It is known that a quadratic equation has either 0, 1, or 2 unique real solutions. Well, look at this equation:
$$\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}=1$$
Without loss of generality, assume a < b < c. Now note that x=a, x=b, and x=c are all...