# Optimization and maximum geometry

What is the side length of the largest square that will fit inside an equilateral triangle with sides of length 1.

I created two equations: Square: Area=$x^2$ Triangle: Area= $\sqrt{3 /4}$

However, how can I find the maximum?? I did $\sqrt{3 /4}=x^2$ ang got fourth root $3 / 2$ which is wrong

• how did u do that? – exchangehelpforuni Jan 19 '17 at 21:36
• is there a way to do that without doing guess and check? – exchangehelpforuni Jan 19 '17 at 21:37
• i did that but nothing helped me.. Just tell me do I need to derive an equation then find its maximum? or no – exchangehelpforuni Jan 19 '17 at 21:37
• If a rectangle has its four vertices on the perimeter of an equilateral triangle, its side lengths fulfill a linear equation. (Figure out which one) If such a rectangle is a square, there is a second linear constraint. – Jack D'Aurizio Jan 19 '17 at 21:39
• – exchangehelpforuni Jan 19 '17 at 21:40

Proof with some words

You can create a rectangle inside your triangle. Let $x_B$ and $x_G$ the $x$-coordinates of the blue and green points, respectively. It is clear that $y_B = y_G = 0$. Moreover, it is easy to see that: $$x_B < 0, x_G >0 ~\text{and}~ x_B = -x_G.$$ The base of the rectangles is $$b = x_G - x_B = 2 x_G.$$

The height of the rectangle is given by the $y$-coordinate of the yellow or red point, let's call them $y_Y$ and $y_R$. Note also that the $x$-coordinates of these two points are:

$$\begin{cases} x_R = x_B \\ x_Y = x_G \end{cases}.$$

Then, using the equation of the diagonal side of your triangle, we have that:

$$\begin{cases} y_R = \frac{\sqrt{3}}{2} + \sqrt{3}x_R = \frac{\sqrt{3}}{2} + \sqrt{3}x_B \\ y_Y = \frac{\sqrt{3}}{2} - \sqrt{3}x_Y = \frac{\sqrt{3}}{2} - \sqrt{3}x_G \end{cases}.$$

As said, $y_R = y_Y$, indeed:

$$y_R = \frac{\sqrt{3}}{2} + \sqrt{3}x_B = \frac{\sqrt{3}}{2} - \sqrt{3}x_G = y_Y.$$

Moreover, $h = y_R$, then:

$$h = \frac{\sqrt{3}}{2} - \sqrt{3}x_G.$$

Finally, we found that the rectangle is defined by:

$$\begin{cases} b = 2x_G\\ h = \frac{\sqrt{3}}{2} - \sqrt{3}x_G \end{cases},$$

while its area is

$$A =bh = 2x_G\left(\frac{\sqrt{3}}{2} - \sqrt{3}x_G\right).$$

Your rectangle is a square only when $b=h$. That is, when:

$$2x_G = \frac{\sqrt{3}}{2} - \sqrt{3}x_G \Rightarrow x_G = \frac{\sqrt{3}}{2(2+\sqrt{3})}.$$

Then:

$$b = h = \frac{\sqrt{3}}{2+\sqrt{3}}$$ and the area is $$A = \left(\frac{\sqrt{3}}{2+\sqrt{3}}\right)^2 = \frac{3}{7+4\sqrt{3}}.$$

It is very important to notice that this square is unique possible by construction. Indeed, we started from a rectangles, and we "proved" that this rectangle is a square only if $b=h= \frac{\sqrt{3}}{2+\sqrt{3}}$. So, the maximum area is

$$A =\frac{3}{7+4\sqrt{3}} = 21 - 12 \sqrt{3} \simeq 0.215390309173472.$$

$1)$ Suppose that $DEFG$ is a square. Let's call $x$ the side of the square.

We also have that $\Delta ADE$ is an equilateral triangle, so $AD=x$.

Now, for the triangle $BDG$ we have:

$$\sin 60°=\frac{DG}{BD}=\frac{x}{BD} \rightarrow BD=\frac{2\sqrt{3}x}{3}$$

$$AB=1=AD+BD=x+\frac{2\sqrt{3}x}{3} \rightarrow x=\frac{3}{3+2\sqrt{3}}=2\sqrt{3}-3$$

It means that doesn't make sense talk about largest square because we have just one and its side is $x=2\sqrt{3}-3$.

$2)$ What could make sense is talk about the biggest rectangle inside the triangle. In order to do that we can call $DE=x$ and then $AD=x$. Let's also call $DG=y$, so

$$\sin 60°=\frac{DG}{BD}=\frac{y}{BD} \rightarrow BD=\frac{2\sqrt{3}y}{3}$$

and then

$$AB=1=AD+BD=x+\frac{2\sqrt{3}y}{3}$$

and we have to maximize:

$$A=xy=\left(1-\frac{2\sqrt{3}y}{3}\right)y$$

and the maximum happens for (using the quadratic standard approach)

$$y=\frac{\sqrt{3}}{4} \rightarrow x=\frac{1}{2}$$

what give us

$$A_{max}=\frac{\sqrt{3}}{8}$$

• Actually infinitely many squares fits insides a triangle, but only one is inscribed (up to rotation). – Lery Jan 19 '17 at 23:40
• @exchangehelpforuni: is it clear? – Arnaldo Jan 20 '17 at 11:23

I think there is a mistake in the proof above. Let's say a square $DFGE$ with side $1/2$ ,such that 2 of its vertices lie on side AB of the given triangle ,can fit in the triangle , but then $EG=\frac{A'C}{2}=\frac{\sqrt{3}}{4}<\frac{1}{2}$ thus if the square has side $\frac{1}{2}$ it can't fit in the triangle.

Well, you actually have to solve two problems:

Firstly, since you want the biggest possible square, it means that it will then be inscribed in the triangle, but you must convinced yourself of this or prove it. It may be proven constructively, I think, or ab absurdo (ie. suppose there is a bigger square in the triangle, that is not inscribed and show that it violates one an hypothesis).

Secondly, once you've convinced yourself of that and that the only way to inscribe a square in a triangle is to have two sides of the triangle touching two of the angles of the square and the last side of the triangle containing the two others points, which means that one of the side of the square is completely contained into the triangle, then you can compute the value you're looking for:

let us, wlog, say that the base of the triangle is containing one complete side of the square. when does $X=Y$ for $X$ and $Y$ as in the following image, for an equilateral triangle with side of length 1:

Now, let us use trigonometry to relate $Y$ to $X$.

when the value $Y$ increase, the value $X$ decrease because it's one side of a right triangle and so we have: $\frac{1-Y}{2}=Z$.

But $Z$ relates to $X$ as follows, since we know the angles are 60° : $\tan(60°)=\frac{X}{Z}$, so $Z=\frac{X}{\tan(60°)}$ and we have: $$\frac{1-Y}{2}=\frac{X}{\tan(60°)} = \frac{X}{\sqrt{3}}$$

So : $$2X = (1-Y)\sqrt{3} = \sqrt{3}-\sqrt{3}Y$$ Now if $X=Y$, then: $$2X = \sqrt{3}-\sqrt{3}X$$ $$(2+\sqrt{3})X = \sqrt{3}$$ and so : $$X = \frac{\sqrt{3}}{(2+\sqrt{3})} = \frac{(2-\sqrt{3})\sqrt{3}}{4-3}=(2-\sqrt{3})\sqrt{3}\approx 0.4641$$

And so the area of the square would be $(2\sqrt{3}-3)^2=4\cdot 3-12\sqrt{3}+9$.

Note that I may have made some computations mistake—even if I think everything is fine—but the general idea is correct.