Difference between revisions of "1986 AIME Problems/Problem 9"
m (Added point P in second solution) |
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D(D--Ea);D(Da--F);D(Fa--E); | D(D--Ea);D(Da--F);D(Fa--E); | ||
MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2); | MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2); | ||
+ | /*P copied from above solution*/ | ||
+ | pair P = IP(D--Ea,E--Fa); dot(MP("P",P,N)); | ||
</asy></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps --> | </asy></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps --> | ||
Revision as of 21:33, 13 March 2015
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 (mass points)
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 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 .
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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