Geometry problem 2 by Dhaval Furia

Let $BP=x, PC = 20-x, AP = y.$ By geometric means, we obtain:

$\frac{y}{x} = \frac{20-x}{y} \Rightarrow y = \sqrt{20x-x^2}$ (i).

Differentiating (i) yields:

$y'(x) = \frac{10-x}{\sqrt{20-x^2}} = 0 \Rightarrow x = 10$ (ii)

$y''(x) = -\frac{(20x-x^2) + (10-x)^2}{(20x-x^2)^{3/2}} \rightarrow y''(10) = -\frac{1}{\sqrt{10}} < 0$ (iii) (hence a maximum).

The maximum length of $AP$ is just $\sqrt{20 \cdot 10 - 10^2} = \boxed{10}.$

By the Pythagoream Theorem:

1. (AB)^2 + (AC)^2 = 400
2. (AP)^2+(BP)^2 = (AB)^2
3. (AP)^2 + (20 - (BP))^2 = (AC)^2

Adding the second and third equations and simplifying yields (AP)^2 = 20 (BP) - (BP)^2. If we call x = (BP), we see that AP will be minimized when (AP)^2 is minimized. This will happen when the function f(x) = 20x - x^2 is minimized. f'(x) = 20 - 2 x = 0 implies x = 10. Further, f''(10) < 0, so that x = 10 maximizes f. f(10) = 200 - 100 = 100. This is the maximum value of (AP)^2, so that the maximum value of (AP) is sqrt(100) = 10.

Since $\triangle ABC$ is right-angled triangle, its hypotenuse $BC$ is the diameter of circumcircle of $\triangle ABC$ . Now any point(in this case $A$ ) on the circle is farthest from a particular diameter only when that point lie on the line form the center of circle and perpendicular to that diameter. Here that point will be at a distance equal to the radius of circle. So maximum height $AP = 10\;cm$

Triangles $\triangle {ABC}$ and $\triangle {APB}$ are similar. So

$\dfrac{|\overline {AP}|}{|\overline {AB}|}=\dfrac{|\overline {AC}|}{|\overline {BC}|}\implies$

$|\overline {AP}|=\dfrac{|\overline {AC}||\overline {AB}|}{20}$

Since side lengths of the triangle are positive definite, therefore

$2|\overline {AC}||\overline {AB}|\leq |\overline {AC}|^2+|\overline {AB}|^2=400\implies |\overline {AP}|\leq \dfrac{200}{20}=\boxed {10}$ .