= Solution
{c}
Label the <polygon vertices> of the nondegenerate <triangle> $K$ as $a_1,a_2,a_3$, and put $m_{ij}=(a_i+a_j)/2$ for its three edge <midpoints>. The six <monomials> $1,x_1,x_2,x_1^2,x_1x_2,x_2^2$ form a <basis> of $\mathcal P_2(K)$, so its <dimension> is six. Nondegeneracy matters: $K$ has nonempty interior, and a <polynomial> vanishing on that interior is the zero <polynomial>.
Let $\lambda_1,\lambda_2,\lambda_3$ be the <affine functions> of the <barycentric coordinate system>, characterized by $\lambda_i(a_j)=\delta_{ij}$ and $\sum_i\lambda_i=1$. Define six <quadratic polynomials>
$$
N_i=\lambda_i(2\lambda_i-1)\quad(i=1,2,3),\qquad
N_{ij}=4\lambda_i\lambda_j\quad(1\leq i<j\leq3).
$$
At $a_k$, $N_i(a_k)=\delta_{ik}$ and all $N_{ij}(a_k)$ vanish. At $m_{kl}$, exactly $\lambda_k$ and $\lambda_l$ equal $1/2$, while the third is zero. Thus every $N_i$ vanishes at all three edge <midpoints>, and $N_{ij}(m_{kl})$ is one for the matching unordered pair and zero otherwise.
These six <polynomials> have the cardinal property for the six <point evaluation functionals>. Evaluating any vanishing <linear combination> at the nodes forces each of its <coefficients> to vanish. They are therefore six <linearly independent> members of a six-dimensional <vector space>, hence a <basis>.
In particular, every $p\in\mathcal P_2(K)$ has the unique expansion
$$
\boxed{p=\sum_{i=1}^3p(a_i)\lambda_i(2\lambda_i-1)+\sum_{1\leq i<j\leq3}p(m_{ij})\,4\lambda_i\lambda_j.}
$$
This proves both existence and uniqueness for arbitrary six prescribed nodal values, which is precisely an <unisolvent finite element>. It also identifies the <finite element degrees of freedom> and shape functions of the <quadratic Lagrange triangle> explicitly.
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