lol

Evan O' Dorney showed me a nice solution: he put the angle in terms of three other angles, and put those angles in terms of inscribed angles, and sinplified it all into 1/2 <OXZ

It's easy to see that $PYQX$ and $SYRZ$ are rectangles, and we know that since $\angle BYA=90$, then $90=\angle YAB+\angle YBA=\angle YZA+\angle YXB=\angle YPQ+\angle YSR$. Hence triangles $YSR$ and $XPQ$ are similar. Thus since $YS||RZ$, we have that if we let the intersection of $PQ$ and $RS$ be $N$, then $\angle XPQ=\angle YSR=\angle ARN$, and quadrilateral $APRN$ is cyclic. Hence $\angle QNR=\angle XAZ=\frac{1}{2}\angle XOZ.$ [asy][asy]unitsize(100); real u=-0.6,v=0.2,w=0.8; pair a=(-1,0),b=(1,0),x=(u,sqrt(1-u^2)),y=(v,sqrt(1-v^2)),z=(w,sqrt(1-w^2)),p=foot(y,a,x),q=foot(y,b,x),r=foot(y,a,z),s=foot(y,b,z),o=extension(p,q,r,s); draw(arc(origin,1,0,180)--cycle, linewidth(0.5)); dot(a); dot(b); dot(o); dot(p); dot(q); dot(r); dot(s); dot(x); dot(y); dot(z); dot(origin); draw(y--z--a--p--y--q,linewidth(0.5)); draw(y--x--b--s--y--r,linewidth(0.5)); draw(p--o--s,linewidth(0.5)); label("$A$",a,W); label("$B$",b,E); label("$X$",x,NW); label("$Y$",y,N); label("$Z$",z,NE); label("$O$",origin,SW); label("$N$",o,SE); label("$P$",p,NW); label("$Q$",q,SW); label("$R$",r,SE); label("$S$",s,NE); [/asy][/asy]

Would we lose points if the intersection of PQ and RS was not on AB in our diagram, but we did not use that fact?

abacadaea wrote: Would we lose points if the intersection of PQ and RS was not on AB in our diagram, but we did not use that fact? You shouldn't. While it's (very) good to include a diagram for readability, the text of your proof should stand on its own without the diagram. So, if your text is solid, you should be fine.

Let C be the intersection of PQ and RS Let D be ft of perp from Y to XZ Then I said (1/2)<XOZ=<XAZ and try to show that PACR is cyclic. I used two cyclic pentagons XPYDQ and ZSYDR (it should be obvious to you. In the actual test, I showed step for such pentagon) and Simson Line from Y to AZB to show that (tiresome to write all steps: try to figure out by urself kk)<APC=<ARC. Then, PACR is cyclic and therefore, the question holds

Although it's an easy problem it's the problem I most enjoyed in the whole contest. I mean I've practically never invoked simson lines in an Olympiad problem.

I was kind of disappointed by this problem... for one thing, I was hoping for a hard geo question so I could have an advantage However, just like last year's #1, a random theorem killed the problem.... That's not the most optimal situation when choosing problems for the USAMO.

Heh, I actually had a little fun with this problem since although I instantly dropped the perpendicular, I completely missed/forgot about (not sure which) Simson lines. Fortunately I ended up proving they were collinear...

I knew what the Simson's line was and proved it. But I spelt it "Simpson's theorem" Is this going to be 1~2 points off? or completely 0 ???? Hope I get 7... T_T

If everything after that was correct, you will get a 7. See this.

lol pretty sure ur fine with that typo =p.

I think $AB$ doesn't actually have to be a diameter (though $O$ has to be the center of the circle.) This solution uses directed angles (not sure if I'm using them correctly or if it's really necessary, though I think it's necessary to account for all different configurations...)

Zhero wrote: I think $AB$ doesn't actually have to be a diameter (though $O$ has to be the center of the circle.) This solution uses directed angles (not sure if I'm using them correctly or if it's really necessary, though I think it's necessary to account for all different configurations...) . [/hide] They are necessary, that is why I had to redraw the diagram in front of Dr.Penev. I was like woops my solution fails darn. Does anyone think that proving it for only a certain configuration of QTRI (where T is the intersection of PQ and RS, and I is the intersection of AZ and BX) loses more than one point? Directed angles kills the problem.

Correct me if I'm wrong, but I don't believe that directed angles are necessary in the original problem; $AXYZB$ was specified to be convex, so the vertices must appear in that order. Could you point out to me any troublesome configurations? (In my generalization, the vertices did not need to appear in that order, hence the need for directed angles.)

OneShotSniper wrote: I knew what the Simson's line was and proved it. But I spelt it "Simpson's theorem" Is this going to be 1~2 points off? or completely 0 ???? Hope I get 7... T_T i'm in the same situation but i spelled it "titu's lemma" do i get points off

redacted, thanks @below

Redacted $ $

to easy for usajmo #3 literally #2 was way more harder then this one.

Post for storage $RS$ and $PQ$ are simson lines from $P$ to $ABZ$ and $ABY$ respectively, so $RS$ and $PQ$ pass through the foot from $Y$ to $AB$ (call it $M$). Now we have $AMRY$ is cyclic so $\measuredangle YMR=\measuredangle YAR= \measuredangle YAZ$. Similarly we have $\measuredangle YMQ=\measuredangle YBX$. Now we have $\measuredangle XAZ=\measuredangle YAZ+\measuredangle XAY=\measuredangle YAZ+\measuredangle XBY=\measuredangle YMR+\measuredangle QMY=\measuredangle QMR$. Which means $\angle QMR=\frac{1}{2}\angle XOZ$

Lol simson lines, solved at g2 problem session Define $W$ as the foot of the perpendicular from $Y$ to $AB$. By definition, $\overline{PQW}$ and $\overline{SRO}$ are Simson lines, which makes $\angle{PWS}$ the desired angle. Note that $\angle{YWA}+\angle{YPA} = 90+90 = 180$ which makes $PYWA$ cyclic. Similarly, $SBWY$ is also cyclic. We split $\angle{PWS}$ into $\angle{PWY}+\angle{YWS}$. Because of the cyclic quads, $\angle{PWY} = \angle{PAY} = \angle{XAY}$ and $\angle{YWS} = \angle{YBS}=\angle{YBZ}$. By inscribed angle theorem, \begin{align*} \angle{XAY}+\angle{YBZ} &= \angle{XAY}+\angle{YAZ} \\ &= \angle{XAZ} \\ &= \frac{\angle{XOZ}}{2} \end{align*}which thus concludes the proof.

So you angle chase. Full proof here. https://infinityintegral.substack.com/p/usajmo-2010-contest-review Angle chase or basic triangle congruency and similarity is the only synthetic way I know to solve stuff (except randomly inverting something and ending up with more circles than I originally have). Usually I just bash using Cartesian Coordinates or Complex Numbers Coordinates or very rarely Barycentric Coordinates. I also tried bashing this by setting A = (-1,0), B = (1,0) and trig expressions for x, y, z. It did not go well.

Solution: Let $\overleftrightarrow{AP} = \ell_1$ and $\overleftrightarrow{BS} = \ell_2$. Let $\ell_1 \cap \ell_2 = D$. By inscribed angle theorem, $\angle{SDP} = \frac{<AOB - <XOZ}{2} = 90^{\circ} - \frac{<XOZ}{2}$. It's easy to see that $$\angle{QPD} + \angle{RSD} = \angle{YPD} + \angle{QPY} + \angle{YSD} + \angle{RSY} = 90^{\circ} + \angle{QPY} + 90^{\circ} + \angle{RSY} = 180^{\circ} + \angle{QPY} + \angle{RSY} = 180^{\circ} + \angle{QXY} + \angle{RZY}$$by considering the cyclic quadrilaterals $PYQX$ and $SYRZ$. Since $A, R, Z$ are collinear and $B, R, X$ are collinear, $\angle{QXY} + \angle{RZY} = \angle{BXY} + \angle{AZY} = 90^{\circ}$ by inscribed angle theorem and the fact that $\overline{AB}$ is a diameter of the circle. Letting the intersection of lines $PQ$ and $RS$ be $C$, and considering quadrilateral $CPDS$, we have that $$\angle{QCR} + \angle{QPD} + \angle{RSD} + \angle{SDP} = 360^{\circ} \Rightarrow \angle{QCR} + 180^{\circ} + 90^{\circ} + \angle{SDP} = 360^{\circ} \Rightarrow \angle{QCR} + \angle{SDP} = 90^{\circ} \Rightarrow \angle{QCR} + 90^{\circ} - \frac{<XOZ}{2} = 90^{\circ} \Rightarrow \angle{QCR} = \frac{<XOZ}{2}$$, as desired. $\blacksquare$

first, we get that $XAZ=XBZ=\frac{XOZ}{2}$ let $PQ$ and $RS$ intersect at $C$ since $PY \perp AX$, $QY \perp BX$, $RY \perp BZ$, $SY \perp AZ$, simson lines yield that $C$ is on $AB$, and $YC \perp AB$ therefore, $YCP=YAP$, $YCR=YBR$, and $YAP=YAX=YBX=YBQ$ we then get $PCR=YCP+YCR=YAP+YBR=YBP+YBR=XBZ=\frac{XOZ}{2}$

Solved with megarnie. Let $PQ$ and $RS$ intersect at $K$. Since $AB$ is the diameter of $(O)$, we obtain $\angle AXB = \angle AZB = 90^{\circ}$, implying that $PXQY$ and $YSZR$ are rectangles. Thus, $$\angle YZR = \angle SRZ = \angle ARK$$However, $XYZB$ is cyclic, so \begin{align*} \angle YZR &= 180^{\circ} - (\angle AZB + \angle YXQ) \\ &= 90^{\circ} - \angle YXQ\\ &= \angle YXP\\ &= \angle QPX \end{align*}Therefore, $\angle ARK = \angle QPX$, which leads to $PRKA$ being a cyclic quadrilateral. It follows that $$\angle PKS = \angle PKR = \angle PAR = \angle PAZ = \dfrac{\angle XOZ}{2}$$as desired.

tedious angle chase! Mark $\measuredangle{XOZ}:=\alpha , \measuredangle{ZAB}=\theta ,PQ \cap SR:=\{J\}$ , so we have $$\measuredangle{PXY}=90^{\circ}-\measuredangle{YXB}=90^{\circ}-\measuredangle{YAB}$$,similarly $$\measuredangle{YZS}=90^{\circ}-\measuredangle{YBA}$$, and since $PXQY$ , $YRZS$ are cyclic we have $$\measuredangle{PQY}=90^{\circ}-\measuredangle{YAB}$$, similarly $$\measuredangle{YRS}=90^{\circ}-\measuredangle{YBA}$$. Also $\measuredangle{XAY}=\measuredangle{AYQ}$ , so $$\measuredangle{QYR}=90^{\circ}-\measuredangle{AYQ}-\measuredangle{BYR}$$, now since $$OZ=OA \implies \measuredangle{AZO}=\theta \implies \measuredangle{XZA}=90^{\circ}-\frac{\alpha}{2}-\theta$$, also $\measuredangle{YBX}=\measuredangle{XAY}$ , so $$\measuredangle{XAY}+\measuredangle{ZBY}=\frac{\alpha}{2}$$, so $\measuredangle{QYR}=\frac{\alpha}{2}$ , now since $YQJR$ is a quadrilateral we get $\measuredangle{QJR}=\frac{\alpha}{2}=\frac{\measuredangle{XOZ}}{2}$. $\square$

[asy][asy]unitsize(100); real u=-0.6,v=0.2,w=0.8; pair a=(-1,0),b=(1,0),x=(u,sqrt(1-u^2)),y=(v,sqrt(1-v^2)),z=(w,sqrt(1-w^2)),p=foot(y,a,x),q=foot(y,b,x),r=foot(y,a,z),s=foot(y,b,z),o=extension(p,q,r,s); draw(arc(origin,1,0,180)--cycle, linewidth(0.5)); dot(a); dot(b); dot(o); dot(p); dot(q); dot(r); dot(s); dot(x); dot(y); dot(z); dot(origin); draw(y--z--a--p--y--q,linewidth(0.5)); draw(y--x--b--s--y--r,linewidth(0.5)); draw(p--o--s,linewidth(0.5)); label("$A$",a,W); label("$B$",b,E); label("$X$",x,NW); label("$Y$",y,N); label("$Z$",z,NE); label("$O$",origin,SW); label("$M$",o,SE); label("$P$",p,NW); label("$Q$",q,SW); label("$R$",r,SE); label("$S$",s,NE); [/asy][/asy] Notice that $P, Q, M$ are collinear as it's the Simson line from $Y$ to $\bigtriangleup{ABX}$. Similarly, $S, R, M$ are collinear as it's the Simson line from $Y$ to $\bigtriangleup{AZB}$. Since $\angle{AMY} = \angle{APY} = 90^\circ$, $APYM$ is cyclic, therefore $\angle{PMY} = \angle{PAY} = \angle{XAY}$. Similarly, $\angle{BMY} = \angle{YSB} = 90^\circ$, so $BMYS$ is cyclic, therefore $\angle{SMY} = \angle{SBY} = \angle{ZBY}$. Since $\angle{ZBY}$, $\angle{YAZ}$ subtend the same arc, they're equal. Finally, $$\angle{PMS} = \angle{PMY} + \angle{SMY} = \angle{XAY} + \angle{ZAY} = \angle{XAZ} = \frac{1}{2}\angle{XOZ}$$So, we are done.

Say $H$ be the foot of Perpendicular from $Y$ to $AB$. Observe $P$,$ Q$ and $H$ are simson lines and they are collinear. Same with $S$, $R$ and $H$. This gurantees $PQ$ and $RS$ are concurrent at $H$. It sufficies To prove: $\angle PHS=\frac{\angle XOZ}{2}$. Observe $YPAH$ and $YPAR$ is concyclic. $\rightarrow$ YPARH is cyclic. $\angle PHS=\angle PHR=\angle PAR$ and $\angle PAR=\angle XAZ$. $\angle XAZ$ is incribed angle of circle with center $O$. $\angle XAZ=\frac{\angle XOZ}{2}$ $\Rightarrow$ $\angle PHS=\angle PHR=\angle PAR=\angle XAZ=\frac{\angle XOZ}{2}$.

Let $W$ be the altitude from $Y$ to $\overline{AB}$. By Simson's line theorem, we have that $PW$ and $RS$ concur at $W$ on $\overline{AB}$. Now make the observation that $PYWA$ and $SYWB$ are separately cyclic. By the inscribed angle theorem, it suffices to show at present that $PW \parallel XC$ and $SW \parallel ZC$. Indeed, \[\angle PWY = \angle PAY = \angle XAY = \angle XCY. \]By implementing another symmetrical argument, we similarly obtain that $SW \parallel ZC$. Hence \begin{align*} \angle PWS &= \angle PWY + \angle YWS \\ &= \angle XCY + \angle YCZ \\ &= \angle XCZ \\ &= \frac{1}{2} \angle XOZ, \end{align*}as desired. $\blacksquare$

Simsion lemma And angle Chase kills this