Second Balkan Mathematical Olympiad

For discussing Olympiad level Geometry Problems
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nafistiham
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Second Balkan Mathematical Olympiad

Unread post by nafistiham » Mon Jan 23, 2012 3:40 pm

let $O$ be the centre of the circle through the points $A,B,C$.Let $D$be the mid point of $AB$.Let $E$ be the centroid of the triangle $ACD$.Prove that, the line $CD$ will be perpendicular to the line $OE$ iff $AB=AC$

the book provides the solution in vector could someone do it in the geometric way?

$7777^{th}$ :lol:
Last edited by nafistiham on Mon Jan 23, 2012 6:23 pm, edited 1 time in total.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Avik Roy
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Re: Second Balkan Mathematical Olympiad

Unread post by Avik Roy » Mon Jan 23, 2012 4:04 pm

প্রশ্নই তো অসম্পূর্ণ মনে হচ্ছে!!!
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

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nafistiham
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Re: Second Balkan Mathematical Olympiad

Unread post by nafistiham » Mon Jan 23, 2012 6:26 pm

nafistiham wrote:Prove that, the line $CD$ will be perpendicular to the line $OE$ iff $AB+AC$
sorry. :oops:
it should have been like this.
nafistiham wrote:Prove that, the line $CD$ will be perpendicular to the line $OE$ iff $AB=AC$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

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Phlembac Adib Hasan
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Re: Second Balkan Mathematical Olympiad

Unread post by Phlembac Adib Hasan » Tue Jan 24, 2012 7:18 pm

Geo 3.pdf
(118.17 KiB) Downloaded 189 times
It's very easy to prove $ OE\perp CD $ when $AB=AC$.So I'm giving its inverse case.
My Proof :
Given that $ OE\perp CD $.We have to prove $AB=AC$.Let $AB\cap CE=F$.(Not shown in the figure).
From Stewart's theorem,\[AF.BC^2+BF.AC^2=AB(CF^2+AF.FB)\]
\[\Rightarrow AB(\frac{BC^2}{4}+\frac{3AC^2}{4})=AB(CF^2+\frac{3AB^2}{16})\]
\[\Leftrightarrow (\frac{BC^2}{4}+\frac{3AC^2}{4})=(CF^2+\frac{3AB^2}{16})\]
\[\Leftrightarrow CF^2=\frac{BC^2}{4}+\frac{3AC^2}{4}-\frac{3AB^2}{16}\]
\[\Leftrightarrow \frac{4CF^2}{9}=\frac{BC^2}{9}+\frac{AC^2}{3}-\frac{AB^2}{12}\]
\[\Rightarrow CE^2=\frac{BC^2}{9}+\frac{AC^2}{3}-\frac{AB^2}{12}\]
\[\Rightarrow CE^2-DE^2=\frac{BC^2}{9}+\frac{AC^2}{3}-\frac{AB^2}{12}-\frac{BC^2}{9}\]
\[\Rightarrow CE^2-DE^2=\frac{AC^2}{3}-\frac{AB^2}{12}\]
Notice that \[OD^2=AO^2-\frac{AB^2}{4}=R^2-\frac{AB^2}{4}\]
\[\Rightarrow CO^2-OD^2=\frac{AB^2}{4}\]
As $ OE\perp CD $, from perpendicular lemma,\[CO^2-OD^2=CE^2-DE^2\]
\[\Rightarrow \frac{AB^2}{4}=\frac{AC^2}{3}-\frac{AB^2}{12}\]
\[\Rightarrow \frac{AB^2}{3}=\frac{AC^2}{3}\]
\[\Rightarrow AB=AC\]
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nafistiham
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Re: Second Balkan Mathematical Olympiad

Unread post by nafistiham » Wed Jan 25, 2012 2:07 pm

thanks for the nice proof. :D (and i am uploading the figure as a jpeg file, so that none has to download it. ;) )
Geo 3.JPG
from a previous post by phlembac adib hasan
Geo 3.JPG (40.55 KiB) Viewed 1835 times
$400^{th}$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

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