Difference between revisions of "2016 AMC 10A Problems/Problem 19"
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==Solution 2(Coordinate Bash)== | ==Solution 2(Coordinate Bash)== | ||
− | 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> are <math>\left(\frac{9}{2},\frac{9}{4}\right)</math>. Furthermore we find that the coordinates of <math>Q</math> are <math>\left(\frac{18}{5}, \frac{9}{5}\right)</math>. Using the [[Pythagorean Theorem]], we get that the length of <math>QD</math> is <math>\sqrt{\left(\frac{18}{5}\right)^2+\left(\frac{9}{5}\right)^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}</math>, and the length of <math>DP</math> is <math>\sqrt{\left(\frac{9}{2}\right)^2+\left(\frac{9}{4}\right)^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}}{4} - \frac{9\sqrt{5}}{5} = \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> are <math>\left(\frac{9}{2},\frac{9}{4}\right)</math>. Furthermore we find that the coordinates of <math>Q</math> are <math>\left(\frac{18}{5}, \frac{9}{5}\right)</math>. Using the [[Pythagorean Theorem]], we get that the length of <math>QD</math> is <math>\sqrt{\left(\frac{18}{5}\right)^2+\left(\frac{9}{5}\right)^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}</math>, and the length of <math>DP</math> is <math>\sqrt{\left(\frac{9}{2}\right)^2+\left(\frac{9}{4}\right)^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}}{4} - \frac{9\sqrt{5}}{5} = \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> ~ minor LaTeX edits by dolphin7 |
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==Solution 3== | ==Solution 3== | ||
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Draw line segment <math>AC</math>, and call the intersection between <math>AC</math> and <math>BD</math> point <math>K</math>. In <math>\delta ABC</math>, observe that <math>BE:EC=1:2</math> and <math>AK:KC=1:1</math>. Using mass points, find that <math>BP:PK=1:1</math>. Again utilizing <math>\delta ABC</math>, observe that <math>BF:FC=2:1</math> and <math>AK:KC=1:1</math>. Use mass points to find that <math>BQ:QK=4:1</math>. Now, draw a line segment with points <math>B</math>,<math>P</math>,<math>Q</math>, and <math>K</math> ordered from left to right. Set the values <math>BP=x</math>,<math>PK=x</math>,<math>BQ=4y</math> and <math>QK=y</math>. Setting both sides segment <math>BK</math> equal, we get <math>y= \frac{2}{5}x</math>. Plugging in and solving gives <math>QK= \frac{2}{5}x</math>, <math>PQ=\frac{3}{5}x</math>,<math>BP=x</math>. The question asks for <math>BP:PQ:QD</math>, so we add <math>2x</math> to <math>QK</math> and multiply the ratio by <math>5</math> to create integers. This creates <math>5(1:\frac{3}{5}:\frac{12}{5})= 5:3:12</math>. This sums up to <math>3+5+12=\boxed{\textbf{(E) }20}</math> | Draw line segment <math>AC</math>, and call the intersection between <math>AC</math> and <math>BD</math> point <math>K</math>. In <math>\delta ABC</math>, observe that <math>BE:EC=1:2</math> and <math>AK:KC=1:1</math>. Using mass points, find that <math>BP:PK=1:1</math>. Again utilizing <math>\delta ABC</math>, observe that <math>BF:FC=2:1</math> and <math>AK:KC=1:1</math>. Use mass points to find that <math>BQ:QK=4:1</math>. Now, draw a line segment with points <math>B</math>,<math>P</math>,<math>Q</math>, and <math>K</math> ordered from left to right. Set the values <math>BP=x</math>,<math>PK=x</math>,<math>BQ=4y</math> and <math>QK=y</math>. Setting both sides segment <math>BK</math> equal, we get <math>y= \frac{2}{5}x</math>. Plugging in and solving gives <math>QK= \frac{2}{5}x</math>, <math>PQ=\frac{3}{5}x</math>,<math>BP=x</math>. The question asks for <math>BP:PQ:QD</math>, so we add <math>2x</math> to <math>QK</math> and multiply the ratio by <math>5</math> to create integers. This creates <math>5(1:\frac{3}{5}:\frac{12}{5})= 5:3:12</math>. This sums up to <math>3+5+12=\boxed{\textbf{(E) }20}</math> | ||
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+ | ==Solution 5 (Easy Coord Bash)== | ||
+ | We set coordinates for the points. Let <math>A=(0,3), B=(6,3), C=(6,0)</math> and <math>D=(0,0)</math>. Then the equation of line <math>AE</math> is <math>y = -\frac{1}{6}x + 3,</math> the equation of line <math>AF</math> is <math>y = -\frac{1}{3}x + 3,</math> and the equation of line <math>BD</math> is <math>y = \frac{1}{2}x</math>. We find that the x-coordinate of point <math>P</math> is <math>\frac 9 2</math> by solving <math> -\frac{1}{6}x + 3=\frac{1}{2}x.</math> Similarly we find that the x-coordinate of point <math>Q</math> is <math>\frac {18} 5</math> by solving <math>-\frac{1}{3}x + 3=\frac{1}{2}x.</math> It follows that <math>BP:PQ:QD=6-\frac 9 2 : \frac 9 2 - \frac {18} 5 : \frac {18} 5= \frac 3 2 : \frac 9 {10} : \frac {18} 5 = 15:9:36.</math> Hence <math>r,s,t=15,9,36</math> and <math>r+s+t=15+9+36=\boxed{\textbf{(E) } 20}.</math> ~ Solution by dolphin7 | ||
==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 20:23, 20 January 2020
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
Use similar triangles. Our goal is to put the ratio in terms of . 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 are . Furthermore we find that the coordinates of are . Using the Pythagorean Theorem, we get that the length of is , and the length of is The length of . Then The ratio Then and is and , respectively. The problem tells us to find , so ~ minor LaTeX edits by dolphin7
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 .
Solution 4 (Mass Points)
Draw line segment , and call the intersection between and point . In , observe that and . Using mass points, find that . Again utilizing , observe that and . Use mass points to find that . Now, draw a line segment with points ,,, and ordered from left to right. Set the values ,, and . Setting both sides segment equal, we get . Plugging in and solving gives , ,. The question asks for , so we add to and multiply the ratio by to create integers. This creates . This sums up to
Solution 5 (Easy Coord Bash)
We set coordinates for the points. Let and . Then the equation of line is the equation of line is and the equation of line is . We find that the x-coordinate of point is by solving Similarly we find that the x-coordinate of point is by solving It follows that Hence and ~ Solution by dolphin7
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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