Quadrilateral $ABCD$ is both cyclic and circumscribed. Its incircle touches its sides $AB$ and $CD$ at points $X$ and $Y$, respectively. The perpendiculars to $AB$ and $CD$ drawn at $A$ and $D$, respectively, meet at point $U$; those drawn at $X$ and $Y$ meet at point $V$, and finally, those drawn at $B$ and $C$ meet at point $W$. Prove that points $U$, $V$ and $W$ are collinear. Proposed by A. Golovanov
Problem
Source: Tuymaada 2012, Problem 6, Day 2, Seniors
Tags: geometry, ratio, trapezoid, trigonometry, trig identities, Law of Sines, geometry proposed
Luis González
22.07.2012 08:02
Let $UV$ cut the perpendicular to $CD$ through $C$ at $W'.$ $P \equiv AD \cap BC.$ WLOG assume that the incircle $(V)$ of $ABCD$ is P-excircle of $\triangle PAB.$ Incircle $(J)$ of $\triangle PAB$ touches $AB$ at $L.$ Since $AB,DC$ are antiparallel WRT $PA,PB,$ then $\triangle PAB \sim \triangle PCD$ are similar with incircles $(J),(V).$ Thus $\frac{UV}{VW'}=\frac{DY}{YC}=\frac{BL}{LA}=\frac{AX}{XB} \Longrightarrow AU \parallel XV \parallel BW' \Longrightarrow W \equiv W'.$
subham1729
22.07.2012 14:35
Hint : Take in radius=$r$, keep $\angle {B},\angle {C}$ fixed .We've $XV=r$ Expressing $AU,BV$ in terms of $r$ and by some trigonometric fixed ratio we are done.
antimonyarsenide
22.07.2012 21:23
We could also get $AX/XB=DY/YC$ by letting $M,N$ be the incircle tangency points with $AD,BC$ respectively and noting that by some basic angle chasing, $\triangle BXV \sim \triangle VMD$, $\triangle BNV\sim \triangle VYD$, so $BXVN\sim VMDY$ and similarly $AXVM\sim VNCY$, so $AX*CY=VM*VN=r^2=BX*YD$. Note that this also shows $BC,XY,AD$ concur I messed around a lot with this kind of idea (except with $P=AB\cap CD$) and also noticed stuff like $\angle WVP=\angle UVP$. I didn't think of the phantom point construction and equal ratios $\iff$ parallelism idea Nice problem though.
Aiscrim
14.06.2014 19:39
A remark to make the problem "computable": $AUBW$ is a trapezoid, hence the collinearity of points $U,V,W$ is equivalent to proving the relation $XV=\dfrac{BX}{AB}\cdot AU+\dfrac{AX}{AB}\cdot BW$ (drop the perpendicular from $W$ onto $AU$, then it is easy..).
Take the radius of the inscribed circle $r$, $\widehat{ABC}=x,\ \widehat{DCB}=y$. By the law of sines in triangles $\triangle{AUD}$ and $\triangle{BWC}$ and in some right triangles, we obtain:
$XV=r, AX=r\tan{\frac{x}{2}},\ BX=\dfrac{r}{\tan{\frac{x}{2}}},\ AU=\dfrac{r\cos{x} (\tan{\frac{x}{2}}+\tan{\frac{y}{2}} )}{\sin{(x+y)}},\ BW=\dfrac{r\cos{y} (\frac{1}{\tan{\frac{x}{2}}}+\frac{1}{\tan{\frac{y}{2}}} )}{\sin{(x+y)}}$
So all that is left to prove is a dull trigonometric relation:
\[ \tan{\frac{y}{2}}+ \frac{1}{\tan{\frac{x}{2}}} =\dfrac{\cos{x} (\tan{\frac{x}{2}}+\tan{\frac{y}{2}} )}{\sin{(x+y)}\tan{\frac{x}{2}}}+\dfrac{\tan{\frac{y}{2}}\cos{y} (\frac{1}{\tan{\frac{x}{2}}}+\frac{1}{\tan{\frac{y}{2}}} )}{\sin{(x+y)}} \Leftrightarrow \]
\[{ \Leftrightarrow \sin{(x+y)} (\tan{\frac{x}{2}}\tan{\frac{y}{2}}+1 ){=} \cos{x} (\tan{\frac{x}{2}}+\tan{\frac{y}{2}} ){+} \tan{\frac{y}{2}} \cos{y} (1+\dfrac{\tan{\frac{x}{2}}}{\tan{\frac{y}{2}}}} ) \]
\[ \Leftrightarrow \sin{(x+y)} \dfrac{\cos{\frac{x-y}{2}}}{\cos{\frac{x}{2}}\cos{\frac{y}{2}}}=\cos{x}
\dfrac{\sin{\dfrac{x+y}{2}}}{\cos{\frac{x}{2}}\cos{\frac{y}{2}}}+\cos{y}\dfrac{\sin{\frac{x+y}{2}}}{\cos{\frac{x}{2}}\cos{\frac{y}{2}}} \]
\[\Leftrightarrow \sin{(x+y)}\cos{\frac{x-y}{2}}=\sin{\dfrac{x+y}{2}}(\cos{x}+\cos{y})\]
which is obviously true$\sin{(x+y)}=2\sin{\dfrac{x+y}{2}}\cos{\dfrac{x-y}{2}},\ \cos{x}+\cos{y}=2\cos{\frac{x+y}{2}}\cos{\frac{x-y}{2}}$.
navi_09220114
07.12.2015 15:46
It is easy to see AX/XB=DY/YC. We prove this fact first. Let F be intersection of AB and CD, reflect segment BC about the angle bisector of BFC to get segment C'B' (B' lies on AB, C' lies on DC), then since AD and BC are antiparallel, then B'C' is parallel to AD, and B'C' is tangent to the incircle of ABCD, say at K. Let the F-excircle of FB'C' tangent to B'C' at K', then if incircle of ABCD tangent to AD at M and tangent to BC at N, then by homothety F, N, K' colinear. So we get: AX/XB=AN/BM=AN/C'K=AN/B'K'=DN/C'K=DN/CM=DY/YC. Now consider the Miquel point P of the quadrilateral ABCD, then F is the center of spiral similarity that maps circumcircle PAD to circumcircle PBC. Since AX/XB=DY/YC, then it also map circumcircle PAD to circumcircle PXY, so PFXY is cyclic. So it is obvious by now, since APU=APV=APW=90. So P, U, V, W are colinear.