IGO 2016 Elementary/2

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Thamim Zahin
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IGO 2016 Elementary/2

Unread post by Thamim Zahin » Tue Jan 10, 2017 2:12 pm

1. Let $ω$ be the circumcircle of triangle $ABC$ with $AC > AB$. Let $X$ be a point on $AC$ and $Y$ be a point on the circle $ω$, such that $CX = CY = AB$.(The points $A$ and $Y$ lie on different sides of the line $BC$). The line $XY$ intersects $ω$ for the second time in point $P$. Show that $PB = PC$.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

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Thamim Zahin
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Re: IGO 2016 Elementary/2

Unread post by Thamim Zahin » Wed Jan 11, 2017 12:06 pm

$\angle CAP = \angle CBP= \angle CYP= \angle CXY= \angle x$
$\angle BAC =\angle BPC =\angle a = \angle YCA$ [Cause, $YCAB$ is a cyclic quad, and $BA=YC$]
$\angle APB=\angle ACB=\angle b$
$\angle PCA=\angle PBA=\angle n$ [All of them by angle chasing]

Now, $180- (\angle a + \angle x)= \angle n+ \angle b$ [from $ \triangle APB$]

Again, $180- (\angle a + \angle x) = \angle x$ [from $ \triangle CXY$]

That means, $\angle n+ \angle b=\angle x$
So,$ \angle PCB=\angle PBC $
Or, $PB=PC$
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I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.

Absur Khan Siam
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Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

Re: IGO 2016 Elementary/2

Unread post by Absur Khan Siam » Thu Feb 23, 2017 1:11 pm

Let ,
$\angle CPY = \angle CBY = \angle p$
$\angle PBC = \angle PYC = \angle q$

$CX = CY \Rightarrow \angle CYX = \angle CXY = \angle q$

$AB = CY , BY = BY , \angle BAY = \angle BCY$
So,$\triangle ABY \cong \triangle BCY$

Thus, $\angle CBY = \angle AYB = \angle ACB = \angle p$

From $\triangle PXC$, $\angle p + \angle PCX = 180 - \angle PXC$
But, $\angle YXC = 180 - \angle PXC = \angle q$.

$\angle p + \angle PCX = \angle q \Rightarrow \angle PCB = \angle PBC$

$\therefore PB = PC$ :)
Last edited by Absur Khan Siam on Thu Feb 23, 2017 1:23 pm, edited 1 time in total.
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

Absur Khan Siam
Posts: 61
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

Re: IGO 2016 Elementary/2

Unread post by Absur Khan Siam » Thu Feb 23, 2017 1:21 pm

Here is the official solution :
We know that $CX = CY$ therefore:
$\angle YXC = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$ [But how?]
Also we have $AB = CY$ therefore $\widehat{AP} + \widehat{CY} = \widehat{AP} + \widehat{AB} = \widehat{PB}$
So $PB = PC$.
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

Absur Khan Siam
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Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

Re: IGO 2016 Elementary/2

Unread post by Absur Khan Siam » Sun Feb 26, 2017 3:21 pm

How we can write $\angle YXC = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$?

joydip
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Re: IGO 2016 Elementary/2

Unread post by joydip » Sun Feb 26, 2017 3:37 pm

As, $ \angle YXC = \angle CPY + \angle PCA$. So ,$ \angle YXC = \angle XYC \Rightarrow \angle CPY + \angle PCA = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$
The more I learn, the more I realize how much I don't know.

- Albert Einstein

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