I haven't thought about a generalization, but here's a solution to this problem. The assertion is equivalent to $BI^2>BS\cdot BR$. It's easy to see that triangles $BSK,BMR$ are similar (in this order of the vertices). We only need to employ a simple angle chase. From here we derive $\frac{BS}{BK}=\frac{BM}{BR}\Rightarrow BS\cdot BR=BM\cdot BK=BM^2$, and since in the right triangle $BMI$ $BM$ is a leg, while $BI$ is the hypothenuse, we get $BI>BM\Rightarrow BI^2>BM^2=BS\cdot BR$, Q.E.D.

grobber wrote: I haven't thought about a generalization, but here's a solution to this problem. What a beautiful proof!! The generalization which I was talking of is that the result holds even if the line through B in not parallel to MK.

Regardless RS//MK, it holds that AS,CR and MK concur at the pole of RS wrt c(ABC). Given that this point is the orthocenter of the traingle RIS and is inside of RIS, we have the claim.

i really dont get your proof please elaborate thanks

marko avila wrote: i really dont get your proof please elaborate thanks First of all I have to correct the mistyping of points of tangency. As K lies on AB and so on I am going to prove that AS, RC and KL concurr at one point wich is the pole of RS wrt c(ABC). I don't remember anymore my previous proof but I con give the following argument. If T and U are the points of tangency from R to c, let's call P and Q the intersection of KL, UT and TL, UM respectively. From Pascal theorem to the hexagon KULMTT we have that P, Q and R are collinear. Since, as is well known, R, Q and C are collinear, then RC cuts KL at P. As P = KL/\UT, then P is the pole of RB. As R, P and C are collinears then the respectiv polars are concurrent. This means that RB, TU and ML concurr at S. As R lies on KM also A lies on TU. Since, as is easy to prove, P is the orthocenter of RIS and P is between I and the foot of I on RS, we have the claim. Let me know if now it is clear enough .

Its clear now thanks

Very nice your proof, Grobber ! I liked specially the characterization of an acute angle, i.e. Lemma. $\triangle ABC,\ D\in BC,\ AD\perp BC\ : \ A<90^{\circ}\Longleftrightarrow AD^2>DB\cdot DC\ .$ Proof. $h_a=AD=c\sin B=b\sin C,\ DB=c\left|\cos B\right|,\ DC=b\left|\cos C\right|\ .$ Therefore, $AD^2>DB\cdot DC\Longleftrightarrow$$bc\sin B\sin C>bc\left|\cos B\cos C\right|\Longleftrightarrow \sin B\sin C>\left|\cos B\cos C\right|\ .$ $\cos B\cos C\le 0\Longrightarrow B\ge 90^{\circ}\ \vee\ C\ge 90^{\circ}\Longrightarrow A<90^{\circ}\ ;$ $\cos B\cos C>0\Longrightarrow \cos (B+C)<0\Longrightarrow \cos A>0\Longrightarrow A<90^{\circ}\ .$ I proved easily the relation $\boxed {\ BR\cdot BS=(p-b)^2\ }$ and from the well-known relation $BI^2=\frac{ac(p-b)}{p}$ results $BI^2>BR\cdot BS\Longleftrightarrow \frac{ac(p-b)}{p}>(p-b)^2\Longleftrightarrow$ $ac>p(p-b)\Longleftrightarrow 4ac>(a+c)^2-b^2\Longleftrightarrow b^2>(a-c)^2\Longleftrightarrow b>\left|a-c\right|\ .$ Remark. If $X\in RS\cap AC$, then the division $(Y,R,B,S)$ is harmonically and the point $I$ belongs to the exterior of the circle with the diameter $RS\ !$

It's easy to see that the triangles $RKB$ and $LSB$ are similar so : $\frac{BK}{BS}=\frac{BR}{BL}=\frac{RK}{SL}=\frac{KM}{LM}$ (last equality by Thales theorem's). Now let $\alpha= <BIS$ and $\beta=<BIR$ then : $$\tan(\alpha)=\frac{BS}{BI}=\frac{BK.LM}{KM.BI}=\frac{BK}{BI}\frac{LM}{KM}$$ and by the sin law and the fact that $$<KLM=<ILM+<ILK=\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}-\frac{A}{2}$$ and by the same way $$<MKL=\frac{\pi}{2}-\frac{C}{2}$$ therefore : $$\tan(\alpha)=\frac{\cos(\frac{B}{2})\cos(\frac{C}{2})}{\cos(\frac{A}{2})}$$ and by the same way $$\tan(\beta)=\frac{\cos(\frac{B}{2})\cos(\frac{A}{2})}{\cos(\frac{C}{2})}$$ so $$\tan(\alpha+\beta)=\frac{\cos(\frac{B}{2})(\cos^2(\frac{C}{2})+\cos^2(\frac{C}{2}))}{\cos^2(\frac{A}{2})\cos^2(\frac{C}{2})\sin^2(\frac{B}{2})} >0$$ and finally $RIS < \frac{\pi}{2}$

Outlines of my bash Apply the law of sines in $\triangle{RKB},\triangle{BLS}$ and express $BR,RK,BS,SL$ in terms of $r$ where $r$ is the inradius.Then apply the law of cosines in $\triangle{IKR},\triangle{ILS}$ to get $IR^2,IS^2$.Also calculate $RS=BR+BS$ in terms of $r$.Then show that $IR^2+IS^2-RS^2 >0$(it wont be too difficult as one would get rid of $r^2$ and only the trigonometrical terms would remain).Thus $cos(\angle{RIS}) >0$ and we are done.

A similar proof to sprmnt21 above can be done with just polar transformation: Applying a polar transformation centered at the incircle, $t$ will be mapped to its pole $T$ on $KL$ (since this is the polar of $A$). With the given parallel condition this point would be the midpoint of $KL$, but that doesn't really matter. Now the polar of $R$ goes through $T$, as does the polar of $S$, and because they are polars, these are perpendicular to $RI$,$RS$ respectively, giving that $T$ is the orthocenter of $RIS$, done. Any time an angle centered on the center of a circle, especially the incenter, is a part of a problem, poles and polars may have a use!

sprmnt21 wrote: Let $I$ be the incenter of triangle $ABC$. Let $K,L$ and $M$ be the points of tangency of the incircle of $ABC$ with $AB,BC$ and $CA$, respectively. The line $t$ passes through $B$ and is parallel to $KL$. The lines $MK$ and $ML$ intersect $t$ at the points $R$ and $S$. Prove that $\angle RIS$ is acute. Quirky solution Let $P$ be the midpoint of $KL$. Lemma. $\angle APC>90^{\circ}$ Proof. Let $N$ be the midpoint of $AC$. Then $$|\overrightarrow{PN}|=\left|\tfrac{1}{2}\left(\overrightarrow{AK}+\overrightarrow{CL}\right)\right| \le \tfrac{1}{2}\left(|AK|+|CL|\right)=\tfrac{1}{2}\left(|AM|+|CM|\right)=\tfrac{1}{2}|AC|$$ so $P$ lies inside the circle with center $N$ and radius $\tfrac{1}{2}|AC|$, proving the claim. $\blacksquare$ Now observe that $\overline{AM}$ is the polar of $R$ and $\overline{CM}$ is the polar of $S$ in the incircle of $\triangle ABC$. Thus, $IR \perp PA$ and $IS \perp PC$. Moreover, we see that $\angle API<90^{\circ}$ and $\angle IPC<90^{\circ}$ since $AKLC$ is convex and $P$ lies on segment $KL$. Thus, $\angle RIS=180^{\circ}-\angle APC$ and the conclusion follows. $\blacksquare$

Angle chasing gives $SBKM$ and $RBLM$ cyclic. Because $\angle SKM + \angle RLM = \angle SBM + \angle RBM = \pi$, $SK$ and $RL$ intersect at a point $Y$ on the incircle. This means $AS$ is the polar of $R$ and $CR$ is the polar of $S$, so $X:=AS\cap CR$ should be the orthocenter of $\triangle RIS$. By Pappus, $X$ lies on $\overline{KL}$ and because $IX\perp RS\parallel KL$, $X$ is in fact the midpoint of $KL$. Therefore $X$ lies strictly inside $\triangle RIS$, and $\angle RIS$ is acute.

Claim 1: $H$ is the orthocenter of $RSI$ if and only if $BI \cdot BX = RB \cdot BS$. Trivial by considering the reflection of the orthocenter over the foot of $I$ to $RS$. Claim 2: $BI \perp RS$. Note that $KILB$ is cyclic, since $KI \perp BK$ and $IL \perp BL$. Therefore, $\angle KBI = \angle IBL = \angle LKI = \angle KLI$, which implies that $\angle BKL = \angle KLB = 90 - \angle IKL$ from which it follows that $BI \perp KL$. Since $KL$ is parallel to $RS$, $RS \perp BI$. Claim 3: $\triangle BRK \sim \triangle BLS$. Note that since $BI \perp RS$ and $\angle KBI = \angle IBL$, $\angle RBK = \angle LBS$. Since $AKIM$ is cyclic: $\angle IAM = \angle MKI$ from which $\angle RKB = 90 - \angle BAI$. In the same vein, $\angle BLS = 90 - \angle ICB$. Since $KRB = 180 - \angle RKB - \angle RBK = 180 - (90 - \angle BAI) - (90 - \angle KBI) = \angle BAI + \angle KBI$, which by symmetry is equal to $\angle LSB$. The desired similarity follows. Claim 4: The midpoint of $KL$ is the orthocenter of $\triangle RSI$. Note that the midpoint of $KL$ (call it $X$) is also the intersection of $BI$ and $KL$. Let $b$ be the length of $AC$ and let $r$ be the inradius of $ABC$. By claim $3$, we know: $$\frac{RB}{KB} = \frac{BL}{BS} \Longleftrightarrow BL^2 = RB \cdot BS \Longleftrightarrow (s-b)^2 = RB \cdot BS$$ It is easy to see that $BX = (s-b) \cos (KBI)$ and $XI = r \cos (KBI)$ from which we get that: $$BI \cdot BX = ((s-b)\cos (KBI) + r \cos (KBI))(s-b)\cos (KBI) = (s-b)^2 - (s-b)^2\sin ^2 (KBI) + r(s-b)\cos^2(KBI)$$ Note that $$\tan (KBI) = \frac{r}{s-b}$$ from which we see that: $$BI \cdot BX = RB \cdot BS = (s-b)^2$$ and by claim $1$ we win.

Solution with awang11. Let $P$ be the midpoint of $MK$. By point-line duality about the incircle, we have $\angle APC=180^\circ-\angle RIS$. By Pappus on $ALC$ and $RBS$, $AS\cap CR=P'$ lies on $MK$. Let $P_\infty$ be the point at infinity along line $MK$. Note that $MK\perp IB\perp RS$. Observe that $$(P',P_\infty;M,K)\stackrel{C}{=} (R,P_\infty;CM\cap RS,B)\stackrel{M}{=}(L,MK\cap AC;C,A)=-1,$$ with the last result following from the Ceva harmonic division configuration. Thus $P'$ is the midpoint of $MK$, so we have $\angle RPS=180^\circ-\angle RIS$. Since $P$ is closer than $I$ to line $RS$, the result follows.

Wow, I actually liked this problem a lot. Solution 1. Note that $RS$ is actually external angle bisector of $\angle ABC$. Trivial angle chase yields that $KBSM$ and $RBLM$ are cyclic. Claim. $AS\perp IR$. Proof. Let $(AKIM)$ and $(IBS)$ intersect at $P$. By the radical axis on $(AKIM)$, $(IBS)$ and $(KBSM)$, we obtain that $KM$, $BS$ and $IP$ are concurrent at $R$. The claim follows. Now all we need to do is to verify that line $AS$ lies between the lines $AI$ and $AR$. Let $F=AI\cap ML$, then $$\measuredangle IFL=\measuredangle AFM=\measuredangle CML+\measuredangle IAC=90^\circ-0.5(\measuredangle A+\measuredangle C)=0.5\measuredangle B=\measuredangle IBL,$$ thus $F$ lies on the circle with diameter $BI$. As $RS$ is tangent to $(BI)$, then $AI$ must intersect $ML$ before it intersects $RS$. Also $ML$ intersects side $BC$, and therefore cannot intersect $BC$ beyond $B$ of the line $BC$. Hence $AS$ lies inside $\angle IAB$ and therefore inside $\angle IAR$ as $MK$ intersects side $AB$, and therefore cannot intersect $AB$ beyond $B$ of the line $AB$. We do exactly the same with $C$ and obtain that orthocentre of $\triangle IRS$ lies inside the triangle, hence $\angle RIS$ is acute. Solution 2. [which I found during the write up of the first solution.] Note that $RS$ is actually external angle bisector of $\angle ABC$. Trivial angle chase yields that $KBSM$ and $RBLM$ are cyclic. Hence as $RK\cdot RM$ is the power of $R$ wrt $(KLM)$ and $SL\cdot SM$ is the power of $S$ wrt $(KLM)$, we get that $$RS^2=RB\cdot RS+SB\cdot SR=RK\cdot RM+SL\cdot SM=RI^2-r^2+SI^2-r^2<RI^2+SI^2,$$ where $r$ denotes the inradius of $ABC$. By law of cosines, we have $RI^2+SI^2>RS^2=RI^2+SI^2-2\cos{\angle RIS} \cdot RI\cdot SI\implies \cos{\angle RIS}>0\implies \angle RIS$ is acute.

Let $X=BI\cap MK$. It is obvious that $X$ lies between $B$ and $I$. Claim: $X$ is the orthocenter of $\triangle RIS$. Proof. Clearly $BI\perp RS$. It is sufficient to prove that $BR\cdot BS=BI\cdot BX$ by PoP. Notice the angles: $$\angle BRM=\angle KML=\angle KLC=\angle LKC=\angle SKB$$ Similarly $\angle RMB=\angle KSB$. That means $\triangle BRM\sim \triangle BKS$. So $$\begin{align*} \frac{BR}{BM}=\frac{BK}{BS}\Longrightarrow BR\cdot BS=BM^2=BI\cdot BX \end{align*}$$ hence $X$ is the orthcenter. $\square$ Since the orthocenter $X$ lies between $B$ and $I$ therefore the angle $\angle RIS$ must be acute.

Let $D$ and $E$ be $CI\cap ML$ and $AI\cap MK$ respectively. They are also midpoints of $ML,MK$ respectively. Since $\angle IDS=\angle IBS=90^\circ$ and $\angle IER=\angle IBR=90^\circ$ we have $IDBS,IBRE$ cyclic. Also, we have $ILBK$ cyclic. We have $$\angle SIR=\angle SMR+\angle MSI+\angle MRI=\angle LMK+\angle DBI+\angle EBI=\angle LMK+\angle DBE.$$ To prove that this is less than $90^\circ$ we only need to show that double of that is less than $180^\circ.$ Conveniently, $$2(\angle LMK+\angle DBE)=\angle LIK+2\angle DBE=180^\circ-\angle LBK+2\angle DBE$$ so it suffices to show that $\angle LBK\ge 2\angle DBE.$ Well, to do this, we dissect $\angle DBE$ into $\angle DBM+\angle EBM$ and we show that $\angle LBD>\angle DBM$ and $\angle KBE>\angle EBM.$ Note that $\angle MKB,\angle MLB$ are obtuse. We claim that this is a general lemma: let $XYZ$ be an obtuse triangle with $\angle X$ obtuse and $P$ be midpoint on $XY$ then $\angle XZP=\angle YZP.$ Clearly, $ZY>ZX$ and that is the most important part. By sine law, $$\frac{\sin(\angle XZP)}{XP}=\frac{\sin(\angle XPZ)}{XZ}>\frac{\sin(\angle YPZ)}{YZ}=\frac{\sin(\angle YZP)}{YP}.$$ Since $XP=YP$ we have $\sin(\angle XZP)>\sin(\angle YZP).$ $\angle X$ is obtuse so both these angles are clearly acute. Thus, we may conclude that $\angle XZP>\angle YZP$ so our proof is complete.

Won't post a solution but this question can also be solved by barycentric coordinates with reference triangle $\triangle RSM$ (after noting that $MB$ is a symmedian you end up with an inequality equivalent to $1>\sin RSM$ which obviously holds).

We show this by showing that the orthocenter lies inside of triangles RIS. We have KRB=MKL=MLC=SLB and similarly RKB=LSB. Then by sim triangles we have BS*BR=BL^2, and letting X be midpoint of RS we have BX=BLcos(b/2) and BI=BLcos(b/2) hence by transitive BI*BX=BS*BR. It’s easily derived that in a triangle ABC by pop BD*DC=AD*DA’=AD*HA’ where D altitude from A, H orthocenter, and A’ reflection of H over D onto circumcircle. Then since B is the foot of altitude (IX perp. KL parallel to RS), we conclude that X is the orthocenter of RIS, to which the problem becomes evident since KML is acute so X is inside MRS, while the line thru B is on the other side of KL than X. (can someone confirm this is rigorous? i was thinking abt proving that the cosine of that angle positive but idk how)

I've gotten much much worse at geometry, but this should work right? The main claim is that $D=IB\cap KL$ is the orthocenter of RIS, which finishes (since it's closer to RS the external angle bisector of ABC than I is). Very easy to deduce (using well known angles to chase) that RBLM, SBKM are cyclic quads, so SKM+RLM=SBM+RBM=180 so $KS\cap RL=P$ lies on the incircle. Finally, by Brokard $KL\cap YM$, R, S, I is an autopolar orthocentric system, so D is the orthocenter as claimed.

The quintessential geometric inequality: so clean to do correctly, so easy to overcomplicate. Here's the nice solution, which took me much too long. We will show that $RI^2+IS^2 > RS^2$. In particular, we have the following claim: Claim. $MLSB$ is cyclic. Proof. $\measuredangle MBS = \measuredangle BMK = \measuredangle MLS$. $\blacksquare$ Similarly, $KBRL$ is cyclic. Hence $$RI^2 + IS^2 > RM \cdot RL + SK \cdot SL = RB \cdot RS + SB \cdot RS = RS^2$$ where we use the deep estimate $r>0$.

HamstPan38825 wrote: The quintessential geometric inequality: so clean to do correctly, so easy to overcomplicate. I overcomplicated