Let $ABC$ a acute triangle. (a) Find the locus of all the points $P$ such that, calling $O_{a}, O_{b}, O_{c}$ the circumcenters of $PBC$, $PAC$, $PAB$: \[\frac{ O_{a}O_{b}}{AB}= \frac{ O_{b}O_{c}}{BC}=\frac{ O_{c}O_{a}}{CA}\] (b) For all points $P$ of the locus in (a), show that the lines $AO_{a}$, $BO_{b}$ , $CO_{c}$ are cuncurrent (in $X$); (c) Show that the power of $X$ wrt the circumcircle of $ABC$ is: \[-\frac{ a^{2}+b^{2}+c^{2}-5R^{2}}4\] Where $a=BC$ , $b=AC$ and $c=AB$.
Problem
Source: Italian TST , day 2, n°2
Tags: geometry, circumcircle, geometry proposed
Ilthigore
26.03.2008 18:04
I made the locus the single point $ H$, which seems wrong (and renders the other two parts very easy).
¬[ƒ(Gabriel)³²¹º]¼
26.03.2008 19:02
a) Call M the midpoint of AP, N of PB and L of PC. So we want to prove that $ \triangle O_aO_bO_c$ and $ \triangle ABC$ are similar and then $ \angle O_bO_cO_a = \angle ACB$ and so on. $ NPMO_c$ is cyclic and so $ \angle O_bO_cO_a = 180 - APB$ and others. Then P stay on the symmetric of the circumcircle of ABC w.r.t. AB, w.r.t BC and w.r.t. AC. But this 3 circle have the common point in the orthocenter H of ABC, so the locus is H. b)cause the orthocenter of $ \triangle ABC$ is the circumcenter of $ \triangle O_aO_bO_c$, then the circle that pass true M,L,N is the Feuerbach circle of $ \triangle ABC$ and of $ \triangle O_aO_bO_c$, so the 2 triangle are equal and then they have the homothetic center on the center of the Feuerbach circle. c) $ \displaystyle pow_\Gamma(P) = R^2 - PO^2 = R^2 - \frac {9R^2 - a^2 - b^2 - c^2}{4} = \frac {a^2 + b^2 + c^2 - 5R^2}{4}$
windrock
26.01.2009 09:34
a) Locus of $ P$ is $ H$, with $ H$ is the orthorcenter of $ \delta ABC$ b) easy to proof that each of lines $ AO_{a}, BO_{b}, C_O{c}$ pass through the midpoint of $ (OP)$ c) $ OX = \frac{OP}{2}$, $ OP$ can calculate from $ a, b, c$ and $ R$