1986 AIME Problems/Problem 9
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, .
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.
Refer to the diagram; 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 4 (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
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