[QUOTE=Lord of the Flies;44345359]Solution 330
Straightforward, minutes.
I put this question here in case any gcse student was lurking around [so they might be able to solve it] [\QUOTE]
Problem 333**
Let ABCD be a fixed convex quadrilateral with BC = DA and BC not parallel with DA. Let two variable points E and F lie of the sides BC and DA, respectively and satisfy BE = DF. The lines AC and BD meet at P, the lines BD and EF meet at Q, the lines EF and AC meet at R. Show that the circumcircles of the triangles PQR , as E and F vary, have a common point other than P.
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Last edited by Arieisit; 03092013 at 23:56. 
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(Original post by Smaug123)
This was on one of our example sheets 
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Solution 332
Assume .
We consider with a given root . By Vieta we obtain and ; whence is a positive integer. If , we get . Hence we can replace with . Thus we have , giving , since ;
If , we get , and so ; .
Solution 331
Since the limit exists and is nonzero, we have converges iff converges.
The proof is routine.
Assume the limit is . Abbreviate . Then for each there exists (both ) such that whenever . This gives us .
Choosing sufficiently small, we are done.
Solution 333
Let the circumcircles of and meet at . We know that is the center of the spiral similarity that sends to and to (wellknown lemma, proved via directed angles ). Our similarity sends also to and we see that the circumcircles of and meet at . Now applying Miquel's theorem to and , we see that the circumcircle of passes through .Last edited by Mladenov; 04092013 at 01:28. 
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(Original post by The Sorcerer)
Ah many commendations my mathematical friend. There was me thinking that you were just going through a mundane mathematical question via conventional methods... but then you just shock your dedicated audience tremendously by unveiling the beautiful esoteric technique that is absorption... ABSORPTION! I ask, WHO IN THE RIGHT MIND WOULD HAVE DARED TO HAVE UTILIZED SUCH AN IMMENSELY DIFFICULT TECHNIQUE?! and what an elegant sponge you chose to use for the job too! Wonderful. Keep up the good work. I will be greatly anticipating and anxiously awaiting your next groundbreaking mathematical display.
Spoiler:ShowNot my problem that's actually what it's calledLast edited by ukdragon37; 04092013 at 00:43. 
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EDIT: I'm satisfied that this reasoning is wrong; please ignore it.
Spoiler:Show
x, y and z are positive integers satisfying . Show that z=3.
I did not see a specific way to eliminate to get z=3 so decided to try to deduce it using inequalities. Firstly, note that . Firstly, consider the following inequality:
(1)
so z > 2.
Next, add 2xy to both sides of (1) to get
Also, note that 2xy > 1, and so:
Now, we have the following two inequalities:
,
Now, if z is 4 or more, we can see that x and y can be chosen so that equality holds, which is a contradiction, so z < 4. This argument is the bit that concerns have been expressed over.
This gives the inequality 2 < z < 4 so z = 3.
Last edited by DJMayes; 04092013 at 22:18. 
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(Original post by DJMayes)
Guys, could you check my logic to my solution of problem 332? I have done it a different way and obtained the solution but there is a concern that my solution is wrong. Could you see what you think?
Solution 332 (Alternate)
Spoiler:Show
x, y and z are positive integers satisfying . Show that z=3.
I did not see a specific way to eliminate to get z=3 so decided to try to deduce it using inequalities. Firstly, note that . Firstly, consider the following inequality:
(1)
so z > 2.
Next, add 2xy to both sides of (1) to get
Also, note that 2xy > 1, and so:
Now, we have the following two inequalities:
,
Now, if z is 4 or more, we can see that x and y can be chosen so that equality holds, which is a contradiction, so z < 4. This argument is the bit that concerns have been expressed over.
This gives the inequality 2 < z < 4 so z = 3.
Last edited by Smaug123; 04092013 at 20:30. Reason: < > <= 
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 05092013 10:50
(Original post by Mladenov)
Solution 332
Assume .
We consider with a given root . By Vieta we obtain and ; whence is a positive integer. If , we get . Hence we can replace with . Thus we have , giving , since ;
If , we get , and so ; .

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(Original post by DJMayes)
Forgive me but I don't see how this method gives a sufficient proof as there are infinitely many solutions without x or y equal to 1. For example, (5,2,3) and (13,5,3) satisfy the equation and using the second root of the quadratic as you did you can find infinitely many larger and larger solutions. I have been bashing my head against this for the better part of a day now and can't see how any of the solutions I've seen so far work as they rely on x or y equalling 1 which is demonstrably not necessary. 
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(Original post by Mladenov)
We replace with till we get with . Throughout the process remains constant, hence the conclusion. 
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I am so out of this business now, but here is one Diophantine equation.
Problem 334**
Find all pairs of positive integers satisfying the equation . 
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Problem 335**/***
Let be a solution of a homogeneous fourth order ODE with real constant coefficients. Determine the general solution and the ODE it solves. 
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(Original post by DJMayes)
Solution 334: (I think)
Spoiler:Show
In order to look for solutions, try to solve for x in terms of y. By applying the quadratic formula, we get:
Now, if integer solutions exist, we must have the bit inside the square root as a perfect square  that is; any possible solutions for x and y must satisfy:
Now, the RHS must be divisible by y. This opens up two possibilities:
1  Each factor of (z2)(z+2) contains a factor of y. As these are four apart, this is only possible if y = 1, 2 or 4. By checking we find that y=1 is a solution but y=2 or 4 are not.
2  One of the factors of (z2)(z+2) is divisible by y^2. Now, if one factor is divisible by y^2, and z2 is greater than 5 ( z > 7) then z^24 > 5y^2 and there are no solutions. This means we are only interested in values of z up to and including 6.
Testing these, we get z=2 and z=3 as solutions with corresponding y values of 0 and 1. As we are interested in integer solutions, the only possible solution is y=1. Now, we are left with solving the following equation:
and so our only positive integer solution is (1, 1)
Spoiler:ShowThere are other solutions, consider (5,8) and (13,21). In fact I think there are an infinite number of solutions. This kind of explains how to solve it along with the rest of that page, it's not very clear though. Any other links on the subject would be useful!Last edited by Pterodactyl; 05092013 at 18:23. 
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 05092013 18:23
(Original post by DJMayes)
Solution 334: (I think)
Spoiler:Show
In order to look for solutions, try to solve for x in terms of y. By applying the quadratic formula, we get:
Now, if integer solutions exist, we must have the bit inside the square root as a perfect square  that is; any possible solutions for x and y must satisfy:
Now, the RHS must be divisible by y. This opens up two possibilities:
1  Each factor of (z2)(z+2) contains a factor of y. As these are four apart, this is only possible if y = 1, 2 or 4. By checking we find that y=1 is a solution but y=2 or 4 are not.
2  One of the factors of (z2)(z+2) is divisible by y^2. Now, if one factor is divisible by y^2, and z2 is greater than 5 ( z > 7) then z^24 > 5y^2 and there are no solutions. This means we are only interested in values of z up to and including 6.
Testing these, we get z=2 and z=3 as solutions with corresponding y values of 0 and 1. As we are interested in integer solutions, the only possible solution is y=1. Now, we are left with solving the following equation:
and so our only positive integer solution is (1, 1)
Spoiler:ShowYour reasoning assumes prime or , which is not necessary. For example, . Also, you have missed the case , which corresponds to . Not a hint, butSpoiler:ShowThere are infinitely many solutions, and one should look for a family of solutions. Think of some recurrence relation and then infinite descent. 
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 05092013 18:31
(Original post by Mladenov)
Spoiler:ShowYour reasoning assumes prime or , which is not necessary. For example, . Also, you have missed the case , which corresponds to . Not a hint, butSpoiler:ShowThere are infinitely many solutions, and one should look for a family of solutions. Think of some recurrence relation and then infinite descent.
OK, now properly done the question.
Solution 334:
Spoiler:Show
Some solutions are (1,1), (2,3), (5,8) and so on. This looks very similar to the Fibonacci sequence. As such, we will guess that, if (x,y) is a solution, (x+y, x+2y) is a solution. This is for obtaining larger solutions. Reversing this, we want to show that if (x,y) is a solution, then we can generate a smaller solution (2xy, yx):
Hence (2xy, yx) is also a solution. We should now prove the validity of this solution.
Firstly, if then and the only solution with x and y equal is (1,1) so for any other solution we have y>x, making yx positive and valid. If then and no solutions exist, and so the x solution is valid.
Hence we have shown that the existence of a solution (x,y) implies the existence of a smaller solution (2xy, yx). We can repeat this process indefinitely until either the new x or y value equals 1, giving the solution (1,1) and so all solutions are generated from these. Working backwards, our family of solutions is Where F_n represents the nth Fibonacci number with F_0 = F_1 = 1.
Last edited by DJMayes; 05092013 at 19:20. 
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 06092013 23:23
(Original post by Mladenov)
Solution 332
Assume .
We consider with a given root . By Vieta we obtain and ; whence is a positive integer. If , we get . Hence we can replace with . Thus we have , giving , since ;
If , we get , and so ; .
Solution 331
Since the limit exists and is nonzero, we have converges iff converges.
The proof is routine.
Assume the limit is . Abbreviate . Then for each there exists (both ) such that whenever . This gives us .
Choosing sufficiently small, we are done.
Solution 333
Let the circumcircles of and meet at . We know that is the center of the spiral similarity that sends to and to (wellknown lemma, proved via directed angles ). Our similarity sends also to and we see that the circumcircles of and meet at . Now applying Miquel's theorem to and , we see that the circumcircle of passes through . 
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 07092013 10:46
(Original post by BlueSam3)
You should probably also justify that explicitly, unless I'm missing something.
To prove it, suppose that the circumscribed circles of and are tangent at . Then, by angle chasing, more specifically, we let be the tangent through ,  the angle between and , and  the angle between and , thus we see , a contradiction. 
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 07092013 18:07
(Original post by Mladenov)
But this follows directly from the condition that .
To prove it, suppose that the circumscribed circles of and are tangent at . Then, by angle chasing, more specifically, we let be the tangent through ,  the angle between and , and  the angle between and , thus we see , a contradiction. 
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LOTF Solution 336
By maths we have the inequality.
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