For the figure, see Linda Gutsche's course, Wednesday afternoon, 25th. By the inscribed angle theorem, $\widehat{S_{A} B C}=\widehat{S_{A} A C}$ and $\widehat{S_{A} C B}=\widehat{S_{A} A B}$. Since $\widehat{S_{A} A B}=\overline{S_{A} A C}$, it follows that $\widehat{S_{A} B C}=\widehat{S_{A} C B}$, and triangle $S_{A} B C$ is isosceles at $S_{A}$. Therefore, $S_{A}$ lies on the perpendicular bisector of $[B C]$.

## IV. Advanced

## 1 Morning: Basic Strategies and Combinatorics

## 1 Tuesday morning, 24th: Martin Rakovsky

When writing a proof by induction, there are important steps to highlight. Let's start with the following example.

Example 35. Show that for any strictly positive integer $n$,

$$1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$$

The first step is to announce that we will use induction: Proceed by induction on $n$. The second step is the base case: If $n=1$, then we have $1^{2}=\frac{1 \cdot(1+1)(2 \cdot 1+1)}{6}$, so the property holds for $n=1$. The third step is the inductive step, which usually starts with: Suppose the property is true for some natural number $n \geq 1$ and show that it is true for $n+1$. We have $1^{2}+2^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$ by the induction hypothesis, so

$$\begin{aligned} 1^{2}+2^{2}+\ldots+n^{2}+(n+1)^{2} & =\frac{n(n+1)(2 n+1)}{6}+(n+1)^{2} \\ & =(n+1) \cdot \frac{n(2 n+1)+6(n+1)}{6} \\ & =(n+1) \cdot \frac{(n+2)(2 n+3)}{6} \end{aligned}$$

Therefore, the property is true for $n+1$, which concludes the proof.

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