Difference between revisions of "2013 AIME I Problems/Problem 9"
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− | ==Problem 9== | + | ==Problem== |
+ | 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 == | ||
+ | Let <math>M</math> and <math>N</math> be the points on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, where the paper is folded. Let <math>D</math> be the point on <math>\overline{BC}</math> where the folded <math>A</math> touches it. | ||
+ | <asy> import cse5; size(8cm); 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", 12*dir(60), dir(90)); pair C = MP("C", (12,0), dir(-20)); pair D = MP("D", (9,0), dir(-80)); pair Y = MP("Y", midpoint(A--D), dir(-50)); pair M = MP("M", extension(A,B,Y,Y+(dir(90)*(D-A))), dir(180)); pair N = MP("N", extension(A,C,M,Y), dir(20)); pair F = MP("F", foot(A,B,C), dir(-90)); pair X = MP("X", extension(A,F,M,N), dir(-120)); | ||
+ | draw(B--A--C--cycle, tpen); draw(M--N^^F--A--D); draw(rightanglemark(D,F,A,15)); draw(rightanglemark(A,Y,M,15)); MA("\theta",F,A,D,1.8); | ||
+ | </asy> | ||
+ | We have <math>AF=6\sqrt{3}</math> and <math>FD=3</math>, so <math>AD=3\sqrt{13}</math>. Denote <math>\angle DAF = \theta </math>; we get <math>\cos\theta = 2\sqrt{3}/\sqrt{13}</math>. | ||
+ | |||
+ | In triangle <math>AXY</math>, <math>AY=\tfrac 12 AD = \tfrac 32 \sqrt{13}</math>, and <math>AX=AY\sec\theta =\tfrac{13}{4}\sqrt{3}</math>. | ||
+ | |||
+ | In triangle <math>AMX</math>, we get <math>\angle AMX=60^\circ-\theta</math> and then use sine-law to get <math>MX=\tfrac 12 AX\csc(60^\circ-\theta)</math>; similarly, from triangle <math>ANX</math> we get <math>NX=\tfrac 12 AX\csc(60^\circ+\theta)</math>. Thus <cmath>MN=\tfrac 12 AX(\csc(60^\circ-\theta) +\csc(60^\circ+\theta)).</cmath> Since <math>\sin(60^\circ\pm \theta) = \tfrac 12 (\sqrt{3}\cos\theta \pm \sin\theta)</math>, we get | ||
+ | <cmath>\begin{align*} | ||
+ | \csc(60^\circ-\theta) +\csc(60^\circ+\theta) &= \frac{\sqrt{3}\cos\theta}{\cos^2\theta - \tfrac 14} = \frac{24 \cdot \sqrt{13}}{35} | ||
+ | \end{align*}</cmath> | ||
+ | Then <cmath>MN = \frac 12 AX \cdot \frac{24 \cdot \sqrt{13}}{35} = \frac{39\sqrt{39}}{35}</cmath> | ||
+ | |||
+ | The answer is <math>39 + 39 + 35 = \boxed{113}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | Let <math>P</math> and <math>Q</math> be the points on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, where the paper is folded. | ||
+ | |||
+ | Let <math>D</math> be the point on <math>\overline{BC}</math> where the folded <math>A</math> touches it. | ||
+ | |||
+ | Let <math>a</math>, <math>b</math>, and <math>x</math> be the lengths <math>AP</math>, <math>AQ</math>, and <math>PQ</math>, respectively. | ||
+ | |||
+ | We have <math>PD = a</math>, <math>QD = b</math>, <math>BP = 12 - a</math>, <math>CQ = 12 - b</math>, <math>BD = 9</math>, and <math>CD = 3</math>. | ||
+ | |||
+ | Using the Law of Cosines on <math>BPD</math>: | ||
+ | |||
+ | <math>a^{2} = (12 - a)^{2} + 9^{2} - 2 \times (12 - a) \times 9 \times \cos{60}</math> | ||
+ | |||
+ | <math>a^{2} = 144 - 24a + a^{2} + 81 - 108 + 9a</math> | ||
+ | |||
+ | <math>a = \frac{39}{5}</math> | ||
+ | |||
+ | Using the Law of Cosines on <math>CQD</math>: | ||
+ | |||
+ | <math>b^{2} = (12 - b)^{2} +3^{2} - 2 \times (12 - b) \times 3 \times \cos{60}</math> | ||
+ | |||
+ | <math>b^{2} = 144 - 24b + b^{2} + 9 - 36 + 3b</math> | ||
+ | |||
+ | <math>b = \frac{39}{7}</math> | ||
+ | |||
+ | Using the Law of Cosines on <math>DPQ</math>: | ||
+ | |||
+ | <math>x^{2} = a^{2} + b^{2} - 2ab \cos{60}</math> | ||
+ | |||
+ | <math>x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})</math> | ||
+ | |||
+ | <math>x = \frac{39 \sqrt{39}}{35}</math> | ||
+ | |||
+ | The solution is <math>39 + 39 + 35 = \boxed{113}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | Proceed with the same labeling as in Solution 1. | ||
+ | |||
+ | <math>\angle B = \angle C = \angle A = \angle PDQ = 60^\circ</math> | ||
+ | |||
+ | <math>\angle PDB + \angle PDQ + \angle QDC = \angle QDC + \angle CQD + \angle C = 180^\circ</math> | ||
+ | |||
+ | Therefore, <math>\angle PDB = \angle CQD</math>. | ||
+ | |||
+ | Similarly, <math>\angle BPD = \angle QDC</math>. | ||
+ | |||
+ | Now, <math>\bigtriangleup BPD</math> and <math>\bigtriangleup CDQ</math> are similar triangles, so | ||
+ | |||
+ | <math>\frac{3}{12-a} = \frac{12-b}{9} = \frac{b}{a}</math>. | ||
+ | |||
+ | Solving this system of equations yields <math>a = \frac{39}{5}</math> and <math>b = \frac{39}{7}</math>. | ||
+ | |||
+ | Using the Law of Cosines on <math>APQ</math>: | ||
+ | |||
+ | <math>x^{2} = a^{2} + b^{2} - 2ab \cos{60}</math> | ||
+ | |||
+ | <math>x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})</math> | ||
+ | |||
+ | <math>x = \frac{39 \sqrt{39}}{35}</math> | ||
+ | |||
+ | The solution is <math>39 + 39 + 35 = \boxed{113}</math>. | ||
+ | |||
+ | ===Note=== | ||
+ | Once you find <math>DP</math> and <math>DQ</math>, you can scale down the triangle by a factor of <math>\frac{39}{35}</math> 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 <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> | ||
+ | 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("$D$", A, S); | ||
+ | pair X = (6,6*sqrt(3)); | ||
+ | draw(B--X--C); | ||
+ | label("$A$",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. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=581ZtcQFCaE&t=98s | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2013|n=I|num-b=8|num-a=10}} | ||
+ | {{MAA Notice}} |
Revision as of 23:05, 3 January 2022
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.
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 | ||
All AIME Problems and Solutions |
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