Euclidea 2.8 3e Fix 📢

Circle ( \omega ) with center ( O ) and point ( A ) on ( \omega ). Goal: Construct all vertices of a square inscribed in ( \omega ) using exactly 3 elementary moves.

Since ( AC ) is a diameter, step 2’s circle has center ( A ) radius ( 2R ). The chord ( PQ ) is the radical axis of ( \omega ) and ( \Gamma ), which lies on the perpendicular bisector of ( AC ), hence passes through ( O ). Therefore ( BD ) is the diameter perpendicular to ( AC ), forming a square with vertices at the endpoints of these perpendicular diameters. euclidea 2.8 3e

I’ll provide a clear, step-by-step solution for — constructing the inscribed square in a given circle using the 3E (three elementary moves) constraint. This is a known minimal-solution challenge in Euclidea. Problem statement (Euclidea 2.8) Given a circle (center ( O ), point ( A ) on circumference). Construct an inscribed square in 3 elementary moves (E). Allowed moves: line (through two points), circle (center + point), perpendicular bisector, parallel line, angle bisector, etc., but each counts as 1E if it's a single construction tool usage. 3E solution outline The key: In a circle, two perpendicular diameters give the 4 vertices of an inscribed square. With just 3 moves, you can’t draw both diameters fully, but you can get their intersection points. Circle ( \omega ) with center ( O

Draw line ( OA ), get ( C ) (other intersection). 2E: Draw circle with center ( A ) through ( C ). This circle intersects the original circle at points ( P ) and ( Q ). 3E: Draw line ( PQ ). Where is ( PQ )? It is the perpendicular bisector of ( AC ) and passes through ( O ). So ( PQ ) is the perpendicular line to ( OA ) through ( O ). The chord ( PQ ) is the radical

Now ( PQ ) meets the original circle at ( B ) and ( D ). Thus points ( A, B, C, D ) are the vertices of the inscribed square, formed by diameters ( AC ) and ( BD ) which are perpendicular.