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#1 2007-08-20 21:16:21
- JaneFairfax
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Geometry problem
Here’s a geometry problem I’ve just made up by myself. 
Let O be a fixed point. Let Q be a variable point such that the length of OQ is less than or equal to 2a, where a is a fixed positive real number. Now let P and R be points satisfying the following conditions:
(i) OQPRO is a rectangle.
(ii) If PQ is shifted parallel to itself to P′Q′ such that P′Q′ and PR mutually bisect each other, then PP′RQ′P is a rhombus with sides of length a.
Prove that the locus of P is a circle of radius 2a.
I’m sure it works. Still, if you find any problem with my problem, do let me know. 
Last edited by JaneFairfax (2007-08-20 21:20:35)
#2 2007-08-22 11:05:32
- John E. Franklin
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Re: Geometry problem
Here are six red points.
Is that what you mean?
I only used up 279 bytes
on that small pic!!
The lower-center red dot
is Q prime, unmarked as such.
Imagine for a moment that even an earthworm may possess a love of self and a love of others.
#3 2007-08-22 21:42:40
- JaneFairfax
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Re: Geometry problem
That’s right. I should have provided a diagram anyway, so here it is.
|P′Q′| = |PQ|
|PP′| = |P′R| = |RQ′| = |Q′P| = a
|OQ| ≤ 2a
Note that P can be below OQ as well, and that Q can also be to the left of O. Indeed, Q can be anywhere within a distance of 2a of O – but for the purpose of analysis, we can assume WLOG that OQ is horizontal.
Last edited by JaneFairfax (2007-08-22 21:52:07)
#4 2007-08-24 21:55:49
#5 2007-08-28 06:59:48
- John E. Franklin
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Re: Geometry problem
What's a locus? Is it a geometry term?
Imagine for a moment that even an earthworm may possess a love of self and a love of others.
#6 2007-09-18 22:41:10
- landof+
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Re: Geometry problem
Definition of a locus
That may be of some use.
I shall be on leave until I say so...
#7 2010-04-06 11:18:24
- JaneFairfax
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Re: Geometry problem
Bumping this because after two and a half years nobody appears to have a solution yet.
#8 2010-11-22 03:51:33
- 123ronnie321
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Re: Geometry problem
I am using your figure to solve this.
let the midpoint of PR be M
Let OQ = t < 2a
Let PQ = k.
PQ = P`Q` = k = 2P'M
PR = OQ = t = 2PM
PM^2 + P`M^2 = P`P^2 = a^2 ... Pythagoras theorem
therefore, k^2 + t^2 = (2a)^2. which is a circle if we make the following assumptions-
Let OQ act like x axis and and OR as Y axis.
Co ordinates of pt P are k,t.
Last edited by 123ronnie321 (2010-11-23 23:49:12)
#9 2010-11-22 09:47:59
- bob bundy
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Re: Geometry problem
Hi Jane,
I haven't been ignoring your problem since 2007. I've only been a member for 6 months or so.
So thanks for bringing back into attention.
See my two diagrams below. I set up what I thought was the right diagram and took the first screen shot. Then I moved Q left a bit and all that happened was moved the same amount (second half of shot).
Would you mind repeating the construction rules.
Mystified Bob 
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
#10 2010-11-23 00:06:07
- bob bundy
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Re: Geometry problem
Hi Jane Fairfax,
Ok, so I hadn't read the problem properly. Now I have.
Bob
Last edited by bob bundy (2010-11-23 00:07:53)
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
#11 2010-11-23 22:28:09
#12 2010-11-24 01:43:02
- bob bundy
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Re: Geometry problem
Hi Jane
I thought I'd already proved it in post 10.
Bob 
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei