Consider $ \rho$ a semicircle of diameter $ AB$. A parallel to $ AB$ cuts the semicircle at $ C, D$ such that $ AD$ separates $ B, C$. The parallel at $ AD$ through $ C$ intersects the semicircle the second time at $ E$. Let $ F$ be the intersection point of the lines $ BE$ and $ CD$. The parallel through $ F$ at $ AD$ cuts $ AB$ in $ P$. Prove that $ PC$ is tangent to $ \rho$. Author: Cosmin Pohoata
Problem
Source: Romanian JBTST V 2007, problem 1
Tags: geometry, parallelogram, geometry proposed
Jan
06.06.2007 21:24
Because $FDAP$ is a parallellogram, we have that $\left|FD\right|=\left|PA\right|$. Now construct $U \in CD$ such that $BU \parallel OD$. We have that \[\left|PO\right|=\left|PA\right|+\left|AO\right|=\left|FD\right|+\left|DU\right|=\left|FU\right| \hspace{1cm}(1)\] and obviously also that \[\left|BU\right|=\left|OD\right|=\left|OC\right| \hspace{1cm}(2)\] and finally also \[\angle DUB = \angle DOB = \angle COP \hspace{1cm}(3)\] From $(1),(2)$ and $(3)$ follows that $\Delta PCO \cong \Delta FBU$ Because $\left|BD\right| = \left|CA\right|=\left|ED\right|$, it follows that $OD \bot EB$. Therefore, $\angle FBU = \frac{\pi}{2}= \angle PCO$. Image not found
pohoatza
06.06.2007 21:41
Very nice solution, Jan! Although the problem is easy, I will post my solution. Because $ AD \| PF$, we have that $ \angle{PFC} = \angle{ADC} = \angle{BCD}$, thus the quadrilateral $ PBFC$ is an issoscel trapezoid. Therefore $ \angle{PCB} = \angle{PFB}\iff 90 + \angle{PCA} = 90 + \angle{DBF}$ $ \iff \angle{PCA} = \angle{DBF}$. But $ \angle{DBF} = \angle{DAE}$, because the quadrilateral $ ABDE$ is cyclic, and $ \angle{DAE} = \angle{ADC}$. Therefore $ \angle{PCA} = \angle{ADC}$, so $ PC$ is tangent to $ \rho$.