I've finished MX2 now, but this thread is too fun lol.
Can I ask how you know to turn that statement into that sum?I've finished MX2 now, but this thread is too fun lol.
Can I ask how you know to turn that statement into that sum?
Does this work?Prove that the tangents are opposite vertices of a cyclic quadrilateral intersect on the secant through the other two verticies IF AND ONLY IF the two products of opposite sides of the cyclic quadrilateral are equal (so prove both ways)
firstly, i don't think you know what if and only if means.Does this work?
Let A, B, C and D be the sides of a cyclic quadrilateral and E be an external point and let two equal tangents from the external point E meet the circle at A and C.
Any secant that passes through E, B and D is a diagonal to the cyclic quadrilateral ABCD.
Therefore triangle ABC is similar to triangle ADC (opposite pairs of triangles in a cyclic quadrilateral are similar).
Hence AB/AD=BC/DC, making (AB)(DC)=(AD)(BC).
Going backwards: assume (AB)(DC)=/=(AD)(BC).
Therefore triangles are not similar (unless (AB)(BC)=(AD)(DC)).
Therefore ABCD is not a cyclic quadrilateral, which is a contradiction.
Lol I just realised what I was doing wrong with the similar triangles :S...firstly, i don't think you know what if and only if means.
secondly, I think you're making wrong assumptions, for example opposite triangles in cyclic quads aren't necessarily similar.
in other words, you've only proven it only one way.
For backwards, you're supposed to explicitly prove that if (AB)(DC)=(AD)(BC), then the inteserction of the tangents at A and C go through the secant of B and D, which you've NOT done. What you've proven is that if tangents at A and C go through secant of B and D, then (AB)(DC)=(AD)(BC)
So your answer is incomplete. You didn't even do the hard bit lol
Sure.Can anyone do the Q?
Cool question! I think I have covered cases correctly:
Yep pretty similar to my solution!Cool question! I think I have covered cases correctly:
First note that there is no solution with either of x,y equal to 0.
Case A: x negative, y positive
Unless y=1, only one side will be an integer. A trivial check shows there is no solution with y=1 and x negative.
Case B: x, y both negative.
By factoring out the (-1)^x and (-1)^y, we get a 1-1 correspondence between negative solution pairs and positive solution pairs with both variables having the same parity. This reduces the problem to...
Case C: x,y positive integers.
Wlog assume x < y.
By inspection there is no solution with x=1.
The condition is then equivalent to x/log(x)=y/log(y).
But differentiating shows that f(t)=t/log(t) decreases from 1 to e and increases thereafter, so we must have 1 < x < e. So x=2 is the only possible source of solutions and there is exactly one real solution to x^2=2^x in the interval (e,infinity).
As it happens, this solution is an integer! It is 4.
Hence the only solution pairs are (2,4),(4,2),(-2,-4),(-4,-2).
Definitely! It feels more like a coincidence than anything actually deep, but still.Yep pretty similar to my solution!
Quite interesting how numbers like 4 and 2 can come out of something that seems so simple
I'll prove the stronger result thatProve that (2^p)-1 is coprime to (2^q)-1 if p and q are distinct primes. Use HSC methods
Let p(n) be the probability of flipping an odd number of heads with n such coins.Here is a nice probability question