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)
![[asy] pointpen = black; pathpen = black +linewidth(0.7); pair A=(0,0),B=(0,25),C=(70/3,25),D=(70/3,0),E=(0,8),F=(70/3,22),G=(15,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(MP("E",E,W)--MP("F",F,(1,0))); D(B--G); D(E--MP("B'",G)--F--B,dashed); MP("8",(A+E)/2,W);MP("17",(B+E)/2,W);MP("22",(D+F)/2,(1,0)); [/asy]](http://latex.artofproblemsolving.com/e/8/4/e8476256c3a6cf7cf8b766055c9c035fb86817da.png)
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
![$l^2 = \frac{25 \cdot 14^2}{9} \Longrightarrow l = \frac{70}{3}$](http://latex.artofproblemsolving.com/7/e/8/7e8ae577a847b2f8283ffcb1b85a61f570702b67.png)
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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