It is obvious that AP=ES If O is center of circle then $ \triangle AOP\equiv\triangle EOS$ and $ \angle OAP\equiv\angle OES$ Because $ \angle APS\equiv\angle PSE$ yields the conclusion.

The condition $ AB+CD=BC+DE$ implies that $ AP=SE$. Denote by $ U$ -the intersection of the segment $ AO$ with the circle $ k$. Denote by $ V$ -the intersection of the segment $ CO$ with the circle $ k$. Suppose that $ R$ is the radius of the circle $ k$. From the power of point $ A$ we get that $ AP^2=AU\cdot (AU+2R)$. From the power of point $ E$ we get that $ SE^2=EV\cdot (EV+2R)$. Since $ AP=SE$ we get that $ AU\cdot (AU+2R)=EV\cdot (EV+2R)\iff EV=AU\iff AO=CO$. So $ \triangle OPA\equiv \triangle OSE\iff \angle AES \equiv \angle EAP$. It's pretty obvious now that $ APSE$ is an issoscelles trapezoid, so $ PS\parallel AE$.

Marius Mainea wrote: It is obvious that AP=ES If O is center of circle then $ \triangle AOP\equiv\triangle EOS$ and $ \angle OAP\equiv\angle OES$ Because $ \angle APS\equiv\angle PSE$ yields the conclusion. My solution is the same. In my opinion this is far too easy problem for such a competition like JBMO.

can somebody please post the figure?

ridgers wrote: can somebody please post the figure? Here is the figure

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Marius Mainea wrote: It is obvious that AP=ES If O is center of circle then $ \triangle AOP\equiv\triangle EOS$ and $ \angle OAP\equiv\angle OES$ Because $ \angle APS\equiv\angle PSE$ yields the conclusion. Can somebody give further explaination about this, especially how did you get AP=ES ?

Dr N0 wrote: Can somebody give further explaination about this, especially how did you get AP=ES ? First Step: Since $AB$, $BC$, $CD$ and $DE$ are tangents and since tangents from the same point are equal, we get: $BP=BQ$ $CQ=CR$ $DR=DS$ Now we add $CR=CQ$ and $DR=DS \Rightarrow CR+DR=CQ+DS \Rightarrow CD=CQ+DS$. Also $AB+CD=BC+DE \Rightarrow CD=BC+DE-AB$. We subtitute the second in the first and get $BC+DE-AB=CQ+DS \Rightarrow AB=BC-CQ+DE-DS \Rightarrow AB=BQ+SE$. Since $BP=BQ \Rightarrow AB=BP+SE \Rightarrow AB-BP=SE \Rightarrow AP=ES$ Second Step: Now we have to prove that $\triangle{AOP}=\triangle{EOS}$ i. $AP=ES$ as we found above ii. $OP=OS$ since they are both radii iii. $\angle{APO}=\angle{ESO}=90$ since the angle between a tangent and a radius is a right angle Now we know that $\triangle{AOP}=\triangle{EOS}$. Let $\angle{AOP}=\angle{EOS}=x \Rightarrow \angle{POS}=180-2x$. $\triangle{POS}$ is isosceles since both $OP$ and $OS$ are radii. $\triangle{POS}:$ $\angle{OPS}=\angle{OSP}=\dfrac{180-\angle{POS}}{2}=\dfrac{180-180+2x}{2}=\dfrac{2x}{2}=x$ Now we have found that both angles $AOP$ and $OPS$ are equal to $x$ and since they are alternate we can now conclude that $ {PS}\parallel{AE}$

The key is to use the condition $$AB+CD = BC+DE \iff AP+BP+CP+RD = BQ+QC+DS+SE \iff AP = S,$$so $\triangle OPA \cong \triangle OSE$, and $\angle A = \angle E$. Thus $PSEA$ is an isosceles trapezoid, and $\overline{PS} \parallel \overline{AE}$.

Use equal tangents to line segments to find that AP=ES. With a well drawn diagram, we try to prove that <EPO=<POA. OP=OS, PA=SE, and angles APO and ESO are both right angles, so we have SAS congruency. So, <POS=180-2<POA (CPCTC), or <SPO=<POA, as desired. $\blacksquare$