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- Nov 18th 2007, 02:01 AM #1

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## (r - 2)^2 + (r - 4)^2 = r^2?

How do I find the value of r?

(r - 2)^2 + (r - 4)^2 = r^2

I know what r is, coincidentally, (it's 10)

But how would I go about finding it if I didn't know?

Wasn't sure where to post this, it's not homework, I got it from a brain teaser (finding the radius of a circle which is inside a square with a 2x4 rectangle in the corner, the radius is r as seen above)

- Nov 18th 2007, 02:18 AM #2

- Nov 18th 2007, 02:31 AM #3

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- Nov 18th 2007, 02:57 AM #4

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Ahh yes. I didn't quite understand at first, how to arrive at

is what trips me up I think. I can grasp everything you just wrote, I was able to test it... but I just don't get my head around the workings of it heh.

I think I'm a lot closer to understanding it than I was before though.

Mind you, I never went much past basic math and some algebra in school, I just do these kinds of things for fun and learning.

I also think my understanding is "contaminated" by knowing the value of r.

What you guys have shown me is useful information though for learning how to do this.

- Nov 18th 2007, 03:30 AM #5
**1.**Initial equation

**2.**Because (when you have parentheses up against eachother as we will, it is understood that they are to be multiplied) you can rewrite the initial equation like this:

**3.**Using the technique called FOIL: first, outer, inner, last. Which basically boils down to (a+b)(c+d)=ac+ad+bc+bd.

So which can be simplified to which can again be simplified to

and using the same technique:

**4.**So applying what we just went over:

**5.**combine like terms such as and so on

**6.**Subtract from both sides (you must do it to both sides, because if you only did it to one side they would no longer be equal. For example, 5=5, if you subtract 1 from either side, you must subtract it from the other, so 4=4, if you do not do it to both sides, they will no longer be equal)

That is how you get to this step.

edit: I can break it down a little more if you are really struggling with any given step here, let me know.

- Nov 18th 2007, 03:49 AM #6

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Oh wow, I actually got it!! Thank you!

I was thinking I was a bit dumb there for a while but I actually understand it now.

Haha, I went back to the same brain teaser, and it happened to have different values this time but I was actually able to figure it out.

- Dec 17th 2007, 08:48 AM #7

- Dec 17th 2007, 08:54 AM #8

- Jan 21st 2008, 02:40 PM #9daddyGuest

- Jan 21st 2008, 07:24 PM #10

- Jan 21st 2008, 07:53 PM #11
it is a nice enough method to note. as there are a lot of problems that it

*would*work for. the technique of substitution is used all over math, so i wouldn't file this approach under the "one trick pony" file.

in this case however, it didn't make life much simpler, so i see no harm in sticking with the straight forward approach...not that Jane was saying anything different

as in a "one trick pony"? (that's not a nice thing to call your ex-wife )...what do you mean?

- Jan 22nd 2008, 11:11 PM #12
The reason for the substitution is that the equation is symmetrical about . For other similar problems, you make similar substitutions involving whatever the equation is symmetrical about. For example …

Solve .

**Your method**

???

<screams in terror and gives up>

**My method**

The equation is symmetrical about so let . Then

or

or

You see the difference between your method and mine? Therefore my method is not a one-trick pony! Next time think before making another similar comment.