Difference between revisions of "2013 AIME I Problems/Problem 9"
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== Solution 3 (Coordinate Bash) == | == Solution 3 (Coordinate Bash) == | ||
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+ | e let the original position of <math>A</math> be <math>A</math>, and the position of <math>A</math> after folding be <math>D</math>. Also, we put the triangle on the coordinate plane such that <math>A=(0,0)</math>, <math>B=(-6,-6\sqrt3)</math>, <math>C=(6,-6\sqrt3)</math>, and <math>D=(3,-6\sqrt3)</math>. | ||
+ | |||
+ | [asy] | ||
+ | import cse5; | ||
+ | size(10cm); | ||
+ | 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 = (9,0); | ||
+ | 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); | ||
+ | draw(M--A--N--cycle); | ||
+ | label("<math>D</math>", A, S); | ||
+ | pair X = (6,6*sqrt(3)); | ||
+ | draw(B--X--C); | ||
+ | label("<math>A</math>",X,dir(90)); | ||
+ | draw(A--X); | ||
+ | [/asy] | ||
+ | |||
+ | Note that since <math>A</math> is reflected over the fold line to <math>D</math>, the fold line is the perpendicular bisector of <math>AD</math>. We know <math>A=(0,0)</math> and <math>D=(3,-6\sqrt3)</math>. The midpoint of <math>AD</math> (which is a point on the fold line) is <math>(\tfrac32, -3\sqrt3)</math>. Also, the slope of <math>AD</math> is <math>\frac{-6\sqrt3}{3}=-2\sqrt3</math>, so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of <math>AD</math>, or <math>\frac{1}{2\sqrt3}=\frac{\sqrt3}{6}</math>. Then, using point slope form, the equation of the fold line is | ||
+ | <cmath>y+3\sqrt3=\frac{\sqrt3}{6}\left(x-\frac32\right)</cmath><cmath>y=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}</cmath> | ||
+ | Note that the equations of lines <math>AB</math> and <math>AC</math> are <math>y=\sqrt3x</math> and <math>y=-\sqrt3x</math>, respectively. We will first find the intersection of <math>AB</math> and the fold line by substituting for <math>y</math>: | ||
+ | <cmath>\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}</cmath><cmath>\frac{5\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=-\frac{39}{10}</cmath> | ||
+ | Therefore, the point of intersection is <math>\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)</math>. Now, lets find the intersection with <math>AC</math>. Substituting for <math>y</math> yields | ||
+ | <cmath>-\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}</cmath><cmath>\frac{-7\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=\frac{39}{14}</cmath> | ||
+ | Therefore, the point of intersection is <math>\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)</math>. Now, we just need to use the distance formula to find the distance between <math>\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)</math> and <math>\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)</math>. | ||
+ | <cmath>\sqrt{\left(\frac{39}{14}+\frac{39}{10}\right)^2+\left(-\frac{39\sqrt3}{14}+\frac{39\sqrt3}{10}\right)^2}</cmath> | ||
+ | The number 39 is in all of the terms, so let's factor it out: | ||
+ | <cmath>39\sqrt{\left(\frac{1}{14}+\frac{1}{10}\right)^2+\left(-\frac{\sqrt3}{14}+\frac{\sqrt3}{10}\right)^2}=39\sqrt{\left(\frac{6}{35}\right)^2+\left(\frac{\sqrt3}{35}\right)^2}</cmath><cmath>\frac{39}{35}\sqrt{6^2+\sqrt3^2}=\frac{39\sqrt{39}}{35}</cmath> | ||
+ | Therefore, our answer is <math>39+39+35=\boxed{113}</math>, and we are done. | ||
+ | |||
+ | Solution by nosaj. | ||
== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=8|num-a=10}} | {{AIME box|year=2013|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:52, 26 May 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
.
[asy]
import cse5;
size(10cm);
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 = (9,0);
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);
draw(M--A--N--cycle);
label("", A, S);
pair X = (6,6*sqrt(3));
draw(B--X--C);
label("
",X,dir(90));
draw(A--X);
[/asy]
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 |
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