# Rectangular WZYX inside the rectangular ACDB

The rectangular WZYX is inside the rectangular ACDB

AB= 8 cm AC= 6 cm WX= 8 cm

What is WZ= ?

Are there an infinite number of possible solutions? we can Imagine sliding point X toward A or B. We can always rotate line XW around X so that W stays on AC. Each angle gives a different solution.

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It may not be possible to keep both $Y$ and $Z$ on the edges of $ABCD$ when you move $X$ around. – Rahul Aug 24 '12 at 23:57
@RahulNarain Sorry, I deleted it because I understood what you were trying to say. I see that only one rectangle is possible where XB is not 0. – mathguy Aug 25 '12 at 2:09

One "algebraically" correct solution is explained by svenkatr, but I found a different solution that doesn't involve degenerate lines.

Please look at the following image for reference.

Haha, sorry it is very poorly drawn. As you can see, $XB$ is $n$, $BY$ is $m$, and $XY$ is $w$.

Now using the Pythagorean theorem, we know that:
$n^2 + m^2 = w^2$ and
$(8-n)^2+(6-m)^2=64$.

Method 1 (the first one I thought of...)

Now please refer to my second picture:

Pretend the lower, left corner of the rectangle is the origin. Now the line with the negative slope contains the points $(0,m)$ and $(n,0)$. This means that our slope is $\frac{m-0}{0-n} = \frac{-m}{n}$. That means that the slope of the perpendicular line is $\frac{n}{m}$. Since the positive slope line intersects $x=0$ at $(0,m)$, then the equation of that line is $y=\frac{n}{m}x+m$. Note that a point on that line is $(8-n,m)$, so let us plug that into the equation and get: $6 = \frac{n(8-n)}{m} + m$

And there you have your three equations:
$n^2 + m^2 = w^2$
$(8-n)^2+(6-m)^2=64$
$6 = \frac{n(8-n)}{m} + m$

My assumption is that you only need help with the algebra, so I let wolfram handle the arithmetic, and I got this result.

We now know that $m=1.89573, n=1.13305$ and most importantly, your answer $w=WZ=2.20853$ =)

Method 2 (probably the fastest)

Triangles AWX and $XBY$ are similar due to the ASA property. Therefore, the sides are proportional, and $\frac{XB}{BY}=\frac{AW}{AX}$. That was the key to this problem.

Method 3 (suggested by svenkatr)

The third equation can involve the sum of the areas. $nm + (8-n)(6-m) + 8z = 6*8 = 48$
http://www.wolframalpha.com/input/?i=n%5E2%2Bm%5E2+%3D+w%5E2%3B+%288-n%29%5E2%2B%286-m%29%5E2%3D64%3B+mn%2B%288-n%29%286-m%29+%2B+8w+%3D+48

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Let $BX = CZ = x$ and let $BY=WC=y$. Let $WZ = XY = z$. Then, we can write the following equations

1) Pythagorean theorem for $\Delta XBY$ gives $x^2 +y^2 = z^2$.

2) Pythagorean theorem for $\Delta WAX$ gives $(8-x)^2 +(6-y^2) = 64$.

3) Area of bigger rectangle = sum of smaller areas gives $xy + (8-x)(6-y)+8z = 48$.

You have 3 equations in 3 unknowns. The quadratic terms might give multiple solutions but you cannot have infinitely many solutions.

In fact, with $x=0$ and $y=z=6$ you get a trivial solution. Wolfram alpha gives 3 other solutions here http://www.wolframalpha.com/input/?i=solve+x%5E2%2By%5E2+%3D+z%5E2%3B+%288-x%29%5E2%2B%286-y%29%5E2%3D64%3B+2xy-6x-8y%2B8z+%3D+0. You know which solutions to discard.

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Is it possible at all for there to be just one more solution? – mathguy Aug 25 '12 at 0:20
@Sidd - if there is one more solution I'm not sure how to find it. – svenkatr Aug 25 '12 at 0:22
Sorry, I only say so because I made a to-scale picture on Microsoft word with the given dimensions, and I got something that looks like it fits, but then again how reliable can that method be...haha – mathguy Aug 25 '12 at 0:24
I feel that there is only one feasible solution because you are constraining the length of WX, which prevents you from rotating and scaling the inner rectangle freely. – svenkatr Aug 25 '12 at 0:25
please look at my answer, just to show you another way to approach this problem for future reference. – mathguy Aug 25 '12 at 2:40