IMO 2009 #2

by Wolstenholme, Nov 5, 2014, 9:36 PM

Let $ ABC$ be a triangle with circumcentre $ O$. The points $ P$ and $ Q$ are interior points of the sides $ CA$ and $ AB$ respectively. Let $ K,L$ and $ M$ be the midpoints of the segments $ BP,CQ$ and $ PQ$. respectively, and let $ \Gamma$ be the circle passing through $ K,L$ and $ M$. Suppose that the line $ PQ$ is tangent to the circle $ \Gamma$. Prove that $ OP = OQ.$

Proof:

Let $ \Omega $ denote the circumcircle of $ \triangle{ABC}. $ Now it is clear that $ MK \parallel AB $ and $ ML \parallel AC $ so we have that $ \angle{KLM} = \angle{PAQ}. $ Moreover since $ PQ $ is tangent to $ \Gamma $ we have that $ \angle{MKL} = \angle{PML} = \angle{APQ} $ so $ \triangle{APQ} \sim \triangle{MKL}. $ This implies that $ AQ \cdot MK = AP \cdot ML \Longrightarrow AQ \cdot BQ = AP \cdot CP $ (since $ BQ = 2MK $ and $ CP = 2ML $) so $ P $ and $ Q $ have the same power with respect to $ \Omega $ which implies that $ OP = OQ $ as desired.

Comment

1 Comment

The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
How did KLM=PAQ angles ?

by A.A.Knightmare, Jul 27, 2020, 10:07 AM

Archives
+ June 2016
+ April 2016
+ March 2016
+ July 2015
+ February 2015
+ June 2014
Shouts
Submit
  • glad to have found a fellow chipotle lover <3

    by nukelauncher, Aug 13, 2020, 6:40 AM

  • the random chinese tst problem is the only thing I read, but I'll assume your blog is nice and give you a shout even though you probably never use aops anymoer

    by fukano_2, Jun 14, 2020, 6:24 AM

  • wolstenholme - op

    by AopsUser101, Jan 29, 2020, 8:27 PM

  • this blog is so hot

    by mathleticguyyy, Jun 5, 2019, 8:26 PM

  • Hi. Nice Blog!

    by User360702, Jan 10, 2019, 6:03 PM

  • helloooooo

    by songssari, Jun 12, 2016, 8:21 AM

  • shouts make blogs happier

    by briantix, Mar 18, 2016, 9:57 PM

  • You were just featured on AoPS's facebook page.

    by mishka1980, Sep 12, 2015, 10:33 PM

  • This is late, but where is the ARML results post?

    by donot, Aug 31, 2015, 11:07 PM

  • "I am Sam"
    "Sam I am"

    by mathwizard888, Aug 12, 2015, 9:13 PM

  • HW$\textcolor{white}{}$

    by Eugenis, Apr 20, 2015, 10:10 PM

  • Uh-oh ARML practice is Thursday... I should start the homework. :P

    by nosaj, Apr 20, 2015, 12:34 AM

  • Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial :P

    by Wolstenholme, Apr 15, 2015, 10:00 PM

  • How are Chebyshev Polynomials trivial? :P

    by nosaj, Apr 13, 2015, 4:10 AM

  • Are you Sam?

    by Eugenis, Apr 4, 2015, 2:05 AM

  • @Brian: yes, yes I did #whoneedsalgskillz?

    @gauss1181; hey!

    by Wolstenholme, Mar 1, 2015, 11:25 PM

  • hello!!! :D

    by gauss1181, Nov 27, 2014, 12:19 AM

  • Hi Wolstenholme did you actually use calc on that tstst problem :o

    by briantix, Aug 2, 2014, 12:25 AM

18 shouts
Contributors
Tags
About Owner
  • Posts: 543
  • Joined: Mar 3, 2013
Blog Stats
  • Blog created: Apr 3, 2013
  • Total entries: 112
  • Total visits: 35003
  • Total comments: 167
Search Blog
a