Difference between revisions of "2014 AIME II Problems/Problem 11"
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+ | ==Solution 2== | ||
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+ | NOTE: Draw your own picture for this one! It requires some tenacity. | ||
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+ | First, we are stuck, right? Nowhere to go. Wait a minute... there's medians and midpoints. Maybe we should draw auxiliary segment <math>MP</math>. For lack of better term, let's call the length <math>x</math>. Meanwhile, because <math>\triangle RPM</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 AE</math> and <math>\overline PM</math> (also using Midsegment Theorem). | ||
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+ | Now we just have to calculate <math>ED</math>. Using the Law of Sines, we get <math>ED = \frac{\sqrt{3}+1}{2}</math>, thus <math>CA = RA</math>, which must be equal to <math>ED - x</math>. (That's minus <math>2x</math> and plus <math>x</math>.) | ||
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+ | Finally, what is <math>RE</math>? It comes out to <math>\frac{\sqrt{6}}{2}</math>. | ||
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+ | We got the three sides. Now all that is left is using the Law of Cosines. | ||
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+ | Taking <math>\triangle AER</math> and using <math>\angle AER</math>, of course, we find out (after some calculation) that <math>AE = \frac{7 - \sqrt{27}}{22}</math>. | ||
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== See also == | == See also == |
Revision as of 17:51, 18 March 2017
Contents
[hide]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
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
NOTE: Draw your own picture for this one! It requires some tenacity.
First, we are stuck, right? Nowhere to go. Wait a minute... there's medians and midpoints. Maybe we should draw auxiliary segment . For lack of better term, let's call the length . Meanwhile, because is similar to (angle, side, and side- and ratio), must be 2. Now, notice that is , because of the parallel segments and (also using Midsegment Theorem).
Now we just have to calculate . Using the Law of Sines, we get , thus , which must be equal to . (That's minus and plus .)
Finally, what is ? It comes out to .
We got the three sides. Now all that is left is using the Law of Cosines.
Taking and using , of course, we find out (after some calculation) that .
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.