Bosnia and Herzegovina TST 1997 Day 1 Problem 2

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Bosnia and Herzegovina TST 1997 Day 1 Problem 2

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Source: Bosnia and Herzegovina Team Selection Test 1997

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#1 h Sep 20, 2018, 11:50 AM • 2 Y

Y by Adventure10, Mango247

In isosceles triangle $ABC$ with base side $AB$ , on side $BC$ it is given point $M$ . Let $O$ be a circumcenter and $S$ incenter of triangle $ABC$ . Prove that $SM \mid \mid AC \Leftrightarrow OM \perp BS$

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#2 h Jul 28, 2023, 12:21 AM

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gobathegreat wrote:

In isosceles triangle $ABC$ with base side $AB$ , on side $BC$ it is given point $M$ . Let $O$ be a circumcenter and $S$ incenter of triangle $ABC$ . Prove that $SM \mid \mid AC \Leftrightarrow OM \perp BS$

$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
First we will do the going and then the return
$\color{red}\boxed{SM||AC}$
$\color{red}\rule{24cm}{0.3pt}$
Let $T\equiv OM\cap BS$
Let $H\equiv OC\cap AB$
$\Rightarrow CM\perp AB$ Since $O$ is circumcenter:
$\Rightarrow \angle ACO=\angle OCB$ Since $AC||SM:$
$\Rightarrow \angle OCB=\angle CSM...(I)$ Since $O$ is circumcenter:
$\Rightarrow \angle OCB=\angle OBM$ By $(I):$
$\Rightarrow \angle CSM=\angle OBM$ $\Rightarrow OSBM \text{ is cyclic}$ $\Rightarrow \angle SBM=\angle TOS...(II)$ Since $S$ is incenter:
$\Rightarrow \angle SBM=\angle SBH$ By $(II):$
$\Rightarrow \angle SBH=\angle TOS=\angle TOH$ $\Rightarrow TOBH \text{ is cyclic}$ $\Rightarrow \angle OTB=\angle OHB=90$ $\Rightarrow OM\perp BS_\blacksquare$ $\color{red}\rule{24cm}{0.3pt}$
$\color{red}\boxed{OM\perp BS}$
$\color{red}\rule{24cm}{0.3pt}$
Let $T\equiv OM\cap BS$
Let $H\equiv OC\cap AB$
Since $\triangle ABC$ is isosceles:
$\Rightarrow OC\perp AB$ $\Rightarrow HBOT \text{ is cyclic}$ $\Rightarrow \angle TOH=\angle HBS=\angle SBM$ $\Rightarrow SBMO \text{ is cyclic}...(I)$ Since $O$ is circumcenter:
$\Rightarrow \angle ACS=\angle OCM=\angle OBC$ By $(I):$
$\Rightarrow \angle ACS=\angle OSM$ $\Rightarrow SM||AC_\blacksquare$ $\color{red}\rule{24cm}{0.3pt}$
$\Rightarrow \boxed{SM||AC\Leftrightarrow OM\perp BS}_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$

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#3 h Jul 28, 2023, 1:30 PM

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Let $D$ be the midpoint of arc $CA$ and let $G$ be such that $BODG$ is a parallelogram and let $N=OG\cap BC$ , then we have to prove that $SN\parallel CA$ .
We use complex numbers. With $(ABC)$ as the unit circle. Let $c=1$ , then $a=d^2$ and $b=\frac{1}{d^2}$ . So $s=d+\frac{1}{d}-1=\frac{d^2-d+1}{d}$ and $g=b+d=\frac{1}{d^2}+d=\frac{d^3+1}{d^2}$ . We have $\begin{align*}n & =\frac{\left (\overline og-o\overline g\right )(c-b)-(o-g)\left (\overline cb-\overline bc\right )}{\left (\overline o-\overline g\right )(c-b)-(o-g)\left (\overline c-\overline b\right )} \\ & =\frac{g\left (\frac{b}{c}-\frac{c}{b}\right )}{\overline g(b-c)+g\left (\frac{1}{c}-\frac{1}{b}\right )} \\ & =\frac{g\frac{b^2-c^2}{bc}}{\overline g(b-c)+g\frac{b-c}{bc}} \\ & =\frac{g\frac{(b+c)(b-c)}{bc}}{\overline g(b-c)+g\frac{b-c}{bc}} \\ & =\frac{g(b+c)}{\overline gbc+g} \\ & =\frac{\frac{d^3+1}{d^2}\left (\frac{1}{d^2}+1\right )}{\frac{\frac{1}{d^3}+1}{\frac{1}{d^2}}\frac{1}{d^2}+\frac{d^3+1}{d^2}} \\ & =\frac{\frac{d^3+1}{d^2}\frac{d^2+1}{d^2}}{\frac{1}{d^3}+1+\frac{d^3+1}{d^2}} \\ & =\frac{\frac{(d^3+1)(d^2+1)}{d^4}}{\frac{d^3+1}{d^3}+\frac{d^3+1}{d^2}} \\ & =\frac{d^2+1}{d(d+1)}, \end{align*}$ so $s-n=\frac{d^2-d+1}{d}-\frac{d^2+1}{d(d+1)}=\frac{d(d-1)}{d+1}$ and $a-c=d^2-1=(d+1)(d-1).$ Therefore we get $\frac{a-c}{s-n}=\frac{(d+1)^2}{d}=d+2+\frac{1}{d}=d+\overline d+2\in \mathbb{R}.$ Done.

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