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Find the area of an equilateral triangle

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anemone

MHB POTW Director
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Feb 14, 2012
3,678
Show that the curve $x^3+3xy+y^3=1$ has only one set of three distinct points, $P$, $Q$, and $R$ which are the vertices of an equilateral triangle, and find its area.
 

MarkFL

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Feb 24, 2012
13,775
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:

\(\displaystyle A=\frac{1}{2}sh=\frac{1}{2}\sqrt{6}\frac{3}{\sqrt{2}}=\frac{3\sqrt{3}}{2}\)
 
Last edited:
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
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:

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

Awesome, MarkFL, awesome!!!(Sun) This is your second time answered to my challenge problem and thank you for participating!:eek:
 

Pranav

Well-known member
Nov 4, 2013
428
Why both of you posted the same solution? :confused:
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,678
Why both of you posted the same solution? :confused:
I am so so sorry for causing the confusion. I was meant to reply to MarkFL by quoting his solution, and then typed beneath it and I accidentally clicked on the "Edit" button(I could edit it since I'm the moderator in the "Challenge Questions and Puzzles" subforum), and when I realized I bungled that I immediately rectified the situation by posting my reply and again, I made another mistake, those solution should be put under quote tag but I didn't. :eek: Sorry again.
 

MarkFL

Administrator
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Feb 24, 2012
13,775
I fixed it. (Hug)
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,678