1. $[\mathbf{2 0}]$ Let $a_{1}, a_{2}, a_{3}, \ldots, a_{100}$ be integers such that
$$
\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+\cdots+a_{100}^{2}}{a_{1}+a_{2}+a_{3}+\cdots+a_{100}}=100
$$

Determine, with proof, the maximum possible value of $a_{1}$.
Answer:
550
Solution 1: We can rearrange the equation as follows:
$$
\begin{array}{c}
a_{1}^{2}+a_{2}^{2}+\cdots+a_{99}^{2}+a_{100}^{2}=100\left(a_{1}+a_{2}+\cdots+a_{99}+a_{100}\right) \\
\left(a_{1}^{2}-100 a_{1}\right)+\left(a_{2}^{2}-100 a_{2}\right)+\left(a_{3}^{2}-100 a_{3}\right)+\cdots+\left(a_{100}^{2}-100 a_{100}\right)=0 \\
\left(a_{1}-50\right)^{2}+\left(a_{2}-50\right)^{2}+\left(a_{3}-50\right)^{2}+\cdots+\left(a_{100}-50\right)^{2}=100 \cdot 50^{2}=500^{2} .
\end{array}
$$

Thus $\left(a_{1}-50\right)^{2} \leq 500^{2}$ and so $a_{1} \leq 550$. Equality holds when $a_{1}=550$ and $a_{i}=50$ for all $i>1$. Therefore, the maximum possible value of $a_{1}$ is 550 .

Solution 2: Let $k=\frac{1}{99}\left(a_{2}+a_{3}+\cdots+a_{100}\right)$. Note, that by the Cauchy-Schwarz inequality, we have:
$$
99 \cdot\left(a_{2}^{2}+a_{3}^{2}+\cdots+a_{100}^{2}\right) \geq\left(a_{2}+a_{3}+\cdots+a_{100}\right)^{2}=(99 k)^{2}
$$
and so:
$$
100=\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+\cdots+a_{100}^{2}}{a_{1}+a_{2}+a_{3}+\cdots+a_{100}} \geq \frac{a_{1}^{2}+99 k^{2}}{a_{1}+99 k}
$$

Now consider the function:
$$
f_{t}(x)=\frac{99 x^{2}+t^{2}}{99 x+t}
$$
for all reals $t$. Note that for $t=a_{1}$, we have that $f_{t}(k) \leq 100$ for some real $k$. Furthermore, because $a_{1}^{2}+a_{2}^{2}+\cdots+a_{100}^{2}>0$, we have $a_{1}+a_{2}+\cdots+a_{100}>0$, and so $f_{t}(x)>0$ for all valid $x$.

We assume $a_{1}>0$, since we are trying to maximize this element and some positive solution exists (i.e. $(100,100, \ldots, 100))$. Hence, the minimum point of $f_{t}(x)$ over $x>0$ needs to be less than 100 . To compute this, set the derivative to be zero as follows:
$$
f_{t}^{\prime}(x)=1-\frac{100 t^{2}}{(t+99 x)^{2}}=0
$$
which has solutions at $x=\frac{t}{11}$ and $x=-\frac{t}{9}$. Taking a second derivative shows that the point $\left(\frac{t}{11}, \frac{2 t}{11}\right)$ is the only valid minimum point in the first quadrant. Hence, we must have $\frac{2 t}{11} \leq 100$ and so $a_{1}=t \leq 550$, as desired. Equality holds at $(550,50, \ldots, 50)$.