Difference between revisions of "2004 AIME II Problems/Problem 7"
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=== Solution 2 (analytic) === | === Solution 2 (analytic) === | ||
Let <math>A = (0,0), B=(0,25)</math>, so <math>E = (0,8)</math> and <math>F = (l,22)</math>, and let <math>l = AD</math> be the length of the rectangle. The [[slope]] of <math>EF</math> is <math>\frac{14}{l}</math> and so the equation of <math>EF</math> is <math>y -8 = \frac{14}{l}x</math>. We know that <math>EF</math> is perpendicular to and bisects <math>BB'</math>. The slope of <math>BB'</math> is thus <math>\frac{-l}{14}</math>, and so the equation of <math>BB'</math> is <math>y -25 = \frac{-l}{14}x</math>. Let the point of intersection of <math>EF, BB'</math> be <math>G</math>. Then the y-coordinate of <math>G</math> is <math>\frac{25}{2}</math>, so | Let <math>A = (0,0), B=(0,25)</math>, so <math>E = (0,8)</math> and <math>F = (l,22)</math>, and let <math>l = AD</math> be the length of the rectangle. The [[slope]] of <math>EF</math> is <math>\frac{14}{l}</math> and so the equation of <math>EF</math> is <math>y -8 = \frac{14}{l}x</math>. We know that <math>EF</math> is perpendicular to and bisects <math>BB'</math>. The slope of <math>BB'</math> is thus <math>\frac{-l}{14}</math>, and so the equation of <math>BB'</math> is <math>y -25 = \frac{-l}{14}x</math>. Let the point of intersection of <math>EF, BB'</math> be <math>G</math>. Then the y-coordinate of <math>G</math> is <math>\frac{25}{2}</math>, so | ||
− | < | + | <cmath> |
\begin{align*} | \begin{align*} | ||
\frac{14}{l}x &= y-8 = \frac{9}{2}\\ | \frac{14}{l}x &= y-8 = \frac{9}{2}\\ | ||
\frac{-l}{14}x &= y-25 = -\frac{25}{2}\\ | \frac{-l}{14}x &= y-25 = -\frac{25}{2}\\ | ||
\end{align*} | \end{align*} | ||
− | </ | + | </cmath> |
Dividing the two equations yields | Dividing the two equations yields | ||
<center><math>l^2 = \frac{25 \cdot 14^2}{9} \Longrightarrow l = \frac{70}{3}</math></center> | <center><math>l^2 = \frac{25 \cdot 14^2}{9} \Longrightarrow l = \frac{70}{3}</math></center> |
Revision as of 19:56, 13 March 2015
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
Contents
Solution
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 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 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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.