2013 AIME I Problems/Problem 9
Contents
Problem
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. We have and , so . Denote ; we get .
In triangle , , and .
In triangle , we get and then use sine-law to get ; similarly, from triangle we get . Thus Since , we get Then
The answer is .
Solution 2
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 3
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 .
Note
Once you find and , you can scale down the triangle by a factor of so that all sides are integers. Applying Law of cosines becomes easier, you just need to remember to scale back up.
Solution 4 (Coordinate Bash)
We 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.
Solution 5
Note: this requires lots of calculations that increase your chance of errors, but it only requires simple understanding of areas, similar triangles, and Heron's formula. I'll just put the strategy here because I am too lazy to calculate it myself right now.
1. notice that the two triangles on the sides of the folded corner are similar. using this, we can find that the side lengths of them are 9,7.8,4.2 and 3, 45/7, 39/7 2. use heron's formula to find the areas of those two triangles. remember that it is sqrt[s(s-a)(s-b)(s-c)] 3. using the area of these triangles, we can find the area of the triangle with the length we need. 4. use heron's formula again, with the unknown length as x, and since we know the area and the other two side lengths, we can just solve for x with this equation.
-EmilyQ
Solution 6(Easy)
Thanks to Solution 1 for the diagram below:
We will use the notation already on the diagram, but our solution is slightly different.
We will only need M and N.
Let NC be length a, which implies NA be 12-a.
Also, AC = 3 because AB = 9
By the Law of Cosines on NCA,
which simplifies to:
Which means that NC = 45/7 and NA = 39/7.
We can do the same thing for MBA.
This time, MB = b.
Which gives:
Which implies that MA = Now, since MAN is 60 degrees, we can apply the Law of Cosines again(I know, I don't like bashy things too) to get:
Which leads us to our answer => 113
~MC
Video Solution
https://www.youtube.com/watch?v=581ZtcQFCaE&t=98s
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 | ||
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