Difference between revisions of "2016 AMC 10A Problems/Problem 19"
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− | In rectangle <math> | + | In rectangle <math>ABCD,</math> <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ratio <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s,</math> and <math>t</math> is <math>1.</math> What is <math>r+s+t</math>? |
<math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math> | <math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math> |
Revision as of 22:39, 3 April 2021
Contents
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 (Similar Triangles)
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
Video Solution
https://www.youtube.com/watch?v=aG9JiBMd0ag
Video Solution 2
~IceMatrix
Video Solution 3
https://youtu.be/4_x1sgcQCp4?t=3406
~ pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.