Difference between revisions of "2014 AIME II Problems/Problem 11"
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==Solution 2== | ==Solution 2== | ||
− | Let <math> | + | Let <math>MN = x.</math> Meanwhile, since <math>\triangle R PM</math> is similar to <math>\triangle RCD</math> (angle, side, and side- <math>RP</math> and <math>RC</math> ratio), <math>CD</math> must be 2<math>x</math>. Now, notice that <math>AE</math> is <math>x</math>, because of the parallel segments <math>\overline A\overline E</math> and <math>\overline P\overline M</math>. |
Now we just have to calculate <math>ED</math>. Using the Law of Sines, or perhaps using altitude <math>\overline R\overline O</math>, we get <math>ED = \frac{\sqrt{3}+1}{2}</math>. <math>CA=RA</math>, which equals <math>ED - x</math> | Now we just have to calculate <math>ED</math>. Using the Law of Sines, or perhaps using altitude <math>\overline R\overline O</math>, we get <math>ED = \frac{\sqrt{3}+1}{2}</math>. <math>CA=RA</math>, which equals <math>ED - x</math> | ||
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We can now use coordinates with <math>D(0,0)</math> as origin and <math>DE</math> along the <math>x</math>-axis. | We can now use coordinates with <math>D(0,0)</math> as origin and <math>DE</math> along the <math>x</math>-axis. | ||
− | Let <math>RD=4</math> instead of <math>1</math> (in the end we | + | Let <math>RD=4</math> instead of <math>1</math> (in the end we will scale down by <math>4</math>). Since <math>\angle D = 60^\circ</math>, we get <math>R(2,2\sqrt{3})</math>, and therefore <math>M(1, \sqrt{3})</math>. |
We use sine-law in <math>\triangle RED</math> to find the coordinates <math>E(2+2\sqrt{3}, 0)</math>:<cmath>DE =4\cdot \frac{\sin 75^\circ}{\sin 45^\circ} = 4(\sin 30^\circ + \cos 30^\circ) = 2+2\sqrt{3}. </cmath> | We use sine-law in <math>\triangle RED</math> to find the coordinates <math>E(2+2\sqrt{3}, 0)</math>:<cmath>DE =4\cdot \frac{\sin 75^\circ}{\sin 45^\circ} = 4(\sin 30^\circ + \cos 30^\circ) = 2+2\sqrt{3}. </cmath> | ||
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Scaling down by <math>4</math>, we get <math>AE=\tfrac 1{22}(7-3\sqrt{3})</math>, so our answer is <math>7+27+22=056</math>. | Scaling down by <math>4</math>, we get <math>AE=\tfrac 1{22}(7-3\sqrt{3})</math>, so our answer is <math>7+27+22=056</math>. | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/muM8UcGKjHo?si=C6o7-C4DgB5i4yKv | ||
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+ | ~MathProblemSolvingSkills.com | ||
+ | |||
== See also == | == See also == |
Revision as of 17:24, 20 January 2024
Problem 11
In , and . . Let be the midpoint of segment . Point lies on side such that . Extend segment through to point such that . Then , where and are relatively prime positive integers, and is a positive integer. Find .
Solution 1
Let be the foot of the perpendicular from to , so . Since triangle is isosceles, is the midpoint of , and . Thus, is a parallelogram and . We can then use coordinates. Let be the foot of altitude and set as the origin. Now we notice special right triangles! In particular, and , so , , and midpoint and the slope of , so the slope of Instead of finding the equation of the line, we use the definition of slope: for every to the left, we go up. Thus, , and , so the answer is .
Solution 2
Let Meanwhile, since is similar to (angle, side, and side- and ratio), must be 2. Now, notice that is , because of the parallel segments and .
Now we just have to calculate . Using the Law of Sines, or perhaps using altitude , we get . , which equals
Using Law of Sine in , we find = .
We got the three sides of . Now using the Law of Cosines on . There we can equate and solve for it. We got . Then rationalize the denominator, we get .
Solution 3
Let be the foot of the perpendicular from to , so . Since is isosceles, is the midpoint of , and by midpoint theorem . Thus, is a parallelogram and therefore . We can now use coordinates with as origin and along the -axis.
Let instead of (in the end we will scale down by ). Since , we get , and therefore .
We use sine-law in to find the coordinates : Since slope, and , it follows that slope. If then we have Now .
Scaling down by , we get , so our answer is .
Video Solution
https://youtu.be/muM8UcGKjHo?si=C6o7-C4DgB5i4yKv
~MathProblemSolvingSkills.com
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.