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Point $X$ is on $\overline{AC}$ such that $AX = 4\cdot CX = 12$. We know $\angle ABC = \angle BXA = 90^\circ.$ What is $BX$? [asy] size(5cm); pair A,B,C,X; A = (0,0); X = (12,0); C = (16,0); B = (12,sqrt(48)); draw(X--B--A--C--B); draw(rightanglemark(A,B,C,20)); draw(rightanglemark(B,X,A,20)); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$X$",X,S); label("$12$",(A+X)/2,S); [/asy]

 Nov 1, 2019
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By similar triangles, the answer is 6 sqrt(2).

 Nov 1, 2019

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