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#2: Post edited
by
Anonymous
·
2022-07-05T16:36:36Z (almost 2 years ago)
We will show that there can be at most three equilateral triangles inscribed in the ellipse, so you must already have all the solutions.- Let $A = (0,-2)$ and let $E$ be the ellipse $x^2 + 3y^2 = 12$. We have $A \in E$.
- Now let us search for equilateral triangles $ABC$ inscribed in $E$, the vertices being described in counterclockwise order. The point $C$ must be the image of $B$ under a $60^{\circ}$ counterclockwise rotation about $A$. Thus a point $B \in E$ will yield a solution to the problem if and only if it also belongs to $E'$, the image of $E$ under a $60^{\circ}$ clockwise rotation about $A$.
- Therefore the number of solutions to the problem is equal to the number of points of intersection of the ellipses $E$ and $E'$, besides $A$ itself. But it is well known that there is at most one conic through any five points, so the number of points of intersection is at most four, including $A$. Thus the problem has at most three solutions.
- We will show that there can be at most three equilateral triangles inscribed in the ellipse passing through $(0,-2)$, so you must already have all the solutions.
- Let $A = (0,-2)$ and let $E$ be the ellipse $x^2 + 3y^2 = 12$. We have $A \in E$.
- Now let us search for equilateral triangles $ABC$ inscribed in $E$, the vertices being described in counterclockwise order. The point $C$ must be the image of $B$ under a $60^{\circ}$ counterclockwise rotation about $A$. Thus a point $B \in E$ will yield a solution to the problem if and only if it also belongs to $E'$, the image of $E$ under a $60^{\circ}$ clockwise rotation about $A$.
- Therefore the number of solutions to the problem is equal to the number of points of intersection of the ellipses $E$ and $E'$, besides $A$ itself. But it is well known that there is at most one conic through any five points, so the number of points of intersection is at most four, including $A$. Thus the problem has at most three solutions.
#1: Initial revision
by
Anonymous
·
2022-07-05T15:57:36Z (almost 2 years ago)
We will show that there can be at most three equilateral triangles inscribed in the ellipse, so you must already have all the solutions.
Let $A = (0,-2)$ and let $E$ be the ellipse $x^2 + 3y^2 = 12$. We have $A \in E$.
Now let us search for equilateral triangles $ABC$ inscribed in $E$, the vertices being described in counterclockwise order. The point $C$ must be the image of $B$ under a $60^{\circ}$ counterclockwise rotation about $A$. Thus a point $B \in E$ will yield a solution to the problem if and only if it also belongs to $E'$, the image of $E$ under a $60^{\circ}$ clockwise rotation about $A$.
Therefore the number of solutions to the problem is equal to the number of points of intersection of the ellipses $E$ and $E'$, besides $A$ itself. But it is well known that there is at most one conic through any five points, so the number of points of intersection is at most four, including $A$. Thus the problem has at most three solutions.