Problem 499

Since $3 < $\sqrt{k^2+4}$ <
4$, then$3^2 < k^2 + 4 <
4^2$. (We can square each part and preserve the direction of the
inequalities since each part is positive.)

Therefore, $9 < k^2 + 4 < 16$
and so $5 < k^2 < 12$.

Since $k$ is a positive integer whose
square is between 5 and 12, then $k =
3$.
Let $S$ be the sum of the 20
smallest odd positive integers.

Then $S = 1 + 3 + 5 + $\cdots$ + 35 + 37 +
39$.

(Note that the smallest odd positive integer is 1 and the 20th integer
in this list must be $19 $\cdot$ 2 =
38$ greater than the 1st integer.)

If we rewrite the terms of $S$ in
reverse order, we obtain $S = 39 + 37 + 35 +
$\cdots$ + 5 + 3 + 1$.

Adding these two representations, we obtain $$2S = 40 + 40 + 40 + \cdots + 40 + 40 + 40$$ There are 20 terms in this sum because there were 20 terms
in each of the sums. Each term in this sum equals 40 because the first
pair adds to 40 and each subsequent pair has one number increased by 2
and one number decreased by 2, which means that the sum does not
change.

Therefore, $2S = 20 \cdot 40 = 800$
and so the sum of the 20 smallest odd integers is $400$.
Since $\triangle ABF$ is
right-angled at $A$, then by the
Pythagorean Theorem, $$BF^2 = AB^2 + FA^2 = 16^2 + 13^2 = 256 + 169 = 425$$ Since $\triangle BDF$ is equilateral, then $BD = DF = BF$ and so $BD^2 = DF^2 = BF^2 = 425$.

Since $\triangle BCD$ is right-angled
at $C$, then $BC^2 + CD^2 = BD^2 = 425$.

Since $BC = 8$, then $8^2 + CD^2 = 425$ which gives $CD^2 = 361$.

Since $CD > 0$, then $CD = \sqrt{361} = 19$.

Since $\triangle DEF$ is right-angled
at $E$, then $DE^2 + EF^2 = DF^2 = 425$.

Since $DE = 5$, then $5^2 + EF^2 = 425$ which gives $EF^2 = 400$.

Since $EF > 0$, then $EF = \sqrt{400} = 20$.

Therefore, the perimeter of $ABCDEF$
is $AB + BC + CD + DE + EF + FA$,
which equals $16 + 8 + 19 + 5 + 20 +
13$or$81$.