A Fascinating Problem I found on a Course in Brilliant

We are finding the distance from the center of the base (which we designate to be the origin of a Cartesian coordinate system) and the right side of the triangle. The equation of that side is $\frac{x}{7}+\frac{y}{24}-1=0$ , and the distance is $\frac{1}{\sqrt{\frac{1}{7^2}+\frac{1}{24^2}}}=\boxed{\frac{168}{25}}$ .

We will focus on the right half of the picture and place a coordinate system with the origin at the center of the circle. Then, we have the equations

$y_1 =24-\frac{24}7x$

for the line and

$y_2 =\sqrt{r^2-x^2}$

for the circle.

For them to touch each other, they have to have a point in common and their derivatives at this point have to be equal. For the derivatives we get

$y_1^\prime = -\frac{24}7$

and

$y_2^\prime = -\frac x{\sqrt{r^2-x^2}}$ .

If we set them equal to each other, we end up with

$24-\frac{24}7x = \sqrt{r^2-x^2}$ $-\frac{24}7 = -\frac x{\sqrt{r^2-x^2}}$

We can substitute the first equation into the second and multiply by $-1$

$\frac{24}7 = \frac x{24-\frac{24}7x}$

$\frac 7{24}x = 24 - \frac{24}7x$

$x = \frac {4032}{625}$

Now, we plug this into the other equation. This yields

$24-\frac{24}7\left(\frac{4032}{625}\right) = \sqrt{r^2-\left(\frac{4032}{625}\right)^2}$

$\frac{1382976}{390625} = r^2 - \frac{16257025}{390625}$

$r^2 = \frac{28224}{625}$

$r = \boxed{\frac {168}{25}}$