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

Revision as of 00:33, 15 November 2022

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 $\dfrac{1}{2} \cdot 24 \cdot 12 = \boxed{144}$ .

Solution 3

The base of $\triangle{ABC}$ is $BC$ . Let the base of $t_1$ be $x$ , the base of $t_2$ be $y$ , and the base of $t_3$ be $z$ . Since $\triangle{ABC}, t_1, t_2,$ and $t_3$ are all similar, the sections in $\triangle{ABC}$ that aren't $t_1,t_2,$ or $t_3$ are all parallelograms. Hence, $BC=x+z+y$ . We can relate $t_1,t_2,$ and $t_3$ by the square root of the ratio of their areas. $\sqrt{\frac{4}{9}}=\frac{2}{3}$ and $\sqrt{\frac{4}{49}}=\frac{2}{7}$ so $y=\frac{3x}{2}$ and $z=\frac{7x}{2}$ . $x+\frac{7x}{2}+\frac{3x}{2}=6x$ , so $\triangle{ABC}$ has a base that is $6$ times $t_1$ . $[\triangle{ABC}]=36[t_1]=36 \cdot 4=\boxed{144}$ .

-PhunsukhWangdu

Solution 4 (Fakesolve)

Basically, we can just assume that $\angle{BAC}$ is a right angle. Due to the parallel lines that we're given

@@ Line 25: / Line 25: @@
 == Solution 4 (Fakesolve) ==
-Basically, we can just assume that $\angle{BAC} is a right angle. Due to the parallel lines that we're given
+Basically, we can just assume that <math>\angle{BAC}</math> is a right angle. Due to the parallel lines that we're given
 == 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

Page

Toolbox

Search