Difference between revisions of "2016 AIME II Problems/Problem 14"
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==Short Simple Solution== | ==Short Simple Solution== | ||
− | Draw a good diagram. Draw <math>CH</math> as an altitude of the triangle. Scale everything down by a factor of <math>100\sqrt{3}</math>, so that <math>AB=2\sqrt{3}</math>. Finally, call the center of the triangle U. Draw a cross-section of the triangle via line <math>CH</math>, which of course includes <math>P, Q</math>. From there, we can call <math>OU=h</math>. There are two crucial equations we can thus generate. WLOG set <math>PU<QU</math>, then we call <math>PU=d-h, QU= | + | Draw a good diagram. Draw <math>CH</math> as an altitude of the triangle. Scale everything down by a factor of <math>100\sqrt{3}</math>, so that <math>AB=2\sqrt{3}</math>. Finally, call the center of the triangle U. Draw a cross-section of the triangle via line <math>CH</math>, which of course includes <math>P, Q</math>. From there, we can call <math>OU=h</math>. There are two crucial equations we can thus generate. WLOG set <math>PU<QU</math>, then we call <math>PU=d-h, QU=d+h</math>. First equation: using the Pythagorean Theorem on <math>\triangle UOB</math>, <math>h^2+2^2=d^2</math>. Next, using the tangent addition formula on angles <math>\angle PHU, \angle UHQ</math> we see that after simplifying <math>-d^2+h^2=-4, 2d=3\sqrt{3}</math> in the numerator, so <math>d=\frac{3\sqrt{3}}{2}</math>. Multiply back the scalar and you get <math>\boxed{450}</math>. Not that hard, was it? |
==Solution 3== | ==Solution 3== |
Revision as of 00:29, 26 February 2018
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 by Shaddoll
Short Simple Solution
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 . An alternate way of proceeding after finding (credit to riemanntensor), was to average the two possible values, you can see for yourself why this would work.
-fatant
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.