The figure below is a trigonal pyramid. △ABC is an equilateral triangle. OA = OB = OC. ∠ AOB = ∠ BOC = ∠COA 30 degrees. Point P, Q, R is fixed respectively on the side OB, OC, OA so that AP + PQ + QR + RB may be minimized. Find PQ : RB.
Answer
1 ： 2
Solution
In the figure below, ∠AOB' = 120 °. ∠OAP = ∠OB'R = 30°.
△OPR is an equilateral triangle.
Answer
1 ： 2
Solution
In the figure below, ∠AOB' = 120 °. ∠OAP = ∠OB'R = 30°.
△OPR is an equilateral triangle.
OP = PR, RB = RB' = PR.
Thus, PQ : RB = PQ : PR = 1 : 2.
