ISL 2009 G4

by utkarshgupta, Apr 25, 2016, 11:51 AM

This proof is for Anant who asked me for a coordinate bash in the shouts :P
I wasn't able to solve it synthetically (at least last year :P)

Problem : (ISL 2009)
Given a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ meet at $E$ and the lines $AD$ and $BC$ meet at $F$. The midpoints of $AB$ and $CD$ are $G$ and $H$, respectively. Show that $EF$ is tangent at $E$ to the circle through the points $E$, $G$ and $H$.

Proposed by David Monk, United Kingdom

Solution :
We will work in the Cartesian plane.
$X=(x,y)$ will mean that the $x$,$y$coordinates of the point $X$ are $x_1,y_1$ respectively.
The slope of a line $l$ will be denoted by $m_l$.

Without loss of Generality,
$$E=(0,0)$$,
$$A=(1,0)$$,
$$C=(-pq,0)$$
Let $m_{BD}=\tan{\theta}$


So set
$$B=(p\cos{\theta},p\sin{\theta})$$
Using $B,D$ lie on the opposite sides of $AC$, and $EB \cdot ED = EA \cdot EC$.

$$D=(-q\cos{\theta},-q\cos{\theta})$$
Now easy calculations yield
$F = AD \cap BC$
$$F = (\frac{q(p+q+\cos{\theta})+pq\cos{\theta}}{q^2-1},\frac{q\sin{\theta}(1+pq)}{q^2-1})$$$$G= (\frac{1+p\cos{\theta}}{2},\frac{p\sin{\theta}}{2})$$$$H= (\frac{-pq-q\cos{\theta}},\frac{-q\sin{\theta}}{2})$$
Observe that the given problem is equivalent to proving directed angles, $\angle FEG = \angle EHG$

That is $\frac{m_{GH}-m_{EH}}{1+m_{GH}m_{EH}} = \frac{m_{GE}-m_{EF}}{1+m_{GE}m_{EF}}$

Now this is just easy calculations (they are actually easy as most of the things cancel out).
This post has been edited 3 times. Last edited by utkarshgupta, Apr 25, 2016, 11:54 AM

Stay insane,Coz it's your will, labour and pain,which takes you to the top of the mountain.

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  • Let's all echo what's written in the blog description - Stay Insane / 'Cause it's your labor, will and pain/ That takes you to the top of soda fountain :D

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  • I have no plans for this blog right now....
    No time here people !
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    I may try some combinatorics :P

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