Question Number 222385 by ajfour last updated on 25/Jun/25
Commented by ajfour last updated on 25/Jun/25
Commented by ajfour last updated on 27/Jun/25
https://youtu.be/Nd0JeAO6fRY?si=87O09KtjNWiH7uXs
Answered by mr W last updated on 25/Jun/25
Commented by mr W last updated on 25/Jun/25
$${x}=\sqrt{{r}^{\mathrm{2}} −\left(\frac{{p}}{\:\sqrt{\mathrm{2}}}−{r}\right)^{\mathrm{2}} }=\sqrt{\sqrt{\mathrm{2}}{pr}−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$${r}^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}{p}+\frac{{p}}{\:\sqrt{\mathrm{2}}}+\sqrt{\sqrt{\mathrm{2}}{pr}−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}}\right)^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} \\ $$$${let}\:\frac{{r}}{{R}}=\lambda=\frac{\mathrm{1}}{\mathrm{5}},\:\xi=\frac{\sqrt{\mathrm{2}}{p}}{{R}} \\ $$$$\lambda^{\mathrm{2}} +\left(\frac{\mathrm{3}\xi}{\mathrm{2}}+\sqrt{\lambda\xi−\frac{\xi^{\mathrm{2}} }{\mathrm{4}}}\right)^{\mathrm{2}} =\left(\mathrm{1}−\lambda\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\xi+\sqrt{\mathrm{4}\lambda\xi−\xi^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{4}\left(\mathrm{1}−\mathrm{2}\lambda\right) \\ $$$$\mathrm{3}\xi+\sqrt{\mathrm{4}\lambda\xi−\xi^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{1}−\mathrm{2}\lambda}=\mu \\ $$$$\sqrt{\mathrm{4}\lambda\xi−\xi^{\mathrm{2}} }=\mu−\mathrm{3}\xi \\ $$$$\mathrm{10}\xi^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}\lambda+\mathrm{3}\mu\right)\xi+\mu^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\xi=\frac{\mathrm{2}\lambda+\mathrm{3}\mu−\sqrt{\mathrm{4}\lambda^{\mathrm{2}} +\mathrm{12}\lambda\mu−\mu^{\mathrm{2}} }}{\mathrm{10}} \\ $$$$\xi=\frac{\lambda+\mathrm{3}\sqrt{\mathrm{1}−\mathrm{2}\lambda}−\sqrt{\lambda^{\mathrm{2}} +\mathrm{2}\lambda−\mathrm{1}+\mathrm{6}\lambda\sqrt{\mathrm{1}−\mathrm{2}\lambda}}}{\mathrm{5}} \\ $$$$\Rightarrow\frac{{p}}{{R}}=\frac{\lambda+\mathrm{3}\sqrt{\mathrm{1}−\mathrm{2}\lambda}−\sqrt{\lambda^{\mathrm{2}} +\mathrm{2}\lambda−\mathrm{1}+\mathrm{6}\lambda\sqrt{\mathrm{1}−\mathrm{2}\lambda}}}{\mathrm{5}\sqrt{\mathrm{2}}} \\ $$$${example}:\:{R}=\mathrm{1},\:{r}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${p}=\frac{\sqrt{\mathrm{2}}+\mathrm{3}\sqrt{\mathrm{30}}−\mathrm{2}\sqrt{\mathrm{3}\sqrt{\mathrm{15}}−\mathrm{7}}}{\mathrm{50}}\approx\mathrm{0}.\mathrm{271} \\ $$
Commented by ajfour last updated on 25/Jun/25
$${thanks}\:{sir},\:{i}\:{reasoned}\:{that}\:{the}\:{other} \\ $$$${root}\:{is}\:{also}\:{valid}:\:{in}\:{this}\:{case}\:{square}'{s} \\ $$$${top}\:{right}\:{corner}\:{creeps}\:{inside}\:{the}\:{circle} \\ $$$${and}\:{goes}\:{to}\:{touch}\:{the}\:{other}\:{side}\:{of} \\ $$$${circle}'{s}\:{circumference}. \\ $$$${here}\:{p}\:\approx\:\mathrm{0}.\mathrm{44} \\ $$
Commented by mr W last updated on 25/Jun/25
$${i}\:{have}\:{rejected}\:{the}\:{other}\:{root}\:{for} \\ $$$${the}\:{hatched}\:{square}: \\ $$
Commented by mr W last updated on 25/Jun/25
Commented by ajfour last updated on 25/Jun/25
$${quite}\:{pragmatic}\:{of}\:{you}. \\ $$
Commented by mr W last updated on 25/Jun/25
![can we find a geometric way to solve following equation system? y^2 +z^2 +λyz=a^2 z^2 +x^2 +λzx=b^2 x^2 +y^2 +λxy=c^2 for contant λ=−2 cos θ ∈(−2, 1] and x, y, z ∈R^+](https://www.tinkutara.com/question/Q222399.png)
$${can}\:{we}\:{find}\:{a}\:{geometric}\:{way}\:{to}\: \\ $$$${solve}\:{following}\:{equation}\:{system}? \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\lambda{yz}={a}^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} +{x}^{\mathrm{2}} +\lambda{zx}={b}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\lambda{xy}={c}^{\mathrm{2}} \\ $$$${for}\:{contant}\:\lambda=−\mathrm{2}\:\mathrm{cos}\:\theta\:\in\left(−\mathrm{2},\:\mathrm{1}\right]\: \\ $$$${and}\:{x},\:{y},\:{z}\:\in{R}^{+} \\ $$
Commented by mr W last updated on 25/Jun/25
Commented by mr W last updated on 25/Jun/25
$${can}\:{we}\:{put}\:{a}\:{triangle}\:{of}\:{any}\:{size} \\ $$$${inside}\:{an}\:{open}\:{regular}\:{pyramid}\:{of} \\ $$$${infinite}\:{size}? \\ $$
Commented by mr W last updated on 25/Jun/25
Commented by ajfour last updated on 25/Jun/25
$${Q}.\:\mathrm{222404} \\ $$