Question-125384

Question Number 125384 by ajfour last updated on 10/Dec/20

Commented by ajfour last updated on 10/Dec/20

The hemisphere has radius 2. The outer circular base has radius 3. Find maximum side length of equilateral triangle with vertices placed as shown.

$${The}\:{hemisphere}\:{has}\:{radius}\:\mathrm{2}. \\ $$$${The}\:{outer}\:{circular}\:{base}\:{has}\:{radius} \\ $$$$\mathrm{3}.\:{Find}\:{maximum}\:{side}\:{length}\:{of} \\ $$$${equilateral}\:{triangle}\:{with}\:{vertices} \\ $$$${placed}\:{as}\:{shown}. \\ $$

Answered by mr W last updated on 10/Dec/20

Commented by mr W last updated on 10/Dec/20

AB=s DC=(((√3)s)/2) DC′′≤DC≤DC′ that means the maximum equilateral ΔABC is ΔABC′, i.e. point C lies also on the base.

$${AB}={s} \\ $$$${DC}=\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}} \\ $$$${DC}''\leqslant{DC}\leqslant{DC}' \\ $$$${that}\:{means}\:{the}\:{maximum}\:{equilateral} \\ $$$$\Delta{ABC}\:{is}\:\Delta{ABC}',\:{i}.{e}.\:{point}\:{C}\:{lies}\:{also} \\ $$$${on}\:{the}\:{base}. \\ $$

Commented by mr W last updated on 10/Dec/20

Commented by mr W last updated on 10/Dec/20

R+(√(r^2 −((s/2))^2 ))=(((√3)s)/2) r^2 −((s/2))^2 =((((√3)s)/2)−R)^2 s^2 −(√3)Rs+R^2 −r^2 ⇒s_(max) =(((√3)R+(√(4r^2 −R^2 )))/2)=((3(√3)+(√7))/2)≈3.921

$${R}+\sqrt{{r}^{\mathrm{2}} −\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} −\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}}−{R}\right)^{\mathrm{2}} \\ $$$${s}^{\mathrm{2}} −\sqrt{\mathrm{3}}{Rs}+{R}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\Rightarrow{s}_{{max}} =\frac{\sqrt{\mathrm{3}}{R}+\sqrt{\mathrm{4}{r}^{\mathrm{2}} −{R}^{\mathrm{2}} }}{\mathrm{2}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}+\sqrt{\mathrm{7}}}{\mathrm{2}}\approx\mathrm{3}.\mathrm{921} \\ $$

Commented by ajfour last updated on 11/Dec/20

$${Great}!\:\:{Thank}\:{you}\:{Sir},\:{sorry}\:{it} \\ $$$${dint}\:{turn}\:{out}\:{much}\:{interesting}. \\ $$

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