Difference between revisions of "1985 AIME Problems/Problem 6"
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== Solution 2 == | == Solution 2 == | ||
This problem can be done using mass points as well. Assign B a weight of 1 and realize that many of the triangles have the same altitude. After continuously using the formulas where the sum of the two weights = the middle weight and the the weight times side = other weight times other side, the problem can be trivialized giving the answer <math>315</math> | This problem can be done using mass points as well. Assign B a weight of 1 and realize that many of the triangles have the same altitude. After continuously using the formulas where the sum of the two weights = the middle weight and the the weight times side = other weight times other side, the problem can be trivialized giving the answer <math>315</math> | ||
| + | |||
| + | Solution by carlos8 | ||
== See also == | == See also == | ||
Revision as of 13:16, 12 October 2016
Contents
Problem
As shown in the figure, triangle 
is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle .
Solution
Let the interior point be 
, let the points on 
, 
and 
be 
, 
and 
, respectively. Let 
be the area of 
and 
be the area of 
. Note that 
and 
share the same altitude from , so the ratio of their areas is the same as the ratio of their bases. Similarly, 
and 
share the same altitude from 
, so the ratio of their areas is the same as the ratio of their bases. Moreover, the two pairs of bases are actually the same, and thus in the same ratio. As a result, we have:
or equivalently 
and so 
.
Applying identical reasoning to the triangles with bases 
and 
, we get 
so that 
and 
. Substituting from this equation into the previous one gives 
, from which we get 
and so the area of 
is 
.
Solution 2
This problem can be done using mass points as well. Assign B a weight of 1 and realize that many of the triangles have the same altitude. After continuously using the formulas where the sum of the two weights = the middle weight and the the weight times side = other weight times other side, the problem can be trivialized giving the answer 
Solution by carlos8
See also
| 1985 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
