Difference between revisions of "1986 AIME Problems/Problem 9"
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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. | ||
− | Since <math> | + | Since <math>PDAF'</math> is a parallelogram, we find <math>PF' = AD</math>, and similarly <math>PE = BD'</math>. So <math>d = PF' + PE = AD + BD' = 425 - DD'</math>. Thus <math>DD' = 425 - d</math>. By the same logic, <math>EE' = 450 - d</math>. |
Since <math>\triangle DPD' \sim \triangle ABC</math>, we have the [[proportion]]: | Since <math>\triangle DPD' \sim \triangle ABC</math>, we have the [[proportion]]: | ||
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<div style="text-align:center;"><math>\frac{425-d}{425} = \frac{PD}{510}</math><br /><math>PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d</math></div> | <div style="text-align:center;"><math>\frac{425-d}{425} = \frac{PD}{510}</math><br /><math>PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d</math></div> | ||
− | Doing the same with <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 = 306</math>. |
=== Solution 2 === | === Solution 2 === |
Revision as of 20:30, 8 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 |