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

1. ## (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)

2. 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.

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

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

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

now continue

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

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

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

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

and just you finished it off from there

edit: Jhevon beat me

4. 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.

5. 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
$\displaystyle (r - 2)^{2} + (r - 4)^{2} = r^{2}$

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

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

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

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

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

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

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

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.

6. 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.

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

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

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

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

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

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

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

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

8. Originally Posted by JaneBennet
Here is a quick way to do it. Let R = r − 2:

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

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

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

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

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

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

Substitute back and you’ll get r = 2 or r = 10.
That is a cute method ^_^

9. Originally Posted by JaneBennet
Here is a quick way to do it. Let R = r − 2:

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

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

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

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

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

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

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.

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.

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?

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

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

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

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

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

<screams in terror and gives up>

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

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

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

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

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

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

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

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.