Let $ABCD$ be a convex quadrilateral. Construct the points $P,Q,R,$ and $S$ on continue of $AB,BC,CD,$ and $DA$, respectively, such that \[BP=CQ=DR=AS.\] Show that if $PQRS$ is a square, then $ABCD$ is also a square.
Problem
Source: Iran Third Round 1996, E2, P2
Tags: inequalities, geometry, similar triangles, geometry proposed
xeroxia
30.03.2011 22:35
Let $\angle ASP = \alpha , \angle BPQ = \beta , \angle CQR = \theta , \angle DRS = \lambda$. Let $X$ and $Y$ be the feet of perpendiculars from $S$ and $Q$ to $AB$, respectively. $\triangle YPQ \sim \triangle XSP$. We have $XS \leq AS$. Thus $Y$ is closer (or equal) to $P$ than $B$. So $\angle PQB \geq \angle PQX$. This yields $90^{\circ} - \theta \geq 90^{\circ} - \beta \Rightarrow \beta \geq \theta$. Similarly draw perpendiculars from $S$ and $Q$ to $CD$. We will have $\lambda \geq \alpha $. Draw perpendiculars from $P$ and $R$ to $BC$. We will have $\theta \geq \lambda$. Similarly draw perpendiculars from $P$ and $R$ to $AD$. We will have $\alpha \geq \beta$. For quadrilateral $ABCD$ we have $\angle A = 90^{\circ} + \beta - \alpha$, $\angle B = 90^{\circ} + \theta - \beta$, $\angle C = 90^{\circ} +\lambda - \theta$, and $\angle D = 90^{\circ} + \alpha - \lambda$. According to above four inequality all four angles $\angle A, \angle B, \angle C, \angle D$ of $ABCD$ are less than or equal to $90^{\circ}$. But their sum is $360^{\circ}$. So they are all $90^{\circ}$. From similar triangles we will get $AP = BP = CR = DS$. So $(AB = BC = CD= DA) \land (\angle A = \angle B = \angle C = \angle D = 90^{\circ}) \Rightarrow ABCD$ is square.
sayantanchakraborty
10.05.2014 09:14
I have found too easy a solution.Have I got the problem right?? $PQRS$ is a square $\Leftrightarrow \triangle{ASP} \cong \triangle{BPQ} \cong \triangle{CQR} \cong \triangle{DRS} \Leftrightarrow QB=RC=DS=AP \Leftrightarrow AB=BC=CD=DA$. If I got the problem right,then it surely was a bad choice for the 3rd round.
joybangla
10.05.2014 14:41
sayantanchakraborty wrote: I have found too easy a solution.Have I got the problem right?? $PQRS$ is a square $\Leftrightarrow \triangle{ASP} \cong \triangle{BPQ} \cong \triangle{CQR} \cong \triangle{DRS} \Leftrightarrow QB=RC=DS=AP \Leftrightarrow AB=BC=CD=DA$. If I got the problem right,then it surely was a bad choice for the 3rd round. No you didn't Iran third rounds are not trivial. You proved that all sides are equal but $ABCD$ could still be a rhombus. See Xeroxia's proof. Getting $AB=BC=CD=DA$ is not hard but proving the angles are equal is hard. So your proof is incomplete.