Difference between revisions of "User:Eznutella888"

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label("$E$", (6,2), dir(0));
 
label("$E$", (6,2), dir(0));
 
</asy>
 
</asy>
 +
 +
We can set coordinates for the points. <math>D=(0,0), C=(6,0),  B=(6,3), A=(0,3)</math>. The line <math>BD</math>'s equation is <math>y = \frac{1}{2}</math>, <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[2]{(\frac{18}{5})^2+(\frac{9}{5})^2} = \sqrt[2]{\frac{405}{25}} = \frac{\sqrt[2]{405}}{5} = \frac{9\sqrt{5}}{5}</math>, and the length of <math>DP</math> = <math>\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}. </math>PQ = DP - DQ = \frac{9\sqrt{5}}{5} - \frac{9\sqrt{5}}{4} = \frac{9\sqrt{5}}{20}. The length of <math>DB = \sqrt[2]{6^2 + 3^2} = \sqrt[2]{45} = 3\sqrt[2]{5}</math>. Then <math>BP= 3\sqrt[2]{5} - \frac{9\sqrt{5}}{4} = frac{3\sqrt{5}}{4}. Then the ratio </math>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 <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>

Revision as of 23:33, 5 February 2016

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

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