Mike's square

Let $B$ be located at $(x, 3\log_a x).$ Because $\overline{AB}$ is parallel to the $x$ -axis, $\overline{BC}$ is parallel to the $y$ -axis, so the coordinates of $C$ are $(x, 5\log_a x).$

Then because the side length of the square is $\sqrt{\frac{5}{64}} = \frac{\sqrt 5}{8},$ we have $5\log_a x - 3\log_a x = 2\log_a x = \frac{\sqrt 5}{8}.$ Dividing by $2$ and rewriting in exponential form, we get $\qquad a^{\sqrt{5}/16} = x. \qquad (1)$

Now consider vertex $A.$ Side $AB$ is parallel to the $x$ -axis, so the coordinates of $A$ are $(x+\frac{\sqrt 5}{8}, \log_a(x+\frac{\sqrt 5}{8})),$ and we have the equation $3\log_a x = \log_a\left(x+\frac{\sqrt 5}{8}\right).$ Raising $a$ to the power of both sides gives $x^3 = x + \frac{\sqrt 5}{8}$ $\qquad x^3-x-\frac{\sqrt 5}{8} = 0. \qquad (2)$ By Descarte's Rule of Signs, there is only one positive solution to this equation (there is only one sign change: between the $x^3$ and $x$ terms); in addition, logarithms are only defined for positive inputs, so we want the positive solution. Now we notice by inspection that $x = \frac{\sqrt 5}{2}$ is a solution to $(2)$ , and is also positive. Thus, $\qquad x = \frac{\sqrt 5}{2}. \qquad (3)$ Now looking back to equation $(1),$ we have $a^{\sqrt 5 / 16} = \frac{\sqrt 5}{2},$ so $a^{\sqrt 5 / 4} = \left(\frac{\sqrt 5}{2}\right)^4 = \frac{25}{16}.$ The answer is $25 + 16 = \boxed{41}.$

Hey, I submitted this problem! I guess I should write a solution then.

There are two possibilities: the square lies below the x-axis or the square lies above the x-axis. It cannot lie half above and half below the axis because all of the graphs touch at the point (1, 0), so if $AB$ is above the x-axis then $C$ cannot be below the x-axis because then it necessarily must be to the left of line $AB$ , and the same applies if $AB$ is below and $C$ is above. We will explore both possibilities in this solution.

Case 1: $ABCD$ is above the x-axis. Then, $C$ must be the top left corner of the square, $B$ the bottom left, and $A$ the bottom right. In addition, the side length of the square is $\frac{\sqrt{5}}{8}$ . We can then set up the following equations:

$log_a(x + \frac{\sqrt{5}}{8}) = 3log_ax$

$3log_ax + \frac{\sqrt{5}}{8} = 5log_ax$

From the second, we get $log_ax = \frac{\sqrt{5}}{16} \rightarrow x^4 = a^{\frac{\sqrt{5}}{4}}$ . So we need to find x.

Getting rid of the log in the first equation, we get $x^3 - x - \frac{\sqrt{5}}{8}$ . With some ingenuity, this can be solved. Let $x = m\sqrt{5}$ . Thus $5m^3\sqrt{5} - m\sqrt{5} - \frac{\sqrt{5}}{8}$ . Clearing fractions and radicals we get $40m^3 - 8m - 1$ = 0. On first glance, we know that $1$ is too high, but $\frac{1}{4}$ is too low. Making use of RRT, we try $m = frac{1}{2}$ , and it works, so $x = \frac{\sqrt{5}}{2}$ .

Let $B$ be the point $(x,3 \log_a(x))$ . We then have two equivalent ways of representing the $y$ coordinate of $C$ : $3 \log_a(x) + \frac{\sqrt{5}}{8} = 5 \log_a(x)$ Through the properties of logs, we can show that this implies: $x^2 = a^{\frac{\sqrt{5}}{8}} \implies x^4 = a^{\frac{\sqrt{5}}{4}}$ Knowing this, we then must solve for $x^4$ .

Similarly using our definition of point $B$ , there are two equivalent ways of writing the $y$ coordinate of point $A$ : $\log_a(x + \frac{\sqrt{5}}{8}) = 3 \log_a(x)$ $\implies x^3 - x - \frac{\sqrt{5}}{8} = 0$ Through magic we can solve this depressed cubic. Two of the solutions are negative, which we may throw out because they are not in the domain of our $\log$ functions. The positive solution is $\frac{\sqrt{5}}{2}$ . This raised to the fourth power is $\frac{25}{16}$ , so our answer is $41$ .

Let the x-coordinates of $A$ be $z$ , then : $A(z,\log_a z),B(z-\frac{\sqrt{5}}{8},\log_a z), C(z-\frac{\sqrt{5}}{8},\log_a z-\frac{\sqrt{5}}{8} ).$ Now, we use the fact that $B,C$ are members of some graphs to get that : $\begin{cases} \log_a z = 3 \log_a\left(z-\frac{\sqrt{5}}{8}\right). \\ \log_a z-\frac{\sqrt{5}}{8}= 5 \log_a\left(z-\frac{\sqrt{5}}{8}\right).\end{cases}$ The first one is a third degree equation easy to verify that $\frac{5\sqrt{5}}{8}$ is it only positive solution, then solve the second one for $\ln a$ to get : $\ln a = \frac{8}{\sqrt{5}} \ln \frac{4}{5}$ .

Now, we can see that that : $a^{\frac{\sqrt{5}}{4}} = \frac{16}{25}$ .

which means that the answer is $41$ .

Let points $A=\left(x_{1},\log_{a}(x_{1})\right),\, B=(x_{2},3\log_{a}(x_{2})),\, C=(x_{2},5\log_{a}(x_{2}))$ . Since side $AB$ is parallel to the x-axis, it is clear that $\log_{a}(x_{1})=3\log_{a}(x_{2})=K$ . In addition, since the sides of a square are equal, we have that $x_{1}-x_{2}=a^{K}-a^{K/3}=5\log_{a}(x_{2})-3\log_{a}(x_{2})=\frac{2K}{3}$ . We know what $K$ is since the area of the square is given. $\left(\frac{2K}{3}\right)^{2}=\frac{5}{64}$ , so $K=\frac{3\sqrt{5}}{16}$ . Substituting this value into the equality above yields: $(x_{2})^{3}-(x_{2})=\frac{\sqrt{5}}{8}$ . It is relatively easy to see that $x_{2}=\frac{\sqrt{5}}{2}$ (the denominator of $x_{2}$ should be 2 to get the 8 in the denominator on the right, and the numerator of $x_{2}$ should have $\sqrt{5}$ in it). Since $x_{2}=a^{K/3}$ , we have that $\frac{\sqrt{5}}{2}=a^{\frac{\sqrt{5}}{16}}$ . Raising both sides to the fourth power gives $a^{\frac{\sqrt{5}}{4}}=\frac{25}{16}$ . $25+16=\boxed{41}$ .