Given five distinct points in the plane, no three collinear, show that four may be chosen which form the vertices of a convex quadrilateral.
Given five distinct points in the plane, no three collinear, show that four may be chosen which form the vertices of a convex quadrilateral.
Given - Five distinct points in the plane with no three collinear.
Now, let there are 5 vertices which form a convex pentagon. then any 4 of them form a convex quadrilateral. Thus, there are 5 convex quadrilaterals.
Now, let there are 4 points(p1,p2,p3,p4) which form a convex quad. in anticlockwise direction and fifth in its interior. Suppose m is the intersection of diagonals p1 p3 and p2p4. But fifth p5 doesn't lie on neither diagonal. thus, quad p1p3p4p5 and quad p2p3p4p5 are convex but other two are not. hence, there are three convex quad's.
Finally, let there are three points (p1,p2,p3) and these points form a triangle with p4 and p5 in its interior. The line p4p5 meets two of three sides, that is p1p2 and p1p3. Thus, p2p3p4p5 is only convex quadrilateral.
Given - Five distinct points in the plane with no three collinear.
Now, let there are 5 vertices which form a convex pentagon. then any 4 of them form a convex quadrilateral. Thus, there are 5 convex quadrilaterals.
Now, let there are 4 points(p1,p2,p3,p4) which form a convex quad. in anticlockwise direction and fifth in its interior. Suppose m is the intersection of diagonals p1 p3 and p2p4. But fifth p5 doesn't lie on neither diagonal. thus, quad p1p3p4p5 and quad p2p3p4p5 are convex but other two are not. hence, there are three convex quad's.
Finally, let there are three points (p1,p2,p3) and these points form a triangle with p4 and p5 in its interior. The line p4p5 meets two of three sides, that is p1p2 and p1p3. Thus, p2p3p4p5 is only convex quadrilateral.
