Problem 10

Since ratio is independent of side length so let vertex as complex 12th roots of unity.This is simple application of complex numbers.

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$We\quad will\quad solve\quad in\quad complex\quad plane,\\ \\ \theta =\cfrac { 2\pi }{ 12 } =\cfrac { \pi }{ 6 } ,let\quad A\quad is\quad at\quad { e }^{ i\left( 0 \right) }\\ \\ B={ e }^{ i\cfrac { \pi }{ 6 } },F={ e }^{ i\cfrac { 5\pi }{ 6 } }\\$ $\\ \Rightarrow AB=\left| B-A \right| =\left| { e }^{ i\cfrac { \pi }{ 6 } }-1 \right| =\sqrt { 2-\sqrt { 3 } } \\ \\ \& \quad AF=\left| F-A \right| =\left| { e }^{ i\cfrac { 5\pi }{ 6 } }-1 \right| =\sqrt { 2+\sqrt { 3 } } \\ \\ \therefore \quad \cfrac { AB }{ AF } +\cfrac { AF }{ AB } =\cfrac { { AB }^{ 2 }+{ AF }^{ 2 } }{ AB.AF } =\cfrac { 2-\sqrt { 3 } +2+\sqrt { 3 } }{ \sqrt { 4-3 } } =4$

Angle half the side makes at the center for AB 15 degrees, and for AF=5 * 15=75.
So AB/AF=(2RSin15) / (2RSin75) +(2RSin15) / (2RSin75)= Tan15 + 1/Tan15=4. Why is it Level 5?