Perpetually Bisected Tangent

If $P = (x, y)$ is the midpoint of the segment between $(a, 0)$ and $(0, b)$ , then clearly $a = 2x$ and $b = 2y$ . Thus the slope of the tangent to $P$ must be $\frac{dy}{dx} = -\frac b a = -\frac y x.$ Thus we solve the differential equation $\frac{dy} y = - \frac{dx} x\ \ \therefore\ \ \ln |y| = -\ln |x| + c\ \ \therefore\ \ y = \frac C x.$ Substituting the given point we find $C = 50$ . Therefore $\left.y\right|_{x=25} = 50/25 = \boxed{2}$ .

Let the tangent to point $P(a, f(a))$ be:

$y - f(a) = f'(a)(x-a) \Rightarrow y = f'(a)x + [f(a) - af'(a)]$

for any real positive value a and which has $x$ and $y-$ intercepts:

$([af'(a) - f(a)]/f'(a), 0)$ and $(0, f(a) - af'(a))$

respectively. If $P$ bisects the tangent line within the first quadrant, then one obtains the equations:

$[af'(a) - f(a)]/[f'(a) + 0] \cdot \frac{1}{2} = a, [ f(a) - af'(a) + 0] \cdot \frac{1}{2} = f(a)$

which both yield the common separable ODE:

$a \cdot f'(a) + f(a) = (af(a))' = 0, f(5) = 10;$

and has the general solution:

$f(a) = \frac{K}{a}$

which after inputting the boundary condition above gives $f(x) = 50/x$ for the required function. Hence, $\boxed{f(25) = 2}$ .