Difference between revisions of "2018 AMC 10A Problems/Problem 13"
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== Problem == | == Problem == | ||
| − | A paper triangle with sides of lengths 3,4, and 5 inches, as | + | A paper triangle with sides of lengths 3,4, and 5 inches, as shown, 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> | ||
draw((0,0)--(4,0)--(4,3)--(0,0)); | draw((0,0)--(4,0)--(4,3)--(0,0)); | ||
Revision as of 18:02, 8 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?
![[asy] draw((0,0)--(4,0)--(4,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,3), NE); label("$C$", (4,0), SE); label("$4$", (2,0), S); label("$3$", (4,1.5), E); label("$5$", (2,1.5), NW); fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray); [/asy]](http://latex.artofproblemsolving.com/2/1/1/21145c9f39a3f3cd34317f8b1f8e3f3541c3c477.png)
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
![\[\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.\]](http://latex.artofproblemsolving.com/1/7/9/1791ad4c4a4f105c0a6a8344d66645913a5abc25.png)
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.
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. 
