# Difference between revisions of "User:Eznutella888"

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label("$E$", (6,2), dir(0)); | label("$E$", (6,2), dir(0)); | ||

</asy> | </asy> | ||

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+ | 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 ! 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).

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 = PQ = DP - DQ = \frac{9\sqrt{5}}{5} - \frac{9\sqrt{5}}{4} = \frac{9\sqrt{5}}{20}. The length of . Then 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