Difference between revisions of "2013 AIME I Problems/Problem 9"
(→Solution 3 (Coordinate Bash)) |
Daniyalqazi2 (talk | contribs) m (→Solution 1) |
||
Line 48: | Line 48: | ||
<math>b = \frac{39}{7}</math> | <math>b = \frac{39}{7}</math> | ||
− | Using the Law of Cosines on <math> | + | Using the Law of Cosines on <math>DPQ</math>: |
<math>x^{2} = a^{2} + b^{2} - 2ab \cos{60}</math> | <math>x^{2} = a^{2} + b^{2} - 2ab \cos{60}</math> |
Revision as of 23:51, 7 August 2017
Problem 9
A paper equilateral triangle has side length . The paper triangle is folded so that vertex touches a point on side a distance from point . The length of the line segment along which the triangle is folded can be written as , where , , and are positive integers, and are relatively prime, and is not divisible by the square of any prime. Find .
Solution 1
Let and be the points on and , respectively, where the paper is folded.
Let be the point on where the folded touches it.
Let , , and be the lengths , , and , respectively.
We have , , , , , and .
Using the Law of Cosines on :
Using the Law of Cosines on :
Using the Law of Cosines on :
The solution is .
Solution 2
Proceed with the same labeling as in Solution 1.
Therefore, .
Similarly, .
Now, and are similar triangles, so
.
Solving this system of equations yields and .
Using the Law of Cosines on :
The solution is .
Solution 3 (Coordinate Bash)
e let the original position of be , and the position of after folding be . Also, we put the triangle on the coordinate plane such that , , , and .
Note that since is reflected over the fold line to , the fold line is the perpendicular bisector of . We know and . The midpoint of (which is a point on the fold line) is . Also, the slope of is , so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of , or . Then, using point slope form, the equation of the fold line is Note that the equations of lines and are and , respectively. We will first find the intersection of and the fold line by substituting for : Therefore, the point of intersection is . Now, lets find the intersection with . Substituting for yields Therefore, the point of intersection is . Now, we just need to use the distance formula to find the distance between and . The number 39 is in all of the terms, so let's factor it out: Therefore, our answer is , and we are done.
Solution by nosaj.
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.