A Problem of Romanian TST

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tanmoy
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A Problem of Romanian TST

Unread post by tanmoy » Mon Dec 14, 2015 11:57 pm

Let $C$ and $D$ be distinct points on a semicircle with diameter $AB$. Denote by $E, F, G$ the midpoints of line segments $AC, CD, DB$, respectively. The perpendicular from $E$ to $AF$ meets the tangent to the semicircle at $A$ at the point $M$, and the perpendicular from $G$ to $BF$ meets the tangent to the semicircle at $B$ at the point $N$. Prove that $MN$ is parallel to $CD$.
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Kazi_Zareer
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Re: A Problem of Romanian TST

Unread post by Kazi_Zareer » Fri May 27, 2016 2:40 am

Here $ECDG$, $MEGN$ and $MCDN$ are cyclics.Now we can conclude with Radical Axis.
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Raiyan Jamil
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Re: A Problem of Romanian TST

Unread post by Raiyan Jamil » Wed Aug 03, 2016 10:46 pm

#kazi zareer... none of the three quads you've mentioned are cyclic quads... :/
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Kazi_Zareer
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Re: A Problem of Romanian TST

Unread post by Kazi_Zareer » Tue Aug 09, 2016 9:22 pm

Raiyan Jamil wrote:#kazi zareer... none of the three quads you've mentioned are cyclic quads... :/
Then tell your whole solution. :?
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tanmoy
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Re: A Problem of Romanian TST

Unread post by tanmoy » Mon Aug 15, 2016 10:28 pm

Kazi_Zareer wrote:
Here $ECDG$, $MEGN$ and $MCDN$ are cyclics.Now we can conclude with Radical Axis.

Totally a wrong solution.
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tanmoy
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Re: A Problem of Romanian TST

Unread post by tanmoy » Mon Aug 15, 2016 11:54 pm

A good probem:
See that $OF \perp CD$(Since $O$ is the center of $(ABCD)$ and $F$ is the midpoint of the chord $CD$.),so if we can pove that $OF \perp MN$,we are done.

Now construct two circles with center $M$,radius $MA$ and center $N$,radius $NB$.It is easy to see that $AB$ is tangent to both circles and as $O$ is the midpoint of $AB$,it is well known that $O$ lies on the radical axis $l$ of those circles $(M,MA)$ and $(N,NB)$.So,our goal is to show that $F$ also lies on $l$ i.e. if $(M,MA) \cap AF=P,(N,NB) \cap BF=Q$,we have to prove that $FP \cdot FA=FQ \cdot FB$.Let $ME \cap AF=S$.

Now,let $AA_{1},BB_{1},CC_{1},DD_{1}$ are the perpendiculars from $A,B,C,D$ to $CD,CD,AF,BF$,respectively..Then as $AA_{1} \parallel OF \parallel BB_{1}$ and $O$ is the midpoint of $AB$,so,$FA_{1}=FB_{1}$.And as $E$ is the midpoint of $AC$ and $MES \perp AF$,$S$ is the midpoint of $AC_{1}$ $\Rightarrow$ $MA=MC_{1}$ $\Rightarrow$ $C_{1} \equiv P$.Similarly,$D_{1} \equiv Q$.

Now,as $AA_{1}CP$ and $BB_{1}DQ$ are cyclic,from Power of a Point Theorem,we get that
$FP \cdot FA=FC \cdot FA_{1}=FD \cdot FB_{1}$(Since $FC=FD,FA_{1}=FB_{1}$)$ =FQ \cdot FB$.

So,from the last line of the 2nd paragraph,we are done.
Last edited by tanmoy on Wed Aug 17, 2016 11:03 am, edited 5 times in total.
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Zawadx
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Re: A Problem of Romanian TST

Unread post by Zawadx » Tue Aug 16, 2016 9:17 pm

tanmoy wrote:
So,our goal is to show that $F$ also lies on $l$ i.e. if $ME \cap AF=P,NG \cap BF=Q$,we have to prove that $FP \cdot FA=FQ \cdot FB$.
Why is this a sufficient condition?

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asif e elahi
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Re: A Problem of Romanian TST

Unread post by asif e elahi » Tue Aug 16, 2016 10:21 pm

tanmoy wrote:So,our goal is to show that $F$ also lies on $l$ i.e. if $ME \cap AF=P,NG \cap BF=Q$,we have to prove that $FP \cdot FA=FQ \cdot FB$.
That' not true actually. But the problem is very easy from here.
Let $\omega=\bigodot (M,MA),\lambda=\bigodot(N,NB)$
Let $\omega\cap AF=A,X$ and $\lambda\cap BF=B,Y$.
\begin{align*}
FX.FA &=FE^2-AE^2\\
&=OG^2-AE^2
\end{align*}
Similarly $FY.FB=OE^2-BG^2$ and $OG^2-AE^2=OE^2-BG^2$ by Pythagoras.
So $FX.FA=FY.FB$ which implies that $F$ lies on the radical axis of $\omega$ and $\lambda$. $O$ has the same property too. So $OF\perp MN$.

tanmoy
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Re: A Problem of Romanian TST

Unread post by tanmoy » Tue Aug 16, 2016 10:51 pm

Sorry,I've made a mistake.Edited
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asif e elahi
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Re: A Problem of Romanian TST

Unread post by asif e elahi » Wed Aug 17, 2016 2:36 am

tanmoy wrote:Because then $OF \perp O_{1}O_{2}$ and as $O_{1}O_{2} \parallel MN$,so we will be done
Why $O_{1}O_{2} \parallel MN$?

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