2)

angle subtended by it is greatest at the centre

half of diameter is greaterst(=radius, for rest chords, it is sqrt(r^2-a^2), where a is not zero)

2)

a=k

b=k(k-a)

where k-a+1 divides k

Keep Smiling

Malay

Results 1 to 7 of 7

- March 19th 2007, 06:02 PM #1

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

## Problem 22

1)Given a function defined on [a,b], we say it is

**null**iff INT(a,b) f^2(x) dx=0.

Note, f(x)=0 is null on any interval however the converse is not true.

Show that if g(x) is any function defined on [a,b] and f(x) is a null function then,

INT(a,b)g(x)f(x) dx=0

2)Show that the diameter is the longest chord. (Not necessarily hard I want to see how many reasons you can come up with).

3)Let a,b be positive integers. Given necessary and sufficient conditions so that,

(ab)/(a+b)

Is a positive integer.

- March 21st 2007, 04:54 AM #2

- April 10th 2007, 08:02 PM #3

- Joined
- Nov 2006
- Posts
- 3

(1)

It seems false the way you have stated the problem. For consider (ix + n). It can easily be shown that

Int(a,b) (ix + n)^2 dx = 0

for

[a,b] = [-sqrt(3n^2), sqrt(3n^2)]

But with (g(x) = x) then

Int(a,b) x(ix + n) dx = 2in^3 * sqrt(3)

(2)

Center the circle on the origin and use a simple max value formula:

(d/dx) chord length = 0 or undefined

(d/dx) 2 * sqrt(r^2 - x^2) = 0 or undefined

-2x * (1 / sqrt(r^2 - x^2)) = 0 or undefined

x = -r, 0, r

Plug in the values

2 * sqrt(r^2 - (-r)^2) = 0

2 * sqrt(r^2 - 0) = 2*r

2 * sqrt(r^2 - r^2) = 0

So, the max value is obviously the line (x = 0) which passes through origin, a.k.a the circle's center.

And finally:

Chord that passes through center of circle = Diameter

Done-and-done.

- April 11th 2007, 04:09 AM #4

- April 11th 2007, 10:47 PM #5

- Joined
- Nov 2006
- Posts
- 3

Oh. Well, if so, then the only valid

*f*(x) would be (*f*(x) = 0). Because the square of any real function is going to be always positive, thus the only way to have zero area under the curve would be for the function to be a constant zero. And (0 * g(x) = 0).

I assumed his question was more complicated than that.

- April 20th 2007, 01:43 AM #6

That's because you assumed the functions involved are continuous! It's pretty trivial in that case

If however they are assumed only integrable (as they should - ps. I love Hacker's problems, coz they are almost never "well posed", and that's exactly what makes a problem interesting ) than we can just apply Cauchy-Schwartz:

Q(uite)E(asily)D(oneth).

For 2, this came to my mind. Consider a tangent to the circle. Rotating this by the touching point, for an angle s ε (0,π) we obtain a chord of the circle, of length say l(s)>0. This function is defined on (0,π), is continuous, and can be extended to a function continuous on [0,π] by the (obvious) l(0)=l(π)=0. So it must attain a maximum in (0,π). We now see it is increasing for s<π/2 and decreasing for s>π/2, so the maximum is attained for s=π/2 - this chord is exactly the diameter.

- April 20th 2007, 03:59 AM #7

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10