Difference between revisions of "2014 AIME II Problems/Problem 11"
Expilncalc (talk | contribs) m (→Solution 2: Fixed parts) |
|||
Line 45: | Line 45: | ||
==Solution 2== | ==Solution 2== | ||
− | + | Call <math>MP</math> <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 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> | ||
Finally, what is <math>RE</math>? It comes out to <math>\frac{\sqrt{6}}{2}</math>. | Finally, what is <math>RE</math>? It comes out to <math>\frac{\sqrt{6}}{2}</math>. | ||
− | We got the three sides. Now all that is left is using the Law of Cosines. | + | We got the three sides. Now all that is left is using the Law of Cosines. There we can equate <math>x</math> and solve for it. |
− | 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>. | + | 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>. The step before? <math>x=\frac{1-\sqrt{3}}{4\sqrt{3}+2}</math>. |
== See also == | == See also == |
Revision as of 22:16, 16 February 2018
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
Call . Meanwhile, because 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
Finally, what is ? It comes out to .
We got the three sides. Now all that is left is using the Law of Cosines. There we can equate and solve for it.
Taking and using , of course, we find out (after some calculation) that . The step before? .
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.