HongKong 2022-23 IMO Selection Tests

The answer is $\left(\genfrac{}{}{0ex}{}{50}{3}\right)$ minus the number of choices of three rods that violate the triangle inequality. The number to be subtracted is the size of the set $$S=\{(a,b,c):1\le a<b<c\le 99,a+b<c,a,b,c\text{ are odd numbers}\}.$$ Note that for $(a,b,c)\in S$, $a+b$ is an even number between $4$ and $98$ (inclusive). For each such even number, we can easily count the number of such pairs $(a,b)$, and for each such $(a,b)$ we can easily count the number of choices of $c$ for which $(a,b,c)\in S$.

As an example, if we fix $a+b=20$, there would be $5$ choices for $(a,b)$, namely, $(1,19)$, $(3,17)$, $(5,15)$, $(7,13)$ and $(9,11)$. For each of these $5$ pairs of $(a,b)$, there are $40$ choices of $c$ for which $(a,b,c)\in S$, namely, $21,23,25,\dots ,97,99$. In the same way, we can see that for each even number $2k$ where $k\in \{2,3,4,\dots ,48,49\}$, there would be $\lfloor \frac{1}{2}k\rfloor$ choices of $(a,b)$ for which $a+b=2k$, and for each such $(a,b)$ there would be $50-k$ choices of $c$ for which $(a,b,c)\in S$. This can be summarized by the following table:

Value of $k$	Value of $a+b$	No. of choices of $(a,b)$	No. of choices of $c$
2	4	$\lfloor \frac{2}{2}\rfloor =1$	$50-2=48$
3	6	$\lfloor \frac{3}{2}\rfloor =1$	$50-3=47$
4	8	$\lfloor \frac{4}{2}\rfloor =2$	$50-4=46$
5	10	$\lfloor \frac{5}{2}\rfloor =2$	$50-5=45$
...	...	...	...
48	96	$\lfloor \frac{48}{2}\rfloor =24$	$50-48=2$
49	98	$\lfloor \frac{49}{2}\rfloor =24$	$50-49=1$

From the table, we see that $$\begin{array}{rl}|S|& =1\times 48+1\times 47+2\times 46+2\times 45+\cdots +24\times 2+24\times 1\\ & =1\times 95+2\times 91+\cdots +24\times 3\\ & =\sum _{k=1}^{24}k(99-4k)\\ & =99(1+2+\cdots +24)-4(1^2+2^2+\cdots +24^2)\\ & =99\cdot \frac{24\cdot 25}{2}-4\cdot \frac{24(24+1)(2\cdot 24+1)}{6}\\ & =10100\end{array}$$ and so the answer is $$\left(\genfrac{}{}{0ex}{}{50}{3}\right)-10100=\frac{50\cdot 49\cdot 48}{6}-10100=9500.$$