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University Problem Solving Thread

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cara.mel:
I found where most of the questions come from =D

I'm very disappointed you didn't include this:
1. You have a compass whose legs are set at a fixed distance apart. How can you draw 2 circles of different radii on paper?

'Answer' - tear out a bit of paper, fold it a few times so its thickness is nonnegligible, and its other dimensions sufficiently small. Now put it on the remaining paper, put the sharp end of the compass on this elevated creation, and draw a circle on the remaining paper. This circle will have a different radius.

Proper answers :D "We received a very creative (sometimes even unphysical) set of solutions from room 610, physics building. It includes: modify the compass or introduce another compass; stretch the paper uniformly in all directions, draw a circle, then unstretch the paper; make a point on the paper for a 0-radius circle; draw a straight line using 1 leg of the compass, creating a circle with infinite radius;
any drawn circle has finite width so there are infinitely many different circles between the inner and outer edges.

We accept their solution of piercing a hole in the paper with the compass and lowering 1 leg into the hole."

kamil9876:
nice thread! just discovered, hope it's not too late to post:

2.)
My proof is rather long, so i'll just give an outline and if interested, ask for the details later:

its true for x=y=z=1/3

Wwe have infinitely many combinations (x,y,z). WE know that (1/3,1/3,1/3) works and does not violate the x+y+z=1 condition. How do we get from that combo to something else that doesn't violate this condition, well we can add c to the x value and add -c to the y value. Now evaluate xy+yz+xz for our new combo (1/3 + c,1/3 - c,1/3) and u find that it is equal to 1/3 - c^2 which is obviously less than 1/3.

But this only proves this property for a subset of the combos we want, as one of the values was assumed to be 1/3. Now we want to find the combos that possibly have no 1/3 for any value. In other words, we're going to change the z value a bit: lets say we add d to the z value and to 'fix up' we subtract d from the y value:

now we have the combo (1/3 + c, 1/3 -c - d, 1/3 + d)

This accounts for all posibilities*

*Proof: if we select an  x-value we want, we then get a c value we must stick to. Now by choosing a z value we want, we're consequently choosing the d value. Now we're limited to only one y value because c and d we already have and this makes sense because by choosing values for x and z there is only one other y value we can have so that x+y+z=1

Now we must work out xy+yz+xz for that general combo presented above and after a fuckload of algebra we get:

xy+yz+x= 1/3 -c^2-cd-d^2
           = 1/3 -((c+d)^2-cd)
And c and d are free paremeters.
Now u must prove that (c+d)^2>cd for all values and ur done. 

Ahmad:
Well done.

Remark:
It's interesting the way symmetry creeps into algebra. In one way a lot of the study of algebra seems to be about data representation and symmetry.

For example given x + y = 1, find the maximum of xy. We can reformulate this problem as find the maximum of x(1-x) by subbing in the y value through the constraint. But it still may not be immediately obvious what the maximum is. However if we let x = 1/2 - c and y = 1/2 + c just as you had we get xy = 1/4 - c^2 <= 1/4 and the maximum becomes obvious. All we did was a simple reformulation of the problem (changed data representation) through a substitution but somehow this substitution seems to be the problem's natural one (because of symmetry).

kamil9876:
Yea. My method was inspired by my solution to a similair problem:

abcd=1

prove that

Here is where my method of showing that its true for some combo and then adjusting algebriacally to account for all combos originated. Basically, it's true when they all equal 1. However to get from one combo to the other u multiply a by x and b by 1/x etc.
You can then generalise this for an n number of variables. The 10 would be n+nC2 and there would be n squares followed by products of two variables(every such pair).

Anyway, to keep up the spirit of the thread, here is one good obscure problem I actually came accross in my yr11 textbook(although the problem was from a chapter we didn't do). Underneath the problem there is the message: "Note: this is a difficult problem". First time I've ever came across such a thing. Very luring indeed.

If a,b and c are positive integers such that no integer greater than 1 divides them all and , prove that a + b is a perfect square

Ahmad:



implies there exists integers m and k such that with

Perhaps I'll post a problem soon, but feel free to continue posting problems and keeping this thread active.

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