1. Suppose $r, s$, and $t$ are nonzero reals such that the polynomial $x^{2}+r x+s$ has $s$ and $t$ as roots, and the polynomial $x^{2}+t x+r$ has 5 as a root. Compute $s$.
Proposed by: Rishabh Das
Answer:
29

Solution: The first equation implies $s t=s$, so $t=1$. Then $x^{2}+x+r$ has 5 as a root, so $r+30=0$, implying $r=-30$. Finally, $x^{2}-30 x+s$ has 1 as a root, so $s=29$.
Remark: We missed the case of $s=t$, so $x^{2}+r x+s$ has $s=t$ as one root, and 1 as the other root (by Vieta's). This means $r=-s-1$. Then
$$
x^{2}+t x+r=x^{2}+s x-(s+1)=(x+(s+1))(x-1)
$$
has 5 as a root, so $s=-6$ is another solution. During the competition, both the answers -6 and 29 (as well as " 29 or -6 ") were accepted.