Question Number 226453 by ajfour last updated on 29/Nov/25
Commented by ajfour last updated on 29/Nov/25
$${Find}\:{radii}\:{of}\:{smaller}\:{circles} \\ $$$${a}\left({top}\right),\:{b}\left({left}\right),\:{c}\left({right}\right). \\ $$
Commented by mr W last updated on 29/Nov/25
$$\frac{{h}}{\mathrm{4}}=\frac{{h}+\mathrm{4}}{\mathrm{10}}\:\Rightarrow{h}=\frac{\mathrm{8}}{\mathrm{3}}\:\Rightarrow{H}=\mathrm{4}+{h}=\frac{\mathrm{20}}{\mathrm{3}}>\mathrm{5} \\ $$$$\Rightarrow{impossible} \\ $$$$\Rightarrow{square}\:{side}\:=\mathrm{4}\:{is}\:{not}\:{possible}! \\ $$
Commented by mr W last updated on 29/Nov/25
$${perhaps}\:{square}\:{side}\:=\mathrm{3}\:? \\ $$
Answered by mr W last updated on 29/Nov/25
![say square side = 3 let AP=x ((QB)/3)=(3/(AP)) ⇒QB=(9/x) x+3+(9/x)=2×5 x^2 −7x+9=0 ⇒x=((7−(√(13)))/2) ⇒QB=10−3−((7−(√(13)))/2)=((7+(√(13)))/2) b=((3+x−(√(3^2 +x^2 )))/( 2))=((3+x−(√(7x)))/2) =((13−(√(13))−(√(14(7−(√(13))))))/4) ≈0.625202 (c/b)=((QB)/3)=((7+(√(13)))/6) ⇒c=(((7+(√(13)))[13−(√(13))−(√(14(7−(√(13)))))])/(24)) ≈1.105102 (a/b)=(3/( (√(3^2 +x^2 ))))=(3/( (√(7x)))) ⇒a=(3/( (√(7×((7−(√(13)))/2)))))×((13−(√(13))−(√(14(7−(√(13))))))/4) ≈0.544155](https://www.tinkutara.com/question/Q226458.png)
$${say}\:{square}\:{side}\:=\:\mathrm{3} \\ $$$${let}\:{AP}={x} \\ $$$$\frac{{QB}}{\mathrm{3}}=\frac{\mathrm{3}}{{AP}}\:\Rightarrow{QB}=\frac{\mathrm{9}}{{x}} \\ $$$${x}+\mathrm{3}+\frac{\mathrm{9}}{{x}}=\mathrm{2}×\mathrm{5} \\ $$$${x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{9}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{7}−\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\Rightarrow{QB}=\mathrm{10}−\mathrm{3}−\frac{\mathrm{7}−\sqrt{\mathrm{13}}}{\mathrm{2}}=\frac{\mathrm{7}+\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$${b}=\frac{\mathrm{3}+{x}−\sqrt{\mathrm{3}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{\:\mathrm{2}}=\frac{\mathrm{3}+{x}−\sqrt{\mathrm{7}{x}}}{\mathrm{2}} \\ $$$$\:\:=\frac{\mathrm{13}−\sqrt{\mathrm{13}}−\sqrt{\mathrm{14}\left(\mathrm{7}−\sqrt{\mathrm{13}}\right)}}{\mathrm{4}} \\ $$$$\:\approx\mathrm{0}.\mathrm{625202} \\ $$$$\frac{{c}}{{b}}=\frac{{QB}}{\mathrm{3}}=\frac{\mathrm{7}+\sqrt{\mathrm{13}}}{\mathrm{6}} \\ $$$$\Rightarrow{c}=\frac{\left(\mathrm{7}+\sqrt{\mathrm{13}}\right)\left[\mathrm{13}−\sqrt{\mathrm{13}}−\sqrt{\mathrm{14}\left(\mathrm{7}−\sqrt{\mathrm{13}}\right)}\right]}{\mathrm{24}} \\ $$$$\:\:\:\:\:\:\:\approx\mathrm{1}.\mathrm{105102} \\ $$$$\frac{{a}}{{b}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +{x}^{\mathrm{2}} }}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{7}{x}}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{7}×\frac{\mathrm{7}−\sqrt{\mathrm{13}}}{\mathrm{2}}}}×\frac{\mathrm{13}−\sqrt{\mathrm{13}}−\sqrt{\mathrm{14}\left(\mathrm{7}−\sqrt{\mathrm{13}}\right)}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\approx\mathrm{0}.\mathrm{544155} \\ $$
Commented by mr W last updated on 29/Nov/25
Commented by ajfour last updated on 29/Nov/25
$${Thank}\:{you}\:{sir}.\:{All}\:{excellent}!{truly}. \\ $$