Question Number 164842 by ajfour last updated on 22/Jan/22 Commented by MJS_new last updated on 22/Jan/22 $$\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{largest}\:{R}\:\mathrm{should}\:\mathrm{be}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle} \\ $$$$\mathrm{with}\:\mathrm{sides}\:\mathrm{2}{a},\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} },\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\Rightarrow\:{R}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}{b}}…
Question Number 33756 by Rasheed.Sindhi last updated on 23/Apr/18 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:\boldsymbol{\mathrm{circum}}-\boldsymbol{\mathrm{circle}}\:\mathrm{and} \\ $$$$\boldsymbol{\mathrm{in}}-\boldsymbol{\mathrm{circle}}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{are}\:\boldsymbol{\mathrm{concentric}}, \\ $$$$\mathrm{the}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{an}\:\boldsymbol{\mathrm{equalateral}}\:\boldsymbol{\mathrm{triangle}}. \\ $$ Answered by MJS last updated on 23/Apr/18 $$\mathrm{draw}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{that}\:“\mathrm{sits}''\:\mathrm{on}\:{O}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\mathrm{with} \\…
Question Number 164748 by ajfour last updated on 21/Jan/22 Commented by mr W last updated on 21/Jan/22 $${is}\:{it}\:{not}\:{always}\:{ok}\:{for}\:{any}\:{a}\:{and}\:{b} \\ $$$${with}\:{b}={a}+\mathrm{1}? \\ $$ Commented by ajfour…
Question Number 164738 by ajfour last updated on 21/Jan/22 Commented by Rasheed.Sindhi last updated on 21/Jan/22 $${Sorry}\:{if}\:{I}\:{don}'{t}\:{understand}\:{sir}! \\ $$$$\mathcal{D}{o}\:{you}\:{mean}\:{construct}\:{it}\:{with} \\ $$$$\mathcal{E}{uclidean}\:{tools}\:{that}\:{is}\:{compass} \\ $$$${and}\:{straightedge}? \\ $$…
Question Number 164627 by amin96 last updated on 19/Jan/22 Answered by mahdipoor last updated on 19/Jan/22 $${a}=\mathrm{2018}+\frac{\mathrm{1}}{\mathrm{2019}}=\mathrm{2019}−\frac{\mathrm{2018}}{\mathrm{2019}} \\ $$$${b}=\mathrm{2019}−\frac{\mathrm{1}}{\mathrm{2018}} \\ $$$${c}=\mathrm{2018}+\frac{\mathrm{2019}}{\mathrm{2018}}=\mathrm{2019}+\frac{\mathrm{1}}{\mathrm{2018}} \\ $$$${d}=\mathrm{2020}−\frac{\mathrm{2020}}{\mathrm{2019}}=\mathrm{2019}−\frac{\mathrm{1}}{\mathrm{2019}} \\ $$$$\Rightarrow{a}<{b}<{d}<{c}…
Question Number 164626 by amin96 last updated on 19/Jan/22 Commented by mr W last updated on 19/Jan/22 $$\left.\Rightarrow\:{C}\right) \\ $$ Answered by mr W last…
Question Number 33452 by ajfour last updated on 16/Apr/18 Commented by ajfour last updated on 16/Apr/18 $${Find}\:{length}\:{of}\:{shadow}\:{s}\:{of}\:{a}\: \\ $$$${stick}\:{of}\:{length}\:\boldsymbol{{l}}\:{when}\:{source} \\ $$$${of}\:{light}\:{is}\:{a}\:{point}\:{source}\:{at} \\ $$$${height}\:{H}\:{above}\:{ground}. \\ $$…
Question Number 164515 by ajfour last updated on 18/Jan/22 Commented by ajfour last updated on 18/Jan/22 $${Find}\:{x}\:{in}\:{terms}\:{of}\:{c},\:{even}\:{when} \\ $$$$\:\:\:\:\:{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$ Commented by mr W…
Question Number 164468 by ajfour last updated on 17/Jan/22 Answered by mr W last updated on 18/Jan/22 Commented by mr W last updated on 18/Jan/22…
Question Number 98675 by Algoritm last updated on 15/Jun/20 Commented by Algoritm last updated on 16/Jun/20 $$\mathrm{someone}\:\mathrm{gave}\:\mathrm{me}\:\mathrm{an}\:\mathrm{original}\:\mathrm{picture}\:\mathrm{and}\:\mathrm{I}'\mathrm{ll}\:\mathrm{translate}\:\mathrm{it} \\ $$ Commented by mr W last updated…