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

Revision as of 13:20, 6 May 2007

Problem

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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Solution

Notice that all four triangles are similar to each other. 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. Since the squares of the lengths of the sides are in proportion with the area, 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 $12^2 = 144$ .

@@ Line 7: / Line 7: @@
 == See also ==
 {{AIME box|year=1984|num-b=2|num-a=4}}
+* [[AIME Problems and Solutions]]
+* [[American Invitational Mathematics Examination]]
+* [[Mathematics competition resources]]
 [[Category:Intermediate Geometry Problems]]

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

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

Revision as of 13:20, 6 May 2007

Problem

Solution

See also