Question-46864

Question Number 46864 by ajfour last updated on 01/Nov/18

Commented by ajfour last updated on 02/Nov/18

If a sphere of radius R touches the walls at A, B, C ; find a,b,c in terms of α,β,γ and R.

$${If}\:{a}\:{sphere}\:{of}\:{radius}\:\boldsymbol{{R}}\:{touches} \\ $$$${the}\:{walls}\:{at}\:{A},\:{B},\:{C}\:;\:{find}\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}} \\ $$$${in}\:{terms}\:{of}\:\alpha,\beta,\gamma\:{and}\:{R}. \\ $$

Commented by ajfour last updated on 02/Nov/18

Answered by MrW3 last updated on 02/Nov/18

Commented by MrW3 last updated on 04/Nov/18

M=center of sphere MA=MB=MC=R let H=OM OA=(√(OM^2 −MA^2 ))=(√(H^2 −R^2 )) OB=(√(OM^2 −MB^2 ))=(√(H^2 −R^2 )) OC=(√(OM^2 −MC^2 ))=(√(H^2 −R^2 )) ⇒OA=OB=OC=h (say) ΔOBQ≡ΔOAQ ⇒∠OQB=∠OQA=φ (say) similarly ⇒∠ORA=∠ORC=ψ (say) ⇒∠OPC=∠OPB=ϕ (say) γ+ϕ+φ=π α+φ+ψ=π β+ψ+ϕ=π ⇒α+β+γ+2(ϕ+φ+ψ)=3π ⇒ϕ+φ+ψ=((3π)/2)−((α+β+γ)/2) ⇒ϕ+π−α=((3π)/2)−((α+β+γ)/2) ⇒ϕ=(π/2)−((−α+β+γ)/2) ⇒φ+π−β=((3π)/2)−((α+β+γ)/2) ⇒φ=(π/2)−((α−β+γ)/2) ⇒ψ+π−γ=((3π)/2)−((α+β+γ)/2) ⇒ψ=(π/2)−((α+β−γ)/2) let QR=u, PQ=v, RP=w (u/(sin α))=((OQ)/(sin ∠ORQ))=((OP)/(sin ψ))=(h/(sin φ sin ψ)) ⇒u=((sin α)/(sin φ sin ψ))h=((sin α)/(cos ((α−β+γ)/2) cos ((α+β−γ)/2)))h=μh (v/(sin γ))=((OP)/(sin ∠OQP))=((OP)/(sin φ))=(h/(sin ϕ sin φ)) ⇒v=((sin γ)/(sin ϕ sin φ))h=((sin γ)/(cos ((−α+β+γ)/2) cos ((α−β+γ)/2)))h=νh (w/(sin β))=((OR)/(sin ∠OPR))=((OR)/(sin ϕ))=(h/(sin ψ sin ϕ)) ⇒w=((sin β)/(sin ψ sin ϕ))h=((sin β)/(cos ((α+β−γ)/2) cos ((−α+β+γ)/2)))h=ωh in ΔPQR: the incircle with radius r touches the triangle at A,B,C. (r/R)=(h/( (√(h^2 +R^2 )))) ⇒r=(R/( (√(1+((R/h))^2 )))) let s=((u+v+w)/2)=((μ+ν+ω)/2)h A_(ΔPQR) =sr=(√(s(s−u)(s−v)(s−w))) ⇒r=(√(((s−u)(s−v)(s−w))/s)) ⇒r=(1/2)(√(((μ+ν−ω)(μ−ν+ω)(−μ+ν+ω))/(μ+ν+ω))) h=λh r=(R/( (√(1+((R/h))^2 ))))=λh ⇒λ^2 [1+((h/R))^2 ]=1 ⇒h=R(√((1/λ^2 )−1)) ⇒r=λh=R(√(1−λ^2 )) u^2 =v^2 +w^2 −2vw cos ∠RPQ ⇒cos ∠RPQ=((v^2 +w^2 −u^2 )/(2vw))=((ν^2 +ω^2 −μ^2 )/(2νω)) a^2 =2r^2 (1+cos ∠RPQ)=r^2 (2+((ν^2 +ω^2 −μ^2 )/(νω))) ⇒a=r(√(2+((ν^2 +ω^2 −μ^2 )/(νω)))) ⇒a=R(√((1−λ^2 )(2+((ν^2 +ω^2 −μ^2 )/(νω))))) similarly ⇒b=R(√((1−λ^2 )(2+((ω^2 +μ^2 −ν^2 )/(ωμ))))) ⇒c=R(√((1−λ^2 )(2+((μ^2 +ν^2 −ω^2 )/(μν))))) with μ=((sin α)/(cos ((α−β+γ)/2) cos ((α+β−γ)/2))) ν=((sin γ)/(cos ((−α+β+γ)/2) cos ((α−β+γ)/2))) ω=((sin β)/(cos ((α+β−γ)/2) cos ((−α+β+γ)/2))) λ=(1/2)(√(((μ+ν−ω)(μ−ν+ω)(−μ+ν+ω))/(μ+ν+ω))) (H/R)=(h/r)=(1/λ) ⇒OM=H=(R/λ) ==================== check with α=β=γ=(π/2): μ=ν=ω=((sin (π/2))/(cos (π/4) cos (π/4)))=2 λ=(1/2)(√((2×2×2)/(2+2+2)))=(1/( (√3))) H=(R/λ)=(√3)R ⇒ correct a=b=c=R(√((1−(1/3))(2+(2^2 /(2×2)))))=(√2)R ⇒ correct

$${M}={center}\:{of}\:{sphere} \\ $$$${MA}={MB}={MC}={R} \\ $$$${let}\:{H}={OM} \\ $$$${OA}=\sqrt{{OM}^{\mathrm{2}} −{MA}^{\mathrm{2}} }=\sqrt{{H}^{\mathrm{2}} −{R}^{\mathrm{2}} } \\ $$$${OB}=\sqrt{{OM}^{\mathrm{2}} −{MB}^{\mathrm{2}} }=\sqrt{{H}^{\mathrm{2}} −{R}^{\mathrm{2}} } \\ $$$${OC}=\sqrt{{OM}^{\mathrm{2}} −{MC}^{\mathrm{2}} }=\sqrt{{H}^{\mathrm{2}} −{R}^{\mathrm{2}} } \\ $$$$\Rightarrow{OA}={OB}={OC}={h}\:\left({say}\right) \\ $$$$ \\ $$$$\Delta{OBQ}\equiv\Delta{OAQ} \\ $$$$\Rightarrow\angle{OQB}=\angle{OQA}=\phi\:\left({say}\right) \\ $$$${similarly} \\ $$$$\Rightarrow\angle{ORA}=\angle{ORC}=\psi\:\left({say}\right) \\ $$$$\Rightarrow\angle{OPC}=\angle{OPB}=\varphi\:\left({say}\right) \\ $$$$ \\ $$$$\gamma+\varphi+\phi=\pi \\ $$$$\alpha+\phi+\psi=\pi \\ $$$$\beta+\psi+\varphi=\pi \\ $$$$\Rightarrow\alpha+\beta+\gamma+\mathrm{2}\left(\varphi+\phi+\psi\right)=\mathrm{3}\pi \\ $$$$\Rightarrow\varphi+\phi+\psi=\frac{\mathrm{3}\pi}{\mathrm{2}}−\frac{\alpha+\beta+\gamma}{\mathrm{2}} \\ $$$$\Rightarrow\varphi+\pi−\alpha=\frac{\mathrm{3}\pi}{\mathrm{2}}−\frac{\alpha+\beta+\gamma}{\mathrm{2}} \\ $$$$\Rightarrow\varphi=\frac{\pi}{\mathrm{2}}−\frac{−\alpha+\beta+\gamma}{\mathrm{2}} \\ $$$$\Rightarrow\phi+\pi−\beta=\frac{\mathrm{3}\pi}{\mathrm{2}}−\frac{\alpha+\beta+\gamma}{\mathrm{2}} \\ $$$$\Rightarrow\phi=\frac{\pi}{\mathrm{2}}−\frac{\alpha−\beta+\gamma}{\mathrm{2}} \\ $$$$\Rightarrow\psi+\pi−\gamma=\frac{\mathrm{3}\pi}{\mathrm{2}}−\frac{\alpha+\beta+\gamma}{\mathrm{2}} \\ $$$$\Rightarrow\psi=\frac{\pi}{\mathrm{2}}−\frac{\alpha+\beta−\gamma}{\mathrm{2}} \\ $$$$ \\ $$$${let}\:{QR}={u},\:{PQ}={v},\:{RP}={w} \\ $$$$\frac{{u}}{\mathrm{sin}\:\alpha}=\frac{{OQ}}{\mathrm{sin}\:\angle{ORQ}}=\frac{{OP}}{\mathrm{sin}\:\psi}=\frac{{h}}{\mathrm{sin}\:\phi\:\mathrm{sin}\:\psi} \\ $$$$\Rightarrow{u}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\phi\:\mathrm{sin}\:\psi}{h}=\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\frac{\alpha−\beta+\gamma}{\mathrm{2}}\:\mathrm{cos}\:\frac{\alpha+\beta−\gamma}{\mathrm{2}}}{h}=\mu{h} \\ $$$$\frac{{v}}{\mathrm{sin}\:\gamma}=\frac{{OP}}{\mathrm{sin}\:\angle{OQP}}=\frac{{OP}}{\mathrm{sin}\:\phi}=\frac{{h}}{\mathrm{sin}\:\varphi\:\mathrm{sin}\:\phi} \\ $$$$\Rightarrow{v}=\frac{\mathrm{sin}\:\gamma}{\mathrm{sin}\:\varphi\:\mathrm{sin}\:\phi}{h}=\frac{\mathrm{sin}\:\gamma}{\mathrm{cos}\:\frac{−\alpha+\beta+\gamma}{\mathrm{2}}\:\mathrm{cos}\:\frac{\alpha−\beta+\gamma}{\mathrm{2}}}{h}=\nu{h} \\ $$$$\frac{{w}}{\mathrm{sin}\:\beta}=\frac{{OR}}{\mathrm{sin}\:\angle{OPR}}=\frac{{OR}}{\mathrm{sin}\:\varphi}=\frac{{h}}{\mathrm{sin}\:\psi\:\mathrm{sin}\:\varphi} \\ $$$$\Rightarrow{w}=\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\psi\:\mathrm{sin}\:\varphi}{h}=\frac{\mathrm{sin}\:\beta}{\mathrm{cos}\:\frac{\alpha+\beta−\gamma}{\mathrm{2}}\:\mathrm{cos}\:\frac{−\alpha+\beta+\gamma}{\mathrm{2}}}{h}=\omega{h} \\ $$$$ \\ $$$${in}\:\Delta{PQR}: \\ $$$${the}\:{incircle}\:{with}\:{radius}\:{r}\:{touches}\:{the} \\ $$$${triangle}\:{at}\:{A},{B},{C}. \\ $$$$\frac{{r}}{{R}}=\frac{{h}}{\:\sqrt{{h}^{\mathrm{2}} +{R}^{\mathrm{2}} }} \\ $$$$\Rightarrow{r}=\frac{{R}}{\:\sqrt{\mathrm{1}+\left(\frac{{R}}{{h}}\right)^{\mathrm{2}} }} \\ $$$${let}\:{s}=\frac{{u}+{v}+{w}}{\mathrm{2}}=\frac{\mu+\nu+\omega}{\mathrm{2}}{h} \\ $$$${A}_{\Delta{PQR}} ={sr}=\sqrt{{s}\left({s}−{u}\right)\left({s}−{v}\right)\left({s}−{w}\right)} \\ $$$$\Rightarrow{r}=\sqrt{\frac{\left({s}−{u}\right)\left({s}−{v}\right)\left({s}−{w}\right)}{{s}}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\left(\mu+\nu−\omega\right)\left(\mu−\nu+\omega\right)\left(−\mu+\nu+\omega\right)}{\mu+\nu+\omega}}\:{h}=\lambda{h} \\ $$$${r}=\frac{{R}}{\:\sqrt{\mathrm{1}+\left(\frac{{R}}{{h}}\right)^{\mathrm{2}} }}=\lambda{h} \\ $$$$\Rightarrow\lambda^{\mathrm{2}} \left[\mathrm{1}+\left(\frac{{h}}{{R}}\right)^{\mathrm{2}} \right]=\mathrm{1} \\ $$$$\Rightarrow{h}={R}\sqrt{\frac{\mathrm{1}}{\lambda^{\mathrm{2}} }−\mathrm{1}} \\ $$$$\Rightarrow{r}=\lambda{h}={R}\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} } \\ $$$$ \\ $$$${u}^{\mathrm{2}} ={v}^{\mathrm{2}} +{w}^{\mathrm{2}} −\mathrm{2}{vw}\:\mathrm{cos}\:\angle{RPQ} \\ $$$$\Rightarrow\mathrm{cos}\:\angle{RPQ}=\frac{{v}^{\mathrm{2}} +{w}^{\mathrm{2}} −{u}^{\mathrm{2}} }{\mathrm{2}{vw}}=\frac{\nu^{\mathrm{2}} +\omega^{\mathrm{2}} −\mu^{\mathrm{2}} }{\mathrm{2}\nu\omega} \\ $$$${a}^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\angle{RPQ}\right)={r}^{\mathrm{2}} \left(\mathrm{2}+\frac{\nu^{\mathrm{2}} +\omega^{\mathrm{2}} −\mu^{\mathrm{2}} }{\nu\omega}\right) \\ $$$$\Rightarrow{a}={r}\sqrt{\mathrm{2}+\frac{\nu^{\mathrm{2}} +\omega^{\mathrm{2}} −\mu^{\mathrm{2}} }{\nu\omega}} \\ $$$$\Rightarrow{a}={R}\sqrt{\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\left(\mathrm{2}+\frac{\nu^{\mathrm{2}} +\omega^{\mathrm{2}} −\mu^{\mathrm{2}} }{\nu\omega}\right)} \\ $$$${similarly} \\ $$$$\Rightarrow{b}={R}\sqrt{\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\left(\mathrm{2}+\frac{\omega^{\mathrm{2}} +\mu^{\mathrm{2}} −\nu^{\mathrm{2}} }{\omega\mu}\right)} \\ $$$$\Rightarrow{c}={R}\sqrt{\left(\mathrm{1}−\lambda^{\mathrm{2}} \right)\left(\mathrm{2}+\frac{\mu^{\mathrm{2}} +\nu^{\mathrm{2}} −\omega^{\mathrm{2}} }{\mu\nu}\right)} \\ $$$${with} \\ $$$$\mu=\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\frac{\alpha−\beta+\gamma}{\mathrm{2}}\:\mathrm{cos}\:\frac{\alpha+\beta−\gamma}{\mathrm{2}}} \\ $$$$\nu=\frac{\mathrm{sin}\:\gamma}{\mathrm{cos}\:\frac{−\alpha+\beta+\gamma}{\mathrm{2}}\:\mathrm{cos}\:\frac{\alpha−\beta+\gamma}{\mathrm{2}}} \\ $$$$\omega=\frac{\mathrm{sin}\:\beta}{\mathrm{cos}\:\frac{\alpha+\beta−\gamma}{\mathrm{2}}\:\mathrm{cos}\:\frac{−\alpha+\beta+\gamma}{\mathrm{2}}} \\ $$$$\lambda=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\left(\mu+\nu−\omega\right)\left(\mu−\nu+\omega\right)\left(−\mu+\nu+\omega\right)}{\mu+\nu+\omega}} \\ $$$$ \\ $$$$\frac{{H}}{{R}}=\frac{{h}}{{r}}=\frac{\mathrm{1}}{\lambda} \\ $$$$\Rightarrow{OM}={H}=\frac{{R}}{\lambda} \\ $$$$==================== \\ $$$${check}\:{with}\:\alpha=\beta=\gamma=\frac{\pi}{\mathrm{2}}: \\ $$$$\mu=\nu=\omega=\frac{\mathrm{sin}\:\frac{\pi}{\mathrm{2}}}{\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\:\mathrm{cos}\:\frac{\pi}{\mathrm{4}}}=\mathrm{2} \\ $$$$\lambda=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{2}×\mathrm{2}×\mathrm{2}}{\mathrm{2}+\mathrm{2}+\mathrm{2}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${H}=\frac{{R}}{\lambda}=\sqrt{\mathrm{3}}{R}\:\Rightarrow\:{correct} \\ $$$${a}={b}={c}={R}\sqrt{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{2}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}}\right)}=\sqrt{\mathrm{2}}{R}\:\Rightarrow\:{correct} \\ $$

Commented by ajfour last updated on 03/Nov/18