Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

Example 9 Let points EE, FF, GG be the midpoints of the edges ABAB, BCBC, CDCD of the regular tetrahedron ABCDABCD. Then the size of the dihedral angle CFGEC-FG-E is ( ).
(A) arcsin63\arcsin \frac{\sqrt{6}}{3}
(B) π2+arccos33\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}
(C) π2arctan2\frac{\pi}{2}-\arctan \sqrt{2}
(1) πarccot22\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}

Multiple choice: answer with the letter of the option you want.

Solution

Analysis: It is easy to know that ACAC is parallel to plane EFGEFG and ACFGAC \perp FG (the intersection line of plane EFGEFG and plane BCDBCD). Therefore, by Basic Conclusion 12, the required dihedral angle is equal to the supplementary angle of the angle formed by ACAC and plane BCDBCD. Draw AOAO \perp plane BCDBCD, with the foot of the perpendicular being OO, then OO is the center of BCD\triangle BCD, and we get ACO\angle ACO =arccot22=\operatorname{arccot} \frac{\sqrt{2}}{2}, so the required dihedral angle is πarccot22\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}. Therefore, the correct choice is (D).

Note: Since the required dihedral angle is greater than π2\frac{\pi}{2}, after obtaining cosACO\cos \angle ACO =33=\frac{\sqrt{3}}{3}, we can directly determine that the correct choice is (D).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.