GEOMETRY MARATHON: SEASON 2

For discussing Olympiad level Geometry Problems
sourav das
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GEOMETRY MARATHON: SEASON 2

Unread post by sourav das » Fri Sep 16, 2011 8:23 pm

I guess it's time to restart Geometry Marathon. 8-)

RULES: Very simple. I'll start this Marathon with a geometry problem. Next who can give the solution of the problem, will get a chance to post a new problem. Different solutions are welcome too. Who will give a different solution also can get a chance to post a new problem. To make things interesting, one have to post solution of a problem within 12 hours. After 12 hours, the problem will be a star problem and the problem will be re-posted in a new topic. But the Geometry Marathon will be continued by posting a new problem from the former solvers....

Hope u will enjoy the Geometry Marathon. :D

Problem 1:
Let $O$ be the circumcentre of an acute-angled triangle $ABC$. A line through $O$ intersects
the sides $CA$ and $CB$ at points $D$ and $E$ respectively, and meets the circumcircle of
triangle $ABO$ again at point $P$ (Not $O$) inside the triangle. A point $Q$ on side $AB$ is such
that $AQ/QB = DP/PE$


Prove that $\angle APQ = 2\angle CAP$
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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Labib
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Re: GEOMETRY MARATHON: SEASON 2

Unread post by Labib » Sat Sep 17, 2011 12:34 am

Is my figure correct?
9.png
Prove that, $\angle APQ = 2 \angle CAP$
9.png (329.23KiB)Viewed 6615 times
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sourav das
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Re: GEOMETRY MARATHON: SEASON 2

Unread post by sourav das » Sat Sep 17, 2011 12:45 am

$Q$ is on $AB$
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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Abdul Muntakim Rafi
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Re: GEOMETRY MARATHON: SEASON 2

Unread post by Abdul Muntakim Rafi » Sat Sep 17, 2011 8:02 pm

Hey guys,what software do you use to draw pictures(of course related to maths) in pc?
Man himself is the master of his fate...

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*Mahi*
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Re: GEOMETRY MARATHON: SEASON 2

Unread post by *Mahi* » Sat Sep 17, 2011 8:43 pm

Please read Forum Guide and Rules before you post.

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FahimFerdous
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Re: GEOMETRY MARATHON: SEASON 2

Unread post by FahimFerdous » Sat Sep 17, 2011 11:07 pm

Hey, I've just started trying the problem. Don't post the solve yet please. :D
Actually, I'm a little busy these days.
Your hot head might dominate your good heart!

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*Mahi*
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Re: GEOMETRY MARATHON: SEASON 2

Unread post by *Mahi* » Sun Sep 18, 2011 12:24 am

$\text{Solution(complex):}$
Let the circumcircle of $ABC$ be the unit circle, let $C$ be $i=\sqrt{-1}$ and $DE$ be the real line. Let $A=a$ and $B=b$.
Then $D=\frac {-a-i} {a-i}$ and $E=\frac {-b-i} {b-i}$.
Now the circumcentre of $\triangle OAB$ will be $o_1=\frac {ab} {a+b}$ and $P$ will be $p=\frac {ab+1} {a+b}$.
We notice that,$D,E,P$ are all real numbers, so we can easily calculate the ratio $DP/PE=\left | \frac {p-d}{p-e} \right |$.
And then we can divide $AB$ with the lemma of $\frac {na+mb} {n+m}$ to get $Q$. After that, calculation yields
$ \frac {a-\frac {ab+1} {a+b}} {\frac {ab+1} {a+b}-\frac {na+mb} {n+m}}=k (\frac {a+b} {b-i})^2$ where $k$ is a real number,which gives $2\angle CAP= \angle APQ$
Last edited by *Mahi* on Sun Sep 18, 2011 9:26 am, edited 1 time in total.
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Abdul Muntakim Rafi
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Re: GEOMETRY MARATHON: SEASON 2

Unread post by Abdul Muntakim Rafi » Sun Sep 18, 2011 1:20 am

Thank you very much... @Mahi
Man himself is the master of his fate...

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Tahmid Hasan
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Re: GEOMETRY MARATHON: SEASON 2

Unread post by Tahmid Hasan » Sun Sep 18, 2011 11:04 am

Abdul Muntakim Rafi wrote:Hey guys,what software do you use to draw pictures(of course related to maths) in pc?
i use cabri
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*Mahi*
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Re: GEOMETRY MARATHON: SEASON 2

Unread post by *Mahi* » Thu Sep 22, 2011 7:28 pm

New Problem!
Let $BE$ and $CF$ be the altitudes in an acute triangle $ABC$. Two circles passing through the points $A$ and $F$ are tangent to the line $BC$ at the points $P$ and $Q$ so that $B$ lies between $C$ and $Q$. Prove that the lines $PE$ and $QF$ intersect on the circumcircle of $\triangle AEF$.
(Intended to those who haven't solved it yet.)
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