Difference between revisions of "1986 AIME Problems/Problem 9"
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Let the length of the segment be <math>x</math> and the area of the triangle be <math>A</math>, using the theorem, we get: | Let the length of the segment be <math>x</math> and the area of the triangle be <math>A</math>, using the theorem, we get: | ||
− | <math>\frac {d + e + f}{A} = \left(\frac {x}{BC}\right)^2</math>, <math>\frac {b + c + d}{A}= \left(\frac {x}{AC}\right)^2</math>, <math>\frac {a + b + f}{A} = \left(\frac {x}{AB}\right)^2</math> | + | <math>\frac {d + e + f}{A} = \left(\frac {x}{BC}\right)^2</math>, <math>\frac {b + c + d}{A}= \left(\frac {x}{AC}\right)^2</math>, <math>\frac {a + b + f}{A} = \left(\frac {x}{AB}\right)^2</math>. |
− | + | Adding all these together and using <math>a + b + c + d + e + f = A</math> we get | |
− | <math>\frac {f + d + b}{A} + 1 = x^2 | + | <math>\frac {f + d + b}{A} + 1 = x^2 \cdot \left(\frac {1}{BC^2} + \frac {1}{AC^2} + \frac {1}{AB^2}\right)</math> |
− | Using [[corresponding angles]] from parallel lines, it is easy to show that <math>\triangle ABC \sim \triangle F'PF</math> | + | Using [[corresponding angles]] from parallel lines, it is easy to show that <math>\triangle ABC \sim \triangle F'PF</math>; since <math>ADPF'</math> and <math>CFPE'</math> are parallelograms, it is easy to show that <math>FF' = AC - x</math> |
Now we have the side length [[ratio]], so we have the area ratio | Now we have the side length [[ratio]], so we have the area ratio | ||
− | <math>\frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2</math> | + | <math>\frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2</math>. By symmetry, we have |
<math>\frac {d}{A} = \left(1 - \frac {x}{AB}\right)^2</math> and <math>\frac {b}{A} = \left(1 - \frac {x}{BC}\right)^2</math> | <math>\frac {d}{A} = \left(1 - \frac {x}{AB}\right)^2</math> and <math>\frac {b}{A} = \left(1 - \frac {x}{BC}\right)^2</math> | ||
Substituting these into our initial equation, we have | Substituting these into our initial equation, we have | ||
<math>1 + \sum_{cyc}\left(1 - \frac {x}{AB}\right) - \frac {x^2}{AB^2} = 0</math> | <math>1 + \sum_{cyc}\left(1 - \frac {x}{AB}\right) - \frac {x^2}{AB^2} = 0</math> | ||
− | <math>1 + \sum_{cyc}1 - 2 | + | <math>\implies 1 + \sum_{cyc}1 - 2 \cdot \frac {x}{AB} = 0</math> |
− | <math>\frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x</math> | + | <math>\implies \frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x</math> |
− | answer follows after some hideous computation. | + | and the answer follows after some hideous computation. |
===Solution 4=== | ===Solution 4=== | ||
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Let <math>F'P = x</math>. Then, because <math>\triangle ABC \sim \triangle F'PF</math>, <math>\frac{AB}{AC}=\frac{F'P}{F'F}</math>, so <math>\frac{425}{510}=\frac{x}{510-d}</math>. Simplifying the LHS and cross-multiplying, we have <math>6x=2550-5d</math>. From the same triangles, we can find that <math>FP=\frac{18}{17}x</math>. | Let <math>F'P = x</math>. Then, because <math>\triangle ABC \sim \triangle F'PF</math>, <math>\frac{AB}{AC}=\frac{F'P}{F'F}</math>, so <math>\frac{425}{510}=\frac{x}{510-d}</math>. Simplifying the LHS and cross-multiplying, we have <math>6x=2550-5d</math>. From the same triangles, we can find that <math>FP=\frac{18}{17}x</math>. | ||
− | <math>\triangle PEE'</math> is also similar to <math>\triangle F'PF</math>. Since <math>EF'=d</math>, <math>EP=d-x</math>. We now have <math>\frac{PE}{EE'}=\frac{F'P}{FP}</math>, and <math>\frac{d-x}{450-d}=\frac{17}{18}</math>. Cross multiplying, we have <math>18d-18x=450 | + | <math>\triangle PEE'</math> is also similar to <math>\triangle F'PF</math>. Since <math>EF'=d</math>, <math>EP=d-x</math>. We now have <math>\frac{PE}{EE'}=\frac{F'P}{FP}</math>, and <math>\frac{d-x}{450-d}=\frac{17}{18}</math>. Cross multiplying, we have <math>18d-18x=450 \cdot 17-17d</math>. Using the previous equation to substitute for <math>x</math>, we have: <cmath>18d-3\cdot2550+15d=450\cdot17-17d</cmath> This is a linear equation in one variable, and we can solve to get <math>d=\boxed{306}</math> |
*I did not show the multiplication in the last equation because most of it cancels out when solving. | *I did not show the multiplication in the last equation because most of it cancels out when solving. |
Revision as of 22:53, 2 April 2018
Problem
In , , , and . An interior point is then drawn, and segments are drawn through parallel to the sides of the triangle. If these three segments are of an equal length , find .
Contents
Solution
Solution 1
Construct cevians , and through . Place masses of on , and respectively; then has mass .
Notice that has mass . On the other hand, by similar triangles, . Hence by mass points we find that Similarly, we obtain Summing these three equations yields
Hence,
Solution 2
Let the points at which the segments hit the triangle be called as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar (). The remaining three sections are parallelograms.
Since is a parallelogram, we find , and similarly . So . Thus . By the same logic, .
Since , we have the proportion:
Doing the same with , we find that . Now, .
Solution 3
Define the points the same as above.
Let , , , , and
The key theorem we apply here is that the ratio of the areas of 2 similar triangles is the ratio of a pair of corresponding sides squared.
Let the length of the segment be and the area of the triangle be , using the theorem, we get:
, , . Adding all these together and using we get
Using corresponding angles from parallel lines, it is easy to show that ; since and are parallelograms, it is easy to show that
Now we have the side length ratio, so we have the area ratio . By symmetry, we have and
Substituting these into our initial equation, we have and the answer follows after some hideous computation.
Solution 4
Refer to the diagram in solution 2; let , , and . Now, note that , , and are similar, so through some similarities we find that . Similarly, we find that and , so . Now, again from similarity, it follows that , , and , so adding these together, simplifying, and solving gives .
Solution 5
Refer to the diagram from Solution 2. Notice that because , , and are parallelograms, , , and .
Let . Then, because , , so . Simplifying the LHS and cross-multiplying, we have . From the same triangles, we can find that .
is also similar to . Since , . We now have , and . Cross multiplying, we have . Using the previous equation to substitute for , we have: This is a linear equation in one variable, and we can solve to get
- I did not show the multiplication in the last equation because most of it cancels out when solving.
(Note: I chose to be only because that is what I had written when originally solving. The solution would work with other choices for .)
See also
1986 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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