2004 AIME II Problems/Problem 7
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
Solution
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
BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \\ BC &= \frac{70}{3}
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)Thus the perimeter of is
, and the 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
\frac{14}{l}x &= y-8 = \frac{9}{2}\\ \frac{-l}{14}x &= y-25 = -\frac{25}{2}\\
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)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.
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 | ||
All AIME Problems and Solutions |