Difference between revisions of "2016 AMC 10A Problems/Problem 19"
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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>. | 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>. | ||
− | ==Solution 4== | + | ==Solution 4 (Mass Points)== |
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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> | ||
Revision as of 11:21, 8 November 2019
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
An alternate solution is to perform the same operations, but only solve for the x-coordinates. By similar triangles, the ratios will be the same.
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 (Cheap Solution)
Use your ruler (it is recommended that you bring a ruler and protractor to AMC10 tests) and accurately draw the diagram as one in solution 1, then measure the length of the segments, you should get a ratio of being , multiplying each side by the result is
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.