124. A nice and easy problem with a right-angled triangle.

by Virgil Nicula, Sep 14, 2010, 2:47 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=366610
Quote:
In $\triangle ABC$ consider the circle $w=C(D,\rho)$ , where $D\in (BC)$ and which is tangent to $AB$ , $AC$ . If are given $A$ , $\rho$ and $s$ , then find the area $S$ of $\triangle ABC$ .
Proof. Observe that the point $D$ belongs to $A$-bisector of given triangle and $AD=\frac {\rho}{\sin\frac A2}$ . Using the well-known relation $AD=\frac {2bc\cdot \cos\frac A2}{b+c}$ obtain

$\boxed{\ \rho (b+c)=bc\cdot \sin A\ }\ (1)$ . Otherwise, $[ABC]=[ABD]+[ACD]$ $\iff$ $\rho (b+c)=bc\cdot\sin A$ . From generalized Pytagoras' relation obtain

$\left[2s-(b+c)\right]^2=b^2+c^2-2bc\cdot\cos A$ $\iff$ $2s^2-2s(b+c)+bc(1+\cos A)=0$ $\iff$ $\boxed {\ s(b+c)=s^2+bc\cdot \cos^2\frac A2\ }\ (2)$ .

Otherwise, from well-known relation $\cos^2\frac A2=\frac {s(s-a)}{bc}$ obtain $bc\cdot\cos^2\frac A2=s[(b+c)-s]$ $\iff$ $s(b+c)=s^2+bc\cdot \cos^2\frac A2$ . Dividing the

relations $(1)$ , $(2)$ obtain $\frac {\rho}{s}=\frac {bc\cdot \sin A}{s^2+bc\cdot\cos^2\frac A2}$ $\iff$ $bc=\frac {\rho s^2}{s\cdot\sin A-\rho\cdot\cos^2\frac A2}$ $\implies$ $S=\frac {bc\cdot \sin A}{2}=$ $\frac {\rho s^2\cdot \sin\frac A2\cos\frac A2}{2s\cdot\sin \frac A2\cos\frac A2-\rho\cdot\cos^2\frac A2}$ $\iff$

$\boxed {S=\frac {\rho s^2\cdot\tan\frac A2}{2s\cdot\tan\frac A2-\rho}}\ (3)$ . Remark. Using the relations $S=sr$ and $r_a=s\cdot\tan\frac A2$ in the relation $(3)$ obtain that $r=\frac {\rho r_a}{2r_a-\rho}$ , i.e. $\boxed{\frac {1}{r}+\frac {1}{r_a}=\frac {2}{\rho}}$ .

Otherwise, denote $\{F,F_a\}\subset AB$ so that $IF\perp AB$ and $I_aF_a\perp AB$ . Since $\left(A,I,D,I_a\right)$ is harmonically obtain $\frac {1}{AI}+\frac {1}{AI_a}=\frac {2}{AD}$ $\iff$ $\frac {1}{r}+\frac {1}{r_a}=\frac {2}{\rho}$ .
This post has been edited 26 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:34 AM

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