Did anyone else construct a counterexample and submit it?

The existence of a counterexample would be very strange, considering the problem asks you to prove the statement.

I know; thats why Im very sure I'm wrong.. did you get this?

No. The problem is correct. man I overlooked something on this problem...ugh

I drew the circle tangent to two sides at P and Q. Then I assumed the circle would intersect the third side. Then I basically said that if the intersections were R and S, everything works out nicely, but otherwise, there's some sort of contradiction. How many points do you think I'll get?

I proved that the circle through P, R, and S is tangent to AB at P and the circle through Q, R, and S is tangent to AC at Q. How many points do you think I will get? (I think around 2 or 3.) EDIT: 500th post

My solution:

Eh, idk if this'll earn points or not Basically, using similar triangles gives $BS \cdot BR=BP^2$, thus the circumcircle of $PSR$ is tangent to $AB$. Similarly, the circumcircle of $QSR$ is tangent to AC. Thus, $AP$ is a tangent of the circumcircle of $PSR$, and $AQ$ is a tangent of the circumcircle of $QSR$. Since these tangents are equal, the circumcircles are the same, QED. Does this get points? I didn't explicity cite radical axes when I noted that the tangents are equal -> same circle.

algebra1337 wrote: My solution:

This was my solution precisely. I proved everything except for the fact that radical axes are always single lines if the two circles are different, which I'm fairly certain is well-known. Also, a diagram for those interested: [asy][asy] import graph; size(10.46cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.9, xmax = 14.56, ymin = -3.95, ymax = 11.77; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); /* draw figures */ draw((2.72,5.22)--(1.18,3.1), zzttqq); draw((1.18,3.1)--(5.42,2.82), zzttqq); draw((5.42,2.82)--(2.72,5.22), zzttqq); draw(circle((2.91,3.51), 1.16)); /* dots and labels */ dot((2.72,5.22),dotstyle); label("$A$", (2.87,5.44), NE * labelscalefactor); dot((1.18,3.1),dotstyle); label("$B$", (1,3.32), NE * labelscalefactor); dot((5.42,2.82),dotstyle); label("$C$", (5.55,3.04), NE * labelscalefactor); dot((1.97,4.19),dotstyle); label("$P$", (1.88,4.6), NE * labelscalefactor); dot((3.67,4.37),dotstyle); label("$Q$", (3.82,4.6), NE * labelscalefactor); dot((1.84,3.06),dotstyle); label("$S$", (1.98,3.25), NE * labelscalefactor); dot((3.9,2.92),dotstyle); label("$R$", (4.03,3.15), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Something slightly different...

Could I get penalized for not taking into account $Q=R'$, or is it trivial? EDIT: Oh then just take Q', and what if we can't because circ(ABR) is also tangent, if circ(ABQ) and circ(ABR) are both tangent, Q=R contradiction. Okay, the patch is trivial probably still a 7. Also what if R' is on the WRONG SIDE of Q, well direct angles whatever. Stupid trivial stuff and dumb orientations, but I should still get a 7.

algebra1337 wrote: My solution:

My first thought when I saw those angles... the configuration of tangent arcs and the final conclusion is a dead giveaway.

BOGTRO wrote: Eh, idk if this'll earn points or not Basically, using similar triangles gives $BS \cdot BR=BP^2$, thus the circumcircle of $PSR$ is tangent to $AB$. Similarly, the circumcircle of $QSR$ is tangent to AC. Thus, $AP$ is a tangent of the circumcircle of $PSR$, and $AQ$ is a tangent of the circumcircle of $QSR$. Since these tangents are equal, the circumcircles are the same, QED. Does this get points? I didn't explicity cite radical axes when I noted that the tangents are equal -> same circle. I hope this works, because I said about the same thing (kinda making an assumption at the end about the equal tangents tho)

JSGandora wrote:

Exactly my solution. You do have to state that it's by radical axis (oops), so I *think* we should get 6/7.

Wait I pretty much said that (the bit about the equal tangents), except after I said that, I tried to "justify" it with an INCORRECT statement... will I get docked points? The incorrect statement was complete BS and was barely relevant to the thing I was proving...

henrypickle wrote: I hope this works, because I said about the same thing (kinda making an assumption at the end about the equal tangents tho) BOGTRO wrote: Eh, idk if this'll earn points or not Basically, using similar triangles gives $BS \cdot BR=BP^2$, thus the circumcircle of $PSR$ is tangent to $AB$. Similarly, the circumcircle of $QSR$ is tangent to AC. Thus, $AP$ is a tangent of the circumcircle of $PSR$, and $AQ$ is a tangent of the circumcircle of $QSR$. Since these tangents are equal, the circumcircles are the same, QED. Does this get points? I didn't explicity cite radical axes when I noted that the tangents are equal -> same circle. I think you might get a 5 for that, since it's not immediately obvious that two circles tangent to the sides at P and Q (with AP = AQ), and going through the same two points, are the same. I'm not sure if this is closer to a 2 or a 5, but it's not complete for sure.

Well, it's basically the definition of radical axes :3

I didn't even know about radical axis but I cited the equal tangents anyways :3

I didn't take the test but here's my solution : Let $\omega$ be the circumcircle of $PRS$, and $\omega'$ the circumcircle of $QRS.$ Then since $\angle BPS = \angle PRS$, we have that $AP$ is a tangent to $\omega$. Similarly $AQ$ is tangent to $\omega'$. And so $A$ belongs to the radical axis of $\omega$ and $\omega'$, which is $RS$ (contradiction) unless $\omega \equiv \omega'$, so done.

Note that $AB$ and $AC$ are tangents to $(PSR)$ and $(QSR)$ at $P$ and $Q$ respectively. Now for the sake of contradiction, assume that these two are different circle. The radical axis of these two circles is $BC$, and as $A$ has equal tangents ($AP$ and $AQ$) to both of these circles. Thus $A$ must lie on the radical axis too, which is a contradiction.

nice short problem

memories of the OTIS application Assume FTSOC that $(PRS)$ and $(QRS)$ are distinct. The angle conditions imply that $AP$ is tangent to $(PRS),$ and $AQ$ is tangent to $(QRS).$ Since $AP^2 = AQ^2,$ $A$ lies on the radical axis $\overline{RS} = \overline{BC}$ of these two circles, which is absurd.

first, the angle equalities give that $BPS$ and $BRP$ are similar this yields $\frac{BP}{BR}=\frac{BS}{BP}$, and $BP^2=BS*BR$ this means that the circle containing $P$, $S$, and $R$ is tangent to $AB$ at $P$ doing the same thing for $C$, we get that the circle containing $Q$, $S$, and $R$ is tangent to $AC$ at $Q$ convieniently, $AP=AQ$ so they are tangents to the same circle, meaning that the circle containing $Q$, $S$, and $R$ is the circle containing $P$, $S$, and $R$, and $PQRS$ are concyclic

Since we know that $\angle{BPS} = \angle{PRS}$ and $\angle{CQR} = \angle{QSR}$, we know that $AP$ is tangent to $(PSR)$ and $AQ$ is tangent to $(QSR)$. We have $$Pow_{(PSR)}(A) = AP^2 = AQ^2 = Pow_{(QSR)}(A),$$so $A$ lies on the radical axis of the circles $(PSR)$ and $(QSR)$. We also have $$Pow_{(PSR)}(B) = BS\cdot BR = Pow_{(QSR)}(B)$$and $$Pow_{(PSR)}(C) = CS\cdot CR = Pow_{(QSR)}(C),$$so both $B$ and $C$ also lie on the radical axis of the two circles. We have that $A$, $B$, and $C$ all lie on the radical axis of the two circles, and that they also aren't collinear, since they are vertices of a non-degenerate triangle. This means that $(PSR)$ and $(QSR)$ must be the same circle, since the radical axis of two distinct circles must be a line. Thus $P, Q, R, S$ are concyclic, and we are done.$\blacksquare$

We assume, for the sake of contradiction that $P$, $Q$, $R$ and $S$ are not cyclic. We also let $\omega_{1}$ and $\omega_{2}$ be the circumcircles of ($PSR$) and ($QSR$) respectively. Our solution is based off the following lemma. $\overline{AP}$ and $\overline{AQ}$ are tangent to $\omega_1$ and $\omega_2$ respectively. Since, $\angle BPS = \angle PRS$, and $\angle CQR = \angle QSR$. So by Tangent Criterion, $\overline{AB}$ and $\overline{AC}$ are tangent to $\omega_1$ and $\omega_2$ respectively. Hence, $\overline{AP}$ and $\overline{AQ}$ are tangent to $\omega_1$ and $\omega_2$ respectively. This gives us \begin{align*} AP^2 = \text{Pow}_{\omega_1}(A)\\ AQ^2 = \text{Pow}_{\omega_2}(A) \end{align*} Since, $AP = AQ$ so, \[\text{Pow}_{\omega_1}(A) = \text{Pow}_{\omega_2}(A)\] This gives us that $A$ lies on the radical axis of $\omega_1$ and $\omega_2$. Hence, a contradiction because the radical axis of $\omega_1$ and $\omega_2$ is line $SR$ and $A$ does not lie on $SR$.

Suppose \((PSR)\) and \((QSR)\) are two distinct circles. Given that \(\measuredangle PRS = \measuredangle BPS\), it follows that \(AB\) is tangent to \((PSR)\). Similarly, since \(\measuredangle QRS = \measuredangle CQS\), \(AC\) is tangent to \((QSR)\). Therefore, \(BC\) serves as the radical axis of these two circles. Since \(AP = AQ\), we have \[ \text{Pow}_{(PSR)} A = AP^2 = AQ^2 = \text{Pow}_{(QSR)} A, \]implying that \(A\) must lie on the radical axis \(BC\). However, this is a contradiction because \(A\) does not lie on \(BC\). Therefore, our initial assumption that \((PSR)\) and \((QSR)\) are separate circles must be incorrect. Consequently, the points \(P\), \(Q\), \(R\), and \(S\) must be concyclic.

We will assume the contrary that $(PSR)$ and $(QSR)$ are different because we need to prove $PQRS$ cyclic we remember theorem 2.9 from EGMO on radical center, by keeping that in the back of our mind. We see that $AB$ and $AC$ are tangent to $(PQRS)$ this is because $\angle BPS =\angle PRS$ and $\angle CQR=\angle QSR$; as they are tangents we can use Power of Point to say that $AP²=AQ² \Rightarrow AP=AQ$, so $A$ lies on the Radical Axis of the two circles $(PSR)$ and $(QSR)$. But $RS$ intersects the two circles that means $RS$ is the radical axis, but $A$ does not intersect $RS$ so our assumption was wrong and $PQRS$ is cyclic.

Note that the circumcircle of $\triangle PSR$ is tangent to $AB$ at $P$ by the first angle condition. Similarly, the circumcircle of $\triangle QRS$ is tangent to $AC$ at $Q$ by the second angle condition. Assume for the sake of contradiction that these two circles are different. Clearly, $A$ has the same power with respect to $(PSR)$ and $(QRS)$ because $AP = AQ\implies AP^2 = AQ^2$. Therefore, $A$ must lie on the radical axis of these two circles, which is just $\overline{BC}$. However, $A$ cannot lie on $\overline{BC}$, so we conclude that $(PSR)$ and $(QRS)$ are the same circles and that $P, Q, R, S$ are concyclic. $\blacksquare$

In triangles $BPS$ and $BRP$ , $\angle BPS = \angle PRS$, and they have a common angle $\angle PBS = \angle RBP$ , Thus using Angle-Angle Similarity Criterion we can say that $$\triangle BPS \sim \triangle BRP$$From this we can conclude that $$\frac{BP}{BS} = \frac{BR}{BP}$$$$BP^2 = BR \cdot BS$$Thus $AB$ is tangent to $(PSR)$. We can use a similar analogy to prove that $AC$ is tangent to $(QSR)$.Therefore $$Pow_{(PSR)}A = AP^2 = AQ^2 = Pow_{(QSR)}A$$and hence $AB$ lies on the radical axis of $(PSR)$ and $(QSR)$, but so does points $S$ and $R$. Clearly $A$ does not lie on $SR$ and since circle can not have two radical axes, we can conclude that the circumcircles $(PSR)$ and $(QSR)$ are same, that is points $P , Q , R , S$ are concyclic

Rewrite the angle conditions as $AB$ and $AC$ being tangent to $(PSR)$ and $(QSR)$, respectively. Since $AP^2=AQ^2$, the point $A$ is on the radical axis of $(PSR)$ and $QSR$. Assuming $(PSR)\neq (QSR)$, the radical axis of two circles are line $BC$ which contradicts $A$ being on the radical axis. Thus, $(PSR)=(QSR)$ which implies points $P$, $Q$, $R$ and $S$ are cyclic as desired.

Assume FTSOC that $(PSR)$ and $(QSR)$ are different. $BPA$ tangent to $(PSR)$, $CQA$ tangent to $(QSR)$ $=>$ A lies on radax but $S$ and $R$ also lie on radax which implies contradiction by ABC degen

Let $S'$ be intersection of $AS$ and $(PSR). $ Since $AS'*AS = AP^2 = AQ^2 \implies S' $ lies on $(QSR) \implies PQRS$ are concylic