Problem:
Equilateral △ABC has side length 600. Points P and Q lie outside the plane of △ABC and are on the opposite sides of the plane. Furthermore, PA=PB=PC, and QA=QB=QC, and the planes of △PAB and △QAB form a 120∘ dihedral angle (the angle between the two planes). There is a point O whose distance from each of A,B,C,P, and Q is d. Find d.
Solution:
Let H be the foot of the perpendicular from P to the plane of △ABC. Because PA=PB=PC, it follows that HA=HB=HC; that is, H is the centroid of the equilateral △ABC. Likewise H is also the foot of the perpendicular from Q to the plane of △ABC. Hence O is the midpoint of PQ and PQ=2d.
Let D be the midpoint of side AB. Hence PD⊥AB and QD⊥AB, from which it follows that ∠PDQ is the angle formed by planes ABP and ABQ, and so ∠PDQ=120∘. Let ∠PDH=x and ∠QDH=y. Then tan(x+y)=−3. Set AB=a. Then DC=23a,DH=31DC=63a, and PH=tanx⋅DH=63atanx. Likewise QH=63atany. Hence 2d=PQ=PH+HQ=63a(tanx+tany), or
tanx+tany=a43d
Because OP=OC=OQ, conclude that O, the midpoint of PQ, is the circumcenter of △PCQ, from which it follows that ∠PCQ=90∘. Then CH is the altitude to the hypotenuse of right △CPQ, implying that CH2=PH⋅QH. Hence
CH2=(32DC)2=3a2=PH⋅HQ=12a2tanxtany
implying that tanxtany=4.
By the Tangent Angle Addition Formula, tan(x+y)=1−tanxtanytanx+tany or
−3=1−4a43d
implying that d=43a. Substituting a=600 in the last equation yields d=450.