Difference between revisions of "2016 AIME II Problems/Problem 14"
Mathtiger6 (talk | contribs) (→Solution 5) |
m (→Solution 3) |
||
Line 9: | Line 9: | ||
==Solution 3== | ==Solution 3== | ||
− | To make numbers more feasible, we'll scale everything down by a factor of <math>100</math> so that <math>\overline{AB}=\overline{BC}=\overline{AC}=6</math>. We should also note that <math>P</math> and <math>Q</math> must lie on the line that is perpendicular to the plane of <math>ABC</math> and also passes through the circumcenter of <math>ABC</math> (due to <math>P</math> and <math>Q</math> being equidistant from <math>A</math>, <math>B</math>, <math>C</math>), let <math>D</math> be the altitude from <math>C</math> to <math>AB</math>. We can draw a vertical cross-section of the figure then: <asy>pair C, D, I, P, Q, O; D=(0,0); C=(5.196152,0); P=(1.732051,7.37228); I=(1.732051,0); Q=(1.732051,-1.62772); O=(1.732051,2.87228); draw(C--Q--D--P--cycle); draw(C--D, dashed); draw(P--Q, dotted); draw(O--C, dotted); label("$C$", C, E); label("$D$", D, W); label("$I$", I, NW); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, SW); dot(O); dot(I);</asy> We let <math>\angle PDI=\alpha</math> so <math>\angle QDI=120^{\circ}-\alpha</math>, also note that <math>\overline{PO}=\overline{QO}=\overline{CO}=d</math>. Because <math>I</math> is the centroid of <math>ABC</math>, we know that ratio of <math>\overline{CI}</math> to <math>\overline{DI}</math> is <math>2:1</math>. Since we've scaled the figure down, the length of <math>CD</math> is <math>3\sqrt{3}</math>, from this it's easy to know that <math>\overline{CI}=2\sqrt{3}</math> and <math>\overline{DI}=\sqrt{3}</math>. The following two equations arise: <cmath> | + | To make numbers more feasible, we'll scale everything down by a factor of <math>100</math> so that <math>\overline{AB}=\overline{BC}=\overline{AC}=6</math>. We should also note that <math>P</math> and <math>Q</math> must lie on the line that is perpendicular to the plane of <math>ABC</math> and also passes through the circumcenter of <math>ABC</math> (due to <math>P</math> and <math>Q</math> being equidistant from <math>A</math>, <math>B</math>, <math>C</math>), let <math>D</math> be the altitude from <math>C</math> to <math>AB</math>. We can draw a vertical cross-section of the figure then: <asy>pair C, D, I, P, Q, O; D=(0,0); C=(5.196152,0); P=(1.732051,7.37228); I=(1.732051,0); Q=(1.732051,-1.62772); O=(1.732051,2.87228); draw(C--Q--D--P--cycle); draw(C--D, dashed); draw(P--Q, dotted); draw(O--C, dotted); label("$C$", C, E); label("$D$", D, W); label("$I$", I, NW); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, SW); dot(O); dot(I);</asy> We let <math>\angle PDI=\alpha</math> so <math>\angle QDI=120^{\circ}-\alpha</math>, also note that <math>\overline{PO}=\overline{QO}=\overline{CO}=d</math>. Because <math>I</math> is the centroid of <math>ABC</math>, we know that ratio of <math>\overline{CI}</math> to <math>\overline{DI}</math> is <math>2:1</math>. Since we've scaled the figure down, the length of <math>CD</math> is <math>3\sqrt{3}</math>, from this it's easy to know that <math>\overline{CI}=2\sqrt{3}</math> and <math>\overline{DI}=\sqrt{3}</math>. The following two equations arise: <cmath> |
− | |||
-fatant | -fatant | ||
Revision as of 21:48, 16 March 2021
Contents
[hide]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 . -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.