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

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== Solution 2 ==
 
== Solution 2 ==
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Alternatively, since the triangles are congruent by <math>AA</math>, then the ratios between the bases and the heights of each of the three triangles would all be equal. The areas of each of the triangles are all perfect squares, so we could assume <math>\dfrac{base}{height} = \dfrac{2}{1}.</math> That means that the base of <math>t_{1}</math> is 4, the base of <math>t_{2}</math> is 6, and the base of <math>t_{3}</math> is 14. Since the quadrilaterals underneath <math>t_{1}</math> and <math>t_{2}</math> are both parallelograms, and opposite sides of a parallelogram are congruent, the base of the large triangle is <math>4 + 14 + 6 = 24</math>. Therefore, the height of the entire triangle would be twelve, so therefore, the area of the large triangle is <math>\dfrac{1}{2} \cdot 24 \cdot 12 = \boxed{144}</math>.
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Alternatively, since the triangles are similar by <math>AA</math>, then the ratios between the bases and the heights of each of the three triangles would all be equal. The areas of each of the triangles are all perfect squares, so we could assume <math>\dfrac{base}{height} = \dfrac{2}{1}.</math> That means that the base of <math>t_{1}</math> is 4, the base of <math>t_{2}</math> is 6, and the base of <math>t_{3}</math> is 14. Since the quadrilaterals underneath <math>t_{1}</math> and <math>t_{2}</math> are both parallelograms, and opposite sides of a parallelogram are congruent, the base of the large triangle is <math>4 + 14 + 6 = 24</math>. Therefore, the height of the entire triangle would be twelve, so therefore, the area of the large triangle is <math>\dfrac{1}{2} \cdot 24 \cdot 12 = \boxed{144}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:16, 5 September 2017

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 1

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 $12^2 = \boxed{144}$.

Solution 2

Alternatively, since the triangles are similar by $AA$, then the ratios between the bases and the heights of each of the three triangles would all be equal. The areas of each of the triangles are all perfect squares, so we could assume $\dfrac{base}{height} = \dfrac{2}{1}.$ That means that the base of $t_{1}$ is 4, the base of $t_{2}$ is 6, and the base of $t_{3}$ is 14. Since the quadrilaterals underneath $t_{1}$ and $t_{2}$ are both parallelograms, and opposite sides of a parallelogram are congruent, the base of the large triangle is $4 + 14 + 6 = 24$. Therefore, the height of the entire triangle would be twelve, so therefore, the area of the large triangle is $\dfrac{1}{2} \cdot 24 \cdot 12 = \boxed{144}$.

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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