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- Feb 14, 2012

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- Thread starter anemone
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- Feb 14, 2012

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Here is my solution:

The first thing I notice is that there is cyclic symmetry between $x$ and $y$, and so setting $y=x$, we find:

\(\displaystyle 2x^3+3x^2-1=(x+1)^2(2x-1)=0\)

Thus, we know the points:

\(\displaystyle (x,y)=(-1,-1),\,\left(\frac{1}{2},\frac{1}{2} \right)\)

are on the given curve. Next, if we begin with the line:

\(\displaystyle y=1-x\)

and cube both sides, we obtain:

\(\displaystyle y^3=1-3x+3x^2-x^3\)

We may arrange this as:

\(\displaystyle x^3+3x(1-x)+y^3=1\)

Since $y=1-x$, we may now write

\(\displaystyle x^3+3xy+y^3=1\)

And since the point \(\displaystyle \left(\frac{1}{2},\frac{1}{2} \right)\) is on the line $y=1-x$, we know the locus of the given curve is the line $y=1-x$ and the point $(-1,-1)$. Hence, there can only be one set of points on the given curve that are the vertices of any triangle, equilateral or otherwise.

Using the formula for the distance between a point and a line, we find the altitude of the equilateral triangle will be:

\(\displaystyle h=\frac{|(-1)(-1)+1-(-1)|}{\sqrt{(-1)^2+1}}=\frac{3}{\sqrt{2}}\)

Using the Pythagorean theorem, we find that the side lengths of the triangle must be:

\(\displaystyle s=\frac{2}{\sqrt{3}}h=\sqrt{6}\)

And so the area of the triangle is:

\(\displaystyle A=\frac{1}{2}sh=\frac{1}{2}\sqrt{6}\frac{3}{\sqrt{2}}=\frac{3\sqrt{3}}{2}\)

\(\displaystyle 2x^3+3x^2-1=(x+1)^2(2x-1)=0\)

Thus, we know the points:

\(\displaystyle (x,y)=(-1,-1),\,\left(\frac{1}{2},\frac{1}{2} \right)\)

are on the given curve. Next, if we begin with the line:

\(\displaystyle y=1-x\)

and cube both sides, we obtain:

\(\displaystyle y^3=1-3x+3x^2-x^3\)

We may arrange this as:

\(\displaystyle x^3+3x(1-x)+y^3=1\)

Since $y=1-x$, we may now write

\(\displaystyle x^3+3xy+y^3=1\)

And since the point \(\displaystyle \left(\frac{1}{2},\frac{1}{2} \right)\) is on the line $y=1-x$, we know the locus of the given curve is the line $y=1-x$ and the point $(-1,-1)$. Hence, there can only be one set of points on the given curve that are the vertices of any triangle, equilateral or otherwise.

Using the formula for the distance between a point and a line, we find the altitude of the equilateral triangle will be:

\(\displaystyle h=\frac{|(-1)(-1)+1-(-1)|}{\sqrt{(-1)^2+1}}=\frac{3}{\sqrt{2}}\)

Using the Pythagorean theorem, we find that the side lengths of the triangle must be:

\(\displaystyle s=\frac{2}{\sqrt{3}}h=\sqrt{6}\)

And so the area of the triangle is:

\(\displaystyle A=\frac{1}{2}sh=\frac{1}{2}\sqrt{6}\frac{3}{\sqrt{2}}=\frac{3\sqrt{3}}{2}\)

Last edited:

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- Feb 14, 2012

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Here is my solution:

\(\displaystyle 2x^3+3x^2-1=(x+1)^2(2x-1)=0\)

Thus, we know the points:

\(\displaystyle (x,y)=(-1,-1),\,\left(\frac{1}{2},\frac{1}{2} \right)\)

are on the given curve. Next, if we begin with the line:

\(\displaystyle y=1-x\)

and cube both sides, we obtain:

\(\displaystyle y^3=1-3x+3x^2-x^3\)

We may arrange this as:

\(\displaystyle x^3+3x(1-x)+y^3=1\)

Since $y=1-x$, we may now write

\(\displaystyle x^3+3xy+y^3=1\)

And since the point \(\displaystyle \left(\frac{1}{2},\frac{1}{2} \right)\) is on the line $y=1-x$, we know the locus of the given curve is the line $y=1-x$ and the point $(-1,-1)$. Hence, there can only be one set of points on the given curve that are the vertices of any triangle, equilateral or otherwise.

Using the formula for the distance between a point and a line, we find the altitude of the equilateral triangle will be:

\(\displaystyle h=\frac{|(-1)(-1)+1-(-1)|}{\sqrt{(-1)^2+1}}=\frac{3}{\sqrt{2}}\)

Using the Pythagorean theorem, we find that the side lengths of the triangle must be:

\(\displaystyle s=\frac{2}{\sqrt{3}}h=\sqrt{6}\)

And so the area of the triangle is:

\(\displaystyle A=\frac{1}{2}sh=\frac{1}{2}\sqrt{6}\frac{3}{\sqrt{2}}=\frac{3\sqrt{3}}{2}\)

Awesome,

- Nov 4, 2013

- 428

Why both of you posted the same solution?

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- #5

- Feb 14, 2012

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I am so so sorry for causing the confusion. I was meant to reply toWhy both of you posted the same solution?

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- #7

- Feb 14, 2012

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Thanks, Mark...you are forever my sweetest admin!I fixed it.