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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Heres a geometry problem Ive 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 2*a*, 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 2 a.**

Im sure it works. Still, if you find any problem with my problem, do let me know.

*Last edited by JaneFairfax (2007-08-19 23:20:35)*

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

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.

**igloo** **myrtilles** **fourmis**

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Thats right. I should have provided a diagram anyway, so here it is.

|P′Q′| = |PQ|

|PP′| = |P′R| = |RQ′| = |Q′P| = *a*

|OQ| ≤ 2*a*

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 2*a* of O but for the purpose of analysis, we can assume WLOG that OQ is horizontal.

*Last edited by JaneFairfax (2007-08-21 23:52:07)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

What's a locus? Is it a geometry term?

**igloo** **myrtilles** **fourmis**

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**landof+****Member**- Registered: 2007-03-24
- Posts: 131

That may be of some use.

I shall be on leave until I say so...

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Bumping this because after two and a half years nobody appears to have a solution yet.

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**123ronnie321****Member**- Registered: 2010-09-28
- Posts: 128

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 00:49:12)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,524

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

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,524

Hi Jane Fairfax,

Ok, so I hadn't read the problem properly. Now I have.

Bob

*Last edited by bob bundy (2010-11-22 01:07:53)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Hi, Ronnie and Bob.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,524

Hi Jane

I thought I'd already proved it in post 10.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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