In a Finnish matriculation examination was the following problem
Let $a_1,a_2,\ldots,a_n$ be reals. For what value the parameter $x$ should be given if one wants to minimize the value of the sum $(x-a_1)^2+\cdots + (x-a_n)^2$. This can be compute easily using derivatives but is there an alternative proof which does not use calculus?
Yes, it can be done without calculus. Just expand and write it as a square like this: $$\sum_{k=1}^n(x-a_k)^2\!=\!nx^2-2x\sum_{k=1}^na_k+\sum_{k=1}^na_k^2\!=\!n\left(x^2-\frac2n\sum_{k=1}^na_k+C\right)\!=\!n\left(x-\frac1n\sum_{k=1}^na_k\right)^2+D$$ where $C$ and $D$ are appropriate constants. You can see the minimum is $\displaystyle\frac{a_1+\ldots+a_n}n$.
just simplify it as a normal quadratic equation?
$$ (x-a_1)^2+\cdots + (x-a_n)^2 = nx^2-(2\sum_{i=1}^{n}a_i)x+\sum_{i=1}^{n}a_i^2 $$
when $x=\sum_{i=1}^{n}a_i/n$ it reaches minimal.
You can do it geometrically. Consider the point $p=(a_1,\dots,a_n)\in\mathbb R^n$ (set $n=3$ for ease of imagination).
Then, using Pythagorean theorem, you see that what you want is to find the point $(x,x,x,\dots,x)$ that minimizes the distance form $p$. This is easily done by projecting $p$ onto the line $L=\{x_1=x_2=\dots=x_n\}$. Which is achieved by intersecting $L$ with its orthogonal through $p$.
The orthogonal to $L$ is clearly $\sum x_i=0$ and the orthogonal to $L$ passing through $P$ is the hyperplane $\pi=\{\sum x_i=\sum a_i\}$.
So $\pi\cap L$ is $(x,\dots,x)$ with $nx=\sum a_i$.
Using orthogonal projection in vector algebra. The essence of this proof is that you want to minimize the Euclidean distance between the vectors $\bar x = (x, ..., x) \in R_n$ and $\bar a = (a_1, ..., a_n) \in R_n$.
You have $f: x \in R \longrightarrow \sum_{\{a_k\}} (a_k - x)^2 \in R$ and want to find $x$ such that $f(x)$ or $\sqrt{f(x)}$ is minimum (it's equivalent because square root is monotonic and $f(x) >= 0 \space \forall x$)
It is trivial to show that $\sqrt{f(x)} = \lVert (a_1, ..., a_n) - (x, ..., x)\rVert_2$ where $\lVert \cdot \rVert_2$ is the Euclidean norm. (It is intended that both vectors have $n$ components.) We can write simply $\lVert \bar a - \bar x \rVert$and note that $\bar x = (x, ..., x)$ is in the vector space $\{(r, ..., r)| r \in R\} = \langle \space \{ \space (1, ..., 1) \space \} \space \rangle \subset R^n$
One can demonstrate that, if $V$ is a vector space and $W = \langle \{w_1, ..., w_m\} \rangle $ is a subspace, and $\lVert \cdot \rVert$ is a norm on $V$, then, $\forall v \in V, w \in W,$
$\lVert v - w \rVert $ is minimum (that is: $\lVert v - w \rVert <= \lVert v - w'\rVert \space \forall w' \in W$) if and only if $w = \sum_{\{w_k\}} \frac{(v \vert w_k)}{\lVert w_k\rVert^2}w_k$
$w$ is then called the orthogonal projection of $v$.
Using the lemma, $\lVert \bar a - \bar x\rVert$ is minimum if and only if $\bar x = \sum_{\{w_k\}}\frac{(\bar a \vert w_k)}{\lVert w_k \rVert_2^2}w_k = \frac{(\bar a \vert (1, ..., 1))}{\lVert (1, ..., 1) \rVert_2^2}(1, ..., 1) = \frac{\sum_{\{a_k\}}a_k}{n}(1, ..., 1)$
or equivalently, $x = \frac{\sum_{\{a_k\}}a_k}{n}$.
