Difference between revisions of "2018 AMC 10A Problems/Problem 13"
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| + | In general, whenever we are asked to make a crease, think about that crease as a line of reflection over which the diagram is reflected. This is why the crease must be the perpendicular bisector of <math>AB</math>, because <math>A</math> must be reflected onto <math>B</math>. (by pulusona) | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 13:35, 1 September 2018
Contents
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
In general, whenever we are asked to make a crease, think about that crease as a line of reflection over which the diagram is reflected. This is why the crease must be the perpendicular bisector of , because must be reflected onto . (by pulusona)
Solution 2
Use the ruler and graph paper you brought to quickly draw a 3-4-5 triangle of any scale (don't trust the diagram in the booklet). Very carefully fold the acute vertices together and make a crease. Measure the crease with the ruler. If you were reasonably careful, you should see that it measures somewhat more than 
units and somewhat less than 
units. The only answer choice in range is .
This is pretty much a cop-out, but it's allowed in the rules technically.
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. 
