Smallest Square Tangent to a Hyperbola

Imgur

Let the upper right corner of the square be $A$ , the upper left corner be $B$ , and the origin to be $O$ . Clearly, the side length of the square is smallest when the square is tangent to the hyperbola at point $A$ . Let $A=(x,y)$ . Clearly, since $A$ is on the top half of the hyperbola, $y=\sqrt{x^2+1}$ so $A=(x,\sqrt{x^2+1})$

---

Finding the side length of the square in terms of $x$

Since the square is rationally symmetric about the origin, the side length of the square is the distance between the points $A=(x,y)$ and $B=(-y,x)$ . This is simply $\begin{aligned}1+S&=\sqrt{\left(x+y\right)^2+\left(x-y\right)^2}\\ &= \sqrt{2x^2+2y^2}\\ &= \sqrt{2x^2+2(x^2+1)}\\ &= \boxed{\sqrt{4x^2+2}}\end{aligned}$

So if we find the x-coordinate of point $A$ , we will be able to find the side length of the square using the above formula.

Now, notice that because the continuation of $AO$ is a diagonal of the square, $\angle BAO=45^{\circ}$ . So if we can find the slope of $AB$ and $AO$ , then we can set the degree measure between the two lines to equal $45^{\circ}$ .

---

Imgur

Finding the slope of $AB$

Let the point of tangency be $(a,\sqrt{a^2+1})$ . The derivative of $\sqrt{x^2+1}$ is $\dfrac{x}{\sqrt{x^2+1}}$ , so the slope at point $(a,\sqrt{a^2+1})$ is $\dfrac{a}{\sqrt{a^2+1}}$ .

Finding the slope of $AO$

Since $AO$ passes through $A=(a,\sqrt{a^2+1})$ and $O=(0,0)$ , the slope is simply $\dfrac{\sqrt{a^2+1}}{a}$ .

---

Now that we know the slopes of $AB$ and $AO$ , we can set the degree measure between them to equal $45^{\circ}$ : $\tan^{-1}\left(\dfrac{\sqrt{a^2+1}}{a}\right)-\tan^{-1}\left(\dfrac{a}{\sqrt{a^2+1}}\right)=45^{\circ}$

Taking the $\tan$ of both sides: $\tan\left(\tan^{-1}\left(\dfrac{\sqrt{a^2+1}}{a}\right)-\tan^{-1}\left(\dfrac{a}{\sqrt{a^2+1}}\right)\right)=1$

Using the tangent angle difference property: $\dfrac{\dfrac{\sqrt{a^2+1}}{a}-\dfrac{a}{\sqrt{a^2+1}}}{1+\left(\dfrac{\sqrt{a^2+1}}{a}\right)\left(\dfrac{a}{\sqrt{a^2+1}}\right)}=1$

Simplifying: $\dfrac{\sqrt{a^2+1}}{a}-\dfrac{a}{\sqrt{a^2+1}}=2$

Let $n=\dfrac{\sqrt{a^2+1}}{a}$ . The equation transforms into the much simpler $n-\dfrac{1}{n}=2$

Rearranging: $n^2-2n-1=0$

Using the quadratic equation, we obtain that $n=1\pm \sqrt{2}$

---

Case 1: $n=1-\sqrt{2}$

We substitute back to get $\dfrac{\sqrt{a^2+1}}{a}=1-\sqrt{2}$

Simplifying and rearranging: $a^2=\dfrac{1}{2-2\sqrt{2}}$

However, since $\dfrac{1}{2-2\sqrt{2}}$ is negative, $a$ is not real. Thus, $n=1-\sqrt{2}$ is an extraneous solution.

Case 2: $n=1+\sqrt{2}$

Doing the same thing as last time, we obtain $a^2=\dfrac{1}{2\sqrt{2}-2}$ $a=\sqrt{\dfrac{1}{2\sqrt{2}-2}}=\boxed{\sqrt{\dfrac{\sqrt{2}-1}{2}}}$

---

Finishing up the problem

We plug $a=\sqrt{\dfrac{\sqrt{2}-1}{2}}$ back into the first formula we derived: $\begin{aligned}AB&=\sqrt{4\left(\sqrt{\dfrac{\sqrt{2}-1}{2}}\right)^2+2}\\ &=\sqrt{4\left(\dfrac{\sqrt{2}-1}{2}\right)+2} \\ &=\sqrt{2\sqrt{2}-2+2}\\ &= \sqrt{2\sqrt{2}}\\ &= \sqrt[4]{8}\end{aligned}$

Thus $1+S=\sqrt[4]{8}\approx 1.681$ , and $\lfloor 1000S\rfloor = \boxed{681}$ .