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Here is a code block:


print("hello world!")

Here's an Asymptote diagram:

Here's a cute diagram from my IMO 2023 G4 solve.

Time for some math! $$ \begin{align*} M_X(t) &= E[e^{tX}] \\ &= \int_{-\infty}^{\infty} e^{tx} \frac{1}{\sigma \sqrt{2\pi}} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2} \right) dx \\ &= \int_{-\infty}^{\infty} \frac{1}{\sigma \sqrt{2\pi}} \exp\left( tx - \frac{(x-\mu)^2}{2\sigma^2} \right) dx \\ &= \int_{-\infty}^{\infty} \frac{1}{\sigma \sqrt{2\pi}} \exp\left( \frac{2\sigma^2 tx - (x^2 - 2\mu x + \mu^2)}{2\sigma^2} \right) dx \\ &= \int_{-\infty}^{\infty} \frac{1}{\sigma \sqrt{2\pi}} \exp\left( \frac{-x^2 + 2(\mu + \sigma^2 t)x - \mu^2}{2\sigma^2} \right) dx \\ &= \int_{-\infty}^{\infty} \frac{1}{\sigma \sqrt{2\pi}} \exp\left( \frac{-[x - (\mu + \sigma^2 t)]^2 + 2\mu\sigma^2 t + \sigma^4 t^2}{2\sigma^2} \right) dx \\ &= \exp\left( \frac{2\mu\sigma^2 t + \sigma^4 t^2}{2\sigma^2} \right) \int_{-\infty}^{\infty} \frac{1}{\sigma \sqrt{2\pi}} \exp\left( -\frac{[x - (\mu + \sigma^2 t)]^2}{2\sigma^2} \right) dx \\ &= \exp\left( \mu t + \frac{1}{2}\sigma^2 t^2 \right) \cdot 1 \\ &= e^{\mu t + \frac{1}{2}\sigma^2 t^2} \end{align*} $$

To confuse your enemy, you must first confuse yourself

sun tzu

Time for some math boxes!

Proposition (meow). some theorem statement.

Lemma (trivial lemma). Consider the following operation on an arbitrary positive integer:

If the number is even, divide it by two.
If the number is odd, triple it and add one.

Prove that this process will eventually reach the number 1, regardless of which positive integer is chosen initially.

Proof. ~~Left as an exercise to the reader~~ Skipped, too easy.□