, 2025

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$81$.