There-are-3-tangent-circumferences-inscribed-in-an-isosceles-right-triangle-Two-of-these-circumferences-have-radius-R-and-are-tangent-to-the-hypotenuse-and-to-the-two-cathetus-The-smaller-circumfere

FacebookTweetPin

Question Number 71819 by Maclaurin Stickker last updated on 20/Oct/19

$$Thereare3tangentcircumferences$$$$inscribedinanisoscelesrighttriangle$$$$Twoofthesecircumferenceshave$$$$radius\mathit{R}andaretangenttothe$$$$hypotenuseandtothetwocathetus.$$$$Thesmallercircumferencehas$$$$radius\mathit{r}andistangenttothetwo$$$$cathetus.HowcanIfindtheradius$$$$ofthesmallercircumferenceasa$$$$functionof\mathit{R}?$$$$(Iwantatiponhowtosolvethe$$$$problem).$$

Commented by mr W last updated on 20/Oct/19

Commented by mr W last updated on 20/Oct/19

$$isthiswhatyoumean?$$

Commented by Maclaurin Stickker last updated on 20/Oct/19

$$Yes!I^{\prime }dlikeatiponhowtosolve.$$

Answered by mr W last updated on 20/Oct/19

Commented by mr W last updated on 20/Oct/19

$$AB=\sqrt{2}r+\sqrt{{(r+R)}^2-R^2}+R=\sqrt{2}r+\sqrt{r(2R+r)}+R$$$$\tan 22.5°=\frac{\sin 45°}{1+\cos 45°}=\frac{\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}=\sqrt{2}-1$$$$DC=\frac{R}{\tan 22.5°}=\frac{R}{\sqrt{2}-1}=(\sqrt{2}+1)R$$$$BC=R+(\sqrt{2}+1)R=(2+\sqrt{2})R$$$$AB=BC$$$$\sqrt{2}r+\sqrt{r(2R+r)}+R=(2+\sqrt{2})R$$$$\sqrt{r(2R+r)}=(1+\sqrt{2})R-\sqrt{2}r$$$$r(2R+r)={(1+\sqrt{2})}^2{R}^2+2r^2-2(1+\sqrt{2})\sqrt{2}Rr$$$$\Rightarrow r^2-2(\sqrt{2}+3)Rr+{(1+\sqrt{2})}^2{R}^2=0$$$$\Rightarrow r=(3+\sqrt{2}-2\sqrt{2+\sqrt{2}})R$$$$\Rightarrow r={(\sqrt{2+\sqrt{2}}-1)}^{2}R\approx 0.7187R$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin