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# BMO Banter watch

1. (Original post by JAKstriked)
The line PQ is tangent to the smaller circle at Q.

Attachment 77276
It says that 'the tangent from P to T touches at Q' - so essentially, it's tangent to S and T, at P and Q respectively. Hence why I drew a tangent from P in my diagram [which is necessarily parallel to the tangent at A]. No?
(Original post by DFranklin)
Don't ask me about BMO geometry (or indeed anything that involves drawing diagrams...)
Fair enough
2. (Original post by GHOSH-5)
It says that 'the tangent from P to T touches at Q' - so essentially, it's tangent to S and T, at P and Q respectively. Hence why I drew a tangent from P in my diagram [which is necessarily parallel to the tangent at A]. No?
I think it is saying that the tangent to T which passes through P touches T at Q.~

3. (Original post by JAKstriked)
I think it is saying that the tangent to T which passes through P touches T at Q.~

I think you're right. Woops!
4. (Original post by GHOSH-5)
I think you're right. Woops!
5. After a bit of rummaging around I've found the BMO primer handbook and it agrees with my diagram which is always a plus.

It offers 2 solutions for this problem, one using standard Phythagoras, the other using inversive geometry. My question is could anyone give me a good explanation of how inversion works? I'm not understanding wikipedia too well and the book offers no theory on it.
6. (Original post by JAKstriked)
After a bit of rummaging around I've found the BMO primer handbook and it agrees with my diagram which is always a plus.

It offers 2 solutions for this problem, one using standard Phythagoras, the other using inversive geometry. My question is could anyone give me a good explanation of how inversion works? I'm not understanding wikipedia too well and the book offers no theory on it.
Take a circle C, centre O and any point P inside the circle (disregarding P as the point O for a min) then the inverse point of P, P' is such that OP * OP' = r^2. Then P and P' are said to be inverse points with respect to C.

There are a few properties of inversion from which you can use to prove things.

(a) A line through O (the center of the circle) becomes a line through O. (If you draw a line through the centre of a circle and find all the inverse points, it will be the line (it will essentially be a different line on top of it I think)).

(b) A line not through O becomes a circle through O (If you draw a line that doesnt go through the center, all the points on this line P' will be inverse to some point P inside the circle, and all the points P form a circle through O).

(c) A circle through O becomes a line not through O (inverse of the property above).

(d) A circle not through O becomes a circle not through O .

These are the 4 main properties (unless theres more )
7. (Original post by refref)
Take a circle C, centre O and any point P inside the circle (disregarding P as the point O for a min) then the inverse point of P, P' is such that OP * OP' = r^2. Then P and P' are said to be inverse points with respect to C.
I might add that O, P and P' must be collinear, in the 'order' O, P, P' or O, P', P (P and P' may also coincide). [As otherwise, P' would represent a set of infinitely many points; e.g. take P to be on the radius of C - then from your definition, P' could be any point on the circle.] Also, P need not be inside the circle (indeed, if it were, then the properties you stated would not hold).
8. Thanks for the help. I'm gonna have to look at a few more examples to get the feel of it but it's a start.

edit: found this: http://whistleralley.com/inversion/inversion.htm which has helped me. just thought id post in case anyone else would like a look.
9. (Original post by GHOSH-5)
I might add that O, P and P' must be collinear, in the 'order' O, P, P' or O, P', P (P and P' may also coincide). [As otherwise, P' would represent a set of infinitely many points; e.g. take P to be on the radius of C - then from your definition, P' could be any point on the circle.] Also, P need not be inside the circle (indeed, if it were, then the properties you stated would not hold).
Well thats what I meant It's kind of a backwards thing, if you had a point P inside the circle and the point P' outside of it, P and P' are invariant, but P is also the invariant point of P', so you can swap them around if you know what I mean
10. From last year's BMO-1.
(Original post by Question 4, BMO-1, 2008)
Find all positive integers n such that both divides and divides .
and similarly ; n+2008 and n+2009 are consecutive positive integers, and are hence coprime, hence the previous two statements are true iff

, where k is some non-negative integer.

If we expand the LHS, we see that it is always larger than the RHS for positive integers n, unless n = 0, in which case k = 0 is necessary and sufficient.

Hence the only positive integer is 1.

Is this solution satisfactory?
11. bit of off the topic : but any ideas what the BMO boundary will be this year?
i know i can pay to take it, but still it's not same as getting 'invited' to take it
12. (Original post by lovmaths)
bit of off the topic : but any ideas what the BMO boundary will be this year?
i know i can pay to take it, but still it's not same as getting 'invited' to take it
Somewhere around 90; we can speculate about it as much as we please, but it should be out within a few weeks.
13. GHOSH I am stealing your thread! (Or hoping you can help me)
http://www.bmoc.maths.org/home/bmo1-2000.pdf
Question 3

Spoiler:
Show
I think its at point P = point A but yet to do a real proof
14. (Original post by refref)
GHOSH I am stealing your thread! (Or hoping you can help me)
http://www.bmoc.maths.org/home/bmo1-2000.pdf
Question 3

Spoiler:
Show
I think its at point P = point A but yet to do a real proof
I've just given it a quick run, and I think it is A as well.
Spoiler:
Show
Clearly, it is either A or the point D on the hypotenuse such that AD is perpendicular to the hypotenuse. So we just need to consider which is larger: the sum of the two legs AB and AC or the sum of the hypotenuse BC and AC.

Any chance you could have a look at post #50?
15. (Original post by GHOSH-5)
I've just given it a quick run, and I think it is A as well.
Spoiler:
Show
Clearly, it is either A or the point D on the hypotenuse such that AD is perpendicular to the hypotenuse. So we just need to consider which is larger: the sum of the two legs AB and AC or the sum of the hypotenuse BC and AC.

Any chance you could have a look at post #50?
Cool, that's what I was thinking (I havent proved that bit yet). Would we still need to prove that it cant be point B or C, or any point on AC or AB (apart from A). I know that bit is an easy proof (any point D on AC would give you AD + CD = AD and BD would be the hypotenuse which is greater than if D was point A, if you know what I mean).

I ordered my olympiad primer today...hopefully it will come by next weekend (If I knew I could enter the BMO without getting ~90 on SMC I would have done this a long time ago , these questions are much more fun than SMC).

(I will have a look at #50....)
16. (Original post by refref)
Cool, that's what I was thinking (I havent proved that bit yet). Would we still need to prove that it cant be point B or C, or any point on AC or AB (apart from A). I know that bit is an easy proof (any point D on AC would give you AD + CD = AD and BD would be the hypotenuse which is greater than if D was point A, if you know what I mean).

I ordered my olympiad primer today...hopefully it will come by next weekend (If I knew I could enter the BMO without getting ~90 on SMC I would have done this a long time ago , these questions are much more fun than SMC).

(I will have a look at #50....)
heyyy i was thinking of ordering Olympiad Primer,
do you know how long the delivery takes??
17. (Original post by refref)
Cool, that's what I was thinking (I havent proved that bit yet).
After a bit of checking, I can confirm that it is A.
Would we still need to prove that it cant be point B or C, or any point on AC or AB (apart from A). I know that bit is an easy proof (any point D on AC would give you AD + CD = AD and BD would be the hypotenuse which is greater than if D was point A, if you know what I mean).
I would give at least a couple of sentences as to why, if the point lies on one of the legs, it must be minimised at A, and similarly, if the point lies on the hypotenuse, it must be at D with AD perp. to BC. Perhaps the odd equation/inequality or two, but that would be sufficient.
18. (Original post by lovmaths)
heyyy i was thinking of ordering Olympiad Primer,
do you know how long the delivery takes??
I'll let you know

(there was no option for delivery, meh, I'll just email them right now to find how long it will take)
19. (Original post by refref)
I'll let you know

(there was no option for delivery, meh, I'll just email them right now to find how long it will take)
thanks!
20. (Original post by refref)
...
Did you get this for Q4 off that 2000 paper?
Spoiler:
Show
k = 2001, 667, 87, 23, 3

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