Problem 864

Let $P_{0}$ be the point on the line through $A M$ that minimizes the distance from $C$.

Then $C P_{0}$ is perpendicular to $A M$.

(Note that any other point $P$ on this line would form $\triangle C P_{0} P$ with right-angle at $P_{0}$, making $C P$ the hypotenuse of the triangle and so the longest side. In particular, any other point $P$ on the line gives $C P>C P_{0}$.)

Consider $\triangle A M C$.

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The area of $\triangle A M C$ equals $\frac{1}{2} A C \cdot 2 c=A C \cdot c$. This is because $M$ lies directly above $A C$, which is a diagonal of the base of the prism, and so the height of $\triangle A M C$ equals the height of the prism, which is $2 c$.

Also, the area of $\triangle A M C$ equals $\frac{1}{2} A M \cdot h$, where $h$ is the perpendicular distance from $C$ to $A M$. (Here, we are thinking of $A M$ as a base of the triangle.)

But $C P_{0}$ is the corresponding height, so $h=C P_{0}$.

In other words, $\frac{1}{2} A M \cdot C P_{0}=A C \cdot c$, and so $C P_{0}=\frac{2 A C \cdot c}{A M}$.

So we need to determine the length of $A C$ and the length of $A M$.

$A C$ is the hypotenuse of right-angled $\triangle A B C$.

Since $A B=2 a$ and $B C=A D=2 b$, then

$$A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{(2 a)^{2}+(2 b)^{2}}=\sqrt{4 a^{2}+4 b^{2}}=2 \sqrt{a^{2}+b^{2}}$$

$A M$ is the hypotenuse of right-angled $\triangle A F M$.

Since $A F=2 c$ and $F M=\frac{1}{2} F H=\frac{1}{2} A C=\sqrt{a^{2}+b^{2}}$, then

$$A M=\sqrt{A F^{2}+F M^{2}}=\sqrt{4 c^{2}+a^{2}+b^{2}}$$

Therefore,

$$\begin{aligned} & C P_{0}=\frac{2\left(2 \sqrt{a^{2}+b^{2}}\right) \cdot c}{\sqrt{4 c^{2}+a^{2}+b^{2}}}=\frac{4 c \sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}+b^{2}+4 c^{2}}} \\ & \text { ANSWER: } \frac{4 c \sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}+b^{2}+4 c^{2}}} \end{aligned}$$