Square and a Point

From the figure above, and calling the square side as $l$ , we have:

$\large \displaystyle \color{#D61F06} (1) \color{#333333} \ x^2 + y^2 = 39$

$\large \displaystyle \color{#D61F06} (2) \color{#333333} \ (l-x)^2 + y^2 = l^2 - 2lx + x^2 + y^2 = 159$

$\large \displaystyle \color{#D61F06} (3) \color{#333333} \ (l-y)^2 + x^2 = l^2 - 2ly + y^2 + x^2 = 219$

Plugging $(1)$ into $(2)$ and $(3)$ :

$\large \displaystyle \color{#D61F06} (4) \color{#333333} \ l^2 - 2lx = 120$

$\large \displaystyle \color{#D61F06} (5) \color{#333333} \ l^2 - 2ly = 180$

Making $(5) - (4)$ :

$\large \displaystyle 2lx - 2ly = 60$

$\large \displaystyle 2l(x - y) = 60$

$\large \displaystyle \color{#D61F06} (6) \color{#333333} \ \large \displaystyle x = \frac{30}{l} + y$

Now substituting $(6)$ into $(1)$ :

$\large \displaystyle \frac{900}{l^2} + \frac{60}{l}y + y^2 + y^2 = 39$

Multiplying by $l^2$ on both sides:

$\large \displaystyle \color{#D61F06} (7) \ \color{#333333} 2y^2 l^2 + 60ly + 900 = 39l^2$

From $(5)$ , muliplying by 30 on both sides we can obtain:

$\large \displaystyle \color{#D61F06} (8) \ \color{#333333} 60ly = 30l^2 - 5400$

And also from $(5)$ :

$\large \displaystyle 2ly = l^2 - 180$

Squaring:

$\large \displaystyle 4l^2y^2 = l^4 - 360l^2 + 32400$

$\large \displaystyle \color{#D61F06} (9) \ \color{#333333} 2l^2y^2 = \frac{l^4 - 360l^2 + 32400}{2}$

Substituting $(8)$ and $(9)$ into $(7)$ :

$\large \displaystyle \frac{l^4 - 360l^2 + 32400}{2} + 30l^2 - 5400 + 900 = 39l^2$

Multiplying by $2$ on both sides:

$\large \displaystyle l^4 - 360l^2 + 32400 + 60l^2 - 9000 = 78l^2$

$\large \displaystyle l^4 - 378l^2 + 23400 = 0$

From this, we have:

$\large \displaystyle l^2 = \frac{378 \pm \sqrt{378^2 - 4\cdot 14400}}{2}$

$\large \displaystyle l^2 = \frac{378 \pm 222}{2}$

Since we want the larger $l$ :

$\color{#20A900} \boxed{ \large \displaystyle l^2 = 300}$

Since $l = \sqrt[4]{k}$ , $k = l^4$ , so:

$\color{#3D99F6} \large \displaystyle k = 90000, \boxed{ \large \displaystyle k - 2020 = 87980}$

Let the center of square $ABCD$ be the origin $O(0,0)$ of the $xy$ -plane, its side length be $2a$ and $Q(x,y)$ . Then $A(-a,-a)$ , $B(a,-a)$ , $C(a,a)$ , and $D(-a,a)$ and by Pythagorean theorem :

$\begin{cases} (x+a)^2 + (y+a)^2 = 159 & ...(1) \\ (x-a)^2 + (y+a)^2 = 39 & ...(2) \\ (x-a)^2 + (y-a)^2 = 219 & ...(3) \end{cases}$

Then $(1)-(2): \ \implies 4ax = 120 \implies ax = 30$ and $(2)-(3): \ \implies 4ay = -180 \implies ay = -45$ and

$\begin{aligned} (1): \quad x^2 + 2ax + a^2 + y^2 + 2ay + a^2 & = 159 \\ x^2 + 60 + a^2 + y^2 - 90 + a^2 & = 159 \\ x^2 + y^2 + 2a^2 & = 189 & \small \blue{\text{Multiply both sides by }a^2} \\ a^2x^2 + a^2y^2 + 2a^4 & = 189a^2 \\ 30^2 + (-45)^2 + 2a^4 & = 189a^2 \\ 2a^4 - 189a^2 + 2925 & = 0 \\ (2a^2-39)(a^2-75) & = 0 & \small \blue{\text{Taking the larger }a} \\ \implies a & = \sqrt{75} \end{aligned}$

Therefore the largest side length of square $ABCD$ , $2a = 2\sqrt{75} = \sqrt{300} = \sqrt[4]{90000}$ $\implies k - 2020 = 90000 -2020 = \boxed{87980}$ .

Coordinate approach: Take $B$ as the origin, let the $x$ -axis run from right to left (or flip the diagram), let the side of the square be $s$ and let the coordinates of $Q$ in this system be $(u,v)$ .

Then the information given (using Pythagoras) becomes: $\begin{aligned} u^2+v^2 &=39 \\ (s-u)^2+v^2 &=159 \\ u^2+(s-v)^2 &=219 \end{aligned}$

We can eliminate $u$ and $v$ by subtracting the first equation from the second and third in turn: $\begin{aligned} s^2-2su=120 &\Rightarrow u=\frac{s^2-120}{2s} \\ s^2-2sv=180 &\Rightarrow v=\frac{s^2-180}{2s} \end{aligned}$

Now we can substitute back into the first equation to get: $\frac{\left(s^2-120 \right)^2+\left(s^2-180 \right)^2}{4s^2} = 39$

Expanding and tidying up gives $s^4 - 378s^2 + 23400=0$

This is easily solved by treating it as a quadratic in $s^2$ . The positive roots we get are $s=10\sqrt3$ and $s=\sqrt{78}$ . We're told to take the larger of these, which is the first; the answer is then $s^4-2020=\boxed{87980}$ . It's worth noting, though, that the smaller positive value of $s$ places the point $Q$ outside the square $ABCD$ ; there is a unique solution inside the square.