IGO 2016 Elementary/2
- Thamim Zahin
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- Joined:Wed Aug 03, 2016 5:42 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.
- Thamim Zahin
- Posts:98
- Joined:Wed Aug 03, 2016 5:42 pm
Re: IGO 2016 Elementary/2
$\angle CAP = \angle CBP= \angle CYP= \angle CXY= \angle x$
$\angle BAC =\angle BPC =\angle a = \angle YCA$ [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, $PB=PC$
$\angle BAC =\angle BPC =\angle a = \angle YCA$ [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.
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Re: IGO 2016 Elementary/2
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, $\angle YXC = 180 - \angle PXC = \angle q$.
$\angle p + \angle PCX = \angle q \Rightarrow \angle PCB = \angle PBC$
$\therefore PB = PC$
$\angle CPY = \angle CBY = \angle p$
$\angle PBC = \angle PYC = \angle q$
$CX = CY \Rightarrow \angle CYX = \angle CXY = \angle q$
$AB = CY , BY = BY , \angle BAY = \angle BCY$
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
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- Joined:Tue Dec 08, 2015 4:25 pm
- Location:Bashaboo , Dhaka
Re: IGO 2016 Elementary/2
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 $PB = PC$.
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
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- Posts:65
- Joined:Tue Dec 08, 2015 4:25 pm
- Location:Bashaboo , Dhaka
Re: IGO 2016 Elementary/2
How we can write $\angle YXC = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$?
Re: IGO 2016 Elementary/2
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 first principle is that you must not fool yourself and you are the easiest person to fool.