(2) is false; the center of mass is always the centroid of a plane figure, so any figure without rotational symmetry is a counterexample. (A scalene triangle for example.)

As for (1) it suffices by the above to show that there is a unique point satisfying the condition you require, that is, that any line through it divides $A$ into areas of equal measure. (We assume that $A$ has positive measure.)

Assume the contrary, and let $a, b$ be two such points. Let $l_a$ be a line passing through $a$ and let $l_b$ be a line passing through $b$, with $l_a, l_b$ parallel, such that the region between $l_a$ and $l_b$ contains a subset $S$ of $A$ with non-zero measure. (Such a pair of lines exists as otherwise $A$ is concentrated on the line connecting $a, b$ and thus has zero measure.) But then in your notation we have $|A|=|A|/2+|A|/2+|S|$ which is impossible. So $a=b$.

This argument does not require any additional hypothesis, and the proof is cute, so I thought I'd include it. I claim: There is at most one point satisfying the condition you require, that is, that any line through it divides $A$ into areas of equal measure. (We assume that $A$ has positive measure.)

Assume the contrary, and let $a, b$ be two such points. Let $l_a$ be a line passing through $a$ and let $l_b$ be a line passing through $b$, with $l_a, l_b$ parallel, such that the region between $l_a$ and $l_b$ contains a subset $S$ of $A$ with non-zero measure. (Such a pair of lines exists as otherwise $A$ is concentrated on the line connecting $a, b$ and thus has zero measure.) But then in your notation we have $|A|=|A|/2+|A|/2+|S|$ which is impossible. So $a=b$.

(2) is false; the center of mass is always the centroid of a plane figure, so any figure without rotational symmetry is a counterexample.

As for (1) it suffices by the above to show that there is a unique point satisfying the condition you require, that is, that any line through it divides $A$ into areas of equal measure. (We assume that $A$ has non-zero measure.)

Assume the contrary, and let $a, b$ be two such points. Let $l_a$ be a line passing through $a$ and let $l_b$ be a line passing through $b$, with $l_a, l_b$ parallel, such that the region between $l_a$ and $l_b$ contains a subset $S$ of $A$ with non-zero measure. (Such a pair of lines exists as otherwise $A$ is concentrated on the line connecting $a, b$ and thus has zero measure.) But then in your notation we have $|A|=|A|/2+|A|/2+|S|$ which is impossible. So $a=b$.

(2) is false; the center of mass is always the centroid of a plane figure, so any figure without rotational symmetry is a counterexample. (A scalene triangle for example.)

As for (1) it suffices by the above to show that there is a unique point satisfying the condition you require, that is, that any line through it divides $A$ into areas of equal measure. (We assume that $A$ has positive measure.)

Assume the contrary, and let $a, b$ be two such points. Let $l_a$ be a line passing through $a$ and let $l_b$ be a line passing through $b$, with $l_a, l_b$ parallel, such that the region between $l_a$ and $l_b$ contains a subset $S$ of $A$ with non-zero measure. (Such a pair of lines exists as otherwise $A$ is concentrated on the line connecting $a, b$ and thus has zero measure.) But then in your notation we have $|A|=|A|/2+|A|/2+|S|$ which is impossible. So $a=b$.