Find the area of an equilateral triangle

In summary, the conversation discusses a challenge problem that involves finding the area of an equilateral triangle formed by three distinct points on a curve, and two users accidentally post the same solution, causing confusion and apologies.
  • #1
anemone
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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.
 
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  • #2
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:
  • #3
MarkFL said:
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:
 
  • #4
Why both of you posted the same solution? :confused:
 
  • #5
Pranav said:
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.
 
  • #6
I fixed it. (Hug)
 
  • #7
MarkFL said:
I fixed it. (Hug)

Thanks, Mark...you are forever my sweetest admin!:eek:
 

FAQ: Find the area of an equilateral triangle

What is an equilateral triangle?

An equilateral triangle is a type of triangle where all three sides are equal in length and all three angles are equal to 60 degrees.

How do you find the area of an equilateral triangle?

The formula for finding the area of an equilateral triangle is A = (√3/4) x s^2, where A is the area and s is the length of one side.

Can you explain the formula for finding the area of an equilateral triangle?

The formula is derived from the fact that an equilateral triangle can be divided into two congruent right triangles with one side being half the length of the base and the other being the height. The formula for the area of a right triangle is 1/2 x base x height, and when we substitute in the values for an equilateral triangle, we get A = (√3/4) x s^2.

Do all equilateral triangles have the same area?

Yes, since all sides and angles of an equilateral triangle are equal, the area will also be the same for all equilateral triangles with the same side length.

How do you measure the side length of an equilateral triangle?

The side length of an equilateral triangle can be measured by using a ruler or measuring tape. Alternatively, if you know the perimeter of the triangle, you can divide it by 3 to get the length of one side.

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