Difference between revisions of "1984 AIME Problems/Problem 3"
m |
(solution, box) |
||
| Line 3: | Line 3: | ||
{{image}} | {{image}} | ||
== Solution == | == Solution == | ||
| − | + | Notice that all four triangles are [[similar triangles|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 <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 = 144</math>. | |
| + | |||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1984|num-b=2|num-a=4}} | |
| − | |||
| − | |||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 21:31, 5 March 2007
Problem
A point 
is chosen in the interior of 
such that when lines are drawn through parallel to the sides of , the resulting smaller triangles 
, 
, and 
in the figure, have areas 
, 
, and 
, respectively. Find the area of .
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 
. Thus, the corresponding side on the large triangle is 
, and the area of the triangle is 
.
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 | ||
