Given a square $ABCD$, let us consider an equilateral triangle $KLM$, whose vertices $K$, $L$ and $M$ belong to the sides $AB$, $BC$ and $CD$ respectively. Find the locus of the midpoints of the sides $KL$ for all possible equilateral triangles $KLM$. Note: The set of points that satisfy a property is called a locus.
Problem
Source: 2022 Argentina OMA Finals L3 p3
Tags: geometry, Locus, Equilateral, midpoint
19.11.2022 15:50
Given the square $ABCD\ :\ A(0,0),B(a,0), \cdots$ Choose $K(\lambda,0)$ and $L(a,\mu)$. $\triangle KLM$ is equilateral if $\mu=2a-\sqrt{3}(a-\lambda)$ Midpoint of $KL\ :\ x=\frac{\lambda+a}{2}, y=\frac{\mu}{2}$. Eliminating the parameter $\lambda$, we find the locus of the midpoints, part of the line $y=\sqrt{3}x+a(1-\sqrt{3})$.
20.11.2022 02:39
Let $I$ the midpoint of $KL$ then $MI \perp KL$ thus $MILC$ is cyclic hence $\angle ICB =\angle ICL=\angle IML =30^{\circ }$ therefore the locus is a part of line passing through $C$ with the slope $\frac{\sqrt 3}{3}$ wrt $BC$
20.11.2022 03:06
an other similar question is to find the locus of their circumcenters
20.11.2022 03:15
vanstraelen wrote: Given the square $ABCD\ :\ A(0,0),B(a,0), \cdots$ Choose $K(\lambda,0)$ and $L(a,\mu)$. $\triangle KLM$ is equilateral if $\mu=2a-\sqrt{3}(a-\lambda)$ Midpoint of $KL\ :\ x=\frac{\lambda+a}{2}, y=\frac{\mu}{2}$. Eliminating the parameter $\lambda$, we find the locus of the midpoints, part of the line $y=\sqrt{3}x+a(1-\sqrt{3})$. PROF65 wrote: Let $I$ the midpoint of $KL$ then $MI \perp KL$ thus $MILC$ is cyclic hence $\angle ICB =\angle ICL=\angle IML =30^{\circ }$ therefore the locus is a part of line passing through $C$ with the slope $\frac{\sqrt 3}{3}$ wrt $BC$ Both posts are correct, but they are not complete solutions. You still have to find the exact locus of the midpoint. It could be the entire line, or the segment contained in the square, or any other possibility that you haven't discarded yet. A crucial part of the problem (and basically the real problem) is to properly bound the locus.