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Difference between revisions of "1985 AIME Problems/Problem 6"

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== Solution ==
 
== Solution ==
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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:
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<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>.
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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>.
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== See also ==
 
== See also ==
 
* [[1985 AIME Problems/Problem 5 | Previous problem]]
 
* [[1985 AIME Problems/Problem 5 | Previous problem]]

Revision as of 17:24, 19 January 2007

Problem

As shown in the figure, triangle $ABC$ 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 $ABC$.

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Solution

Let the interior point be $P$, let the points on $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$ be $D$, $E$ and $F$, respectively. Let $x$ be the area of $\triangle APE$ and $y$ be the area of $\triangle CPD$. Note that $\triangle APF$ and $\triangle BPF$ share the same altitude from $P$, so the ratio of their areas is the same as the ratio of their bases. Similarly, $\triangle ACF$ and $\triangle BCF$ share the same altitude from $C$, 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: $\frac{40}{30} = \frac{124 + x}{65 + y}$ or equivalently $372 + 3x = 260 + 4y$ and so $4y = 3x+ 112$.

Applying identical reasoning to the triangles with bases $\overline{CD}$ and $\overline{BD}$, we get $\frac{y}{35} = \frac{x+y+84}{105}$ so that $3y = x + y + 84$ and $2y = x + 84$. Substituting from this equation into the previous one gives $x = 56$, from which we get $y = 70$ and so the area of $\triangle ABC$ is $56 + 40 + 30 + 35 + 70 + 84 = 315$.

See also

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