Difference between revisions of "User:Eznutella888"

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Hello fellow users of AOPS, my name is <math>\mathbb{EZNUTELLA} \text{888}</math>! 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>
 
 
We can set coordinates for the points. <math>D=(0,0), C=(6,0),  B=(6,3),</math> and <math>A=(0,3)</math>. The line <math>BD</math>'s equation is <math>y = \frac{1}{2}x</math>, line <math>AE</math>'s equation is <math>y = -\frac{1}{6}x + 3</math>, and line <math>AF</math>'s equation is <math>y = -\frac{1}{3}x + 3</math>. Adding the equations of lines <math>BD</math> and <math>AE</math>, the coordinates of <math>P</math> is <math>(\frac{9}{2},\frac{9}{4})</math>. Furthermore the coordinates <math>Q</math> is <math>(\frac{18}{5}, \frac{9}{5})</math>. Using the Pythagorean Theorem, the length of <math>QD</math> is <math>\sqrt{(\frac{18}{5})^2+(\frac{9}{5})^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}</math>, and the length of <math>DP</math> = <math>\sqrt{(\frac{9}{2})^2+(\frac{9}{4})^2} = \sqrt{\frac{81}{4} + \frac{81}{16}} = \sqrt{\frac{405}{16}} = \frac{\sqrt{405}}{4} = \frac{9\sqrt{5}}{4}.</math> <math>PQ = DP - DQ = \frac{9\sqrt{5}}{5} - \frac{9\sqrt{5}}{4} = \frac{9\sqrt{5}}{20}.</math> The length of <math>DB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}</math>. Then <math>BP= 3\sqrt{5} - \frac{9\sqrt{5}}{4} = \frac{3\sqrt{5}}{4}.</math> Then the ratio <math>BP : PQ : QD = \frac{3\sqrt{5}}{4} : \frac{9\sqrt{5}}{20} : \frac{9\sqrt{5}}{5} = 15\sqrt{5} : 9\sqrt{5} : 36\sqrt{5} = 15 : 9 : 36 = 5 : 3 : 12.</math> Then <math>r, s,</math> and <math>t</math> is <math>5, 3,</math> and <math>12</math>, respectively. The problem tells us to find <math>r + s + t</math>, so <math>5 + 3 + 12 = \boxed{\textbf{(E) }20}</math>
 

Latest revision as of 23:10, 13 January 2017

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