Solution with Danielle Wang (well, mostly by her): Let $\delta$ and $\epsilon$ denote $\angle MNB$ and $\angle CPQ$. Also, assume $AMNPQ$ has side length $1$. In what follows, assume $AB < AC$. (The case $AB=AC$ is trivial, although technically the lines coincide and so they are not parallel, oops.) First, we note that \begin{align*} BN &= (c-1) \cos B + \cos \delta \\ CP &= (b-1) \cos C + \cos \epsilon \\ \implies a &= 1 + BN + CP \\ \implies \cos \delta + \cos \epsilon &= \cos B + \cos C - 1. \end{align*}Also, by Law of Sines, we have $\frac{c-1}{\sin\delta} = \frac{1}{\sin B}$ and similarly on triangle $CPQ$, and from this we deduce \[ \sin \epsilon - \sin \delta = \sin B - \sin C. \]Using sum-to-product formulas on our relations implies that \[ \tan \left( \frac{\epsilon-\delta}{2} \right) = \frac{\sin B - \sin C}{\cos B - \cos C + 1}. \]Now note that $\ell$ makes an angle of $\frac12(\pi+\epsilon-\delta)$ with line $BC$. Moreover, if line $OI$ intersects line $BC$ with angle $\varphi$ then \[ \tan\varphi = \frac{r - R \cos A}{\frac12(b-c)}. \]So in order to prove the result, we only need to check that \[ \frac{r - R \cos A}{\frac12(b-c)} = \frac{\cos B - \cos C + 1}{\sin B - \sin C}. \]Using the fact that $b = 2R\sin B$, $c = 2R\sin C$, this just reduces to the fact that $r/R + 1 = \cos A + \cos B + \cos C$, which is the so-called Carnot theorem. ADDENDUM April 28: Here is a very short solution with complex numbers, from the official solutions file. In fact, it shows that we only need $AM = AQ = NP$ and $MN = QP$. We use complex numbers with $ABC$ the unit circle, assuming WLOG that $A$, $B$, $C$ are labeled counterclockwise. Let $x$, $y$, $z$ be the complex numbers corresponding to the arc midpoints of $BC$, $CA$, $AB$, respectively; thus $x+y+z$ is the incenter of $\triangle ABC$. Finally, let $s > 0$ be the side length of $AM = AQ = NP$. Then, since $MA = s$ and $MA \perp OX$, it follows that \[ m - a = i \cdot sx. \]Similarly, $n-p = i \cdot sy$ and $a-q = i \cdot sz$, so summing these up gives \[ i \cdot s(x+y+z) = (p-q) + (m-n) = (m-n) - (q-p). \]Since $MN = PQ$, the argument of $(m-n) - (q-p)$ is along the external angle bisector of the angle formed, which is perpendicular to $\ell$. On the other hand, $x+y+z$ is oriented in the same direction as $OI$, as desired.

Lemma : The locus of the point $ P $, whose sum of the oriented distances $ \text{dist}(P,BC) $ $ + $ $ \text{dist}(P,CA) $ $ + $ $ \text{dist}(P,AB) $ is fixed, is a line which is perpendicular to the line connecting the incenter and the circumcenter of $ \triangle ABC. $ Proof : See Distance sum (post #4 or post #5). ____________________________________________________________ Back to the main problem : Let $ T $ be a point varies on the external bisector of $ \angle MSQ. $ From Lemma $ \Longrightarrow $ it suffices to prove $ \text{dist}(T,BC) $ $ + $ $ \text{dist}(T,CA) $ $ + $ $ \text{dist}(T,AB) $ (oriented distance) is constant. Using oriented areas we get $$ [TNP]+[TQA]+[TAM]=[AMNPQ]-[TMN]-[TPQ]=[AMNPQ]=\text{constant}, $$so combining $ AM $ $ = $ $ NP $ $ = $ $ QA $ we conclude that $ \text{dist}(T,BC) $ $ + $ $ \text{dist}(T,CA) $ $ + $ $ \text{dist}(T,AB) $ is fixed.

Me trying to do a geometry problem is like trying to eat soup with a toothpick.

This reminded me of this problem; I wrote the parallel stuff down but couldn't get much after that.

Let $X$ be the midpoint of $MP$ and let $Y$ be the midpoint of $NQ$. It is well known that $XY\perp \ell$. Using TelvCohl's lemma and linearity of distance, it suffices to show \[ d(N,AB)+d(N,AC)+d(Q,BC)=d(P,AB)+d(P,AC)+d(M,BC) \] Or equivalently, that \[ [NAQ]+[NAM]+[QPN]=[PAM]+[PAQ]+[MNP] \] But both sides are equal to the area of the pentagon, so we're done.

adihaya wrote: lmao "Angel" Bisector Yeah, the diagram is 3/5 of an inverted pentagram

Solution: Notice that this is geo, so you probably can't solve it legit. Then you notice that if ABC is equilateral, then MN and PQ are parallel. Thus, there is no point S and no line l. Thus, we found a counterexample and the problem is flawed.

Not quite. What you have instead is a https://en.wikipedia.org/wiki/Principle_of_explosion; which just means you don't need to prove the problem in the edge cases where the premise is wrong. The actual flaw is that $\ell$ and $IO$ are sometimes the same line, and thus are not parallel.

Wait coincident lines are parallel? (This is off topic but they have to be parallel, since parallelism is an equivalence relation, and for various other reasons.)

Ack can't we just say in that case that $MN$ and $PQ$ intersect at the point of infinity and hence the angle bisector is essentially a line also parallel to $MN$ and $PQ$ which should also be parallel to $OI$ since the triangle is isosceles and stuff (This still doesn't account for v_Enhance's remark of $O\equiv I$ but I imagine in that case we can just take limits and continuity or something and be like gg)

pi37 wrote: Wait coincident lines are parallel? (This is off topic but they have to be parallel, since parallelism is an equivalence relation, and for various other reasons.) Yeah they "should" be parallel, but if you look at the definition usually given of "not intersecting" they aren't because they meet at every point. Anyways, that's just a silly remark; I can't imagine anyone cares.

Simple synthetic solution that doesn't use Telv's Lemma (didn't find during the test sadly ): Define $K=(SPN)\cap (SQM)$ and let $X,Y$ denote the midpoints of $MQ,NP$ respectively. We will show that $XY||IO$, which proves the problem since it's well known in configurations pertaining to $MN=PQ$ that $XY||\ell$. By spiral similarity and $MN=PQ$, we have $KNM\cong KPQ$. Thus $KM=KQ\implies K\in \overline {AXI}$. Let $AI\cap (ABC)=L\ne A$. Note that it suffices to show that $\frac {OL}{IL}=\frac {YK}{XK}$, as that will imply $YXK, OIL$ are homothetic and consequently $IO||XY$. The ratios are easy to chase. Notice that the spiral similarity gives $\frac {YK}{XK}=\frac {PN}{MQ}=\frac {AQ}{MQ}$, while $\frac {OL}{IL}=\frac {OL}{ BL}$. These two ratios are equal due to $AQM\sim OLB$ and we are done. And with this post, I am officially done with my 4 years of math olympiad career.

Here's a proof of the lemma used by pi37 and XmL. Lemma: Let $\triangle ABC$ be a triangle and let $E, F$ lie on $\overline{AC}, \overline{AB}$ respectively such that $BF = CE.$ Denote by $P, Q$ the respective midpoints of $\overline{BE}, \overline{CF}$ and let $\ell$ be the bisector of $\angle BAC.$ Then $\ell \perp PQ.$ Proof: Let $X$ be the point such that $ABXC$ is a parallelogram and let $P', Q'$ be the reflections of $A$ in $P, Q$ respectively. It's sufficient to show that $\ell \perp P'Q'.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(20cm); 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 = 82.45491139714903, xmax = 159.96835444316292, ymin = 52.64257438914551, ymax = 104.43349292123433; /* image dimensions */ draw((95.92783993159496,100.3167722802034)--(90.16049630182664,78.93337381879812)--(118.67944203587702,78.93337381879812)--cycle, red); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 5., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 5., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((95.92783993159496,100.3167722802034)--(90.16049630182664,78.93337381879812), red); draw((90.16049630182664,78.93337381879812)--(118.67944203587702,78.93337381879812), red); draw((118.67944203587702,78.93337381879812)--(95.92783993159496,100.3167722802034), red); draw((90.16049630182664,78.93337381879812)--(108.23167655030545,88.75284637221652)); draw((118.67944203587702,78.93337381879812)--(93.89419417452434,92.7766879113214)); draw((112.91209840610868,57.54997535739284)--(90.16049630182664,78.93337381879812), red); draw((112.91209840610868,57.54997535739284)--(118.67944203587702,78.93337381879812), red); draw((95.92783993159496,100.3167722802034)--(102.46433292053713,67.36944791081123), linetype("2 2") + blue); draw((95.92783993159496,100.3167722802034)--(116.6457962788064,71.39328944991613), linetype("2 2") + blue); /* dots and labels */ dot((95.92783993159496,100.3167722802034),linewidth(3.pt) + dotstyle); label("$A$", (95.05644657442731,100.97646264673196), NW * labelscalefactor); dot((90.16049630182664,78.93337381879812),linewidth(3.pt) + dotstyle); label("$B$", (89.28774380438436,77.90165156663275), SW * labelscalefactor); dot((118.67944203587702,78.93337381879812),linewidth(3.pt) + dotstyle); label("$C$", (119.13938046878134,77.90165156663275), SE * labelscalefactor); dot((108.23167655030545,88.75284637221652),linewidth(3.pt) + dotstyle); label("$E$", (108.89013185792834,88.99100252260276), NE * labelscalefactor); dot((93.89419417452434,92.7766879113214),linewidth(3.pt) + dotstyle); label("$F$", (92.8721804770324,92.79946648727932), NW * labelscalefactor); dot((99.19608642606605,83.84311009550731),linewidth(3.pt) + dotstyle); label("$P$", (98.19282866299434,84.0624020977272), NW * labelscalefactor); dot((106.28681810520068,85.85503086505976),linewidth(3.pt) + dotstyle); label("$Q$", (106.76187258354356,86.19066137210528), NE * labelscalefactor); dot((112.91209840610868,57.54997535739284),linewidth(3.pt) + dotstyle); label("$X$", (112.81060946863714,56.45103835382208), SE * labelscalefactor); dot((102.46433292053713,67.36944791081123),linewidth(3.pt) + dotstyle); label("$P'$", (101.55323804360188,66.47625967260305), SW * labelscalefactor); dot((116.6457962788064,71.39328944991613),linewidth(3.pt) + dotstyle); label("$Q'$", (117.29115530944719,70.956805513399), SE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that $AFQ'C$ is a parallelogram since its diagonals bisect one another. Hence, $Q' \in \overline{XC}$ and $CQ' = AF.$ Using $CX = AB$, it follows that $XQ' = BF.$ Similarly, $P' \in \overline{XB}$ and $XP' = CE.$ Because $BF = CE$, it follows that $\triangle XP'Q'$ is isosceles. Hence, the bisector $\ell'$ of $\angle BXC$ is perpendicular to $P'Q'.$ But since $\triangle ABC$ and $\triangle XCB$ are homothetic, $\ell' \parallel \ell$, and the desired result follows.

whatshisbucket wrote: Me trying to do a geometry problem is like trying to eat soup with a toothpick.

@above: That's what is called "bashing."

mathguy5041 wrote: whatshisbucket wrote: Me trying to do a geometry problem is like trying to eat soup with a toothpick.

Interestingly, this technique does not work when the soup has chunks in it, at which point the toothpick becomes a very useful consumption tool. I wonder what the mathematical analogy of this is.

"Some geo is easy to bash and some geo is hard to bash"

All geo can be bashed if you try hard enough

wait I'm pretty sure this geo is much harder to coord/complex/bary bash than problem 3 due to the awkward condition

Um, I'm pretty sure bisecting an angel wouldn't look too good. Also, is trig a viable option?

Thank you so much for explaining this for me, v_Enhance! The complex number solution is very clever indeed, I still wonder how they came up with it though...

What an EXCELLENT problem! Since $MN=QP$, the angle bisector is perpendicular to \[\vec{NM}-\vec{PQ}=\vec{M}-\vec{N}-\vec{Q}+\vec{P}=\vec{AM}+\vec{NP}+\vec{QA}.\]Note that the final expression is parallel to the sum of the unit vectors along the sides of $ABC$, so if we rotate it by $\pi/2$, we will get a vector parallel to \[\vec{OL_A}+\vec{OL_B}+\vec{OL_C}\]where $L_A,L_B,L_C$ are the arc-midpoints on $(ABC)$. It is well known that this is $\vec{OI}$, so we see that the angle bisector of $\angle MSQ$ is parallel to $\vec{OI}$, as desired.

Consider point $R$ such that $NPQR$ is a rhombus. Then, the angle bisector of $MNR$ is parallel to $\ell$. As this line is perpendicular to $MR$, we wish to show that $OI$ is perpendicular to $MR$. Now, suppose $QR$ hits $AB$ at $K$. Then, $AQK$ is homothetic to $ACB$, so we wish to show that $MR$ is perpendicular to the line connecting the incenter and circumcenter of $AQK$. So, we can restate the question as follows: Quote: Given triangle $ABC$, construct $D$ and $E$ on $AB$, $BC$ respectively such that $AD=CE=AC$. Then, $DE$ is perpendicular to $OI$. The new question is very easy, though. $DO^2-EO^2=Pow_O(D)-Pow_O(E)=b(b-c)-b(b-a)=b(a-c)$, whereas $DI^2-EI^2=CI^2-AI^2=(s-c)^2-(s-a)^2=c^2-a^2+(a+b+c)(a-c)=b(a-c)$. So, $DO^2-EO^2=DI^2-EI^2$, and $DE\perp OI$ as desired.

Woah. This problem follows from two lemmas: Lemma 1: If $ABC$ is a triangle, then the sum of the unit vectors in the directions of $AB$, $BC$, and $CA$ is perpendicular to $OI$. Proof: Easy complex bash Lemma 2: If $ABCD$ is a quadrilateral with $AB=BC=CD$, then the angle bisector of lines $AB,CD$ and (using points as vectors) $A+C-B-D$ are perpendicular. Proof: Easy complex bash with $B$ at the origin, using the fact that the angle bisector of $AB,CD$ is parallel through the line between the midpoints of $AD$ and $BC$. Now, $P-N+A-Q+M-A=P+M-N-Q$ is perpendicular to $OI$ by lemma $1$, and perpendicular to $\ell$ by lemma $2$, so we are done. $\blacksquare$

Remark: Beautiful problem! This synthetic solution took me a while to find. Motivated by a certain IMO 2005 # 1, we construct $D$ such that $DQ \parallel NP$ and $DN \parallel QP$. By definition $DNPQ$ is a parallelogram. Note that by these parallelisms, $\ell$ is parallel to the angle bisector of $\angle MND$, and $\triangle MND$ is isosceles, so it suffices to show $OI \perp MD$. This is actually remarkably easy. I claim that if we let $AI \cap (ABC)$ again at point $K$, then $\triangle IOK \sim \triangle MDQ$. This would in fact finish the problem since we already know that $OK \perp DQ$ and $KI \perp QM$, and spiral similarity would imply the final desired perpendicularity $IO \perp MD$. Through $OK \perp DQ$ and $KI \perp QM$ we already know that $\angle MQD = \angle IKO$. It remains to show that $\tfrac{MQ}{DQ} = \tfrac{IK}{KO}$. Note that by Incenter-Excenter Lemma $BK = IK$ and by definition $DQ = AQ$. Note that since $BO = OK$ and $AM = AQ$, and by circular arcs which gives $\angle BOK = \angle MAQ$, the triangles $\triangle MAQ \sim \triangle BOK$. Hence,\[\frac{MQ}{DQ} = \frac{MQ}{QA} = \frac{BK}{KO} = \frac{IK}{KO} \]as desired. $\blacksquare$ Diagram is shown below:

Wait does this just work??

If so, this is another great example of spiral similarities destroying weird geometry problems...

oh wait i think this was like one of the first posts. I'll still leave this here for posterity reasons though.

What in the heck is an equilateral pentagon?

yofro wrote: What in the heck is an equilateral pentagon? It's a pentagon with equal lengths, but not necessarily equal angles.

Pentagon with angles of 108 degrees, but the sides don't have to be equal. Regular pentagons are equiangular pentagons with the same side lengths.

ok so i think i mustve accidentally deleted part of my solution while typing cuz this makes no sense... i just have some random points that i didn't define

Equilateral pentagon... never seen that before. The motivation is that the line $OI$ is completely unrelated to the equilateral pentagon at the moment, so we find a nice characterisation of the direction of the line.

Now using that claim, and also the fact that the angle bisector of $\angle MSQ$ is perpendicular to the vector sum $\overrightarrow{NM}+\overrightarrow{QP}$ (as NM=QP), we have \[\ell\perp(\overrightarrow{NM}+\overrightarrow{QP})=(\overrightarrow{AM}+\overrightarrow{NP}+\overrightarrow{QA})\perp OI\quad\implies\quad\ell\parallel OI.\]

Let $\angle MNB = x$ and $\angle CPQ = y$. Also let AM = MN = NP = PQ = QA = 1. By this we have that BM = c - 1, CQ = b - 1. Now $BN = (c - 1) \cdot \cos \beta + \cos x$ and $CP = (b - 1) \cdot \cos \gamma + \cos y$ $\Rightarrow$ a = 1 + BN + CP $\Rightarrow$ $\cos x + \cos y = \cos \beta + \cos \gamma - 1$. Now by law of sines on $\triangle BNM$, we have that $\frac{BM}{\sin x} = \frac{MN}{\sin \beta}$ $\Leftrightarrow$ $\frac{c - 1}{\sin x} = \frac{1}{\sin \beta}$. Also by law of sines on $\triangle CPQ$ we have that $\frac{CQ}{\sin y} = \frac{PQ}{\sin \gamma}$ $\Leftrightarrow$ $\frac{b - 1}{\sin y} = \frac{1}{\sin \gamma}$ $\Rightarrow$ $\sin y - \sin x = \sin \beta - \sin \gamma$. Now $\tan (\frac{y - x}{2}) = \frac{\sin (\frac{y - x}{2})}{\cos (\frac{y - x}{2})} = \frac{\frac{\sin \beta - \sin \gamma}{2\cos (\frac{y + x}{2}})}{\cos (\frac{y - x}{2})} = \frac{\sin \beta - \sin \gamma}{2\cos (\frac{y + x}{2}) \cdot \cos (\frac{y - x}{2})} = \frac{\sin \beta - \sin \gamma}{\cos x + \cos y} = \frac{\sin \beta - \sin \gamma}{\cos \beta - \cos \gamma + 1}$. Now let $\ell \cap BC = L$ $\Rightarrow$ $\angle BLN = 90 + \frac{y}{2} - \frac{x}{2}$. Also let $OI \cap BC = Z$ and $\angle IZB = \varphi$. Then we have that $\tan\varphi = \frac{r - R \cos \alpha}{\frac{1}{2} \cdot (b - c)}$. So now we only need to prove that $\frac{r - R \cos A}{\frac{1}{2}\cdot (b - c)} = \frac{\cos \beta - \cos \gamma + 1}{\sin \beta - \sin \gamma}$. Since $b = 2R\sin \beta$ and $c = 2R\sin \gamma$, this just reduces to the fact that $\frac{r}{R} + 1 = \cos \alpha + \cos \beta + \cos \gamma$, which is Carnot theorem $\Rightarrow$ $\angle BLN = \angle OZB$ $\Rightarrow$ $OI \parallel \ell$ and we are ready.

Suppose triangle $ABC$ has circumcenter $O$, incenter $I$ and circummidarc triangle $M_A M_B M_C$; now \begin{align*}\ell \perp \overrightarrow{MN} + \overrightarrow{PQ} = -(\overrightarrow{NP} + \overrightarrow{QA} + \overrightarrow{AM}) \perp -(\overrightarrow{OM_A} + \overrightarrow{OM_B} + \overrightarrow{OM_C}) \parallel \overrightarrow{OI}\end{align*}as $I$ is the orthocenter of triangle $M_A M_B M_C$, and we're done.