Side $\overline{AB}$ of $\triangle ABC$ has length $21$. The bisector of angle $A$ meets $\overline{BC}$ at $D$, and $CD = 4$. The set of all possible values of $AC$ is an open interval $(m,n)$. What is $m+n$?
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Solution
Let $AC=x$. By the angle bisector theorem, $BD=\dfrac{CD\cdot AB}{AC}=\dfrac{4(21)}x=\dfrac{84}x$, so $BC=\dfrac{84}x+4$.
From $AB+AC>BC$: $21+x>\dfrac{84}x+4 \Rightarrow x^2+17x-84>0 \Rightarrow (x-4)(x+21)>0 \Rightarrow x>4$.
From $AB+BC>AC$: $21+\dfrac{84}x+4>x \Rightarrow x^2-25x-84<0 \Rightarrow (x-28)(x+3)<0 \Rightarrow x<28$.
The third inequality $AC+BC>AB$, i.e. $x+\frac{84}x+4>21$, always holds because $x+\frac{84}x\ge2\sqrt{84}>17$.
So $(m,n)=(4,28)$ and $m+n=\boxed{32}$.
