BMO1 questions

No idea on the geometry (I am not good at geometry!).

For the other one:

Spoiler

Reply 2

For showing area of AMCP = 1/3 area of ABCD, draw the diagonal AC and see where it gets you. For this one, you don't need the lines MQ, NP, and indeed, they might be distracting, so try to just draw the lines you've got and see if you can spot something.

For the MNPQ one, you could try something similar. (Basically, with these kind of questions, usually just draw lots of diagonals making everything into smaller triangles and the solution will pop up eventually.)

I'm not sure if I've got the best solution, but a hint is that it might be easier showing area of MNPQ = area of AMCP, rather than MNPQ ＝ 1/3 area of ABCD. Experiment a little and see what you get.

Reply 3

So would this be correct for Q1?

Spoiler

I think that's right :O It should be anyway :P

And...Oh, wow...Can't believe proving AMCP = 1/3 ABCD is that simple :O

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Reply 4

Zygroth

So would this be correct for Q1?

Spoiler

I can't follow what you're doing here at all. You will need to put more explanation if you want to make sure you get the marks.

Reply 5

Darshan

I did the 2nd one last year in the BMO and I got 7, but only because of a slip up.

Your explanation's not making much sense though.

Reply 6

Zygroth

And...Oh, wow...Can't believe proving AMCP = 1/3 ABCD is that simple :O

Spoiler

I don't know the BMO marking particularly, but as long as you write that argument out clearly, I can't see why it shouldn't get full marks.

Spoiler

I'm not sure what you meant, but once you drawn MP, MQ, AP, it's only a matter of spotting that certain triangles are the same size... Happy hunting!

Reply 7

Oh really? Meh :frown:

I'll write it out

Spoiler

Ugh...So yeah, that's how bad I am at explaining :P I thought a load of numbers would make sense :s-smilie:

Oh well...I'll learn :P

Ugh, I still can't find the triangles...Does the line down MP (Right in the middle of the 2 quadrilaterals) represent a line of symmetry? I think it does, but I have no idea how to prove it :frown:

Reply 8

Zygroth

Oh really? Meh :frown:

I'll write it out

Spoiler

Ugh...So yeah, that's how bad I am at explaining :P I thought a load of numbers would make sense :s-smilie:

Oh well...I'll learn :PWhat you've done now is a lot better and should get full marks. I know it's maths, but there's no reason you can't use words and explain things!

Incidentally, with your approach, there's no reason to work it out for the case where you don't care where 6 goes first. The reason I suggested that is that the most common way people did this question was:

Spoiler

Reply 9

Oh wow, that's a much neater way :O No idea what xPy is though :frown:

Reply 10

Zygroth

Oh wow, that's a much neater way :O No idea what xPy is though :frown:

It's not really any neater than yours. xPy is the number of ways of choosing y elements from x where order matters. So xPy = x!/y!, as opposed to xCy = x!/(y!(x-y)!)

Reply 11

Oh, I see...
Anyway, on to more questions I can't do, I'm fine with those 2 now :smile:

Two touching circles S and T share a common tangent which meets S at A and T at B. Let AP be a diameter of S and let the tangent from P to T touch it at Q. Show that AP = PQ.
Ugh, I'm not too good with circles. I don't think I'd even attempt a circle question if I saw one in the BMO, but maybe I'll learn something :smile:

For positive real numbers a, b, c, prove that
(a^2 + b^2)^2 >/= (a + b + c) (a + b - c) (b + c - a) (c + a - b)
Er...I have absolutely no idea how. I'm sure this kind of thing must come up a lot, but I've never done it.

Let n be an integer. Show that, if 2 + 2 rt(1 + 12n^2) is an integer, then it is a perfect square.

And also this one from the 06 paper:
Let n be an integer greater than 6. Prove that if n + 1 and n - 1 are both prime, then n^2(n^2 + 16) is divisible by 720. Is the converse also true?
Yeah, I'm as far as "It needs to be a multiple of 2 and 3, so a multiple of 6" with the reasons for it, but not sure what else I need to do. Probably something along the lines of it needs to be a multiple of 12 or something, but I can't see why :s-smilie:

Reply 12

The geometry question: Do you know inversion? If so, what happens when you invert with centre P?

Otherwise, it's not hard (though not too easy either) to work out the lengths PQ and AP in terms of the radii of the two circles. Try to draw some radii of the circles, and then work out the distance from P to the centre of T first.

(a^2+b^2)^2 \geq (a+b+c)(a+b-c)(c+a-b)(b+c-a)

Do you know Heron's formula? If you do, does the RHS seem familiar somehow?

If you don't, it's fully possible to multiply everything out (remember

(x+y)(x-y)=x^2-y^2

!)

Other two questions to follow.

Reply 13

2+2\sqrt{1+12n^2}

is an integer, then it is a perfect square:

Again, there are probably loads of solutions, but here's hints towards one:

Do you recognise this as the quadratic formula perhaps? If

x = 2+2\sqrt{1+12n^2}

, could you perhaps construct a quadratic for which x is a root?

When you're there, you'll realise that there is a particular rather small integer that has to divide x, dividing that away. After that, there's some manipulation, and then you can show your result by consider prime factorisations and stuff. Let me know if you get to this stage but are still stuck.

Final question:

Note that

720=2^4\cdot 3^2 \cdot 5

.

So if you can prove that

n^2(n^2+16)

is divisible by 16, 9 and 5, you are done. None of these cases is hard, and you basically use the same kind of ideas as you used for showing that it is divisible by 6.

Reply 14

For the geometry problem again, if you prefer a geometrical approach (which you should):

Let X be the common point (the point of tangency) between S and T. Draw the line PT. (Can you see that if the problem is true, then you will have to have that PT passes through X? Use the power of a point theorem.) Now, show that PT indeed passes through X, and from there solve the problem.

Reply 15

Erm, I'm not too sure what multiplying it all out would do...I did it, I think, then I'm left with some weird inequality which I'm still unable to solve :frown:

I have no idea how to get rid of any of the unknowns :s-smilie:

I did find that when multiplied out, part of it included: -a^4 -b^4 -2a^2b^2 which is -(a^2 + b^2)^2. It looked promising, but there's such a load of other terms as well I can't find anything else.

Then the quadratic thing...
2 + 2 rt(1 + 12n^2)
= 2 + rt(4) x rt(1 + 12n^2)
= 2 + rt(4 + 48n^2)
If I then use it as part of [-b +/- rt(b^2 - 4ac)] / 2a
b = -2, ac = -12n^2
So it would be something like
ax^2 - 2x + c
Finding it hard to see where I'm supposed to go with this :frown:

For the divisible by 6 thing, I can't remember what I was thinking of when I said divisible by 3 :O I knew that it needed to be a multiple of 2 because odd numbers + 1 = divisible by 2. Then I said something along the lines of dividing by 3, or either n+1 or n-1 will be even...Must've been wrong, because it's 6 :s-smilie:

Or something like that.

As for geometry, I think I wont do any of those Qs :s-smilie:

There's so much I don't know about it... Inversion? Power of a point? :s-smilie:

Even after reading it on Wikipedia, I still don't understand those things.

Reply 16

Zygroth: These questions are supposed to be hard. Maybe it's just the way it's coming across in your posts, but I think you are giving up much too quickly without spending time on them. If you've not tried these kinds of problems before, spending a full day (10 hours or so) on a problem would not be at all unreasonable.

The solutions are usually going to be a lot more complicated than you're used to as well. So again, don't give up just because you've tried something and you've ended up with a long complicated expression.

If you do want help with any of these questions, you're going to have to post more detailed workings than you have so far.

Have you had a look at http://www.mathcomp.leeds.ac.uk/News%20from%20UKMT.htm? It has some links to material to help you prepare for BMO1.

Reply 17

Inequality:

I get

2a^4+2b^4 + c^4 \geq 2a^2c^2 + 2b^2c^2

when I multiply everything out. Is this what you get?

Then, do you know the AM-GM inequality? It's pretty essential that you know it for BMO. Can you see how to apply it here?

2+2\sqrt{1+12n^2}

I get

x^2 - 4x - 48n^2

, simply by spotting that if

x = 2 + 2\sqrt{1 + 12n^2}

,

then the other root has to be (not technically has to be, but it makes a nice quadratic if it is)

2 - 2\sqrt{1+12n^2}

.

Then we can write the polynomial as

\left(t-(2+2\sqrt{1+12n^2})\right)\left(t - (2-2\sqrt{1+n^2})\right) = 0

and multiplying stuff out, we get the polynomial I wrote above.

From here, you can spot that x has to be divisible by 4 if it is an integer. Can you see why for yourself? (This can also be seen directly from

x = 2 + 2\sqrt{1+12n^2}

)Then, we can write x = 4k, and manipulate a little to get

k(k-1) = 48n^2

.

Now, note that you only need to show that k is a square, because if it is, then x = 4k is automatically also a square since 4 is a square. Can you see how we can deduce, from the above equation by consider prime factorisations, that k must be a square?

Hint:

Spoiler

Divisible by 720 question:

Do you know modular arithmetic? Don't worry if you don't, the question is still very solvable, only it will take more work.

It is true that it must be divisible by 3, in fact it must be divisible. In fact, it follows straight from n > 6, n - 1 prime, n + 1 prime, that n is divisible by 3, and thus n^2 is divislbe by 9, and thus n^2(n^2 + 16) divisible by 9.

Hint:

Spoiler

.

The other ones (divisible by 16, by 5) use the same basic ideas. the 16 case is about the same difficultly, but the 5 one will take more work, especially if you do not know any modular arithmetic.

As for the geometry:

Do not worry about inversion, it is by no means required knowledge (although it does lead to neat solutions to some problems). Power of point, also known as intersecting chords theorem, is however very useful when you're dealing with circle geometry, and might be a good idea to learn. Of course, if you do not want to do the geometry questions on BMO, by all means, it's your choice.

Reply 18