Easy in form

The hypotenuse of any right triangle is greater than any of its legs, hence the side of the equilateral triangle is $>4$ , or i.e. $>\sqrt{16}$

But the only test option that had the number under the root larger than $16$ was $\boxed{\sqrt{\frac{52}{3}}}$ . No need to use a calculator.

Let $d_1$ be the distance between base point of the taller pole and flag's ground touch point, and let $d_2$ be the analogous distance for the other pole. Moreover let $l$ be the length of a flag side. We get the system $\left\{ \begin{array}{l} l^{2} - 1 = d_1^{2} + d_2^{2} + 2 d_1 d_2\\ l^{2} -16 = d_1^{2}\\ l^{2} - 9 = d_2^{2} \end{array} \right.$ Then one has $\begin{array}{l} l^{2}-1= l^{2} - 16 + l^{2} - 9 + 2 \sqrt{(l^{2} - 16)(l^{2}-9)}\\ 24-l^2=2\sqrt{(l^{2} - 16)(l^{2}-9)}\\ 576+l^{4} - 48l^{2} = 4(l^{2} - 25 l^{2} + 144) = 4l^{2} -100 l^{2} + 576\\ l^{2} - 48 = 4 l^{2} -100\\ 3l^2 = 52\\ l=\sqrt{52/3} \end{array}$

Here comes a long, long solution, based on Hector Flores' idea:

Call x and y the distances from the bases of the poles to the bottom vertex of the flag. Using pythagoras theorem we get three different equations, and they all describe the side of the triangle. These are:

x^2 + 16 and y^2 + 9 and (x+y)^2 + 1 We see that x^2 + 7 = y^2 Firstly, we want to break out y.

(x+y)^2 + 1 = X^2 + 16 // evolve and subtract x^2 2xy + y^2 + 1 = 16 // Substitute in x^2 + 7 and subtract 1 2xy + (x^2 + 7) = 15 // Subtract (x^2+ 7) 2xy = 8 - x^2 y =( 8 - x^2)/2x

Now, we shall substitute this into: x^2 + 16 = y^2 + 9

That gives us the following:

x^2 + 16 = [( 8 - x^2)/2x]^2 + 9 // now, clean up a bit

x^2 + 7 = (64 - 16x^2 + x^4)/4x^2 // and continue by multiplication with 4x^2

Prepare for pq

3x^4 + 44x^2 - 64 = 0

We now set x^2 = n

So:

3n^2 + 44n - 64 = 0

n^2 +( 44n/3) - 64/3 = 0

n = -22/3 + (484/9 + 252/9)^0.5) n = -22/3 + (676/9)^0.5 n = -22/3 + 26/3 = 4/3 x^2 = n = 4/3 x = 2/(3)^0.5

Put this into x^2 + 16 = y^2 + 9 4/3 + 48/3 = y^2 + 27/3 25/3 = y^2 y = 5/(3)^0.5

x + y = 7/(3)^0.5

Put this into (x+y)^2 + 1 =( triangel side)^2 And you shall get that the triangel side is (52/3)^0.5

I really need to learn to formate...