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Imagine having a piece of paper with two different shapes on it, each at a random location. Can we always draw a straight line through the piece of paper, in a manner that divides both objects in half?

Keep in mind that I am an 8th grader, and most probably unfamiliar with technical terms. This just came to my mind. But if it's necessary to use technical terms, I guess that is my problem and I'll have to look them up.

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Yes, this is the Ham Sandwich Theorem. Taking $n=2$, the statement says that given two measurable objects in $2$-dimensional space (e.g., two "nice" shapes drawn on a piece of paper), it is possible to divide both in half using a single line.

I'll sketch a proof of the case $n=2$ below.

Let $A$ and $B$ be the two shapes. Let $\ell$ be any line. I'll show that we can translate $\ell$ in some way so that it cuts $A$ in half. Let $f(x)$ be the area of $A$ to the right of $\ell$ when $\ell$ it has been translated $x$ units to the right. (If $\ell$ is a horizontal line, replace "right" with "upward/up.") If $A$ is a measurable set (almost any shape you could imagine drawing will be measurable), then $f(x)$ is a continuous function of $x$. That is, shifting the line $\ell$ by a very small amount will result in a very small change in how much of $A$ is to the right of $\ell$. If you move $\ell$ really far to the left (say by $N$ units), all of $A$ is to the right of $\ell$, so $f(-N)=\text{area}(A)$. If you move $\ell$ really far to the right (say by $M$ units), all of $A$ is to the left of $\ell$, so $f(M)=0$. So by the intermediate value theorem, there is some value of $x$ between $-N$ and $M$ such that $f(x)=\frac{1}{2}\cdot \text{area}(A)$. That is, $\ell$ translated to the right $x$ units cuts $A$ in half.

Now let $\theta$ be some angle between $0$ and $180$ degrees. Choose a line that makes angle $\theta$ with the positive $x$-axis. By the above paragraph, we can translate this line in some way so that it cuts $A$ in half. Let $\ell_\theta$ be this translate. That is, $\ell_\theta$ makes angle $\theta$ with the horizontal and cuts $A$ into two pieces of equal area. Let $g(\theta)$ be the area of $B$ to the right of $\ell_\theta$ minus the area of $B$ to the left of $\ell_\theta$. (When $\theta=0$, replace "right" with "below" and "left" with "above," and when $\theta=180$, replace "right" with "above" and "left" with "below.") If $B$ is measurable, then $g(\theta)$ is a continuous function of $\theta$. That is, changing the angle of the line slightly will only result in a small change in $g(\theta)$. Note that $\ell_0$ and $\ell_{180}$ are the same line (one is just rotated $180$ degrees), so \begin{align*} g(0)&=\text{area of $B$ below $\ell_0$}-\text{area of $B$ above $\ell_0$}\\ &=-(\text{area of $B$ above $\ell_0$}-\text{area of $B$ below $\ell_0$})\\ &=-(\text{area of $B$ above $\ell_{180}$}-\text{area of $B$ below $\ell_{180}$})\\ &=-g(180). \end{align*} Using the intermediate value theorem again, there exists $\theta$ between $0$ and $180$ such that $g(\theta)=0$. That is \begin{align*} \text{area of $B$ right of $\ell_\theta$}-\text{area of $B$ left of $\ell_\theta$}&=0\\ \text{area of $B$ right of $\ell_\theta$}&=\text{area of $B$ left of $\ell_\theta$} \end{align*} So $\ell_\theta$ cuts $B$ in half. But $\ell_\theta$ was also chosen so that it cuts $A$ in half. So $\ell_\theta$ is the line we want.

Proof adapted from Theorem III here: https://faculty.missouri.edu/~casazzap/pdf/teach/pizza.pdf

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