2004 AIME II Problems/Problem 7
Contents
Problem
is a rectangular sheet of paper that has been folded so that corner is matched with point on edge The crease is where is on and is on The dimensions and are given. The perimeter of rectangle is where and are relatively prime positive integers. Find
Solutions
Solution 1 (Synthetic)
Since is the perpendicular bisector of , it follows that (by SAS). By the Pythagorean Theorem, we have . Similarly, from , we have Thus the perimeter of is , and our answer is .
Solution 2 (analytic)
Let , so and , and let be the length of the rectangle. The slope of is and so the equation of is . We know that is perpendicular to and bisects . The slope of is thus , and so the equation of is . Let the point of intersection of be . Then the y-coordinate of is , so Dividing the two equations yields
The answer is as above.
Solution 3 (Coordinate Bashing)
Firstly, observe that if we are given that and , the length of the triangle is given and the height depends solely on the length of . Let Point . Since , point E is at (8,0). Next, point is at since and point is at since by symmetry. Draw line segment . Notice that this is perpendicular to by symmetry. Next, find the slope of EB, which is . Then, the slope of is -.
Line EF can be written as y=. Plug in the point , and we get the equation of EF to be y=. Since the length of =25, a point on line lies on when . Plug in into our equation to get . . Therefore, our answer is .
Solution 4 (Trig)
Firstly, note that , so . Then let , so . Then , or
using supplementary and double angle identities. Multiplying though and factoring yields
It is clear from the problem setup that , so the correct value is . Next, extend rays and to intersect at . Then , so . By similar triangles, , so . The perimeter is
An even faster way to finish is, to draw a line segment where is a point on such that is perpendicular to . This makes right triangle , Also, note that has length of (draw the diagram out, and note the ). From here, through , we can note that . is parallel and congrurent to and , and hence we can use this to calculate the perimeter. The perimeter is simply
Solution 5 (Fast, Pythagorean)
Use the prepared diagram for this solution.
Call the intersection of DF and B'C' G. AB'E is an 8-15-17 right triangle, and so are B'DG and C'FG. Since C'F is 3, then using the properties of similar triangles GF is 51/8. DF is 22, so DG is 125/8. Finally, DB can to calculated to be 25/3. Add all the sides together to get .
-jackshi2006
Solution 6(fast as wind[rufeng])
Call the intersection of , , and . Since and are congruent, we know that the three lines intersect. We already know so we just need to find , call it . Drop an altitude from to and call it . . Using Pythagorean Theorem, we have . Triangles and are similar (AA), so we get Simplify and we get .
We find the area of by using the fact that it is a trapezoid.
A different way to find the area: height of with as base
Since and are congruent(SAS), their height from is the same. .
The perimeter is so our answer is .
Solution 7 (Similar to solution 5, more in depth)
Let the endpoint of the intersection of the fold near be . Since trapezoid is folded, it is congruent to trapezoid . Therefore, . Since is a right triangle, from the pythagorean theorem. From here, we can see that triangles by AA similarity. From here, we find from a lot of similarities. Let .
Since :
Since ,
from which we get .
Finally, our answer is , which is .
~ Wesserwessey7254
See also
2004 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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