A geometry problem by A Former Brilliant Member

By pythagorean theorem

$PC=\sqrt{3^2+2^2}=\sqrt{9+4}=\sqrt{13}$

$AC=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

Since $\triangle AQP \sim \triangle ABC$ , then

$\dfrac{PQ}{CB}=\dfrac{PA}{AC}$ $\large \implies$ $\dfrac{PQ}{3}=\dfrac{2}{5}$ $\large \implies$ $PQ=\dfrac{6}{5}$

Applying pythagorean theorem on $\triangle PQC:$ $\large \implies$ $(PQ)^2+(CQ)^2=(CP)^2$ $\large \implies$ $\left(\dfrac{6}{5} \right)^2+(CQ)^2=\left(\sqrt{13}\right)^2$ $\large \implies$ $CQ=\dfrac{17}{5}$

Since $m \angle CBP \cong m \angle CQP = 90^\circ$ , $BPQC$ is a cyclic quadrilateral.

Applying Ptolemy's Theorem , we have

$(BQ)(CP)=(PQ)(BC)+(BP)(QC)$ $\large \implies$ $(BQ)(\sqrt{13})=\left(\dfrac{6}{5} \right)(3)+(2)\left(\dfrac{17}{5} \right)$ $\large \implies$ $BQ=\dfrac{4}{5}\sqrt{13}$

Finally,

$a+b+c=4+5+13=$ $\boxed{22}$

$\triangle CBA$ and $\triangle QPA$ are similar, therefore $\cos(\angle QAB)=\frac 45$ and $\overline{QA}=2\times\frac45=\frac85$

Applying law of cosines to $\triangle QAB$

$\overline {BQ}=\sqrt{4^2+\left(\frac85\right)^2-2\times4\times\frac85\times\frac45}=\dfrac45\sqrt{13}$

Answer is $4+5+13=\boxed{22}$