Right ABC has AB=3, BC=4, and AC=5. Square XYZW is inscribed in ABC with X and Y on AC, W on AB, and Z on BC. What is the side length of the square?

Using similar triangles twice, and  calling the length of the side of the square x,

in the triangle CYZ,  CY/x = 4/3, so CY = 4x/3,

in the triangle WXA,  x/XA = 3/4, so XA = 3x/4.

The hypotenuse of the triangle is

CY + YX + XA  = 4x/3 + x + 3x/4 = 37x/12 = 5,

so x = 60/37.

There may be other methods of solving this one.....but here's my attempt.....

Locate  point D  on BC and let it lie x units from the origin......and let a be the side of the square

Locate point E on  AC  and  F on AB  such that  DE = DF  = a   and DE,DF are perpendicular to each other

Now angle ACB  = angle FDB = atan(3/4)

 So, cos [atan(3/4)] = DB/ FD = x / a    →  x = ( a)cos[ atan(3/4)]

And sin ACB = sin [atan (3/4)]

So, sin [atan (3/4) ]  = a / [ 4 - x]

And substituting for x, we have

sin [atan(3/4)] = a / [ 4 - a* cos [atan(3/4)]]

And with a little help from WolframAlpha,  a = 60/37.........and this is the side  of the square

And x = BD = (60/37)cos[atan(3/4)] = 48/37

Here's the [ approximate] pic :

Right ABC has AB=3, BC=4, and AC=5. Square XYZW is inscribed in ABC with X and Y on AC, W on AB, and Z on BC. What is the side length of the square?

\(\small{ \begin{array}{lcl} XW = WZ = x \qquad u = ZB \qquad v = WB\\ \end{array}\\ \begin{array}{lrcl} (1) & \frac{v}{x} &=& \frac{3}{5} \quad \rightarrow \quad v = \frac{3}{5} x\\ (2) & \frac{u}{x} &=& \frac{4}{5} \quad \rightarrow \quad u = \frac{4}{5} x\\\\ & \sin{(A)} &=& \frac{x}{3-v} \\ & \sin{(B)} = \sin{(90^{\circ}-A)} = \cos{(A)} &=& \frac{x}{4-u} \\ (3) & [\sin{(A)}]^2 + [\cos{(A)}]^2 =1 &=& \left(\frac{x}{3-v}\right)^2 + \left(\frac{x}{4-u}\right)^2 \\\\ \\ \hline \\ &\left( \dfrac{x}{3-v} \right)^2 + \left(\dfrac{x}{4-u} \right)^2 &=& 1 \\\\ & \dfrac{x^2}{ (3-v)^2 } + \dfrac{x^2}{ (4-u)^2 } &=& 1 \qquad v = \frac{3}{5} x \qquad u = \frac{4}{5} x\\\\ & \dfrac{x^2}{ \left(3-\dfrac{3}{5} x \right)^2 } + \dfrac{x^2}{ \left(4-\dfrac{4}{5} x \right)^2 } &=& 1 \\\\ & \dfrac{x^2}{ 3^2 \left(1-\dfrac{1}{5} x \right)^2 } + \dfrac{x^2}{ 4^2\left(1-\dfrac{1}{5} x \right)^2 } &=& 1 \\\\ & \dfrac{x^2}{ 3^2 } + \dfrac{x^2}{ 4^2 } &=& \left(1-\dfrac{1}{5} x \right)^2 \\\\ & \dfrac{x^2}{ 3^2 } + \dfrac{x^2}{ 4^2 } &=& \dfrac{ (5-x)^2 }{5^2}\\\\ & x^2 \left(\dfrac{1} { 3^2 }+ \dfrac{1}{ 4^2 } \right) &=& \dfrac{ (5-x)^2 }{5^2}\\\\ & x^2 \left(\dfrac{4^2+3^2} { 3^24^2 } \right) &=& \dfrac{ (5-x)^2 }{5^2} \qquad | \quad 3^2+4^2=5^2\\\\ & x^2 \left(\dfrac{5^2} { 3^24^2 } \right) &=& \dfrac{ (5-x)^2 }{5^2} \qquad | \quad 5^25^2=5^4\\\\ & x^2 \left( \dfrac{5^4} { 3^24^2 } \right) &=& (5-x)^2 \\\\ & x^2 \left( \dfrac{5^4} { 3^24^2 } \right) &=& 5^2-2\cdot 5 x + x^2 \\\\ & x^2 \left( \dfrac{5^4} { 3^24^2 } \right) -x^2 &=& 5^2-2\cdot 5 x \\\\ & x^2 \left( \dfrac{5^4} { 3^24^2 } -1 \right) &=& 5^2-2\cdot 5 x \\\\ & x^2 \left( \dfrac{5^4-3^24^2} { 3^24^2 } \right) &=& 5^2-2\cdot 5 x \\\\ & \mathbf{ x^2 \left( \dfrac{5^4-3^24^2} { 3^24^2 } \right) +2\cdot 5 x - 5^2 } & \mathbf{=}& \mathbf{ 0 }\\\\ \\ \hline \\ & x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \quad a= \frac{5^4-3^24^2} { 3^24^2 } \quad b=2\cdot 5 \quad c=-5^2 \\\\ & x^2 \left( \frac{5^4-3^24^2} { 3^24^2 } \right) +2\cdot 5 x - 5^2 &=& 0 \\\\ & x_{1,2} &=& {-2\cdot 5 \pm \sqrt{(-2\cdot 5)^2 - 4\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right) \cdot(-5^2) } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm \sqrt{2^25^2 + 2^25^2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right) } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm 2\cdot 5 \sqrt{1 + \left( \frac{5^4-3^24^2} { 3^24^2 }\right) } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm 2\cdot 5 \sqrt{\frac{3^24^2+ 5^4-3^24^2} { 3^24^2 } } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm 2\cdot 5 \sqrt{\frac{5^4} { 3^24^2 } } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm 2\cdot 5 \left( \frac{5^2} { 3\cdot 4 } \right) \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-10 \pm \frac{250} { 12 } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-10 \pm \frac{250} { 12 } \over \frac{962} { 144 } } \\\\ & x &=& {-10 + \frac{250} { 12 } \over \frac{962} { 144 } } \\\\ & x &=& {\frac{130} { 12 } \over \frac{962} { 144 } } \\\\ & x &=& \left( \frac{130} { 12 } \right) \cdot \left( \frac{ 144 }{962} \right) \\\\ & x &=& \left( \frac{130} { 12 } \right) \cdot \left( \frac{ 12^2 }{962} \right) \\\\ & x &=& \frac{130\cdot 12} { 962 } \\\\ & x &=& \frac{1560} { 962 } \\\\ & x &=& \frac{780} { 481 } \\\\ & x &=& \frac{60} { 37 } \\\\ & \mathbf{x} &\mathbf{=}& \mathbf{ 1.621\overline{621} } \end{array}\\ \text{The side length of the square is } \mathbf{1.621\overline{621}} }\)

"Right ABC has AB=3, BC=4, and AC=5. Square XYZW is inscribed in ABC with X and Y on AC, W on AB, and Z on BC. What is the side length of the square?"

Using similar triangles twice, and  calling the length of the side of the square x,

in the triangle CYZ,  CY/x = 4/3, so CY = 4x/3,

in the triangle WXA,  x/XA = 3/4, so XA = 3x/4.

The hypotenuse of the triangle is

CY + YX + XA  = 4x/3 + x + 3x/4 = 37x/12 = 5,

so x = 60/37.

Irritating, Alan's solution appeared during the time I was typing mine in.