aspect ratio AB: BC: CA wanted, <AIO=90^o, <CIO=45^o, incenter, circumcenter

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aspect ratio AB: BC: CA wanted, <AIO=90^o, <CIO=45^o, incenter, circumcenter

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Source: All-Siberian Open School Olympiad 2019-20 11.4

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#1 h Jul 6, 2020, 7:14 AM

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Let points $O$ and $I$ be the center of the circumscribed and inscribed circles triangle $ABC$ , respectively. It is known that the angle $AIO$ is right, and $\angle CIO =45^o$ . Find the aspect ratio $AB: BC: CA$ .

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#2 h Jul 6, 2020, 4:19 PM • 1 Y

Y by Mango247

We have :
$135^{\circ} =\angle AIO + \angle OIC = \angle AIC = 90^{\circ} + \frac{\angle ABC}{2} \Leftrightarrow \angle ABC=90^{\circ}$ Thus $O$ lies on $AC$ , let $R$ and $r$ be the circumradius and the inradius of $ABC$ , respectively, and let $\alpha = \angle BAC$ . From Euler's Theorem, we get :
$R(R-2r)=OI^2=(R \sin \frac{\alpha}{2})^2=R^2(\frac{1- \cos \alpha}{2}) \Leftrightarrow \cos \alpha = \frac{4r-R}{R}$ From the law of Sine, we have :
$\frac{OC}{\sin \angle OIC} = \frac{OI}{\sin \angle ICO}$ $R\sqrt{2} = \frac{OI}{\sin (45^{\circ}-\frac{\alpha}{2})} \Leftrightarrow R^2(1- \sin \alpha)=R(R-2r) \Leftrightarrow \sin \alpha = \frac{2r}{R}$ Thus we have :
$1 = \cos^2 \alpha + \sin^2 \alpha = \frac{R^2 - 8Rr + 20r^2}{R^2} \Leftrightarrow R = \frac{5r}{2}$ Therefore,
$AB : BC : CA = 2R \cos \alpha : 2R \sin \alpha : 2R = 3 : 4 : 5$

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#3 h Jul 6, 2020, 4:50 PM • 1 Y

Y by Mango247

Same as above $O$ lies on $AC$ and $\angle ABC=90$
Let $AD$ is angle bisector. Then $\triangle IOC= \triangle IDC$ so $CD=OC=\frac{AC}{2}$
$\frac{BC}{AC} = \frac{BC}{2CD}=\frac{AB+AC}{2AC} \to 2BC=AB+AC$
$AC^2=AB^2+BC^2=BC^2+(2BC-AC)^2 \to 5BC=4AC \to AB:BC:AC=3:4:5$

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