A Problem of Romanian TST
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 $CD$.
"Questions we can't answer are far better than answers we can't question"
- Kazi_Zareer
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Re: A Problem of Romanian TST
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We cannot solve our problems with the same thinking we used when we create them.
- Raiyan Jamil
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Re: A Problem of Romanian TST
#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
Then tell your whole solution.Raiyan Jamil wrote:#kazi zareer... none of the three quads you've mentioned are cyclic quads... :/
We cannot solve our problems with the same thinking we used when we create them.
Re: A Problem of Romanian TST
Kazi_Zareer wrote:
Totally a wrong solution.
"Questions we can't answer are far better than answers we can't question"
Re: A Problem of Romanian TST
A good probem:
Last edited by tanmoy on Wed Aug 17, 2016 11:03 am, edited 5 times in total.
"Questions we can't answer are far better than answers we can't question"
Re: A Problem of Romanian TST
Why is this a sufficient condition?tanmoy wrote:
- asif e elahi
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Re: A Problem of Romanian TST
That' not true actually. But the problem is very easy from here.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 $FP \cdot FA=FQ \cdot FB$.
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 $OF\perp MN$.
Re: A Problem of Romanian TST
Sorry,I've made a mistake.Edited
"Questions we can't answer are far better than answers we can't question"
- asif e elahi
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Re: A Problem of Romanian TST
Why $O_{1}O_{2} \parallel MN$?tanmoy wrote:Because then $OF \perp O_{1}O_{2}$ and as $O_{1}O_{2} \parallel MN$,so we will be done