Difference between revisions of "2018 AMC 10A Problems/Problem 13"
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Simply make a 3x4x5 inch triangle and then cut it out (using fine rips). Then, make the fold and measure. It will be <math>\boxed{D} \frac{15}{8}</math> inches in length. | Simply make a 3x4x5 inch triangle and then cut it out (using fine rips). Then, make the fold and measure. It will be <math>\boxed{D} \frac{15}{8}</math> inches in length. | ||
+ | Warning: <math>\boxed{C}\frac{7}{4}=\frac{14}{8}</math> which is extremely close to <math>\boxed{D}\frac{15}{8}</math>, so make sure you're precise and have a good ruler. | ||
== See Also == | == See Also == |
Revision as of 20:32, 9 February 2018
Problem
A paper triangle with sides of lengths 3,4, and 5 inches, as shown, 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 measure. It will be inches in length.
Warning:
which is extremely close to
, so make sure you're precise and have a good ruler.
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.