Question Number 201037 by mr W last updated on 28/Nov/23
Commented by mr W last updated on 28/Nov/23
$${a}\:{triangle}\:{has}\:{sides}\:{a},\:{b},\:{c}.\:{find}\:{the} \\ $$$${fraction}\:{of}\:{its}\:{area}\:{covered}\:{by}\:{all} \\ $$$$\left({infinite}\right)\:{inscribed}\:{circles}\:{as}\:{shown}. \\ $$
Answered by mr W last updated on 29/Nov/23
Commented by mr W last updated on 29/Nov/23
$$\frac{{r}_{{n}} −{r}_{{n}+\mathrm{1}} }{{r}_{{n}} +{r}_{{n}+\mathrm{1}} }=\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\lambda \\ $$$$\Rightarrow\frac{{r}_{{n}+\mathrm{1}} }{{r}_{{n}} }=\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda}={constant}\:\Rightarrow\:{G}.{P}. \\ $$
Commented by mr W last updated on 29/Nov/23
Commented by mr W last updated on 29/Nov/23
![let λ_A =sin (A/2) (r_(A,n+1) /r_(A,n) )=...=(r_(A,1) /r_0 )=((1−λ_A )/(1+λ_A ))=k_A area of all incribed circles: A_(IC) =π(Σ_(n=0) ^∞ r_(A,n) ^2 +Σ_(n=0) ^∞ r_(B,n) ^2 +Σ_(n=0) ^∞ r_(C,n) ^2 )−2πr_0 ^2 =πr_0 ^2 ((1/(1−k_A ^2 ))+(1/(1−k_B ^2 ))+(1/(1−k_C ^2 ))−2) =πr_0 ^2 [(1/(1−(((1−sin (A/2))/(1+sin (A/2))))^2 ))+(1/(1−(((1−sin (B/2))/(1+sin (B/2))))^2 ))+(1/(1−(((1−sin (C/2))/(1+sin (C/2))))^2 ))−2] =((4πΔ^2 )/((a+b+c)^2 ))[(1/(1−(((1−sin (A/2))/(1+sin (A/2))))^2 ))+(1/(1−(((1−sin (B/2))/(1+sin (B/2))))^2 ))+(1/(1−(((1−sin (C/2))/(1+sin (C/2))))^2 ))−2] (A_(IC) /Δ)=((4πΔ)/((a+b+c)^2 ))[(1/(1−(((1−sin (A/2))/(1+sin (A/2))))^2 ))+(1/(1−(((1−sin (B/2))/(1+sin (B/2))))^2 ))+(1/(1−(((1−sin (C/2))/(1+sin (C/2))))^2 ))−2] =((8π sin A sin B sin C)/((sin A+sin B+sin C)^2 ))[(1/(1−(((1−sin (A/2))/(1+sin (A/2))))^2 ))+(1/(1−(((1−sin (B/2))/(1+sin (B/2))))^2 ))+(1/(1−(((1−sin (C/2))/(1+sin (C/2))))^2 ))−2]](https://www.tinkutara.com/question/Q201105.png)
$${let}\:\lambda_{{A}} =\mathrm{sin}\:\frac{{A}}{\mathrm{2}} \\ $$$$\frac{{r}_{{A},{n}+\mathrm{1}} }{{r}_{{A},{n}} }=…=\frac{{r}_{{A},\mathrm{1}} }{{r}_{\mathrm{0}} }=\frac{\mathrm{1}−\lambda_{{A}} }{\mathrm{1}+\lambda_{{A}} }={k}_{{A}} \\ $$$${area}\:{of}\:{all}\:{incribed}\:{circles}: \\ $$$${A}_{{IC}} =\pi\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{r}_{{A},{n}} ^{\mathrm{2}} +\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{r}_{{B},{n}} ^{\mathrm{2}} +\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{r}_{{C},{n}} ^{\mathrm{2}} \right)−\mathrm{2}\pi{r}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\:\:=\pi{r}_{\mathrm{0}} ^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{1}−{k}_{{A}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−{k}_{{B}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−{k}_{{C}} ^{\mathrm{2}} }−\mathrm{2}\right) \\ $$$$\:\:=\pi{r}_{\mathrm{0}} ^{\mathrm{2}} \left[\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{sin}\:\frac{{A}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{A}}{\mathrm{2}}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{sin}\:\frac{{B}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{B}}{\mathrm{2}}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{sin}\:\frac{{C}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{C}}{\mathrm{2}}}\right)^{\mathrm{2}} }−\mathrm{2}\right] \\ $$$$\:\:=\frac{\mathrm{4}\pi\Delta^{\mathrm{2}} }{\left({a}+{b}+{c}\right)^{\mathrm{2}} }\left[\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{sin}\:\frac{{A}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{A}}{\mathrm{2}}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{sin}\:\frac{{B}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{B}}{\mathrm{2}}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{sin}\:\frac{{C}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{C}}{\mathrm{2}}}\right)^{\mathrm{2}} }−\mathrm{2}\right] \\ $$$$\frac{{A}_{{IC}} }{\Delta}=\frac{\mathrm{4}\pi\Delta}{\left({a}+{b}+{c}\right)^{\mathrm{2}} }\left[\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{sin}\:\frac{{A}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{A}}{\mathrm{2}}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{sin}\:\frac{{B}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{B}}{\mathrm{2}}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{sin}\:\frac{{C}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{C}}{\mathrm{2}}}\right)^{\mathrm{2}} }−\mathrm{2}\right] \\ $$$$\:=\frac{\mathrm{8}\pi\:\mathrm{sin}\:{A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C}}{\left(\mathrm{sin}\:{A}+\mathrm{sin}\:{B}+\mathrm{sin}\:{C}\right)^{\mathrm{2}} }\left[\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{sin}\:\frac{{A}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{A}}{\mathrm{2}}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{sin}\:\frac{{B}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{B}}{\mathrm{2}}}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{sin}\:\frac{{C}}{\mathrm{2}}}{\mathrm{1}+\mathrm{sin}\:\frac{{C}}{\mathrm{2}}}\right)^{\mathrm{2}} }−\mathrm{2}\right] \\ $$