24. The lengths of the sides of pentagon $A B C D E$ are as follows: $A B=16 \mathrm{~cm}, B C=14 \mathrm{~cm}$, $C D=17 \mathrm{~cm}, D E=13 \mathrm{~cm}, A E=14 \mathrm{~cm}$. Five circles with centres at the points $A$, $B, C, D, E$ are drawn so that each circle touches both of its immediate neighbours. Which point is the centre of the largest circle?
A
B
C
D
E
Solution
A
Let $R_{A}, R_{B}, R_{C}, R_{D}, R_{E}$ be the radii, in $\mathrm{cm}$, of the circles with centres at $A, B, C, D, E$ respectively. Each side of the pentagon is equal to the sum of the radii of the circles whose centres are at its endpoints. That is:
$$
\begin{array}{l}
A B=R_{A}+R_{B}=16 \\
B C=R_{B}+R_{C}=14 \\
C D=R_{C}+R_{D}=17 \\
D E=R_{D}+R_{E}=13 \\
E A=R_{E}+R_{A}=14 .
\end{array}
$$

Adding these gives
$$
\begin{aligned}
2\left(R_{A}+R_{B}+R_{c}+R_{D}+R_{E}\right) & =74 \\
\text { so } \quad R_{A}+R_{B}+R_{c}+R_{D}+R_{E} & =37 .
\end{aligned}
$$

Adding [1] and [3] gives
$$
R_{A}+R_{B}+R_{c}+R_{D}=33 .
$$

Subtracting [7] from [6] gives $R_{E}=4$.
Substituting $R_{E}$ into [5] gives $R_{A}=10$.
Substituting $R_{A}$ into [1] gives $R_{B}=6$.
Substituting $R_{B}$ into [2] gives $R_{C}=8$.
Substituting $R_{C}$ into [3] gives $R_{D}=9$. Hence the largest radius is $R_{A}$.