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Subject: "Seemingly simple problem"     Previous Topic | Next Topic
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Bractals
Member since Jun-9-03
Aug-31-04, 06:26 PM (EST)
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"Seemingly simple problem"
 
   Here is a problem that I found in the September issue of Crux Mathematicorum. I thought it would be simple when I first read it, but I seem to have hit a stonewall.

Given two triangles ABC and PQR such that A and P are the midpoints of sides QR and BC and lines QR and BC bisect angles BAC and QPR. Prove that AB + AC = PQ + PR.


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  Subject     Author     Message Date     ID  
Seemingly simple problem Bractals Aug-31-04 TOP
  RE: Seemingly simple problem darij Sep-04-04 1
     RE: Seemingly simple problem grobber Sep-05-04 2
         RE: Seemingly simple problem Bractals Sep-07-04 3
             RE: Seemingly simple problem grobber Sep-09-04 4
                 RE: Seemingly simple problem darij Sep-11-04 5
                     RE: Seemingly simple problem alexb Sep-12-04 6
                 RE: Seemingly simple problem Bractals Sep-16-04 7
  RE: Seemingly simple problem Bractals Sep-16-04 8

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darij
guest
Sep-04-04, 08:57 AM (EST)
 
1. "RE: Seemingly simple problem"
In response to message #0
 
   This is a problem from an Indian mathematical olympiad of the year 2002. It's actually one of those op02 geometry problems I didn't solve. Anyway, I'll give it another try. It actually does NOT look simple to me, as usually such problems with implicite conditions aren't too simple.

Darij


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grobber
Member since Jul-27-04
Sep-05-04, 10:51 AM (EST)
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2. "RE: Seemingly simple problem"
In response to message #1
 
   Here's something with all the good stuuf cut out, because I don't really have time to get into it.

First of all, try to show that BRCQ is cyclic. Then let R' be the symmetric of R wrt the external bisector of QPR, and let B' be the symmetric of B wrt the external bisector of BAC. B' and R' lie on the same circle as B,R,C,Q. All we need to do in order to show what we want is show that QR'=CB', and this is done by showing that the angles QRR' and CBB' are equal (this is a trivial angle chase).

Terribly sorry I can't give any more details, but maybe I'll come back.


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Bractals
Member since Jun-9-03
Sep-07-04, 08:54 PM (EST)
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3. "RE: Seemingly simple problem"
In response to message #2
 
   Thanks for the insight grobber. I checked it out using Sketchpad and BRCQ definitely looks cyclic. Proving that it is cyclic still has me stumped.


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grobber
Member since Jul-27-04
Sep-09-04, 03:35 PM (EST)
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4. "RE: Seemingly simple problem"
In response to message #3
 
   You might want to try to prove this first:

Given a cyclic quadrilateral ABCD, let M, N be the midpoints of AC, BD respectively. Show that AC bisects angle BMD iff BD bisects angle ANC.

This can be used to show that the quadrilateral we're concerned with is cyclic (some sort of "converse", although not exactly a converse). This, I think, gives further insight. Moreover, the cyclic quadrilateral satisfying this is a special kind: a harmonic quadrilateral, i.e. one for which the tangents to the circumcircle in A, C concur on BD (and those in B, D concur on AC).


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darij
guest
Sep-11-04, 03:42 PM (EST)
 
5. "RE: Seemingly simple problem"
In response to message #4
 
   >You might want to try to prove this first:
>
>Given a cyclic quadrilateral ABCD, let M, N be the midpoints
>of AC, BD respectively. Show that AC bisects angle BMD iff
>BD bisects angle ANC.

This has been proved on MathLinks:

https://www.mathlinks.ro/Forum/viewtopic.php?p=22938

Actually, the proof is due to Treegoner; I had just rewritten
it in a more detailed way.

Darij


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alexb
Charter Member
1337 posts
Sep-12-04, 10:37 AM (EST)
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6. "RE: Seemingly simple problem"
In response to message #5
 
   >This has been proved on MathLinks:
>
>https://www.mathlinks.ro/Forum/viewtopic.php?p=22938
>
>Actually, the proof is due to Treegoner; I had just
>rewritten
>it in a more detailed way.

Lemmas 2 and 3 may be simply combined into a reference to La Hire's theorem:

https://www.cut-the-knot.org/Curriculum/Geometry/TangentsAndDiagonals.shtml


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Bractals
Member since Jun-9-03
Sep-16-04, 06:56 AM (EST)
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7. "RE: Seemingly simple problem"
In response to message #4
 
   I was able to prove to myself that BRCQ is a cyclic quadrilateral.

Question: If line AB intersects the circumcircle of BRCQ at points B and C', does AC' = AC ? It'seems obvious that it does, but I would like to see a proof.


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Bractals
Member since Jun-9-03
Sep-16-04, 00:14 AM (EST)
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8. "RE: Seemingly simple problem"
In response to message #0
 
   Hi,

Here is my first cut at a
proof. Any comments appreciated.

Bractals


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