circles through fixed point,OA+OB=c (2009 All-Siberian Correspondence 10.5)

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circles through fixed point,OA+OB=c (2009 All-Siberian Correspondence 10.5)

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#1 h May 3, 2021, 7:22 PM • 3 Y

Y by Mango247, Mango247, Mango247

On different sides of the angle with the vertex $O$ , points $A$ and $B$ are taken such that the sum of the lengths $OA$ and $OB$ is constant. Prove that the circumcircles of all possible triangles $AOB$ pass, bedises point $O$ , through one more common point.

This post has been edited 4 times. Last edited by parmenides51, May 3, 2021, 10:59 PM

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#2 h May 4, 2021, 6:56 AM • 1 Y

Y by Mango247

Given $O(0,0),A(\lambda,0),B(b\cos \alpha,b\sin \alpha)$ and $OA+OB=t$ , a fixed value; then $b=t-\lambda$ .

Midpoint of the circumcircle $(\frac{\lambda}{2},\frac{t-\lambda-\lambda\cos \alpha}{2\sin \alpha})$ .

Equation of this circumcircle $x^{2}-\lambda x+y^{2}-\frac{t-\lambda-\lambda\cos \alpha}{\sin \alpha} \cdot y =0$ .
$x^{2}+y^{2}-\frac{t}{\sin \alpha} \cdot y +(-x+\frac{1+\cos \alpha}{\sin \alpha} \cdot y)\lambda = 0$ .

We solve the system $\left\{\begin{array}{ll} x^{2}+y^{2}-\frac{t}{\sin \alpha} \cdot y =0 \\ -x+\frac{1+\cos \alpha}{\sin \alpha} \cdot y =0 \end{array}\right.$
The second equation $y=\tan \frac{\alpha}{2} \cdot x$ is the bisector of the angle $\alpha$ .
This line cuts the circle $x^{2}+y^{2}-\frac{t}{\sin \alpha} \cdot y =0$ twice, $x=0$ and $x=\frac{t}{2}$ .

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#3 h May 4, 2021, 7:56 AM

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Let $T$ be the intersection of $\odot [AOB]$ and the bisector of $\angle AOB$ . Then apply Ptolemy Theorem we get $AT\cdot OB+BT\cdot OA=AB\cdot OT$ And $AT=BT$ , so, $OT=\frac {AT\cdot (AO+BO)}{AB}$ And $\frac {AT}{AB}=\frac 1{2\cos \angle TAB}=\frac 1{2\cos \frac 12 \angle AOB}$ is constant, $AO+BO$ is also constant. So, $OT$ is constant, which means $\odot [AOB]$ passes another common point.
$[asy] size(300); pair a,b,o; o=(0,0); real an=58; pair m=(6,0); draw(o--m^^o--(rotate(an,o)*m)); real t=5,r=1,s; for(r=1; r<t; r+=0.6) {b=(r,0); pair n=(t-r,0); a=rotate(an,o)*n; draw(a--b); pair oo; oo=circumcenter(a,o,b); draw(circle(oo,length(oo-o)));} b=(2.2,0); pair n=(2.8,0); a=rotate(an,o)*n; pair q=(2.5,1.38); draw(o--q^^q--a^^q--b,dotted); dot("$A$",a,NNW); dot("$B$",b,SE); dot("$O$",o,WSW); dot("$T$",q,ENE); [/asy]$

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