Problem 4

Let

A,B

be adjacent vertices of a regular

n

-gon (

n\ge5

) with center

O

. A triangle

XYZ

, congruent to and initially coinciding with

OAB

, moves so that

Y

and

Z

each trace the whole boundary of the polygon, while

X

remains inside the polygon. Find the locus of

X

Step 3 of 5: Fix the line containing

X

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In plain words

Equal subtended angles mean that $BX$ is the bisector of the fixed polygon angle at $B$ .

\angle XBY=\angle XBZ

Detailed analysis

Thus $BX$ bisects $\angle YBZ=\angle ABC$ . The bisector of the regular polygon angle at $B$ is the fixed line $BO$ , so every position of $X$ in this phase lies on the same line through $B$ and $O$ .