Difference between revisions of "1985 AIME Problems/Problem 6"
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Applying identical reasoning to the triangles with bases <math>\overline{CD}</math> and <math>\overline{BD}</math>, we get <math>\frac{y}{35} = \frac{x+y+84}{105}</math> so that <math>3y = x + y + 84</math> and <math>2y = x + 84</math>. Substituting from this equation into the previous one gives <math>x = 56</math>, from which we get <math>y = 70</math> and so the area of <math>\triangle ABC</math> is <math>56 + 40 + 30 + 35 + 70 + 84 = 315</math>. | Applying identical reasoning to the triangles with bases <math>\overline{CD}</math> and <math>\overline{BD}</math>, we get <math>\frac{y}{35} = \frac{x+y+84}{105}</math> so that <math>3y = x + y + 84</math> and <math>2y = x + 84</math>. Substituting from this equation into the previous one gives <math>x = 56</math>, from which we get <math>y = 70</math> and so the area of <math>\triangle ABC</math> is <math>56 + 40 + 30 + 35 + 70 + 84 = 315</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | This problem can be done using mass points as well. | ||
== See also == | == See also == |
Revision as of 13:14, 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.
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 |