LOGIN/REGISTER
Please Wait...
It is currently Sep 02, 2010, 11:24 am
Post new topic Reply to topic  Page 1 of 1
  [ 3 posts ]  Share: Facebook
Author Message
PhilAndrew
Poincare Conjecture

Posts: 119
Location: Bucharest
Romania    European Union
Post Posted: Jun 06, 2007, 11:00 am • # 1 


Consider \rho a semicircle of diameter AB. A parallel to AB cuts the semicircle at C, D such that AD separates B, C. The parallel at AD through C intersects the semicircle the second time at E. Let F be the intersection point of the lines BE and CD. The parallel through F at AD cuts AB in P. Prove that PC is tangent to \rho.

Author: Cosmin Pohoata

_________________
Life is complex: it has both real and imaginary components.
 
 
Jan
Yang-Mills Theory
User avatar

Posts: 606
Location: Belgium
Belgium
Post Posted: Jun 06, 2007, 11:24 am • # 2 


Because FDAP is a parallellogram, we have that \left|FD\right|=\left|PA\right|.
Now construct U \in CD such that BU \parallel OD.

We have that \left|PO\right|=\left|PA\right|+\left|AO\right|=\left|FD\right|+\left|DU\right|=\left|FU\right| \hspace{1cm}(1)
and obviously also that \left|BU\right|=\left|OD\right|=\left|OC\right| \hspace{1cm}(2)
and finally also \angle DUB = \angle DOB = \angle COP \hspace{1cm}(3)

From (1),(2) and (3) follows that \Delta PCO \cong \Delta FBU

Because \left|BD\right| = \left|CA\right|=\left|ED\right|, it follows that OD \bot EB.

Therefore, \angle FBU = \frac{\pi}{2}= \angle PCO.

Image
 
 
pohoatza
Navier-Stokes Equations

Posts: 1100
Location: Bucharest
Romania
Post Posted: Jun 06, 2007, 11:41 am • # 3 


Very nice solution, Jan!

Although the problem is easy, I will post my solution.

Because AD \| PF, we have that \angle{PFC} = \angle{ADC} = \angle{BCD}, thus the quadrilateral PBFC is an issoscel trapezoid.

Therefore \angle{PCB} = \angle{PFB}\iff 90 + \angle{PCA} = 90 + \angle{DBF}

\iff \angle{PCA} = \angle{DBF}.

But \angle{DBF} = \angle{DAE}, because the quadrilateral ABDE is cyclic, and \angle{DAE} = \angle{ADC}.

Therefore \angle{PCA} = \andle{ADC}, so PC is tangent to \rho.

_________________
Cosmin Pohoata, Bucharest, Romania
 
 
Display posts from previous:  Sort by  
  Share on Facebook Previous topic | Next topic 

Community » Olympiad Section » Geometry » Geometry Proposed & Own Problems

All times are UTC - 8 hours [ DST ]
Share: Facebook

Moderators: yetti, MithsApprentice, N.T.TUAN, Peter, darij grinberg, orl, pohoatza, pbornsztein, High School Olympiad Moderators

Post new topic Reply to topic  Page 1 of 1
  [ 3 posts ] 

Login
Username:   Password:   Log me on automatically each visit  

You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Introduction to Counting & Probability
All the basics of counting and probability. Over 400 problems! Great for MATHCOUNTS and AMC preparation.
Online School
Learn math and problem solving from the best, with the best. Many classes covering different subjects and skill levels.