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Points $A$, $B$, and $C$ are on sides $\overline{WX}$, $\overline{YZ}$, and $\overline{XY}$ of rectangle $WXYZ$ as shown below such that $XA = 4$, $YB = 18$, $\angle ACB= 90^\circ$, and $YC = 2XC$. Find $AB$. [asy] pair A,B,C,D,P,Q,R; A = (0,0); P = (0,0.7); B = (0,1); R = (0.4,1); C = (1,1); Q = (1,0.2); D = (1,0); draw(P--B--C--D--A--P--R--Q--P); draw(rightanglemark(P,R,Q,2)); label("$W$", A, SW); label("$Z$",D,SE); label("$X$",B,NW); label("$Y$",C,NE); label("$C$",R,N); label("$A$",P,W); label("$B$",Q,E); label("$4$", (B+P)/2,W); label("$18$", (C+Q)/2,E); [/asy]

 

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 Feb 16, 2020
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Note that triangles XCA and CYB are similar.  Since YC = 2XC, the ratio of similarity is 2, which means that XCA and CYB are 30-60-90 triangles.  Then CA = 2*4 = 8 and BC = 18*2/sqrt(3) = 12*sqrt(3), so by the Pythagorean Theorem, AB = sqrt(8^2 + (12*sqrt(3))^2) = 4*sqrt(31).

 Feb 16, 2020

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