Difference between revisions of "1985 AIME Problems/Problem 6"
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[[Image:AIME 1985 Problem 6.png]] | [[Image:AIME 1985 Problem 6.png]] | ||
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Let the interior point be <math>P</math>, let the points on <math>\overline{BC}</math>, <math>\overline{CA}</math> and <math>\overline{AB}</math> be <math>D</math>, <math>E</math> and <math>F</math>, respectively. Let <math>x</math> be the area of <math>\triangle APE</math> and <math>y</math> be the area of <math>\triangle CPD</math>. Note that <math>\triangle APF</math> and <math>\triangle BPF</math> share the same [[altitude]] from <math>P</math>, so the [[ratio]] of their areas is the same as the ratio of their bases. Similarly, <math>\triangle ACF</math> and <math>\triangle BCF</math> share the same altitude from <math>C</math>, 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: | Let the interior point be <math>P</math>, let the points on <math>\overline{BC}</math>, <math>\overline{CA}</math> and <math>\overline{AB}</math> be <math>D</math>, <math>E</math> and <math>F</math>, respectively. Let <math>x</math> be the area of <math>\triangle APE</math> and <math>y</math> be the area of <math>\triangle CPD</math>. Note that <math>\triangle APF</math> and <math>\triangle BPF</math> share the same [[altitude]] from <math>P</math>, so the [[ratio]] of their areas is the same as the ratio of their bases. Similarly, <math>\triangle ACF</math> and <math>\triangle BCF</math> share the same altitude from <math>C</math>, 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: | ||
<math>\frac{40}{30} = \frac{124 + x}{65 + y}</math> or equivalently <math>372 + 3x = 260 + 4y</math> and so <math>4y = 3x+ 112</math>. | <math>\frac{40}{30} = \frac{124 + x}{65 + y}</math> or equivalently <math>372 + 3x = 260 + 4y</math> and so <math>4y = 3x+ 112</math>. |
Latest revision as of 14:18, 9 May 2020
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 1
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. Assign B a weight of 1 and realize that many of the triangles have the same altitude. After continuously using the formulas that state (The sum of the two weights) = (The middle weight), and (The weight side) = (Other weight) (The other side), the problem yields the answer
Solution 3
Let the interior point be and let the points on , and be , and , respectively. Also, let Then notice that by Ceva's, However, we can deduce from the fact that and share the same height. Similarly, and Plug and chug and you get Then notice by the same height reasoning, Clear the fractions and combine like terms to get We know so subtraction yields or Plugging this in to our previous ratio statement yields so Basic addition gives us
-dchen
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 |