Consider the angle between two rays $l_1$ and $l_2$ with a vertex O and point A in this angle. Now Consider all possible triangles with vertex O such that two sides of them belong to $l_1$ and $l_2$ and the third side $l$ passes through A. Where should we put line $l$ so that the area of the triangle is minimized?
2026-05-05 03:35:50.1777952150
How should we minimize the area of this triangle?
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Let $PQ$ be the side of the optimal triangle (the existence of which we take for granted for the moment). Then an infinitesimal rotation of $\ell$ around $A$ changes the area of the triangle by something proportional to $PA^2-PB^2$. So $\ell$ is optimal (or at least extremal) iff $PA=PB$.
To constrct $\ell$, reflect the figure at $A$ (i.e., rotate by $180^\circ$). That gives us a parallelogram with $A$ the intersection (and hence midpoint) of diagonals.
Once we have this result, we can prove correctned without handwaving infinitesimals around: In the figure with the parallelogram, add another line $\ell'$ and compare the small triangles that make up the difference for the triangles bounded by $\ell, \ell'$. Reflect one of them at $A$ to verify that the added triangle is larger than the subtracted triangle