2001 AMC 10 Problems/Problem 15

Revision as of 19:33, 31 March 2020 by Penguinninja2008 (talk | contribs) (Solution 1)


A street has parallel curbs $40$ feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is $15$ feet and each stripe is $50$ feet long. Find the distance, in feet, between the stripes.

$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25$

Solution 1

Drawing the problem out, we see we get a parallelogram with a height of $40$ and a base of $15$, giving an area of $600$.

If we look at it the other way, we see the distance between the stripes is the height and the base is $50$.

[asy] draw((0,0)--(5,0)); draw((2.5,5)--(7.5,5)); draw((0,0)--(2.5,5)); draw((5,0)--(7.5,5),linewidth(2)); draw((2,4)--(6,2),dashed);[/asy]

The area is still the same, so the distance between the stripes is $600/50 = \boxed{\textbf{(C)}\ 12}$.

Solution 2

Alternatively, we could use similar triangles--the $30-40-50$ triangle (created by the length of the bordering stripe and the difference between the two curbs) is similar to the $x-y-15$ triangle, where we are trying to find $y$ (the shortest distance between the two stripes). Therefore, $y$ would have to be $\boxed{\textbf{(C)}\ 12}$.

Solution 3 (basically solution one in words)

The area of the parallelogram in the first picture is base * height which is 15 * 40 = 600. The goal of the problem is to find the minimum distance between the stripes (it's not really said in the problem but that's the goal). If you look at the parallelogram like in diagram two, the base would be 50. Since it's still the same parallelogram, and the shortest distance between two parallel lines is the line that is perpendicular to both lines, then the height would be 50*h = 600, so h = 12.

It's easy to see why the minimum height would be at its perpendicular line. If you try and move the point away from the perpendicular line, you would see that the only thing changing is the distance between the original point and the new point, which would increase distance, not minimize it. The point at the perpendicular line is the only place where you would have zero distance between it and the original point (because it is the original point), and simple observation or Pythagorean theorem would tell you that that is the min distance.

Solution by IronicNinja~

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS