A geometry problem by Shamin Yeaser

Let $|AR| = |QP| = y$ and $|QA| = |PR| = x$ . Then $|RC| = \dfrac{12}{y}$ and $|BQ| = \dfrac{20}{x}$ , and so by Pythagoras

$|PC|^{2} = |RC|^{2} + |PR|^{2} = \left(\dfrac{12}{y}\right)^{2} + x^{2}$ and $|BP|^{2} = |QP|^{2} + |BQ|^{2} = y^{2} + \left(\dfrac{20}{x}\right)^{2}$ .

Then $|BP| \times |PC| = \sqrt{\left(y^{2} + \dfrac{20^{2}}{x^{2}}\right)\left(\dfrac{12^{2}}{y^{2}} + x^{2}\right)} = \sqrt{144 + x^{2}y^{2} + \dfrac{20^{2} \times 12^{2}}{x^{2}y^{2}} + 400}$ . Let this be equation (A).

Now as triangles $\Delta PRC$ and $\Delta BQP$ are similar we know that

$\dfrac{|PR|}{|RC|} = \dfrac{|BQ|}{|QP|} \Longrightarrow \dfrac{x}{\dfrac{12}{y}} = \dfrac{\dfrac{20}{x}}{y} \Longrightarrow \dfrac{xy}{12} = \dfrac{20}{xy} \Longrightarrow x^{2}y^{2} = 12 \times 20$ .

Plugging this result into equation (A) gives us that

$|BP| \times |PC| = \sqrt{144 + 12 \times 20 + \dfrac{20^{2} \times 12^{2}}{12 \times 20} + 400} = \sqrt{144 + 240 + 20 \times 12 + 400} = \sqrt{1024} = \boxed{32}$ .