25. Geometric equations.

by Virgil Nicula, Apr 22, 2010, 10:29 PM

PP1. Let $ABC$ be triangle. Denote the distance $h_{a}$ from the point $A$ to the side $BC$ and $2s=a+b+c$ . Prove that :

$\left\{\begin{array}{cc}1.\blacktriangleright & \boxed{\ h_{a}=\frac{2s(s-a)x}{x^{2}+s(s-a)}\Longleftrightarrow x\in\{\ r_{b}\ ,\ r_{c}\ \}\ }\ .\\\\ 2.\blacktriangleright & \boxed{\ h_{a}=\frac{2(s-b)(s-c)x}{|x^{2}-(s-b)(s-c)|}\Longleftrightarrow x\in \{\ r\ ,\ r_{a}\ \}\ }\ .\end{array}\right\|$



PP2. Let $x$ a real number and the triangle $ABC$ . Prove that there is the following identity: $4R\sin\frac{A+x}{2}\sin\frac{B+x}{2}\sin\frac{C+x}{2}=2R\sin x\sin\frac{x}{2}+r\cos\frac{x}{2}+s\sin\frac{x}{2}$ .

Study some particular cases, for example $x\in\left\{0,\frac {\pi}{2}, \pi\right\}$ . Solve the ecuation: $s=2R\sin x+r\cot\frac{x}{2}$ , where $x\in (0, \pi)$ .


PP3. Let $ABC$ be an equilateral triangle with $AB=2$ . Consider two points $E\in (AC)$ , $F\in (AB)$ so that $AE=AF$ and $AELF$ is a tangential

quadrilateral, where $L\in BE\cap CF$ . Prove that if the $\triangle BLC$ and $\square AELF$ have same inradius $r$ , then $r=\sqrt 3-\sqrt 2$ . Nice !


Proof 1. Denote the incenters $P$ and $R$ of $\triangle BLC\ ,\ \square AELF$ respectively. Prove easily that $AR=2r$ , $PR=d=\sqrt 3-3r$ , $BP^2=1+r^2$ and $BR^2=1+(d+r)^2$ .

Observe that $m\left(\widehat{PBR}\right)=30^{\circ}$ and apply the theorem of Cosinus in $\triangle PBR$ . Obtain that $PR^2=BP^2+BR^2-2\cdot BP\cdot BR\cdot\cos \widehat {PBR}\iff$

$d^2=\left(1+r^2\right)+\left[1+(r+d)^2\right]-\sqrt{3\left(1+r^2\right)\left[1+(r+d)^2\right]}\iff$ $2\left(1+r^2+rd\right)=\sqrt{3\left(1+r^2\right)\left[1+(r+d)^2\right]}\iff$

$4\left(1+r^4+r^2d^2+2r^2+2rd+2r^3d\right)=$ $3\left(1+r^2+2rd+d^2+r^2+r^4+2r^3d+r^2d^2\right)\iff$ $1+r^4+r^2d^2+2r^2+2rd+2r^3d=3d^2\iff$

$\left(1+r^2+rd\right)^2=3d^2\iff$ $1+r^2+rd=d\sqrt{3}\iff$ Now plug in $(*)\ :\ 1+r^2+r\left(\sqrt{3}-3r\right)=\sqrt 3\cdot \left(\sqrt{3}-3r\right)\iff$ $1+r\sqrt 3-2r^2=3-3r\sqrt 3\iff$

$r^2-2r\sqrt{3}+1=0\iff$ $r_{1,2}\in\left\{\sqrt{3}\pm\sqrt{2}\right\}$ . Obviously $r$ can't exceed the altitude of the triangle, which is $a\sqrt{3}$ . Hence $\boxed{r=\sqrt{3}-\sqrt{2}}$ .

Proof 2. Denote the midpoint $D$ of $[BC]$ , the incenters $P$ and $R$ of $\triangle BLC\ ,\ \square AELF$ respectively and $m\left(\widehat{ABR}\right)=x$ . Prove easily that $AR=2r$ , $PR=d=\sqrt 3-3r$

and $RD=d+r=\sqrt 3-2r$ . Thus, $\left\{\begin{array}{ccc}
m\left(\widehat{RBD}\right)=60^{\circ}-x & \implies & \tan\left(60^{\circ}-x\right)=\frac {RD}{BD}=\sqrt 3-2r\\\\
m\left(\widehat{PBD}\right)=30^{\circ}-x & \implies & \tan\left(30^{\circ}-x\right)=r\end{array}\right|\implies$ $\frac {1}{\sqrt 3}=\tan 30^{\circ}=$

$\tan\left[\left(60^{\circ}-x\right)-\left(30^{\circ}-x\right)\right]=\frac {\left(\sqrt 3-2r\right)-r}{1+r\left(\sqrt 3-2r\right)}\implies$ $\sqrt 3\left(\sqrt 3-3r\right)=1+r\sqrt 3-2r^2\implies$ $r^2-2r\sqrt 3+1=0\implies$ $r=\sqrt 3-\sqrt 2$ .


An easy extension. Let an $A$-isosceles $\triangle ABC$ with $AB=2$ and $E\in (AC)$ , $F\in (AB)$ so that $AE=AF$ and $AELF$ is a tangential

quadrilateral, where $L\in BE\cap CF$ . Prove that $\triangle BLC$ and $\square AELF$ have same inradius $r\implies$ $\boxed{\tan\frac B2=\frac {2r}{r^2+1}}=\sin\widehat{LBC}$ .
This post has been edited 31 times. Last edited by Virgil Nicula, May 6, 2016, 11:26 PM

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