MathLabs

Problem 5

Let ABCDEF ABCDEF be a convex hexagon such that AB∥DE AB\parallel DE , BC∥EF BC\parallel EF , and CD∥FA CD\parallel FA . Let RA,RC,RE R_A,R_C,R_E be the circumradii of triangles FAB FAB , BCD BCD , DEF DEF , respectively, and let p p be the perimeter of the hexagon. Prove that RA+RC+RE≥p/2 R_A+R_C+R_E\ge p/2.
Step 1 of 4: Write the circumradius formula
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In plain words

Write the circumradius formula

2RA=BFsin⁡A2R_A=\dfrac{BF}{\sin A}
Detailed analysis

For a triangle, 2R=(opposite side)/sin⁡(opposite angle)2R=\text{(opposite side)}/\sin(\text{opposite angle}). Therefore 2RA=BF/sin⁡A2R_A=BF/\sin A , 2RC=BD/sin⁡C2R_C=BD/\sin C , and 2RE=FD/sin⁡E2R_E=FD/\sin E , where A,B,C,D,E,F A,B,C,D,E,F are the hexagon angles.