95. Three circles in an angle.

by Virgil Nicula, Aug 31, 2010, 3:51 AM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=54897

PP1. Let $\Gamma (J,\rho )$ be a circle which is tangent externally to the incircle $w=C(I,r)$ of $\triangle ABC$ and to the its sidelines $AB$ , $AC$ . Find the ratio $\frac {\rho}{r}$ .

Proof. In the interior of the angle $\widehat {BAC}$ exist two circles, $\Gamma_1\left(J_1,\rho_1 \right)$ and $\Gamma_2\left(J_2,\rho_2 \right)$ with mentioned properties, where $\rho_1<r<\rho_2$ and $\rho_1\rho_2=r^2$ .

Prove easily that $\boxed{\frac {s-a}{r}=\frac {2\sqrt {r\rho}}{|\rho -r|}}\ (1)$ , where $a+b+c=2s$ . The relation $(1)$ is equivalently with $(s-a)^2\cdot\rho^2-2r\left[(s-a)^2+2r^2\right]\cdot\rho+r^2(s-a)^2=0$ $\iff$

$\frac {\rho}{r}\in\left\{\frac {(s-a)^2+2r^2\pm 2r\sqrt{(s-a)^2+r^2}}{(s-a)^2}\right\}$ . Since $IA^2=(s-a)^2+r^2$ obtain that $\frac {\rho}{r}\in \left\{\frac {IA^2+r^2\pm 2r\cdot IA}{(s-a)^2}\right\}$ $\iff$ $\frac {\rho}{r}\in\left\{\left(\frac {IA\pm r}{s-a}\right)^2\right\}$ .

Since $\frac {IA\pm r}{s-a}=\frac {1\pm \frac {r}{IA}}{\frac {s-a}{IA}}=$ $\frac {1\pm \sin\frac A2}{\cos\frac A2}=$ $\frac {1\pm \cos\frac {\pi -A}{2}}{\sin \frac {\pi -A}{2}}$. In conclusion, $\left\{\begin{array}{c}
\rho_1=r\cdot \tan^2\frac {\pi -A}{4}\\\\
\rho_2=r\cdot\cot^2\frac {\pi -A}{4}\end{array}\right\|$ $\iff$ $\frac {\rho}{r}\in \left\{\tan^2\frac {\pi -A}{4}\ ,\ \cot^2\frac {\pi -A}{4}\right\}$ .


Extension. Let $w=C(I,r)$ , $w_1=C\left(I_1,r_1\right)$ , $w_2=C\left(I_2,r_2\right)$ be three circles which are internally to the angle $\widehat{BAC}$ , are tangent to

rays $(AB$ , $(AC$ and the circle $w$ is externally tangent to the circles $w_1$ , $w_2$ . Then $\left\{r_1,r_2\right\}=\left\{r\cdot \tan^2\frac {\pi -A}{4}\ ,\ r\cdot\cot^2\frac {\pi -A}{4}\right\}$ .


Remark. Let $a_1$ , $a_2$ , $a_3$ , $\ldots$ , $a_{n-1}$ , $a_n$ , $\ldots$ be a geometrical progression, where $a_1>0$ , $a_2>0$ . Prove that

$4\cdot\left(\sum_{k=1}^{n-1}\sqrt {a_ka_{k+1}}\right)^2+\left(a_n-a_1\right)^2=$ $\left(a_1+a_n+2\cdot \sum_{k=2}^{n-1} a_k\right)^2$ . Give a geometrical interpretation.
This post has been edited 10 times. Last edited by Virgil Nicula, Nov 23, 2015, 1:51 PM

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