Deg Graph

Graph Theory... Hamiltonian Proof?

I'm terrible at proof, any help would be great! Thanks

Theorem: Let G be a simple graph with p vertices, p >= 2, such that for each vertex v of G, deg v >= 1/2(p-1). Then G contains a Hamiltonian path.

Hint: Construct a graph H from G by adding a new vertex w to G and joining w to each vertex of G. Then show H is Hamiltonian and use a Hamiltonian cycle of H to construct the Hamiltonian path in G.

I have not done graph theory in years some of my terminology is probably wrong; but the proof should be correct.

Use ore's theorem

"Let G be a simple graph of order n>=3 such that, for every pair of distinct non adjacent vertices and u and v, deg(u) + deg(v) >= n . Then G is a Hamiltonian graph."

If we construct H, then each vertex increases in degree by 1. Hence now deg >= 1/2(p-1) + 1 =.5p + .5 this is equivelent to saying deg>=.5 (p+1). Hence for any two verticies the sum of they're degree will be >= (p+1) our graph has p+1 verticies now so this satisfies the conditions for Ore's theorem. So H is hamiltonian. Take a cycle, and take away w. This is now a path in G.

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