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

**Variant** · November 18th 2007, 01:01 AM

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)

**Jhevon** · November 18th 2007, 01:18 AM

Originally Posted by Variant

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)

expand what's in the brackets and simplify to get a quadratic.

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

$\Rightarrow r^2 - 4r + 4 + r^2 - 8r + 16 = r^2$

$\Rightarrow r^2 - 12r + 20 = 0$

now continue

**bobak** · November 18th 2007, 01:31 AM

Do you know how to expand a $(a+b)^2$ ?
$(a+b)^2 = a^2 +2ab + b^2$

so $(r - 2)^2 = r^2 - 4r + 4$
and $(r - 4)^2 = r^2 - 8r + 16$

so your equation becomes $r^2 - 4r + 4 + r^2 - 8r + 16 = r^2$

then you collect like terms to get
$r^2-12r+20 = 0$
now you factor
$(r-10)(r-2) = 0$

and just you finished it off from there

edit: Jhevon beat me

**Variant** · November 18th 2007, 01:57 AM

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.

**angel.white** · November 18th 2007, 02:30 AM

Originally Posted by Variant

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.

1. Initial equation
$(r - 2)^{2} + (r - 4)^{2} = r^{2}$

2. Because $a^{2}=a*a$ (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:
$(r - 2)(r-2) + (r - 4)(r-4) = r^2$

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

So $(r-2)(r-2)=r*r+(-2)r+(-2)r+(-2)(-2)$ which can be simplified to $r*r-2r-2r+4$ which can again be simplified to $r^{2}-4r+4$

and using the same technique: $(r-4)(r-4)= r^{2}-8r+16$

4. So applying what we just went over:
$r^{2} -4r+4 +r^{2}-8r+16 = r^2$

5. combine like terms such as $r^{2}+r^{2}=2r^{2}$ and so on
$2r^{2} -12r+20= r^2$

6. Subtract $r^{2}$ 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)
$r^{2} -12r+20=0$

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.

**Variant** · November 18th 2007, 02:49 AM

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.

**JaneBennet** · December 17th 2007, 07:48 AM

Here is a quick way to do it. Let R = r − 2:

$\mbox{Then }(r-2)^2+(r-4)^2\ =\ r^2$

$\Rightarrow\ R^2+(R-2)^2 =\ (R+2)^2$

$\begin{array}{rcl} \Rightarrow\ R^2 &=& (R+2)^2-(R-2)^2\\\\ {} &=& [(R+2)+(R-2)][(R+2)-(R-2)]\\\\ {} &=& 2R\cdot4\ =\ 8R \end{array}$

$\Rightarrow\ R^2-8R\ =\ 0$

$\Rightarrow\ R(R-8)\ =\ 0$

$\Rightarrow\ R =\ 0\ \mbox{or }R\ =\ 8$

Substitute back and you’ll get r = 2 or r = 10.

**angel.white** · December 17th 2007, 07:54 AM

Originally Posted by JaneBennet

Here is a quick way to do it. Let R = r − 2:

$\mbox{Then }(r-2)^2+(r-4)^2\ =\ r^2$

$\Rightarrow\ R^2+(R-2)^2 =\ (R+2)^2$

$\begin{array}{rcl} \Rightarrow\ R^2 &=& (R+2)^2-(R-2)^2\\\\ {} &=& [(R+2)+(R-2)][(R+2)-(R-2)]\\\\ {} &=& 2R\cdot4\ =\ 8R \end{array}$

$\Rightarrow\ R^2-8R\ =\ 0$

$\Rightarrow\ R(R-8)\ =\ 0$

$\Rightarrow\ R =\ 0\mbox{ or }R\ =\ 8$

Substitute back and you’ll get r = 2 or r = 10.

That is a cute method ^_^

daddy · January 21st 2008, 01:40 PM

Originally Posted by JaneBennet

Here is a quick way to do it. Let R = r − 2:

$\mbox{Then }(r-2)^2+(r-4)^2\ =\ r^2$

$\Rightarrow\ R^2+(R-2)^2 =\ (R+2)^2$

$\begin{array}{rcl} \Rightarrow\ R^2 &=& (R+2)^2-(R-2)^2\\\\ {} &=& [(R+2)+(R-2)][(R+2)-(R-2)]\\\\ {} &=& 2R\cdot4\ =\ 8R \end{array}$

$\Rightarrow\ R^2-8R\ =\ 0$

$\Rightarrow\ R(R-8)\ =\ 0$

$\Rightarrow\ R =\ 0\ \mbox{or }R\ =\ 8$

Substitute back and you’ll get r = 2 or r = 10.

this won't be neat once you try to solve other similar problems however.
hence this seems to be a "one trick pony" if you know what i mean.

**angel.white** · January 21st 2008, 06:24 PM

Originally Posted by daddy

this seems to be a "one trick pony" if you know what i mean.

Yeah, I know what you mean, my ex-wife was the same way.

**Jhevon** · January 21st 2008, 06:53 PM

Originally Posted by daddy

this won't be neat once you try to solve other similar problems however.
hence this seems to be a "one trick pony" if you know what i mean.

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

Originally Posted by angel.white

Yeah, I know what you mean, my ex-wife was the same way.

as in a "one trick pony"? (that's not a nice thing to call your ex-wife

)...what do you mean?

**JaneBennet** · January 22nd 2008, 10:11 PM

Originally Posted by daddy

this won't be neat once you try to solve other similar problems however.
hence this seems to be a "one trick pony" if you know what i mean.

The reason for the substitution is that the equation $(r-2)^2+(r-4)^2\ =\ r^2$ is symmetrical about $r-2$ . For other similar problems, you make similar substitutions involving whatever the equation is symmetrical about. For example …

$\color{white}.\quad.$ Solve $(x+123)^2+(x+456)^2=(x+789)^2$ .

Your method
$\color{white}.\quad.$ $(x+123)^2+(x+456)^2=(x+789)^2$

$\Rightarrow\ x^2+246x+15129+x^2+912x+207936=x^2+1578x+622521$

$\Rightarrow\ x^2+\ldots$ ???

<screams in terror and gives up>

My method
The equation is symmetrical about $x+456$ so let $X=x+456$ . Then

$\color{white}.\quad.$ $(x+123)^2+(x+456)^2=(x+789)^2$

$\Rightarrow\ (X-333)^2+X^2=(X+333)^2$

$\Rightarrow\ X^2-666X+333^2+X^2=X^2+666X+333^2$

$\Rightarrow\ X^2-1332X=0$

$\Rightarrow\ X=0$ or $X=1332$

$\Rightarrow\ x=-456$ or $x=876$

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.

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