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Question Number 129244 by ajfour last updated on 14/Jan/21

Commented by ajfour last updated on 14/Jan/21

$$Intermsofthesidesof△ABC,$$$$findlargestradiusspherethat$$$$canbeplacedagainsttheroom$$$$cornerandtriangle.$$

Answered by mr W last updated on 22/Feb/21

Commented by mr W last updated on 22/Feb/21

Commented by mr W last updated on 22/Feb/21

Commented by mr W last updated on 22/Feb/21

$$saytheplanecontainingthetriangle$$$$ABCintersectsthecoordinateaxes$$$$atP,Q,R.$$$$\mathrm{\Delta }PQRisacircumtriangleofthe$$$$giventriangle\mathrm{\Delta }ABC.$$$$saythesidesof\mathrm{\Delta }ABCarea,b,c.$$$$itsareais{\mathrm{\Delta }}_{ABC}.$$$${\mathrm{\Delta }}_{ABC}=\sqrt{s(s-a)(s-b)(s-c)}$$$$withs=\frac{a+b+c}{2}$$$$$$$$saytheradiusofthesmallsphere$$$$undertheplaneisRand$$$$itscenterisG(R,R,R).thedistance$$$$fromGtotheplane\mathrm{\Delta }PQRisR.$$$$$$$$sayP(p,0,0),Q(0,q,0),R(0,0,r).$$$$theequationoftheplanecontaining$$$$\mathrm{\Delta }ABCaswellas\mathrm{\Delta }PQRis$$$$\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1$$$$wehave$$$$R=\frac{\mid \frac{R}{p}+\frac{R}{q}+\frac{R}{r}-1\mid }{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}+\frac{1}{r^2}}}=\frac{1-(\frac{R}{p}+\frac{R}{q}+\frac{R}{r})}{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}+\frac{1}{r^2}}}$$$$\Rightarrow R=\frac{1}{\frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\sqrt{\frac{1}{p^2}+\frac{1}{q^2}+\frac{1}{r^2}}}$$$$duetosymmetryofrw.r.t.p,q,rwe$$$$knowRismaximumwhenp=q=r.$$$$$$$$PQ=\sqrt{p^2+q^2}$$$$QR=\sqrt{q^2+r^2}$$$$RP=\sqrt{r^2+p^2}$$$$p=q=rmeansPQ=QR=RP,i.e.$$$$\mathrm{\Delta }PQRisequilateral.$$$$$$$$thelargestequilateraltriangle\mathrm{\Delta }PQR$$$$circumscribing\mathrm{\Delta }ABChassidelength$$$$PQ=QR=RP=l=\sqrt{\frac{2(a^2+b^2+c^2)}{3}+\frac{8{\mathrm{\Delta }}_{ABC}}{\sqrt{3}}}$$$$with{\mathrm{\Delta }}_{ABC}=areaof\mathrm{\Delta }ABC.$$$$\left(seeQ60313formoredetails\right)$$$$inthiscase:p=q=r=\frac{l}{\sqrt{2}}$$$$R_{max}=\frac{1}{\frac{3}{p}+\frac{\sqrt{3}}{p}}=\frac{p}{3+\sqrt{3}}=\frac{l}{(3+\sqrt{3})\sqrt{2}}$$$$R_{max}=\frac{1}{(3+\sqrt{3})\sqrt{2}}\times \sqrt{\frac{2(a^2+b^2+c^2)}{3}+\frac{8{\mathrm{\Delta }}_{ABC}}{\sqrt{3}}}$$$$R_{max}=\frac{\sqrt{a^2+b^2+c^2+4\sqrt{3}{\mathrm{\Delta }}_{ABC}}}{3(\sqrt{3}+1)}$$$$$$$$btw,forthebigsphereovertheplane$$$$containing\mathrm{\Delta }ABCwehave$$$$R=\frac{\mid \frac{R}{p}+\frac{R}{q}+\frac{R}{r}-1\mid }{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}+\frac{1}{r^2}}}=\frac{(\frac{R}{p}+\frac{R}{q}+\frac{R}{r})-1}{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}+\frac{1}{r^2}}}$$$$\Rightarrow R_{max}=\frac{\sqrt{a^2+b^2+c^2+4\sqrt{3}{\mathrm{\Delta }}_{ABC}}}{3(\sqrt{3}-1)}$$

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