2016 AIME II Problems/Problem 14
Contents
Problem
Equilateral has side length . Points and lie outside the plane of and are on opposite sides of the plane. Furthermore, , and , and the planes of and form a dihedral angle (the angle between the two planes). There is a point whose distance from each of and is . Find .
Solution 1
The inradius of is and the circumradius is . Now, consider the line perpendicular to plane through the circumcenter of . Note that must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since are collinear, and , we must have is the midpoint of . Now, Let be the circumcenter of , and be the foot of the altitude from to . We must have . Setting and , assuming WLOG , we must have . Therefore, we must have . Also, we must have by the Pythagorean theorem, so we have , so substituting into the other equation we have , or . Since we want , the desired answer is .
Solution 2 (Short & Simple)
Draw a good diagram. Draw as an altitude of the triangle. Scale everything down by a factor of , so that . Finally, call the center of the triangle U. Draw a cross-section of the triangle via line , which of course includes . From there, we can call . There are two crucial equations we can thus generate. WLOG set , then we call . First equation: using the Pythagorean Theorem on , . Next, using the tangent addition formula on angles we see that after simplifying in the numerator, so . Multiply back the scalar and you get . Not that hard, was it?
Solution 3
To make numbers more feasible, we'll scale everything down by a factor of so that . We should also note that and must lie on the line that is perpendicular to the plane of and also passes through the circumcenter of (due to and being equidistant from , , ), let be the altitude from to . We can draw a vertical cross-section of the figure then: We let so , also note that . Because is the centroid of , we know that ratio of to is . Since we've scaled the figure down, the length of is , from this it's easy to know that and . The following two equations arise: Using trig identities for the tangent, we find that Okay, now we can plug this into to get: Notice that only appears in the above system of equations in the form of , we can set for convenience since we really only care about . Now we have Looking at , it's tempting to square it to get rid of the square-root so now we have: See the sneaky in the above equation? That we means we can substitute it for : Use the quadratic formula, we find that - the two solutions were expected because can be or . We can plug this into : I'll use because both values should give the same answer for . Wait! Before you get excited, remember that we scaled the entire figure by ?? That means that the answer is . -fatant
Solution 4
We use the diagram from solution 3. From basic angle chasing, so triangle QCP is a right triangle. This means that triangles and are similar. If we let and , then we know and We also know that
-EZmath2006
Solution 5
We use the diagram from solution 3.
Let and . Then, by Stewart's on , we find
The altitude from to is so
Furthermore, the altitude from to is , so, by LoC and the dihedral condition,
Squaring the equation for and substituting yields
Substituting into the other equation,
Squaring both of these gives
Substituting and solving for gives , as desired.
-mathtiger6
Solution 6 (Geometry)
Let be midpoint be the center of equilateral be the center of sphere Then (See upper diagram).
We construct the circle PQMD, use the formulas for intersecting chords and get (See lower diagram).
We apply the Law of Sine to and get We apply the Pythagorean Law on and and get vladimir.shelomovskii@gmail.com, vvsss
Solution 7
Let be the midpoint of and the center of . Then all lie in the same vertical plane. We can make the following observations:
- The equilateral triangle has side length , so and divides so that and ;
- is the midpoint of since is equidistant from – it is also the circumcenter of ;
- , the dihedral angle.
To make calculations easier, we will denote , so that and .
Denote and , where the tangent addition formula on yields Using and , we have After multiplying both numerator and denominator by we have But note that by power of a point at , where we deduce by symmetry that on the diagram below:
Thus Earlier we assigned the variable to the length which implies . Thus the distance is equal to .
Solution 8 (Law of Cosines)
Let be the center of . Let be the midpoint of . Let and . Let and . We will be working in the plane that contains the points: , , , , , and .
Since , , and are collinear and , is a right triangle with . Since , .
, , , and . By Law of Cosines . Substituting for and simplifying, we get . Squaring and simplifying, we get . Adding to both sides we get . Since is the midpoint of ,
~numerophile
See also
2016 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.