Difference between revisions of "2016 AMC 10A Problems/Problem 19"
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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>, we find that the coordinates of <math>P</math> is <math>(\frac{9}{2},\frac{9}{4})</math>. Furthermore we find that 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> 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> | 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>, we find that the coordinates of <math>P</math> is <math>(\frac{9}{2},\frac{9}{4})</math>. Furthermore we find that 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> 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> | ||
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+ | ==Solution 3== | ||
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+ | Extend <math>AF</math> to meet <math>CD</math> at point <math>T</math>. Since <math>FC=1</math> and <math>BF=2</math>, <math>TC=3</math> by similar triangles <math>\triangle TFC</math> and <math>\triangle AFB</math>. It follows that <math>\frac{BQ}{QD}=\frac{BP+PQ}{QD}=\frac{2}{3}</math>. Now, using similar triangles <math>\triangle BEP</math> and <math>\triangle DAP</math>, <math>\frac{BP}{PD}=\frac{BP}{PQ+QD}=\frac{1}{3}</math>. WLOG let <math>BP=1</math>. Solving for <math>PQ, QD</math> gives <math>PQ=\frac{3}{5}</math> and <math>QD=\frac{12}{5}</math>. So our desired ratio is <math>5:3:12</math> and <math>5+3+12=\boxed{\textbf{(E) } 20}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}} | {{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:47, 2 February 2017
Contents
[hide]Problem
In rectangle and . Point between and , and point between and are such that . Segments and intersect at and , respectively. The ratio can be written as where the greatest common factor of and is What is ?
Solution 1
Since Similarly, . This means that . As and are similar, we see that . Thus . Therefore, so
Solution 2
Coordinate Bash: We can set coordinates for the points. and . The line 's equation is , line 's equation is , and line 's equation is . Adding the equations of lines and , we find that the coordinates of is . Furthermore we find that the coordinates is . Using the Pythagorean Theorem, the length of is , and the length of = The length of . Then The ratio Then and is and , respectively. The problem tells us to find , so
Solution 3
Extend to meet at point . Since and , by similar triangles and . It follows that . Now, using similar triangles and , . WLOG let . Solving for gives and . So our desired ratio is and .
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AMC 10 Problems and Solutions |
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