Math Is Fun Forum

I guess your right.

irspow  I think all you have to do is declare a variable

unsigned int x;

If you were to assign a float or decimal value to it, it would automaticly be converted to an int and change the sign if its negative I think.... But I believe, the drawback is it doesn't round. Thus if you assigned a value of 3.9999999999 to an int it would have a value of 3. A rounding function should be relativly easy to write, though.

My apologees!

Anyway, I was thinking about this problem a lot last night.

At any rate, maybe we should just keep cranking out rearragned forms and see if we find anything.

Pgj * Tgc = Pgc * Tgj   = Rg

Pbj * Tbc = Pbc * Tbj   = Rb

Pgj = 20 + Pbj

Tgj = Tbj + 10/3

hmm...

Oops. The position where the boy saw the cyclist on the graph I drew is obviously wrong. But fortunatly, that should have no effect on the truthfullness of the equations. (if I did them right in the first place)

Well I managed to crank 11 equations out of this problem. There are 11 variables to consider so it should just be a matter of rearranging now. But I'm too exhausted to do it at this point so I'll just post what I found for now.

We'll let A = 0, and t = 0 be the time when they first begin. Also I changed jack and jill to boy and girl, so we can represent them with different letters.

1. The position (or distance traveled from A) when the boy saw the cyclist equals the time in hours from when he first started, times his rate. Thus Pbc = Tbc * Rb

2. Supposedly, the boy passed the cyclest two hours after he passed the jogger. So Tbc = Tbj + 2

3. When the boy saw the jogger, he was 50 miles from point b. So Pbj = B - 50

4. The position of the boy when he saw the jogger equals his rate times the time it took to get there. Thus Pbj = Tbj * Rb

5. Supposedly, the girl saw the cyclist 20 minutes before she got 16 miles from point B. How long would it take her to get 16 miles frome point B? Well, Rg * T = (B - 16)  T = (B - 16)/Rg. But she saw the cyclest 20 minutes before this time. We're working the problem in hours so thats 1/3 hour before. Thus Tgc = (B - 16)/Rg  -1/3

6. Position of the girl when she saw the cyclist equals her rate, times the time it took her to get there. Thus Pgc = Rg * Tgc

7. We are told the girl met the jogger when she was 30 miles from be, thus Pgj = B - 30.

8. Position when girl saw the jogger equals her rate times the time it took to get there. Thus Pgj = Rg * Tgj

9. The cyclist drove from the position where he passed the boy, to the position where he passed the girl. So the distance he traveled between the 2 was (Pgc - Pbc). He did this from the time he passed the boy, to the time he passed the girl, and was driving at 8 mph. Thus the distance between the two also 8 * (Tgc - Tbc). These two distances are equal so (Pgc - Pbc) = 8 * (Tgc - Tbc)

10. The boy met the jogger 50 miles from B and the girl met the jogger 30 miles from B. Thus the positions where they met the jogger were 20 miles apart. Thus Pgj - Pbj = 20. (this is provable using equations 3 and 7)

11. The jogger traveled 20 miles from the position where he met the boy to the position where he met the girl. We don't know how long this was but we do know his rate was 6. So 6 * T = 20. T = 10/3. Thus The time from when the boy saw him to when the girl saw him was 10/3 hours. So Tgj - Tbj = 10/3. Rearraned Tgj = 10/3 + Tbj

12. Oops! I wrote this already. My bad!

Ok so thats 10 equations then. Not 11. 10 is just a combination of equations 3 and 7 (ironicly) so it doesn't count. And 12 I wrote already (in a rearranged form)

The equation may still be solvable at this point, with a lot of rearranging. But I'm going to see if I can find one last equation.

nice! your doing calculus!

Wish I had discovered math earlier. I could have gone very far by now.

Does everyone in your school take calculus in 11 and 12th year? Here usually you only get calculus if your extra good.

Ah I see it now.

I saw that "zort in a mirror" episode in Pinky and The Brain. Brain roxxors! He should be a calculus teacher.

darn! Clearly america's math education system is too slow. (not that I ever used it)

I bet your from athens greece.

So your in trig then? Or did you get AP and get calculus? I've rarely seen such tough geometry problems, and to think your getting them in highschool. Thats wierd...(98.6% of all americans are IDIOTS! at least mathwise...)

Whats the name of your current math/geometry book? I'd like to see if I can get my hands on it.

incase your not sure where I got the requirements from. If you have an expression such as |x + 1| if the value of x + 1 is less then zero, it will be negative. Therefore, if x + 1 is less then 1, the expression will haves its sign changed. Thus |x + 1| equals -(x + 1) when the expression is less then 1. To find when x + 1 is less then zero: x + 1 < 0  solved x < -1. Lets try a value less then -1 as a test. |-2 + 1| = 1. But -2 is less then 1 so instead of | x + 1| we can write -( x + 1) x = -2 so

-(-2 + 1)

-(-1)

1.

This gave us the same answer as the absolute value expression. Why? Because thats what the absolute value symbols mean. When the value is negative (less then 0) the sign is reversed to make if positve.

Thus   if (x < 0) |x| = -x

Maybe. I'm not sure where you got that formula.

I think the reason I didn't see the obvious solution was I did not assume the line drawn from the small circle to the edge of the large, had the angle as the line from the center of the large circle to the point of tangency.

Lets see if I can prove it.

(edit) dang. It seems obvious enough but I can't seem to prove it. I'll keep working on it but maybe you guys can get it before me.

Prove this!

If a small circle B is inside a large circle A, and circle B is tangent to circle A at point C, prove the line drawn through point A to point B passes through the point C. (the point of tangency). Point A and point B are the centerpoints of circles A and B.

heres a drawing of how graeme did it. I take it he found the center coordinates of the small circle same as I did. Then just used the pythagorean theorem plus the radius.