Difference between revisions of "User:Eznutella888"

Revision as of 22:54, 31 October 2016

Hello fellow users of AOPS, my name is $\mathbb{EZNUTELLA} \text{888}$ .

I have taken many math competitions, including the Gauss, Pascal and Cayley by the Centre for Education in Mathematics of Computing, University of Waterloo. 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, 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. $D=(0,0), C=(6,0), B=(6,3),$ and $A=(0,3)$ . The line $BD$ 's equation is $y = \frac{1}{2}x$ , line $AE$ '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 $AE$ , we find that the coordinates of $P$ is $(\frac{9}{2},\frac{9}{4})$ . Furthermore we find that the coordinates $Q$ is $(\frac{18}{5}, \frac{9}{5})$ . Using the Pythagorean Theorem, the length of $QD$ is $\sqrt{(\frac{18}{5})^2+(\frac{9}{5})^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}$ , and the length of $DP$ = $\sqrt{(\frac{9}{2})^2+(\frac{9}{4})^2} = \sqrt{\frac{81}{4} + \frac{81}{16}} = \sqrt{\frac{405}{16}} = \frac{\sqrt{405}}{4} = \frac{9\sqrt{5}}{4}.$ $PQ = DP - DQ = \frac{9\sqrt{5}}{5} - \frac{9\sqrt{5}}{4} = \frac{9\sqrt{5}}{20}.$ The length of $DB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}$ . Then $BP= 3\sqrt{5} - \frac{9\sqrt{5}}{4} = \frac{3\sqrt{5}}{4}.$ Then the ratio $BP : PQ : QD = \frac{3\sqrt{5}}{4} : \frac{9\sqrt{5}}{20} : \frac{9\sqrt{5}}{5} = 15\sqrt{5} : 9\sqrt{5} : 36\sqrt{5} = 15 : 9 : 36 = 5 : 3 : 12.$ Then $r, s,$ and $t$ is $5, 3,$ and $12$ , respectively. The problem tells us to find $r + s + t$ , so $5 + 3 + 12 = \boxed{\textbf{(E) }20}$ $\boxed{\textbf{(B)}}$

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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.
+Hello fellow users of AOPS, my name is <math>\mathbb{EZNUTELLA} \text{888}</math>.
-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).
+I have taken many math competitions, including the Gauss, Pascal and Cayley by the Centre for Education in Mathematics of Computing, University of Waterloo. 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, the AMC 10, as well as the COMC (Canadian Open Mathematics Challenge).
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Difference between revisions of "User:Eznutella888"

Revision as of 22:54, 31 October 2016