Difference between revisions of "2016 AIME II Problems/Problem 14"
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<cmath>180=\angle{QOC}+\angle{CO}P=2\angle{OCP}+2\angle{OCQ}=2\angle{QCP}</cmath> | <cmath>180=\angle{QOC}+\angle{CO}P=2\angle{OCP}+2\angle{OCQ}=2\angle{QCP}</cmath> | ||
so triangle QCP is a right triangle. This means that triangles <math>CQI</math> and <math>CPI</math> are similar. If we let <math>\angle{IDQ}=x</math> and <math>\angle{PDI}=y</math>, then we know <math>x+y=120</math> and <cmath>\frac{PG}{GC}=\frac{GC}{GQ}\Rightarrow\frac{100\sqrt{3}\tan{y}}{200\sqrt{3}}=\frac{200\sqrt{3}}{100\sqrt{3}\tan{x}}\Rightarrow\tan{x}\tan{y}=4</cmath> We also know that <cmath>PQ=2d=100\sqrt{3}(\tan{x}+\tan{y})</cmath> <cmath>d=50\sqrt{3}(\tan{x}+\tan{y})</cmath> <cmath>\frac{d}{1-\tan{x}\tan{y}}=50\sqrt{3}\cdot\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}</cmath> <cmath>\frac{d}{-3}=50\sqrt{3}\tan{(x+y)}</cmath> <cmath>d=-150\sqrt{3}\tan{120}=-150\sqrt{3}(-\sqrt{3})=\boxed{450}</cmath> | so triangle QCP is a right triangle. This means that triangles <math>CQI</math> and <math>CPI</math> are similar. If we let <math>\angle{IDQ}=x</math> and <math>\angle{PDI}=y</math>, then we know <math>x+y=120</math> and <cmath>\frac{PG}{GC}=\frac{GC}{GQ}\Rightarrow\frac{100\sqrt{3}\tan{y}}{200\sqrt{3}}=\frac{200\sqrt{3}}{100\sqrt{3}\tan{x}}\Rightarrow\tan{x}\tan{y}=4</cmath> We also know that <cmath>PQ=2d=100\sqrt{3}(\tan{x}+\tan{y})</cmath> <cmath>d=50\sqrt{3}(\tan{x}+\tan{y})</cmath> <cmath>\frac{d}{1-\tan{x}\tan{y}}=50\sqrt{3}\cdot\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}</cmath> <cmath>\frac{d}{-3}=50\sqrt{3}\tan{(x+y)}</cmath> <cmath>d=-150\sqrt{3}\tan{120}=-150\sqrt{3}(-\sqrt{3})=\boxed{450}</cmath> | ||
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+ | -EXmath2006 | ||
== See also == | == See also == |
Revision as of 14:29, 29 December 2019
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 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
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
-EXmath2006
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.