2003 AMC 10B Problems/Problem 20
- The following problem is from both the 2003 AMC 12B #14 and 2003 AMC 10B #20, so both problems redirect to this page.
Problem
In rectangle and . Points and are on so that and . Lines and intersect at . Find the area of .
Solution 1
because The ratio of to is since and from subtraction. If we let be the height of
The height is so the area of is .
Solution 2
We can look at this diagram as if it were a coordinate plane with point being . This means that the equation of the line is and the equation of the line is . From this we can set of the follow equation to find the coordinate of point :
We can plug this into one of our original equations to find that the coordinate is , meaning the area of is
Solution 3
At points and , segment is 5 units from segment . At points and , the segments are 2 units from each other. This means that collectively, the two lines closed the distance between them by 3 units over a height of 3 units. Therefore, to close the next two units of distance, they will have to travel a height of 2 units.
Then calculate the area of trapezoid and triangle separately and add them. The area of the trapezoid is and the area of the triangle is .
Solution 4
Since then , where and are ponts on and respectivley which make the areas similar. This process can be done over and over again multiple times by the ratio of , or something like this we have to find the ratio of the areas when the sides have shrunk by length
Let be the area of the shape whose length is Now comparing the ratios of to we get By applying an infinite summation
Solution 5
Drop a perpendicular from to and call the intersection point and are similar and so are and (You can prove this with by observing that is congruent to and so on). This means that so which means that Then,
See Also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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