Difference between revisions of "2018 AMC 10A Problems/Problem 13"
(→Solution 4) |
Harsha12345 (talk | contribs) m (→Solution 2) |
||
Line 24: | Line 24: | ||
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) | 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(draw it)== |
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 <math>\frac{7}{4}</math> units and somewhat less than <math>2</math> units. The only answer choice in range is <math>\boxed{\textbf{D) } \frac{15}{8}}</math>. | 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 <math>\frac{7}{4}</math> units and somewhat less than <math>2</math> units. The only answer choice in range is <math>\boxed{\textbf{D) } \frac{15}{8}}</math>. | ||
This is pretty much a cop-out, but it's allowed in the rules technically. | This is pretty much a cop-out, but it's allowed in the rules technically. | ||
+ | |||
==Solution 3== | ==Solution 3== | ||
Revision as of 20:28, 27 January 2019
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
Note
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(draw it)
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.
Solution 3
Since is a right triangle, we can see that the slope of line is = . We know that if we fold so that point meets point the crease line will be perpendicular to and we also know that the slopes of perpendicular lines are negative reciprocals of one another. Then, we can see that the slope of our crease line is . Let us call the midpoint of point , the midpoint of point , and the crease line . We know that is parallel to and that 's length is . Using our slope calculation from earlier, we can see that. With this information, we can solve for : We can then use the Pythagorean Theorem to find . Thus, our answer is .
Solution 4
Make use of the diagram in Solution 3. It can be deduced that . Let . In , . Then also would be .
In , . After some quick math, we get . Solving for will give .
.
~OlutosinNGA
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.