Problem
GEO-B1-M02-P027 Nine Unit Segments
Inside a convex pentagon \(ABCDE\), point \(O\) is chosen and connected to all vertices. Consider the five sides of the pentagon and the five segments \(OA,OB,OC,OD,OE\). What is the largest possible number of these ten segments that can be equal to \(1\)?
C. Hint 1. First prove that all ten segments cannot be unit segments.
D. Hint 2. If two neighboring radial segments and the side between them are all equal to 1, an equilateral triangle appears.
E. Full solution.
We first show that ten unit segments are impossible. If all ten segments had length \(1\), then triangles \(OAB\), \(OBC\), \(OCD\), \(ODE\), and \(OEA\) would all be equilateral.
Hence each of the angles \(\angle AOB\), \(\angle BOC\), \(\angle COD\), \(\angle DOE\), and \(\angle EOA\) would be \(60^\circ\). Their sum around point \(O\) must be \(360^\circ\), but it would be \(5\cdot60^\circ=300^\circ\), a contradiction. Thus at most nine are possible.
Now construct an example with nine. Take a regular hexagon of side \(1\) with center \(O\). Choose five consecutive vertices as \(A,B,C,D,E\). Then \(O\) lies inside the pentagon.
In this example all five segments \(OA,OB,OC,OD,OE\) are equal to \(1\), and four sides \(AB,BC,CD,DE\) are also equal to \(1\). The remaining side \(EA\) is not \(1\). Therefore the maximum is exactly nine.
A. Source analysis. Main objects: a pentagon, an interior point, and ten segments. The obvious approach is to try to make all ten unit segments, but the hidden obstruction is the angle sum around the interior point. The construction for equality uses five consecutive vertices of a regular hexagon. Number of key ideas: 2.
F. Difficulty justification. This is Level 7: a regional-style extremal problem requiring both an upper bound and an example.
G. Check. This is not a one-step exercise: the maximum is proved by forbidding ten segments and then giving a separate sharp construction for nine.