Let $ABCD$ be a parallelogram. The circle with diameter $AC$ intersects the line $BD$ at points $P$ and $Q$. The perpendicular to the line $AC$ passing through the point $C$ intersects the lines $AB$ and $AD$ at points $X$ and $Y$, respectively. Prove that the points $P,Q,X$ and $Y$ lie on the same circle.
Problem
Source: Baltic Way 2008
Tags: geometry, parallelogram, rhombus, geometry unsolved
Nemion
13.11.2008 04:02
Prolongate $ PQ$ until it meets $ XY$ at $ O$ then it is clear that $ OP\cdot OQ = OC^2$. On the other hand since $ \triangle{DYC}$ similar to $ \triangle{BCX}$ we get that: $ \frac {DY}{BC} = \frac{DC}{BX}$, also from $ \triangle{ODC}$ similar to $ \triangle{OBX}$ and $ \triangle{ODY}$ similar to $ \triangle{OBC}$ $ \rightarrow$ $ \frac {OY}{OC} = \frac {DY}{BC}$ and $ \frac {OC}{OX} = \frac {BC}{BX}$ then we get $ OC^2 = OY\cdot OX$ $ \Rightarrow$ $ PQXY$ is cyclic.
limes123
13.11.2008 19:55
+ case when ABCD is rhombus (trivial). Nice and easy solution any other ideas?
Agr_94_Math
14.11.2008 03:38
@Nemion Can u elaborate please?
The QuattoMaster 6000
14.11.2008 05:24
Here's another (less nice) solution:
Lemma: Consider right $ \triangle ABC$ with $ \angle BAC = 90$. Let $ P$ and $ Q$ lie on $ AB$ and $ AC$ respectively such that $ PQ\parallel BC$. Let the line through $ B$ perpendicular to $ BC$ be $ b$ and define the line through $ C$ perpendicular to $ BC$ as $ c$. Let $ CP$ meet $ b$ at $ X$ and $ BQ$ meet $ c$ at $ Y$. Then, $ \angle XAB = \angle YAC$.
Proof: Let $ XC$ and $ BY$ meet at $ Z$, $ XB$ and $ AC$ meet at $ L$, $ YC$ and $ AU$ meet at $ M$, the feet of the perpendiculars from $ X$ to $ AB$ and $ AC$ be $ R$ and $ T$ respectively, and the feet of the perpendiculars from $ Y$ to $ AC$ and $ AB$ be $ S$ and $ T$ respectively. Now, it suffices to show that $ \angle XAT = \angle YAU$, so we want for $ \triangle XAT\sim \triangle YAU$. Thus, it suffices to have $ \frac {YU}{XT} = \frac {AU}{AT}$. Now, from $ UY\perp AM$, we have that $ UY\parallel AC$, so $ \frac {AU}{YC} = \frac {AM}{MC}$. Since $ \angle BCM = 90 = \angle BAC$, we have that $ \frac {AM}{MC} = \frac {AC}{BC}$, so $ \frac {AU}{YC} = \frac {AC}{BC}$. Similarly, $ \frac {TA}{XB} = \frac {AB}{BC}$. Dividing, we get that $ \frac {AU}{TA} = \frac {YC}{XB}\cdot \frac {AC}{AB}$.
Furthermore, $ \frac {PA}{XT} = \frac {PC}{XC}$ and $ \frac {AQ}{YU} = \frac {BQ}{BY}$. Dividing, we get that $ \frac {UY}{XT} = \frac {AQ}{AP}\cdot \frac {BY}{BQ}\cdot \frac {PC}{XC}$. Notice that $ \frac {PA}{QA} = \frac {AB}{AC}$. Now, recall that we want for $ \frac {AU}{TA} = \frac {YU}{XT}$, so $ \frac {BY}{BQ}\cdot \frac {PC}{XC} = \frac {YC}{XB}$. Notice that $ \frac {PC}{BQ} = \frac {ZC}{ZB}$ from $ PQ\parallel BC$. Furthermore, we have that $ \frac {YC}{XB} = \frac {ZC}{ZX}$ and $ \frac {BY}{XC} = \frac {BZ}{ZX}$ both from $ YC\parallel BX$. This gives us that $ \frac {BY}{XC}\cdot \frac {PC}{BQ} = \frac {BZ}{ZX}\cdot \frac {ZC}{ZB} = \frac {ZC}{XZ} = \frac {YC}{XB}$, as desired.
Now for the problem. Allow for $ AP$ to meet $ XY$ at $ T$, $ AQ$ to meet $ XY$ at $ U$, let the line parallel to $ XY$ through $ A$ meet $ CP$ at $ Z$, let the midpoint of $ AC$ be $ M$, allow for $ CZ$ to meet $ AP$ at $ L$, and $ AX$ to meet $ PC$ at $ K$. Since $ M$ is the midpoint of $ AC$, we have that\[ 1 = \frac {PL}{AL}\cdot \frac {AM}{MC}\cdot \frac {KC}{PK}\implies \frac {PL}{LA} = \frac {PK}{KC}\implies LK\parallel AC\]by Ceva's Theorem. Apply the lemma to right $ \triangle APC$ to get that $ \angle APZ = \angle XPC$. Reflecting everything into $ M$, we see that the $ XY$ maps to $ AZ$, $ P$ maps to $ Q$, $ A$ maps to $ C$, and $ B$ maps to $ D$, so $ Y$ maps to $ Z$, so $ \angle XPC = \angle APZ = \angle CQY$. Now, since $ \angle QUC = 90 - \angle QAC = \angle PAC = \angle APQ$, we have htat $ QPTU$ is cyclic. Then, $ \angle QPX = \angle QPT - \angle TPX = 180 - \angle TUQ - \angle UQY = 180 - \angle QYC$, establishing that $ PQYX$ is cyclic.
jevgeniy
24.05.2009 19:43
Quote: Prolongate $ PQ$ until it meets $ XY$ at $ O$ then it is clear that $ OP\cdot OQ = OC^2$. After that it can be easily noticed that $ ODCY$ figure is homothetic to $ OBXC$ figure with homothetic center $ O$ because $ BC\|\|DY$ and $ BX\|\|DC$. Thus $ \frac{XO}{CO}=\frac{CO}{YO}$ and $ XO\cdot YO=CO^2$, from which $ OP\cdot OQ=XO\cdot OY$ follows. But that means $ P$, $ Q$, $ X$, $ Y$ are concyclic, q.e.d. I think that could be a little bit shorter, however, expressing the same idea.