Difference between revisions of "2013 AIME I Problems/Problem 9"
m (→Solution 1) |
(→Problem 9) |
||
Line 1: | Line 1: | ||
==Problem 9== | ==Problem 9== | ||
− | A paper equilateral triangle <math>ABC</math> has side length 12. The paper triangle is folded so that vertex <math>A</math> touches a point on side <math>\overline{BC}</math> a distance 9 from point <math>B</math>. The length of the line segment along which the triangle is folded can be written as <math>\frac{m\sqrt{p}}{n}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m+n+p</math>. | + | A paper equilateral triangle <math>ABC</math> has side length <math>12</math>. The paper triangle is folded so that vertex <math>A</math> touches a point on side <math>\overline{BC}</math> a distance <math>9</math> from point <math>B</math>. The length of the line segment along which the triangle is folded can be written as <math>\frac{m\sqrt{p}}{n}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m+n+p</math>. |
+ | <asy> | ||
+ | import cse5; | ||
+ | size(12cm); | ||
+ | pen tpen = defaultpen + 1.337; | ||
+ | real a = 39/5.0; | ||
+ | real b = 39/7.0; | ||
+ | pair B = MP("B", (0,0), dir(200)); | ||
+ | pair A = MP("A", (9,0), dir(-80)); | ||
+ | pair C = MP("C", (12,0), dir(-20)); | ||
+ | pair K = (6,10.392); | ||
+ | pair M = (a*B+(12-a)*K) / 12; | ||
+ | pair N = (b*C+(12-b)*K) / 12; | ||
+ | draw(B--M--N--C--cycle, tpen); | ||
+ | draw(M--A--N--cycle); | ||
+ | fill(M--A--N--cycle, mediumgrey); | ||
+ | pair shift = (-20.13, 0); | ||
+ | pair B1 = MP("B", B+shift, dir(200)); | ||
+ | pair A1 = MP("A", K+shift, dir(90)); | ||
+ | pair C1 = MP("C", C+shift, dir(-20)); | ||
+ | draw(A1--B1--C1--cycle, tpen);</asy> | ||
== Solution 1 == | == Solution 1 == |
Revision as of 12:44, 30 December 2013
Contents
[hide]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 .
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.