A circle $k$ with center $O$ and radius $r$ and a line $p$ which has no common points with $k$, are given. Let $E$ be the foot of the perpendicular from $O$ to $p$. Let $M$ be an arbitrary point on $p$, distinct from $E$. The tangents from the point $M$ to the circle $k$ are $MA$ and $MB$. If $H$ is the intersection of $AB$ and $OE$, then prove that $OH=\frac{r^2}{OE}$.
Problem
Source: 2015 JBMO TST - Macedonia, Problem 2
Tags: circles, tangent, cyclic quadrilateral, geometry
FabrizioFelen
31.12.2015 17:44
My solution: It is easy to see that: $AB$ is the polar of $M$ respect to $k$ since $H$ lies on $AB$ $\Longrightarrow$ the polar of $H$ through of $M$ and the polar of $H$ is a line perpendicular of $OH$ $\Longrightarrow$ the polar of $H$ is $p$ let $EC$ and $ED$ the tangents of $E$ to $k$ $\Longrightarrow$ $C$,$D$,$H$ are collinear and $r^2=OC^2=OH.OE$ since $\angle CEO=\angle HCO$ $\Longrightarrow$ $OH=\frac{r^2}{OE}$...
SP0SkopjeMK
16.01.2016 05:43
The problem has a quite simple elementary solution, as well. Since $OAEM$ and $OBEM$ are cyclic, the points $O,A,E,M,B$ are concyclic. Now, $\angle OBA=\angle OAB=\angle OEB$, so $\triangle OHB\sim\triangle OBE$, thus we get the desired relation.
Devastator
22.05.2018 09:34
Simple yet nice in my opinion. Can be done in 5 mins lel. Here's my sol:
Let OM and AN intersect at K
As angles HKM and HEM are both right, then KMEH cyclic.
Therefore, by Power of a Point, $OH\cdot OE= OK\cdot OM$
But as OKA and OAM are both right, triangles OKA and OAM are similar.
Hence, $OK\cdot OM = OA^2=r^2$
Inspiration:
I know that it's trivial when M=E, so I thought of drawing OM to get that same configuration also.
Nice solution too SPO