Problems and Solutions

krassi_holmz · 2006-01-07 23:18:08

x^2+y^2=x^2+y^2+2xy-2xy=(x+y)^2-2xy=1-2xy.
x^4+y^4=17=x^2^2+y^2^2+2x^2y^2-2x^2y^2=(x^2+y^2)^2-2x^2y^2=(1-2xy)^2-2x^2y^2 = 1-4xy+4x^2y^2-2x^2y^2=1-4xy+2x^2y^2 =>
2x^2y^2-4xy=17-1=16
x^2y^2-2xy=16/2=8

irspow · 2006-01-08 04:51:53

I might be missing something obvious, but...

1/2 + 1/3 + 1/4 = 13/12? Doesn't one of the partners in k+78 own a 1/12 imaginary share?

I think that this problem could only be solved if we knew who was holding the fake share.

Last edited by irspow (2006-01-08 04:54:55)

Jai Ganesh · 2006-01-08 16:22:43

krassi_holmz is right!
To irspow : In problem # k + 78, when it is stated that the shares are in the ration a:b:c, it need not be true that a+b+c be equal to 1. a's share of the total would be a/(a+b+c), b's share would be b/(a+b+c) and c's c/(a+b+c).

Jai Ganesh · 2006-01-08 17:28:19

Problem # k + 83

Four horses are tethered at 4 corners of a square field of side 70 metres so that they just cannot reach one another. What is the area left ungrazed by the horses?

krassi_holmz · 2006-01-08 17:45:49

70^2 - 45^2 Pi ?

Jai Ganesh · 2006-01-08 22:50:51

krassi_holmz, you are given another chance.

Ricky · 2006-01-09 03:36:33

Last edited by Ricky (2006-01-09 03:40:26)

irspow · 2006-01-09 13:06:26

Last edited by irspow (2006-01-09 13:39:29)

Jai Ganesh · 2006-01-09 16:16:22

Well done, irspow

Jai Ganesh · 2006-01-09 16:39:06

Problem # k + 84

Two spheres of radii 6 cm and 1 cm are inscribed in a right circular cone. The bigger sphere touches the smaller sphere and also the base of the cone. What is the height of the cone?

irspow · 2006-01-10 11:21:02

I'll take an ill-attempted stab.

Jai Ganesh · 2006-01-10 21:54:50

irspow is correct! Although the actual solution is arrived at differently!

Jai Ganesh · 2006-01-10 21:58:32

Probelm # k + 85

If u and v are the roots of the equation x² + ax + b = 0, what are roots of the equation x² -ax + b = 0?

mathsyperson · 2006-01-11 03:02:37

That's an interesting one. I think it's something like this:

Ricky · 2006-01-11 04:16:41

I'm not sure, this is what I get:

Last edited by Ricky (2006-01-11 04:16:58)

mathsyperson · 2006-01-11 05:09:16

I think our answers are the same, and I've just taken a long way round.

Ricky · 2006-01-11 08:34:24

Yea, I guess they are. They looked completely different at first glance.

irspow · 2006-01-11 11:32:20

irspow · 2006-01-14 03:44:50

irspow · 2006-01-14 06:33:56

irspow · 2006-01-14 08:12:05

irspow · 2006-01-14 12:15:28

darn that k+42, I just cant figure it. Please someone, put me out of my misery. I think that you have to incorporate a geometric series somehow, but everything that I try turns to nonsense.

mathsyperson · 2006-01-14 12:47:28

Yes, k+42 is an incredibly difficult one. I think it could probably be solved brutally by getting excel to do all the calculations for you, but it's still tough.

Also, I think your answer to k+40 is wrong. I remember it being much smaller.

ryos · 2006-01-14 16:12:47

I didn't check if this one has already been solved, but since irspow asked about it, I gave it a go.

Last edited by ryos (2006-01-14 16:15:25)

Jai Ganesh · 2006-01-14 17:13:17

Four days Pongal break, and so many solutions posted! I shall reply to all of them after I return from Holiday on Jan 16. mathsyperson is right, irspow's solution to problem # k + 40 isn't correct. It is much smaller. . Same about ryos' solution to problem # k + 42.

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