Given any triangle $ABC.$ Let $O_1$ be it's circumcircle, $O_2$ be it's nine point circle, $O_3$ is a circle with orthocenter of $ABC$, $H$, and centroid $G$, be it's diameter. Prove that: $O_1,O_2,O_3$ share axis. (i.e. chose any two of them, their axis will be the same one, if $ABC$ is an obtuse triangle, the three circle share two points.)
Problem
Source: 2015 Taiwan TST Round 1 Quiz2 P2
Tags: geometry
14.04.2015 21:46
Let $D,E,F$ be the feet of the altitudes on $BC,CA,AB.$ $M$ is the midpoint of $BC$ and $EF,FD,DE,$ cut $BC,CA,AB$ at $P,Q,R.$ Since $(B,C,D,P)=-1$ $\Longrightarrow$ $PB \cdot PC=PM \cdot PD$ $\Longrightarrow$ $P$ has equal power WRT $(O_1)$ and 9-point circle $\odot(DEF) \equiv (O_2)$ and similarly $Q$ and $R$ $\Longrightarrow$ $\overline{PQR}$ is radical axis of $(O_1),(O_2).$ Since $PH$ is the polar of $A$ WRT the circle $(M)$ with diameter $BC,$ then $PH$ is perpendicular to $AM$ at $T,$ the 2nd intersection of $\odot(AEHF)$ and $(O_3).$ Hence $PH \cdot PT=PE \cdot PF$ $\Longrightarrow$ $P$ has equal power WRT $(O_2)$ and $(O_3)$ and similarly $Q,R$ $\Longrightarrow$ $\overline{PQR}$ is radical axis of $(O_2),(O_3).$ The conclusion follows.
15.04.2015 03:27
In general, given two circles $ \odot (O_1, R_1) $ and $ \odot (O_2, R_2) $ . Let $ P, Q $ be the insimilicenter, exsimilicenter of $ \odot (O_1) \sim \odot (O_2) $, respectively . Then $ \odot (O_1), \odot (O_2), \odot (O_3) \equiv \odot (PQ) $ are coaxial . ____________________________________________________________ Proof : Since $ \odot (O_3) $ is the locus of the point $ T $ such that $ TO_1:TO_2=R_1:R_2 $ , so for any point $ T $ on $ \odot (O_3) $ we have $ \frac{{TO_1}^2-{R_1}^2}{{TO_2}^2-{R_2}^2}=\left ( \frac{R_1}{R_2} \right )^2=\text{Const} \Longrightarrow \odot (O_1), \odot (O_2), \odot (O_3) $ are coaxial . Done
15.11.2015 20:46
Does this work? Note that the centres of the circles are collinear on some line $l$. Then, their axis' are perpendicular to $l$ and they are concurrent by the Radical Axis Theorem. Thus, they must coincide i.e. They have the same axis.
16.11.2015 15:00
Does the solution above work? It seems a little short for the problem...
16.11.2015 16:09
MathPanda1 wrote: Does the solution above work? It seems a little short for the problem... All you did was prove that the radical axii are parallel
16.11.2015 18:47
aops wrote: Theorem 3: (Radical Axis Concurrence Theorem) The three pairwise radical axes of three circles concur at a point, called the radical center. See http://www.artofproblemsolving.com/wiki/index.php/Radical_axis Is parallel radical axii a possibility?
16.11.2015 19:00
MathPanda1 wrote: aops wrote: Theorem 3: (Radical Axis Concurrence Theorem) The three pairwise radical axes of three circles concur at a point, called the radical center. See http://www.artofproblemsolving.com/wiki/index.php/Radical_axis Is parallel radical axii a possibility? Yes, but just because they're parallel doesn't mean that they're the same line...
16.11.2015 23:31
Well, don't they have to meet? The only way parallel lines meet is if they coincide. Unless point of infinity exists...
17.11.2015 01:50
MathPanda1 wrote: Well, don't they have to meet? The only way parallel lines meet is if they coincide. Unless point of infinity exists... Parallel radical axii coincide at the point at infinity; a valid way to do this problem would be to show that there exists some point with equal power wrt to all three circles, which would force the conclusion.
17.11.2015 04:58
Thank you for pointing out the errors in my proof, EulerMacaroni. I really appreciate it!
09.05.2020 20:52
here's my solution let $D,E,F$ be the feet of altitudes and define $X_a=EF \cap BC$ and so define $X_b,X_c$ I'll claim that $X_aX_bX_c$ is the radical axis of $O_1,O_2,O_3$ claim(1): $X_a$ is on the radical axis of $O_1,O_2$ proof: easily $X_aE.X_aF=X_aB.X_aC$ $\blacksquare$ claim(2): $X_a$ is on the radical axis of $O_1,O_3$ proof: we have that $A_{HM}$ point , $H$and $X_a$ are collinear (since the image of $X_a$ is the $A_{HM}$ point after inversion around $H$ with radius $-\sqrt{AH.HD}$ ) and of course the $A_{HM}$ point is on $O_3$ so $X_aA_{HM}.X_aH=X_aB.X_aC$ since $HBCA_{HM}$ is cyclic $\blacksquare$ and we win
30.12.2023 15:14
Clearly centers of three circles are lies on Euler line ,so it suffice to find a point $P$ with equal power with respect to all three circles. ( Then the line perpendicular to Euler line and passing from $P$ would be their radical axis. ) Take $P$ on Euler line such that $H$ be between $P$ and $O$ and $PH = x$. Let $l=HO$ , $N =$ center of nine point circle and $K$ be the center of $O_3$. we want to choose x such that these three numbers are equal : \[ (I): P^P_{O_1} = PO^2 - R^2 = (x+l)^2 - R^2 = x^2 + 2xl + l^2 - R^2 \]\[ (II): P^P_{O_2} = PN^2 - \frac{R^2}{4} = (x + \frac{l}{2})^2 - \frac{R^2}{4} = x^2 + xl + \frac{l^2}{4} - \frac{R^2}{4}\]\[ (III): P^P_{O_3} = PK^2 - KH^2 = (x+ \frac{l}{3})^2 - \frac{l^2}{9} = x^2 + \frac{2l}{3}x\]Easily you could check that $x= \frac{\frac{3}{4}(R^2-l^2)}{l}$ works.
12.06.2024 14:31
Let $D$ be the midpoint of $BC$ and $H_A$ be the $A$-Humpty point. Then \[ \frac{\text{Pow}(D,O_1)}{\text{Pow}(D,O_3)} = \frac{ - DB^2}{DH_A \cdot DG} = \frac{-3 DB^2}{DH_A \cdot DA} = -3 \]hence we finish by coaxiality lemma.