Difference between revisions of "2016 AIME II Problems/Problem 14"
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The altitude from <math>P</math> to <math>ABC</math> is <math>\sqrt{a^2 - (200\sqrt{3})^2}</math> so | The altitude from <math>P</math> to <math>ABC</math> is <math>\sqrt{a^2 - (200\sqrt{3})^2}</math> so | ||
<cmath>PQ = 2x = \sqrt{a^2 - (200\sqrt{3})^2} + \sqrt{b^2 - (200\sqrt{3})^2}.</cmath> | <cmath>PQ = 2x = \sqrt{a^2 - (200\sqrt{3})^2} + \sqrt{b^2 - (200\sqrt{3})^2}.</cmath> | ||
+ | |||
Furthermore, the altitude from <math>P</math> to <math>AB</math> is <math>\sqrt{a^2 - 300^2}</math>, so, by LoC and the dihedral condition, | Furthermore, the altitude from <math>P</math> to <math>AB</math> is <math>\sqrt{a^2 - 300^2}</math>, so, by LoC and the dihedral condition, | ||
<cmath>a^2 - 300^2 + b^2 - 300^2 + \sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 4x^2.</cmath> | <cmath>a^2 - 300^2 + b^2 - 300^2 + \sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 4x^2.</cmath> | ||
+ | |||
Squaring the equation for <math>PQ</math> and substituting <math>a^2 + b^2 = 4x^2</math> yields | Squaring the equation for <math>PQ</math> and substituting <math>a^2 + b^2 = 4x^2</math> yields | ||
<cmath>2\sqrt{a^2 - (200\sqrt{3})^2}\sqrt{b^2 - (200\sqrt{3})^2} = 6\cdot 200^2.</cmath> | <cmath>2\sqrt{a^2 - (200\sqrt{3})^2}\sqrt{b^2 - (200\sqrt{3})^2} = 6\cdot 200^2.</cmath> | ||
+ | |||
Substituting <math>a^2 + b^2 = 4x^2</math> into the other equation, | Substituting <math>a^2 + b^2 = 4x^2</math> into the other equation, | ||
<cmath>\sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 2\cdot 300^2.</cmath> | <cmath>\sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 2\cdot 300^2.</cmath> |
Revision as of 18:28, 7 March 2021
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 .
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
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
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.