Math Is Fun Forum

Thats what I thought irspower, but with a little care and a nifty observation, it can be solved in 2d.

Take a look at my pic. Its pretty much self explanatory.

The key to this problem is the slopes of the two cirlces are equal at the intersection. This gives us an extra equation that does not contain R. Then we have two equations for x and y that do not contain R. We can then solved for y in terms of x. Substitute this value of x in equation a to solve for x, then solve for y. Once we have the values of x and y we can use equation b to find the radius.

In the end I got the same answer Greame did, so I guess thats the correct answer.

The slope observation I think is an interesting one that I never made before. Its very usefull in finding the intersection of tangent circles and the size of tangent circles (as in this case). A circle of an unkown radius, with a given centerpoint, can intersect a given circle, at an infinite number of points. But there are only two points for which they intersect, and are tangent, the points of intersection where the slopes are equal. That gives us an extra equation to find the intersection. :-)

Neat! Thanks for the explanation, MathsIsFun! Fractals were mentioned in num3rs the other week but I'd never heard of them. Now I have!

It is an interesting problem. I think I could solve it but not in a formal way. (naming the postulates and axioms, etc.)

I really need to review geometry. Did algebra 2 so fast it was like a fleeting dream...

"What am I talking about? I failed at that so dramatically that I'm not even sure if it's possible." lol. Thats pretty cool you dropped out and taught yourself.

A lot of people who are serious about learning seem to prefer college to highschool. In college, kids are there because they chose to go there. In high school, they're there because they're required by law. Of course theres always the idiot who goes to college to party it up and all but they usually don't last long since there grades are bad.  At least thats what I heard.

"High school will force you to learn how some crock-brained educational theorist thinks you should learn."

Eek! Crock brained teachers? Sometimes I'm really glad I was homeschooled....

.....

Okay..lol. I come here asking you guys to help me do this problem, and I end up helping you do it.

In this problem it helps to substitute the constant (300/pi)^2 with C and the expression inside the radical with u.

Here we are:

Phi huh? Is that pronounced fie? (ryhmes with pi) or fee? (ryhmes with tea)

phi pi, pho phum!

I'd give it a √60 out of 10

Totally "rad"! ;-)

lol! Great joke!

Here we are! The solution to this nightmare of a problem is really quite simple, it just requires a little trick.

As you can see, the trick is to extend the beam out just enough till it touches the uper left corner and the required lengths and angles can be found as shown. Once you have these, the solution to the problem is easily attainable.

Well I"ve been thinking about it, but unless I'm mistaken, it appears that only produces another unknown dimension, so that doesn't help.

I shall reveal the solution to the problem momentarily.