User:Eznutella888
Hello fellow users of AOPS, my name is 
! As you can see I like math. That's why I'm here.
I have taken many math competitions, including the Canadian Gauss, Pascal and Cayley. I have also taken Canadian Intermediate Mathematics Examination, and the Math Challengers competition sponsored by the Canadian Math Challengers Society. I also have taken AMC 8, and this year I'm taking the AMC 10, as well as the COMC (Canadian Open Mathematics Challenge).
![[asy] size(6cm); pair D=(0,0), C=(6,0), B=(6,3), A=(0,3); draw(A--B--C--D--cycle); draw(B--D); draw(A--(6,2)); draw(A--(6,1)); label("$A$", A, dir(135)); label("$B$", B, dir(45)); label("$C$", C, dir(-45)); label("$D$", D, dir(-135)); label("$Q$", extension(A,(6,1),B,D),dir(-90)); label("$P$", extension(A,(6,2),B,D), dir(90)); label("$F$", (6,1), dir(0)); label("$E$", (6,2), dir(0)); [/asy]](http://latex.artofproblemsolving.com/1/5/8/15893c761d311ed13254fdea73b5a3079e88884f.png)
We can set coordinates for the points. 
. The line 
's equation is 
, 
's equation is 
, and line 
's equation is 
. Adding the equations of lines and , the coordinates of 
is 
. Furthermore the coordinates 
is 
. Using the Pythagorean Theorem, the length of 
is ![$\sqrt[2]{(\frac{18}{5})^2+(\frac{9}{5})^2} = \sqrt[2]{\frac{405}{25}} = \frac{\sqrt[2]{405}}{5} = \frac{9\sqrt{5}}{5}$](http://latex.artofproblemsolving.com/c/c/3/cc3e1f1b36d193fe4bd136c7dfdc7cf71de55274.png)
, and the length of 
= ![$\sqrt[2]{(\frac{9}{2})^2+(\frac{9}{4})^2} = \sqrt[2]{\frac{81}{4} + \frac{81}{16}} = \sqrt[2]{\frac{405}{16}} = \frac{\sqrt[2]{405}}{4} = \frac{9\sqrt{5}}{4}.$](http://latex.artofproblemsolving.com/7/a/5/7a561917144e396317efbb42361c9d084c71a89a.png)
PQ = DP - DQ = \frac{9\sqrt{5}}{5} - \frac{9\sqrt{5}}{4} = \frac{9\sqrt{5}}{20}. The length of ![$DB = \sqrt[2]{6^2 + 3^2} = \sqrt[2]{45} = 3\sqrt[2]{5}$](http://latex.artofproblemsolving.com/a/c/1/ac1e3f31a5568a55fb0db3c0b4022c1a18786c97.png)
. Then ![$BP= 3\sqrt[2]{5} - \frac{9\sqrt{5}}{4} = frac{3\sqrt{5}}{4}. Then the ratio$](http://latex.artofproblemsolving.com/e/f/9/ef93800c3791e9571e45f3d0ea89b76a235e4589.png)
BP : PQ : QD: = frac{3\sqrt{5}}{4} : \frac{9\sqrt{5}}{20} : \frac{9\sqrt{5}}{5} = 15\sqrt[2]{5} : 9\sqrt[2]{5} : 36\sqrt[2]{5} = 15 : 9 : 36 = 5 : 3 : 12. Then 
and 
is 
and 
, respectively. The problem tells us to find 
, so 
