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

Bob

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

Bob

]]>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

]]>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.

That may be of some use.

]]>|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.

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.]]>

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.

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