Is finding a clique containing at least half of the vertices in NP-complete?

Question

I guess one can answer with yes, but I couldn't come with a reduction.

I tried to reduce from the normal clique-problem with an input $k$.

Is there a clique with a minimum of $k$ vertices?

Case 1: $k \geq \frac{|V(G))|}{2}$ one can add m vertices with $m := 2k - |V(G)|$

Case 2: $k < \frac{|V(G))|}{2}$ is not clear to me

Answer 1

The answer is yes.

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Given a graph $G=(V, E)$ and a number $k$, I'll describe a reduction from $$L:=\{(G,k)|G \text{ contains a clique of size } \geq k\}$$ to the language

$$HC:=\{(G)|G \text{ contains a clique of size } \geq\vert\frac{V(G)}{2}\vert\}$$

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Build a new graph $H=G\cup K_m$ were $m=\vert V(G)\vert - k$.
Also, add edges by connecting every vertex in $K_m$ with every vertex in $G$

Prop: $H$ has a clique of size $\geq \vert\frac{V(H)}{2}\vert\iff G$ has a clique of size $\geq k$