Difference between revisions of "2014 AIME I Problems/Problem 1"
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== Problem 1 == | == Problem 1 == | ||
| − | The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the | + | The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters. |
<asy> | <asy> | ||
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== Solution == | == Solution == | ||
| − | The rectangle is divided into three smaller | + | The rectangle is divided into three smaller rectangles with a width of 50 mm and a length of <math>\dfrac{80}{3}</math>mm. According to the Pythagorean Theorem (or by noticing the 8-15-17 Pythagorean triple), the diagonal of the rectangle is |
| − | <math>\sqrt{50^2+\frac{80}{3}^2}=\frac{170}{3}</math>mm. Since that on the lace, there are 6 of these diagonals, a width, and at least 200 mm | + | <math>\sqrt{50^2+\left(\frac{80}{3}\right)^2}=\frac{170}{3}</math>mm. Since that on the lace, there are 6 of these diagonals, a width, and an extension of at least 200 mm on each side. Therefore, the minimum of the lace in millimeters is |
<cmath>6\times \dfrac{170}{3}+50+200\times 2=\boxed{790}.</cmath> | <cmath>6\times \dfrac{170}{3}+50+200\times 2=\boxed{790}.</cmath> | ||
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{{AIME box|year=2014|n=I|before=First Problem|num-a=2}} | {{AIME box|year=2014|n=I|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 17:35, 18 July 2022
Problem 1
The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.
![[asy] size(200); defaultpen(linewidth(0.7)); path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,-4)..tension 0.8 ..origin; path laceR=reflect((75,0),(75,-240))*laceL; draw(origin--(0,-240)--(150,-240)--(150,0)--cycle,gray); for(int i=0;i<=3;i=i+1) { path circ1=circle((0,-80*i),5),circ2=circle((150,-80*i),5); unfill(circ1); draw(circ1); unfill(circ2); draw(circ2); } draw(laceL--(150,-80)--(0,-160)--(150,-240)--(0,-240)--(150,-160)--(0,-80)--(150,0)^^laceR,linewidth(1));[/asy]](http://latex.artofproblemsolving.com/a/6/1/a61b1059117868ce9afa4391360bfa9424211553.png)
Solution
The rectangle is divided into three smaller rectangles with a width of 50 mm and a length of 
mm. According to the Pythagorean Theorem (or by noticing the 8-15-17 Pythagorean triple), the diagonal of the rectangle is
mm. Since that on the lace, there are 6 of these diagonals, a width, and an extension of at least 200 mm on each side. Therefore, the minimum of the lace in millimeters is
![\[6\times \dfrac{170}{3}+50+200\times 2=\boxed{790}.\]](http://latex.artofproblemsolving.com/8/2/6/8267e933205c7df72fe0f51938f7c776bce8be80.png)
See also
| 2014 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
