Difference between revisions of "2018 AMC 10A Problems/Problem 13"
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A paper triangle with sides of lengths 3,4, and 5 inches, as shon, is folded so that point <math>A</math> falls on point <math>B</math>. What is the length in inches of the crease? | A paper triangle with sides of lengths 3,4, and 5 inches, as shon, is folded so that point <math>A</math> falls on point <math>B</math>. What is the length in inches of the crease? | ||
<asy> | <asy> | ||
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{{AMC10 box|year=2018|ab=A|num-b=12|num-a=14}} | {{AMC10 box|year=2018|ab=A|num-b=12|num-a=14}} | ||
+ | {{AMC12 box|year=2018|ab=A|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:31, 8 February 2018
Problem
A paper triangle with sides of lengths 3,4, and 5 inches, as shon, is folded so that point falls on point . What is the length in inches of the crease?
Solution 1
First, we need to realize that the crease line is just the perpendicular bisector of side , the hypotenuse of right triangle . Call the midpoint of point . Draw this line and call the intersection point with as . Now, is similar to by similarity. Setting up the ratios, we find that Thus, our answer is .
~Nivek
Solution 2 (if you are already out of time)
Simply make a 3x4x5 inch triangle and then cut it out (using fine rips). Then, make the fold and mesure. It will be inches in length.
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.