Urn with probability one half

Statement

An urn holds $21$ balls, of which $r$ are red and the rest white. Two balls are drawn at random without replacement, and the probability that both are red is exactly $\dfrac12$. How many red balls are there?

Solution

The probability that both are red is $\frac{\binom{r}{2}}{\binom{21}{2}}=\frac{r(r-1)}{21\cdot 20}=\frac12.$ So $r(r-1)=\dfrac{21\cdot 20}{2}=210$. We seek two consecutive integers with product $210$: $15\cdot 14=210$, hence $r=\boxed{15}.$ Check: $\dfrac{15\cdot 14}{21\cdot 20}=\dfrac{210}{420}=\dfrac12$.