Difference between revisions of "1986 AIME Problems/Problem 9"
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== Problem == | == Problem == | ||
− | In <math> | + | In <math>\triangle ABC</math>, <math>AB= 425</math>, <math>BC=450</math>, and <math>AC=510</math>. An interior [[point]] <math>P</math> is then drawn, and [[segment]]s are drawn through <math>P</math> [[parallel]] to the sides of the [[triangle]]. If these three segments are of an equal length <math>d</math>, find <math>d</math>. |
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | [[Image:1986_AIME-9.png]] | + | [[Image:1986_AIME-9.png|center]] |
Let the points at which the segments hit the triangle be called <math>D, D', E, E', F, F'</math> as shown above. All three smaller triangles and the larger triangle are [[similar triangles|similar]] (<math>\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF</math>). This is easy to find using repeated [[alternate interior angles]]. The remaining three sections are [[parallelogram]]s, which is also simple to see by the parallel lines. | Let the points at which the segments hit the triangle be called <math>D, D', E, E', F, F'</math> as shown above. All three smaller triangles and the larger triangle are [[similar triangles|similar]] (<math>\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF</math>). This is easy to find using repeated [[alternate interior angles]]. The remaining three sections are [[parallelogram]]s, which is also simple to see by the parallel lines. | ||
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Since <math>\triangle DPD' \sim \triangle ABC</math>, we have the [[proportion]]: | Since <math>\triangle DPD' \sim \triangle ABC</math>, we have the [[proportion]]: | ||
− | < | + | <cmath>\frac{425-d}{425} = \frac{PD}{510} \Longrightarrow PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d</cmath> |
− | Doing the same with <math>\triangle PEE'</math>, we find that <math>PE' =510 - \frac{17}{15}d</math>. Now, <math>d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = 306</math>. | + | Doing the same with <math>\triangle PEE'</math>, we find that <math>PE' =510 - \frac{17}{15}d</math>. Now, <math>d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}</math>. |
=== Solution 2 === | === Solution 2 === | ||
Define the points the same as above. | Define the points the same as above. | ||
− | Let <math>[CE'PF] = a</math>, <math>[E'EP] = b</math>, <math> | + | Let <math>[CE'PF] = a</math>, <math>[E'EP] = b</math>, <math>[BEPD'] = c</math>, <math>[D'PD] = d</math>, <math>[DAF'P] = e</math> and <math>[F'D'P] = f</math> |
Key theorem: the ratio of the areas of 2 similar triangles is the ratio of a pair of corresponding sides squared. | Key theorem: the ratio of the areas of 2 similar triangles is the ratio of a pair of corresponding sides squared. | ||
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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 {c + e + d}{A} = \left(\frac {x}{BC}\right)^2</math>, <math> | + | <math>\frac {c + e + d}{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 | 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*\left(\frac {1}{BC^2} + \frac {1}{AC^2} + \frac {1}{AB^2}\right)</math> | <math>\frac {f + d + b}{A} + 1 = x^2*\left(\frac {1}{BC^2} + \frac {1}{AC^2} + \frac {1}{AB^2}\right)</math> | ||
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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> | + | <math>\frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2</math>, by symmetry, we have |
− | <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*\frac {x}{AB} = 0</math> | <math>1 + \sum_{cyc}1 - 2*\frac {x}{AB} = 0</math> | ||
− | <math> | + | <math>\frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x</math> |
− | answer follows after some hideous computation | + | answer follows after some hideous computation. |
== See also == | == See also == | ||
{{AIME box|year=1986|num-b=8|num-a=10}} | {{AIME box|year=1986|num-b=8|num-a=10}} | ||
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+ | [[Category:Asymptote needed]] | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 19:34, 9 April 2008
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
[hide]Solution
Solution 1
Let the points at which the segments hit the triangle be called as shown above. All three smaller triangles and the larger triangle are similar (). This is easy to find using repeated alternate interior angles. The remaining three sections are parallelograms, which is also simple to see by the parallel lines.
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 2
Define the points the same as above.
Let , , , , and
Key theorem: 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 answer follows after some hideous computation.
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 |