Students in 2009-2010 WOOT should have found this problem very familiar....

What was the woot problems.

It was: WOOT wrote: Points P and Q are on $ \odot O$ and Z is outside $ \odot O$ such that ZP and ZQ are tangent to the circle. Points A and B are on the circle such that PA is parallel to ZB. Let line ZB meet the circle again at point C. Show that QA passes through the midpoint of segment BC.

2011 Practice AIME III problem 2. WOOT students must be having a good year.

yeah. 2 aime problems? really?

WOW, WOOT is just too good at predicting problems. Or maybe there's some AMC problem writer in the committee disguised as a WOOT student. And of course, luck had it that I just HAD to be too young and inexperienced for WOOT 2009-2010, and I just HAD to take AIME 1 instead of AIME 2. Yay.

Will there be any classes on the same level of WOOT over the summer?

Anyways,

@professordad: lol same here. @hrithikguy: Mathematical Tapas Though it's not competition-oriented, I believe it's around the same audience.

I did it by using a phantom point, by showing that if it's not the bisector, then there's gonna be a contradiction with angle measurements. Will that work?

lolwut I think that usually, AIME 2 =/= USAJMO 2 . So I did some angle chasing...

Not only has it appeared in WOOT, this problem also appeared on last year's Sharygin geometry olympiad: http://www.geometry.ru/olimp/sharygin/2010/zaochsol-e.pdf (problem 16) http://www.artofproblemsolving.com/Forum/viewtopic.php?t=374577

I thought this one was easy, thanks to the Olympiad Geometry class.

Although my solution was really long and rather crude(it took me almost 2 hours), I was able to angle chase and prove that angle MCW, where W is the center of the circle and M is the point of intersection between AC and BE, is a right angle. It involved drawing an insane amount of lines( you need to extend a bunch of lines and drop an altitude), but I am quite happy I was able to solve the problem.

Let the intersection of $BE$ and $AC$ be $M$. Let the intersecion of $BD$ and $AC$ be $N$. Let $PA= a, AN=b, NM=c$ and $MC=d$. We wish to prove $b+c = d$. It is clear that $P$ is harmonic conjugate of $N$ with respecto to $A$ and $C$. We therefore have: $\frac{PA}{PC}= \frac{AN}{NC} => \frac{a}{a+b+c+d}=\frac{b}{c+d} => ac+ad= ab+b^2+bc+bd$ Since $\angle PDB =\angle BED=\angle BMN$ we have that $PBMD$ is cyclic. Hence: $PN*NM = NB*ND = AN*NC => (a+b)c = b(c+d) => ac = bd => a= \frac{bd}{c}$ Plugging this on our other ecuation we get: $bd+\frac{bd^2}{c} = \frac{b^2d}{c}+b^2+bc+cd$ $=> bdc + bd^2 = b^2d+b^2c+bc^2 + bcd$ $=> b(d-c)(d+c) = b^2(d+c) => b(d-c) = b^2 => d = b+c$ which is what we wanted to prove. Obvioulsy $d+c$ is not $0$.

This is also 4-40 in Challenging Problems in Geometry.

Let $BE$ intersect $AC$ at $M$ and let the center of the circle be $O$. Since $DE \| AC$ and $A$, $B$, $C$, $D$ and $E$ lie on a circle, it follows that $DECA$ is an isosceles trapezoid and $\widehat{DA}=\widehat{EC}$. By Chord-Intersection Angle Theorem, \[\angle{PMB} = \frac{\widehat{EC} + \widehat{AB}}{2} = \frac{\widehat{DA} + \widehat{AB}}{2}=\frac{\widehat{DB}}{2}=\angle{POB}\] Hence $POMB$ is cyclic and $\angle{OMP}=\angle{OBP}=90^\circ$. This implies that $M$ is the midpoint of $AC$.

Hard to believe it's only been about a year since I was struggling with this problem..

RedFireTruck wrote: ahxun2006 wrote: Why bump a month old? because its hard problem i wanna put my solution in thread for storage oh okay. ;-;

Solution: Let $O$ be the center of the circle. Now, introduce a point $X'$ on $\overline{AC}$ such that $\overline{OX'} \perp \overline{AC}$. In other words, $X'$ will be the midpoint of $\overline{AC}$. We wish to show that $\overline{BE}$ intersects $\overline{AC}$ at a point $X = X'$. It suffices to show that $\angle{BED} = \angle{BX'P}$, as this implies $B, X', E$ are collinear. Firstly, we have that $PBOD$ is cyclic which gives us that $\angle{BDP} = \angle{BOP}$. Similarly, by $PBOX'$ cyclic, we have that $\angle{BX'P} = \angle{BOP}$. Finally notice that $\angle{BDP} = \angle{BED}$ because both angles subtend arc $BD$ (inscribed angle theorem). Altogether, this gives us $\angle{BED} = \angle{BDP} = \angle{BOP} = \angle{BX'P}$, as desired. $\blacksquare$

Let $O$ be the center of $\omega$ and let $M=BE\cap AC$. Note that we have \[\angle BMP = \angle BED,\]by the condition $AC\parallel DE$, and \[\angle BED = \frac{1}{2}\angle BOD=\angle BOP,\]by circle and tangent properties. Since $\angle BMP =\angle BOP$, this implies that $BOMP$ is cyclic, meaning that \[\angle OMP=180-\angle OBP=90,\]by tangent properties. This means that since $OM\perp AC$, $M$ must be the midpoint of $AC$ by circle properties, finishing the problem.

v_Enhance wrote: 2011 Practice AIME III problem 2. WOOT students must be having a good year. bro I wasn't even born yet lol

Let $O$ be the center of $\omega$ and $K = \overline{AC} \cap \overline{BE}$. Claim: $OKBP$ is cyclic. Proof. Let $N = \overline{OP} \cap \overline{BD}$. Since $P$ is the intersection of the two tangents $PB$, $PD$ to $\omega$, we have $PB=PD$, or $P$ is on the perpendicular bisector of $BD$. Thus, $OP \perp BD$, or $\angle OND = 90^{\circ}$. By angle chasing, we have \begin{align*} \angle BED &= \angle BEC + \angle CED \\ &= \angle BDC + \angle CDP \, (\text{since} \, BCDE \, \text{is cyclic}) \\ &= \angle BDP \\ &= 90^{\circ} - \angle ODB \\ &= \angle DON \\ &= \angle BON = \angle BOP \end{align*}On the other hand, since $AC \parallel DE$ and $P,A,C$ are collinear, we have $$\angle BED = \angle BKC = \angle BKP$$It follows that $$\angle BKP = \angle BED = \angle BOP$$which is equivalent to $OKBP$ being cyclic. $\square$ Then, we have $\angle OKP = \angle OBP = 90^{\circ}$, which leads to $OK \perp AC$. Since $AC$ is a chord of $\omega$, $K$ is the midpoint of $AC$, as desired. $\square$ Edit: Oops, missed a simpler angle chase for $\angle BED = \angle BOP$.

Let the intersection of $AC$ and $BE$ be $Q$. We want to show $AQ=CQ$. Let $\angle{PBC}=\alpha$ and $\angle{PDC}=\beta$. We have $\angle{DAC}=\alpha,\angle{BAC}=\beta,\angle{BED}=\alpha+\beta=\angle{AQE}=\angle{BQC}=\angle{BDP}$. Hence $BPDQ$ is a cyclic quadrilateral. This means that $\angle{DQP}=\angle{DBP}=\alpha+\beta$, so $\triangle{AEQ} \simeq \triangle{CDQ}$, implying that $AQ=CQ$ and we're done.

By the tangent and collinearity condition we get that ABCD is a harmonic quadrilateral. Now from $DE\parallel AC$, $-1 = (A,C;B,D)\stackrel{E}{=}(A,C;AC\cap BE, P_\infty)$ by projecting onto $AC$ ($P_\infty$ is the point at infinity along line $AC$). Since $(A,C;AC\cap BE, P_\infty) = -1$ it is well known that $AC\cap BE$ must be the midpoint of AC, which is what we wanted to prove, so we are ready.

Let $M = \overline{AC} \cap \overline{BE}$. By definition, $ABCD$ is a harmonic quadrilateral, and hence, \[-1 = (B,D;A,C) \overset{E}{=} (M,\infty;A,C).\] The desired result follows by EGMO 9.8. $\square$

A proof without taking $O$ into account. First of all, let: $\bullet$ $AC\cap BD=T$ $\bullet$ $AC\cap BE=Q$ Since $ADEC$ is a isosceles trapezoid, we wish to show $AQ=QC\Longleftrightarrow QD=QE$. Let's prove it: $$\angle BED=\angle AQB=\angle PDB$$which implies $PDQB$ is a cyclic quadrilateral. Thus, $$\angle BED=\angle PBD=\angle PQD=\angle QDE\Longleftrightarrow DQ=DE$$as desired.

this is an egmo problem I just did a few days ago

Let $F = BE \cap AC$, so we just need to prove that $OF \perp AP$. This is equivalent to $BFOP$ cyclic. Now let $\angle BED = \theta$. By the parallel condition we have $\angle BFP = \theta$ and clearly $\angle BOD = 2 \theta$ where $O$ is the center of $\omega$. Now we have $\angle POB = \frac{\angle BOD}{2} = \theta$, so $\angle POB = \angle BFP$ and thus $BFOP$ cyclic, done.

Noice harmonic