Let $C$ be a circunference, $O$ is your circumcenter, $AB$ is your diameter and $R$ is any point in $C$ ($R$ is different of $A$ and $B$) Let $P$ be the foot of perpendicular by $O$ to $AR$, in the line $OP$ we match a point $Q$, where $QP$ is $\frac{OP}{2}$ and the point $Q$ isn't in the segment $OP$. In $Q$, we will do a parallel line to $AB$ that cut the line $AR$ in $T$. Denote $H$ the point of intersections of the line $AQ$ and $OT$. Show that $H$, $B$ and $R$ are collinears.
Problem
Source: Cono Sur Olympiad 1997 P2
Tags: geometry, geometry unsolved, cono sur
ACGNmath
02.11.2017 17:23
Let $B = (1,0)$, $A = (-1,0)$, $R = (r_1, r_2)$. Then $P = \left(\frac{r_1-1}{2}, \frac{r_2}{2}\right)$ $Q = \left(\frac{3r_1-3}{4}, \frac{3r_2}{4}\right)$ $T = \left(\frac{3r_1-1}{4},\frac{3r_2}{4}\right)$ Equation of $AQ$: $y=\frac{3r_2}{3r_1+1}x+\frac{3r_2}{3r_1+1}$ $OT: y = \frac{3r_2}{3r_1-1}x$ To find $H$, intersect $AQ$ and $OT$. $\frac{3r_2}{3r_1+1}x + \frac{3r_2}{3r_1+1} = \frac{3r_2}{3r_1-1} x$ $x = \frac{3r_1-1}{2}$ $y = \frac{3r_2}{2}$ Thus $H$ = $\left(\frac{3r_1-1}{2}, \frac{3r_2}{2}\right)$. Equation of $HB$: $y = \frac{r_2}{r_1-1}x - \frac{r_2}{r_1-1}$ This passes through the point $(r_1, r_2)$ Conclusion follows.
manifoldaora
13.06.2019 21:48
Observe that triangles $TPQ$ and $APO$ are similar. This implies that $2TP=PA$ and therefore $RT=TP$. Also, by thales' Theorem it follows that $TH=OT$. By the $SSS$ criterion, triangles $RTH$ and $TPO$ are similar. It follows that the angle $TRH=90^{\circ}$, and therefore the result follows.
NiltonCesar
26.06.2019 21:24
A solution with Menelaus's Theorem (when I didn't know how to use Search Function): https://artofproblemsolving.com/community/u402236h1641697p10345939