Difference between revisions of "1986 AIME Problems/Problem 9"
m (→Solution 1: <center>, minor) |
Dgreenb801 (talk | contribs) m (→Solution 1) |
||
Line 9: | Line 9: | ||
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]]: | ||
Line 15: | Line 15: | ||
<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 |