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galactus · Feb 17, 2008, 06:49 AM

I am posing this problem for kicks. For those out there who like a nice math challenge problem. Since this site rarely gets them, I thought I would pose one if anyone would like a go.

"Looking at the graph, find the value of k so that areas of region A and region B are equal"

xphelper · Feb 17, 2008, 07:59 AM

Originally Posted by galactus

I am posing this problem for kicks. For those out there who like a nice math challenge problem. Since this site rarely gets them, I thought I would pose one if anyone would like a go.

"Looking at the graph, find the value of k so that areas of region A and region B are equal"

0.2

galactus · Feb 17, 2008, 01:03 PM

No, I am sorry to say that is incorrect. I will leave it up until tomorrow evening to see if anyone else wants a go.

ebaines · Feb 18, 2008, 01:02 PM

How about: k=0.724611?

galactus · Feb 18, 2008, 01:57 PM

Yes, indeed, ebaines. I knew you could get it. :)

ebaines · Feb 18, 2008, 02:19 PM

Originally Posted by galactus

Yes, indeed, ebaines. I knew you could get it. :)

Thanks! However, I must admit that I was unable to find a closed-form solution for this problem, but rather used an approximation method to solve a nasty equation:

1 = b*sin(b) + cos(b)

where b = arcsin(k)-pi.

Is there a more elegant way?

galactus · Feb 18, 2008, 02:59 PM

Yes, that's exactly what I done. I used Newton's method.

galactus · Feb 23, 2008, 10:17 AM

SOLUTION:

Let (a,k) be the point of intersection of y=k with y=sin(x).

Then, k=sin(a) and if the areas are equal:

$\int_{0}^{a}(k-sin(x))dx=\int_{0}^{a}(sin(a)-sin(x))dx=a\cdot{sin(a)}+cos(a)-1=0$

Solve for 'a' using Newton's method or some other estimation techniques and get: $a=2.331122$ , so $k=sin(a)=0.724611$

The horizontal line which intersects sin(x) creating two equal regions is y=0.724611