2014 AIME II Problems/Problem 14
Contents
Problem
In , and . Let and be points on the line such that , , and . Point is the midpoint of the segment , and point is on ray such that . Then , where and are relatively prime positive integers. Find .
Diagram
Solution 1
Let us just drop the perpendicular from to and label the point of intersection . We will use this point later in the problem. As we can see, is the midpoint of and is the midpoint of is a triangle, so . is triangle.
and are parallel lines so is triangle also. Then if we use those informations we get and and or . Now we know that , we can find for which is simpler to find. We can use point to split it up as , We can chase those lengths and we would get , so , so , so We can also use Law of Sines: Then using right triangle , we have So . And we know that . Finally if we calculate . . So our final answer is .
-Gamjawon -edited by srisainandan6 to clarify and correct a small mistake
Solution 2
Here's a solution that doesn't need .
As above, get to . As in the figure, let be the foot of the perpendicular from to . Then is a 45-45-90 triangle, and is a 30-60-90 triangle. So and ; also, , , and . But and are parallel, both being orthogonal to . Therefore , or , and we're done.
Solution 3
Break our diagram into 2 special right triangle by dropping an altitude from to we then get that Since is a 45-45-90,
We know that and are 30-60-90. Thus,
. So our final answer is .
Solution 4
Draw the . Now, take the perpendicular bisector of to intersect the circumcircle of and at as shown, and denote to be the circumcenter of . It is not difficult to see by angle chasing that is cyclic, namely with diameter . Then, by symmetry, and as are both subtended by equal arcs they are equal. Hence, . Now, draw line and intersect it at at point in the diagram. It is not hard to use angle chase to arrive at a parallelogram, and from our length condition derived earlier, . From here, it is clear that ; that is, is just the intersection of the perpendicular from down to and ! After this point, note that . It is easily derived that the circumradius of is . Now, is a triangle, and from here it is easy to arrive at the final answer of . ~awang11's sol
Solution 5
Let
Let
points are collinear.
In
In vladimir.shelomovskii@gmail.com, vvsss
Video solution
https://www.youtube.com/watch?v=SvJ0wDJphdU
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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