{sin}\:\left(\mathrm{18}°\:+\:{x}\right)}{\mathrm{sin}\:\mathrm{18}°}\: \\ $$$$\mathrm{sin}^{−\mathrm{1}}…
Question Number 185619 by Mingma last updated on 24/Jan/23 Commented by mr W last updated on 24/Jan/23 $${a}={BD},\:{b}={EC} \\ $$$${generally}\:{x}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab} \\ $$$${x}=\sqrt{\mathrm{3}^{\mathrm{2}}…
Question Number 185602 by Mingma last updated on 24/Jan/23 Answered by mr W last updated on 24/Jan/23 $${say}\:{AB}={DE}=\mathrm{1} \\ $$$${BD}={DC}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{20}°} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{20}°\:\mathrm{sin}\:\left(\alpha+\mathrm{20}°\right)}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{20}°} \\ $$$$\mathrm{sin}\:\left(\alpha+\mathrm{20}°\right)=\frac{\mathrm{1}}{\mathrm{2}} \\…