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Difference between revisions of "1984 AIME Problems/Problem 3"

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== Solution ==
 
== Solution ==
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By the transversals that go through <math>P</math>, all four triangles are [[similar triangles|similar]] to each other by the <math>AA</math> postulate. Also, note that the length of any one side of the larger triangle is equation to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity <math>K = \dfrac{ab\sin C}{2}</math> to show that the areas are proportional (the sides are proportional and the angles are equal) Hence, we can write the lengths of corresponding sides of the triangle as <math>2x,\ 3x,\ 7x</math>. Thus, the corresponding side on the large triangle is <math>12x</math>, and the area of the triangle is <math>12^2 = \boxed{144}</math>.
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By the transversals that go through <math>P</math>, all four triangles are [[similar triangles|similar]] to each other by the <math>AA</math> postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity <math>K = \dfrac{ab\sin C}{2}</math> to show that the areas are proportional (the sides are proportional and the angles are equal) Hence, we can write the lengths of corresponding sides of the triangle as <math>2x,\ 3x,\ 7x</math>. Thus, the corresponding side on the large triangle is <math>12x</math>, and the area of the triangle is <math>12^2 = \boxed{144}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:16, 14 February 2010

Problem

A point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$, the resulting smaller triangles $t_{1}$, $t_{2}$, and $t_{3}$ in the figure, have areas $4$, $9$, and $49$, respectively. Find the area of $\triangle ABC$.

[asy] size(200); pathpen=black;pointpen=black; pair A=(0,0),B=(12,0),C=(4,5); D(A--B--C--cycle); D(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12); MP("A",C,N);MP("B",A,SW);MP("C",B,SE); /* sorry mixed up points according to resources diagram. */ MP("t_3",(A+B+(B-A)*3/4+(A-B)*5/6)/2+(-1,0.8),N); MP("t_2",(B+C+(B-C)*5/12+(C-B)*5/6)/2+(-0.3,0.1),WSW); MP("t_1",(A+C+(C-A)*3/4+(A-C)*5/12)/2+(0,0.15),ESE); [/asy]

Solution

By the transversals that go through $P$, all four triangles are similar to each other by the $AA$ postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity $K = \dfrac{ab\sin C}{2}$ to show that the areas are proportional (the sides are proportional and the angles are equal) Hence, we can write the lengths of corresponding sides of the triangle as $2x,\ 3x,\ 7x$. Thus, the corresponding side on the large triangle is $12x$, and the area of the triangle is $12^2 = \boxed{144}$.

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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