equilateral triangle inscribed in an ellipse
A high-schooler I know was given the following problem:
In the ellipse $x^2+3y^2=12$ is inscribed an equilateral triangle. One of the triangle's vertices is at the point $(0,-2)$. Find the triangle's other vertices.
The book has one answer: $(\pm1.2\sqrt3,1.6)$. But I know of two more answers: $(0,2)$ and $(\pm\sqrt{12},0)$. Are there any more?
2 answers
The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| msh210 | (no comment) | Mar 25, 2022 at 04:33 |
First of all, let's suppose that the points on the ellipse $(x_1,y_1),(x_2,y_2)$ are of the same distance from $(0,-2)$. We will get three equations from this assumption:
$$ \begin{cases} x_1^2+3y_1^2=12 & (1)\\\\ x_2^2+3y_2^2=12 & (2)\\\\ x_1^2+(y_1+2)^2=x_2^2+(y_2+2)^2 & (3) \end{cases} $$
After substituting $x_1^2,x_2^2$ with $12-3y_1^2,12-3y_2^2$ in equation $(3)$ respectively, we will get $-2y_1^2+4y_1+16=-2y_2^2+4y_2+16$, which is equivalent to $(y_1-1)^2=(y_2-1)^2$. Now we will divide to two separate cases.
Case 1:
If $(y_1-1),(y_2-1)$ are of the same sign (positive or negative), we can infer that $y_1=y_2$, and after equating $(1),(2)$ we will get $x_1^2+3y_1^2=x_2^2+3y_2^2$, Hence $x_1^2=x_2^2$. So there are two options again
Subcase 1:
if $x_1=x_2$, then the two points are that same point. In this case the answer will depend on your definition of an equilateral triangle. if you consider a line as an equilateral triangle, then every point on the ellipse will be a solution. If you will demand that the distance between all three (if we will consider the same point we got as two solutions) points will be the same, there is only one solution, that all the points will be $(0,-2)$. If you consider only triangles with three distinct points, we will have no solution in this case.
Subcase 2:
if $x_1=-x_2$, for simplicity i will denote $x_1=x\ y_1=y,\ x_2=-x,\ y_2=y$. The distance between the points $(x,y)$ and $(-x,y)$ is $2x$, and the distance between those two points to $(0,-2)$ is $\sqrt{x^2+(y+2)^2}$, so if we will require that this two length will be equal we will get the equation $4x^2=x^2+(y+2)^2$, and after substituting $x^2$ with $12-3y^2$ we will get
$$ 0=9y^2-36+y^2+4y+4=10y^2+4y-32=2(y+2)(5y-8) $$
If $y=2$, then we will be at the last case, other wise we will have $y=\frac{8}{5}=1.6$, $|x|=\sqrt{12-3y^2}=\frac{6}{5}\sqrt{3}=1.2\sqrt{3}$, so the two points will be $(\pm 1.2\sqrt{3},1.6)$ as in the book.
Case 2:
If the signs of $(y_1-1),(y_2-1)$ are opposite, we will get $y_1=2-y_2$. From equations $(1),(2)$ we will get $x_1^2-x_2^2=3(y_2^2-y_1^2)=3(y_2+y_1)(y_2-y_1)=12(y_2-1)=12(1-y_1)$. Now, if we will require that the distance between $(x_1,y_1),(x_2,y_2)$ equals to the distance between $(x_1,y_1),(0,-2)$ we could write
$$ \begin{align} (x_1-x_2)^2+(y_1-y_2)^2 & =x_1^2+(y_1+2)^2\\ \\ x_1^2+x_2^2-2x_1x_2+(2y_1-2)^2 & =x_1^2+y_1^2+4y_1+4\\ \\ x_2^2-2x_1x_2+4y_1^2-8y_1+4 & =y_1^2+4y_1+4\\ \\ -2x_1x_2+x_2^2+3y_1^2-12y_1 & =0\ / {\color{gray}\text{because}\ 3y_1^2=12-x_1^2\ \text{we will get} }\\ \\ -2x_1x_2+12+x_2^2-x_1^2-12y_1 & =0\ / {\color{gray}\text{remember that}\ x_1^2-x_2^2=12(1-y_1)}\\ \\ -2x_1x_2-12(1-y_1)+12(1-y_1) & =0\\ \\ -2x_1x_2 & =0 \end{align} $$
Thus $x_1$ or $x_2$ are vanishing. Without loss of generality i will assume that $x_1=0$. From that we will get that $y_1=\pm 2$. Because $y_1+y_2=2$ and $y_2\leq 2$, we will conclude that $y_1\geq 0$ so $y_1=2$. Because $y_1+y_2=2$ we will get that $y_2=0$, so $x_2=\pm\sqrt{12}$, as you got as well.
Conclusion: All the solutions are $\left[(1.2\sqrt{3},1.6),(-1.2\sqrt{3},1.6)\right],\ \left[(0,2),(\sqrt{12},0)\right],\ \left[(0,2),(-\sqrt{12},0)\right]$.
0 comment threads
The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| msh210 | (no comment) | Jul 17, 2022 at 02:53 |
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.
0 comment threads
1 comment thread