In rectangle $WXYZ$, $A$ is on side $\overline{WX}$, $B$ is on side $\overline{YZ}$, and $C$ is on side $\overline{XY}$. If $AX = 2$, $BY = 24$, $\angle ACB= 90^\circ$, and $CY = 3 \cdot CX$, then find $AB$.
[asy] pointpen = black; pathpen = black + linewidth(0.7); pair W = (0,0), X = (2,0), Y = (2,5), Z = (0,5), C = relpoint(Z--X,5), B = intersectionpoint(Y--Y+10*(rotate(90)*(Z-X)),Z--C), A = intersectionpoint(W--W+10*(rotate(90)*(X-Y)),Z--C), D = foot(C,A,B); D(MP("W",W)--MP("X",X)--MP("Y",Y,N)--MP("Z",Z,N)--cycle); D(C--MP("C",C,NE)); D((0,1.7)--MP("A",A,W)--X); D(B--MP("B",B,N)); MP("2",(0,1),W); MP("3",(0,1.3),W); MP("24",WAYNE,NE); D(rightanglemark(Y,C,X,9)); [/asy]
Let $CX=y$. Since $\triangle ACX\sim\triangle DCY$, we have $\frac y{CD}=\frac 23$, so $CD=\frac32y$ and $XY=\frac52y$. (We can also find $CD$ by noting that $[ACD]+[BCD]=12$, so $\frac32y\cdot\frac52y = 12$.) Applying the Pythagorean Theorem on $\triangle XCD$, we have $2^2+\left(\frac32y\right)^2=\left(\frac52y\right)^2$:
\begin{align*}4+(\tfrac 34)^2 y^2&=\left(\tfrac 54\right)^2 y^2\\ 16&=9y^2\\ y^2&=\frac{16}{9}\\ y&=\frac 43. \end{align*}
Thus, $XY = AC = \frac{5}{2}\cdot \frac 43 = \frac{20}{3}$, so $AB=\boxed{026}$.
what the
since when did point d exist
Okay, let's clarify.
Draw a perpendicular from $C$ to $\overline{AB}$, and let that point be $D$.
[asy] pointpen = black; pathpen = black + linewidth(0.7); pair W = (0,0), X = (2,0), Y = (2,5), Z = (0,5), C = relpoint(Z--X,5), B = intersectionpoint(Y--Y+10*(rotate(90)*(Z-X)),Z--C), A = intersectionpoint(W--W+10*(rotate(90)*(X-Y)),Z--C), D = foot(C,A,B); D(MP("W",W)--MP("X",X)--MP("Y",Y,N)--MP("Z",Z,N)--cycle); D(C--MP("C",C,NE)); D((0,1.7)--MP("A",A,W)--X); D(B--MP("B",B,N)); MP("2",(0,1),W); MP("3",(0,1.3),W); MP("24",WAYNE,NE); D(rightanglemark(Y,C,X,9)); D(rightanglemark(C,D,A,9)); MP("D",D,S); [/asy]
$AX = 2$, $CY = 3CX$, then $XY = 2 \cdot \frac{3}{2}CX = 3CX$
Sidenote: In general, it is better not to assign lengths on the diagram, because they sometimes lead to confusion.
Applying the Pythagorean Theorem to $\triangle XCD$, $(CX)^2 + (CD)^2 = (XY)^2$
$(CX)^2 + (3CX)^2 = (3CX)^2$ (Notice that $CD=3CX$)
$1 + 3^2 = 3^2$
The equation is satisfied. We have $XY = 3CX = \frac{20}{3}$.
From here, you can proceed with $\triangle ACB$.
We have $\triangle ACB$ with $AC = \frac{20}{3}$ and $CB = 24$.
Applying Pythagorean Theorem, $AB^2 + (\frac{20}{3})^2 = 24^2$
$AB^2 = 24^2 -(\frac{20}{3})^2 = 576-\frac{400}{9}$ = $\frac{576 \cdot 9}{9}-\frac{400}{9} = \frac{5184-400}{9}=\frac{4784}{9}$
$AB = \frac{\sqrt{4784}}{\sqrt{9}} = \frac{4}{3}\sqrt{299}$. Multiply numerator and denominator by 3 to get $\boxed{[026]}$