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 
, and the length of 
= 
DB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}
BP= 3\sqrt{5} - \frac{9\sqrt{5}}{4} = frac{3\sqrt{5}}{4}. Then the ratio 
r, s,
t
5, 3,12
r + s + t
5 + 3 + 12 = \boxed{\textbf{(E) }20}$
