EDIT Sorry guys, there was a flaw in my solution.

We have \[ KL\bot CI \] $I$ is incenter of the $ \triangle ABC $. Hence, $MKLP$ - cyclic.

Just $AI \parallel CL$ and the rest is easily angle chasing.

Define $S$ as the intersection of $MN$ and $PQ$. Then try to use Ceva in triangle $ MPS $ with point C in its interior : you will easily get $ \angle MSC = \angle ABC/2 $ Then $ \angle PLK + \angle PMK = 90 + \angle ABC/2 + 90 - \angle ABC/2 = 180 $

Let $X$ and $Y$ be the feet of the perpendicular from $M$ and $P$ to $PQ$ and $MN$ respectively. Clearly $MX\perp PQ$ and $PY\perp MN$. Let $H$ be the intersection point of $MX$ and $PY$ and let $S$ be the intersection point of $PQ$ and $MN$. From here, we see $H$ is the orthocenter of triangle $MPS$. Let \[AB=a+b\\ BC=b+c\\ CA=c+a\] and \[\angle BAC=2\alpha\\ \angle ABC=2\beta\\ \angle BCA=2\gamma.\] We have that $\alpha+\beta+\gamma=90^{\circ}$ and that \[CK=a\cos{\beta}\\ CL=b\cos{\alpha}.\] We also have that \[\frac{YH}{XH}=\frac{MH}{PH}=\frac{\sin{\beta}}{\sin{\alpha}}\] because af Sines Law in $MHP$. After that, we see that, if $I_{C}$ is the center of the excircle $\omega_{c}$ of $ABC$ opposite to $C$, and if $\omega_{c}$ touches $AB$ at $T$, then \[\tan{\alpha}=\frac{b}{TI_{C}} \text{ and } \tan{\beta}=\frac{a}{TI_{C}}\] so we must have that \[a\tan{\alpha}=b\tan{\beta}\Rightarrow \frac{a\cos{\beta}}{b\cos{\alpha}}=\frac{\sin{\beta}}{\sin{\alpha}}\Rightarrow \frac{CK}{CL}=\frac{YH}{XH}\] but we can see that $HX\parallel CL$ and $HY\parallel CK$ so triangles $YHX$ and $KCL$ are under an homothetic transformation with center in $S$, and therefore $\boxed{S, C, H \text{ are collinear}}$. We clearly see $MPXY$ is cyclic, and since $XY\parallel KL$ then \[\boxed{MKLP \text{ is cyclic}}\] as desired.

Let $MN\cap PQ=S$. Then we prove that $SC$ is perpendicular to $AB$. Let $SC\cap AB=D, w_{a}\cap BC=X, w_{b}\cap AC=Y$. If $PC\cap w_{a}=C_{1}$ then $(P,C_{1};X,Q)=-1$ so pencil $P(P,C_{1};X,Q)=-1$, if we denote $H_{1}=PX\cap SC$ then $(S,H_{1};C,D)=-1$. Similarly if $MY\cap SC=H_{2}$ we have $(S,H_{2};C,D)=-1$ sooo $H_{1}=H_{2}=H$, but $MY$ is perpendicular to $PQ$ (since 90=$\angle(APQ)+\angle(PAQ/2)=\angle(APQ)+\angle(AMY))$ as so $PX$ is perpendicular to $MN$ $\longrightarrow$ $H$ is orthocenter of $\triangle SPM$ so $SC\perp AB$. Then power of point $SK*SN=SC*SD=SL*SP$ id $SC\cap AB=D$ and we are done.

We wish to prove that the perpendicular bisectors of $KM, PL$ intersect on the bisector of $MN$ (or $AB$). Now the perpendicular bisector of PQ intersects AB at A and as $CQ=s-b$, so the bisector of PL intersects AB $\frac{s-b}{2}$ along from A to B, and this is $\frac{s-a}{2}$ away from the midpoint of AB. Do the same for the other bisector, and to prove they intersect on the bisector of AB, we need that $(s-a)\tan(A/2)=(s-b)\tan(B/2)$ but this is true as both are equal to the inradius.

@school5 why ABA'B' is cyclic and KL||A'B' ?

A mistake ...

@mavropnevma you mean the solution is wrong?

Adam9509 wrote: Let $X$ and $Y$ be the midpoints of $PQ$ and $MN$ respectively. We may see that $MX\perp PQ$ and $PY\perp MN$. If $X$ is the midpoint of $PQ$ then $AX\perp PQ$, not $MX$... Actually if $w_{a}\cap BC=X, w_{b}\cap AC=Y$, then $PX\cap MY=H$, where $H$ is the orthocenter of triangle $PMS$...

Thanks monsterrr, I meant to write the feet of the altitudes, not the midpoints.

In $\triangle ABC$ , the $A$-exincircle $w_a$ is tangent to $AB$ and $AC$ in $P$ and $Q$ respectively. The $B$-exincircle $w_b$ is tangent to $BA$ and $BC$ in $M$ and $N$ respectively. Denote the projection $K$ of $C$ on $MN$ and the projection $L$ of $C$ on $PQ$ . Prove that $MKLP$ is cyclically. Proof. Denote $S\in KM\cap LP$ . Observe that $AM=BP=s-c$ , $MP=a+b$ and $\left\{\begin{array}{c} m\left(\widehat{PMS}\right)=90^{\circ}-\frac B2\\\\ m\left(\widehat{MPS}\right)=90^{\circ}-\frac A2\\\\ m\left(\widehat{MSP}\right)=90^{\circ}-\frac C2\end{array}\right\|$ . Apply the theorem of Sines in $\triangle MPS\ :\ \frac {a+b}{\cos\frac C2}=$ $\frac {SM}{\cos \frac A2}=\frac {SP}{\cos \frac B2}\implies$ $SM^2-SP^2=\frac {(a+b)^2}{\cos^2\frac C2}\cdot\left(\cos ^2\frac A2-\cos^2\frac B2\right)=$ $\frac {ab(a+b)^2}{2s(s-c)}\cdot(\cos A-\cos B)=$ $\frac {ab(a+b)^2}{2s(s-c)}\cdot\frac {2(b-a)s(s-c)}{abc}\implies$ $\boxed{SM^2-SP^2=\frac {(a+b)(b^2-a^2)}{c}}$ . Apply the theorem of Cosines in the triangles $CAM$ and $CBM\ :\ \left\{\begin{array}{c} CM^2=(s-c)^2+b^2+2b(s-c)\cos A\\\\ CP^2=(s-c)^2+a^2+2a(s-c)\cos B\end{array}\right\|\implies$ $CM^2-CP^2=b^2-a^2+2(s-c)(b\cos A-a\cos B)$ . Observe that $b\cos A-a\cos B=$ $\frac {b^2-a^2}{c}$ and $CM^2-CP^2=$ $b^2-a^2+2(s-c)\cdot \frac {b^2-a^2}{c}=\left(b^2-a^2\right)\cdot\left(1+\frac {a+b-c}{c}\right)\implies$ $\boxed{CM^2-CP^2=\frac {(a+b)(b^2-a^2)}{c}}$ . In conclusion, $SM^2-SP^2=CM^2-CP^2=\frac {(b-a)(a+b)^2}{c}\implies$ $SC\perp AB$ . Denote $R\in SC\cap AB$ . Since $RMKC$ and $RPLC$ are cyclically obtain that $\left\{\begin{array}{c} SC\cdot SR=SK\cdot SM\\\\ SC\cdot SR=SL\cdot SP\end{array}\right\|\implies$ $SK\cdot SM=SL\cdot SP\implies$ $MKLP$ is cyclically.

mavropnevma wrote: In a triangle $ABC$, the excircle $\omega_a$ opposite $A$ touches $AB$ at $P$ and $AC$ at $Q$, while the excircle $\omega_b$ opposite $B$ touches $BA$ at $M$ and $BC$ at $N$. Let $K$ be the projection of $C$ onto $MN$ and let $L$ be the projection of $C$ onto $PQ$. Show that the quadrilateral $MKLP$ is cyclic. (Bulgaria) It suffice to prove that triangles $CKL$ and $AIB$ is similar, where $I$ is the incenter of $ABC$. It is obvious that $ \angle KCL\ = 90^{\circ}+\angle\frac{C}{2} = \angle AIB\ $. Also, we can easily count that $ \frac{CK}{CL}=\frac{(p-a){\cos{\beta}}}{(p-b){\cos{\alpha}}}\ =\frac{AI}{BI}$, where $\angle A\ = 2\alpha$, $\angle B\ = 2\beta$, $p$ is the semi-perimetr of $ABC$.

apluscactus wrote: Define $S$ as the intersection of $MN$ and $PQ$. Then try to use Ceva in triangle $ MPS $ with point C in its interior : you will easily get $ \angle MSC = \angle ABC/2 $ Then $ \angle PLK + \angle PMK = 90 + \angle ABC/2 + 90 - \angle ABC/2 = 180 $ Don't you mean using the triangle NQS (the one not containing C) ? I see how to use Ceva with this one (it just needs the law of sines in CQN), but not with yours... and if there was no mistake, I'm keen to see the details.

JPP wrote: Don't you mean using the triangle NQS (the one not containing C) ? I see how to use Ceva with this one (it just needs the law of sines in CQN), but not with yours... and if there was no mistake, I'm keen to see the details. I give a complete detailed proof : Use Ceva in triangle $MSP$ with point $C$ and let $x=\angle MSC$ : $\frac{\sin\angle NMC}{\sin\angle CMB} \frac{\sin\angle APC}{\sin\angle CPQ}=\frac{\sin x}{\sin (\angle ABC/2 + \angle BAC/2 - x)}$ But the law of sines gives us : $\frac{\sin\angle NMC}{\sin\angle CMB}= \frac{CN}{BC} \frac{MB}{MN}$ and similarly : $\frac{\sin\angle APC}{\sin\angle CPQ}= \frac{AC}{CQ} \frac{PQ}{AP}$ Hence, since $MB=AP$ : $\frac{\sin x}{\sin (\angle ABC/2 + \angle BAC/2 - x)}=\frac{CN*AC*PQ}{BC*MN*CQ} (*)$ Now, $f(x)=\frac{\sin x}{\sin (\angle ABC/2 + \angle BAC/2 - x)}$ is an increasing function of $ x$ between $0$ and $\angle ABC/2 + \angle BAC/2$. Therefore, there exists a unique $x$ that satisfies the identity (*). But for $x=\angle ABC/2$, the identity (*) works ! (To show this, you have to express all the lengths $CN$, $AC$, $PQ$, $BC$, $MN$, $CQ$ in terms of the sides of triangle ABC.) This last part is just a long computation, not really interesting !

Oh, yeah, I should have seen that. Thank you !

Take $D\in PQ\cap MN$. It suffices (from trivial angle chasing reasons) to prove that $AB\perp CD$. Let $I, I_a$ and $I_b$ be the centers of the incircle and $\omega_A,\omega_B$, respectively. Denote by $E$ and $F$ the intersections of $I_aI_b$ with $MN$ and $PQ$, respectively. The quadrilateral $BFI_aP$ is cyclic since $\widehat{BPF}=\widehat{BI_aF}=90^\circ-A/2$, hence $BF\perp I_aI_b$. Therefore points $M,N,F$ lie on the circle with diameter $BI_b$. We also have $\widehat{ENC}= 90^\circ-B/2 = \widehat{I_aBP} = \widehat{I_aFP} = \widehat{DFC}$, so the quadrilateral $CFDN$ is cyclic. Finally, we get $\widehat{NDC} = \widehat{NFC} = \widehat{NMI_b}$, then it follows that $CD\parallel MI_b$ and we are done.

Do you know who is the author of the problem (I guess probably Stoyan Boev) ? Very beautiful result supposing to use different approaches.

Can someone post a full synthetic solution? I do not understand how $ABA'B'$ is cyclic and how to arrive at $A'B'||KL$ in school5's solution.

Even though I'm being so late , as I skimmed through solutions I didn't notice anyone mentioning this properties of the figure and I think it would be such a pity if nobody said it..

$(I_B) \cap AC = E$ etc... If $X = PQ \cap MN$ and assuming the problem is true, since the lines adjoining the projections of $C$ on those lines is antiparallel to $AB$, it suffice to show $CX \perp AB$. Now, to use excircles, we have $M(M, MC \cap (I_A); E, N) = -1 \implies (CX \cap AB, ME \cap CX; C, X) = -1$. Using a symmetrical argument, this means $PD \cap ME = T \in CX$. Now, we find things about $PD$ etc... Well, $PD \perp BI_A$ which is the external bisector at $B$, which is perpendicular to the angle bisector. But, of course $BI_B \perp MN$ so $MN \perp PD$. Hence, considering $MCP$, we have $T$ is the orthocentre so done.

Dear Mathlinkers, 1. C' the points of intersection of MK and PL 2. According to Hadamard theorem , CC' is the C'-altitude of C'MP ; this is the difficult point 3. C'' the foot of the C'-altitude of C'MP 4. C''MKC and C""PLK are concyclic 5. with the three chords theorem, we are done Sincerely Jean-Louis

We note that $MP=a+b$.Let $PQ \cap MN=X$.Applying the sine rule in $\triangle{XPM}$ we get $\boxed{XM=\frac{(a+b)cos\frac{A}{2}}{cos\frac{C}{2}}}$.We also note that $BM=BN=s$ so sine rule in $\triangle{BNM}$ gives $NM=(a+b+c)sin\frac{B}{2}$.So $XK=MX-MN+NK or \boxed {XK=\frac{(a+b)cos\frac{A}{2}}{cos\frac{C}{2}}-(a+b+c)sin\frac{B}{2}+\frac{b+c-a}{2}sin\frac{B}{2}}$.Similarly $XL$ can be found trigonometrically.Now all we have to verify is that $\frac{XK}{XL}=\frac{XP}{XM}=\frac{cos\frac{B}{2}}{cos\frac{A}{2}}$ which is easy.

The method of Trigonometrical bashing was foolish though.Though I wrote it was easy I don't really think so(maybe it is easy for those who bash trigonometry really well).The more methodical bash is that of JS Gandora.

I post a more general problem (the crux of this one) and some more properties of this configuration. Consider triangle $ABC$ and let $\omega_b,\omega_c$ be the excircles opposite vertices $B,C$. Let these touch the sides $BC,CA,AB$ at points $B_c,B_b,B_a$ and $C_b,C_a,C_c$ respectively. Let $\ell$ denote the exterior angle bisector of $\angle BAC$ and $I_b,I_c$ be the centres of these excircles. Let $X=B_aB_c \cap C_aC_b$ and $Y=B_bB_c \cap C_bC_c$ and $M=C_aC_b \cap B_bB_c$ and $N=B_aB_c \cap C_bC_c$. Then we make the following observations: 1.) Points $A,X,Y$ are collinear with $XY \perp BC$. 2.) $MN=\ell$ Proof:- We consider the following Lemma: "There exists a conic $\mathcal{H}$ which passes through all six points $B_x,C_y$ with $x,y \in \{a,b,c\}$" Proof of the Lemma is as follows: We fix the points $B_a,C_a,B_c,C_b$ and prove that the corresponding Ray's subtend the same cross ratio on $B_b,C_c$. This is quite simple to compute: $C_c(B_a,C_b;C_a,B_c)=(B,C_b;C_cC_a \cap BC,B_c)=\frac{a(2a+b+c)}{(b+c-a)^2}$ which is symmetric in $b,c$ and so these two cross ratios are equal. Therefore, by the projective definition of a conic, we get the validity of our Lemma. For part 1, we apply Pascal's theorem on the sextuple $(B_a,B_c,B_b,C_a,C_b,C_c)$ which lies on $\mathcal{H}$ to get that points $A,X,Y$ are collinear. Now, some angle chasing gives that $X$ is the orthocenter of triangle $YB_cC_b$ (by showing it lies on the other two altitudes) we get that $XY \perp BC$. For part 2, we apply Pascal's theorem on the sextuple $(B_a,B_c,B_b,C_c,C_b,C_a)$ which lies on $\mathcal{H}$ to get that points $M,N,B_bC_c \cap B_a,C_a=T$ are collinear. Notice that it is clear that $MN \parallel \ell$ by simple angle chasing. Now, we have $B_b,B_a$ are symmetric in $\ell$ and $C_cC_a$ are also symmetric in $\ell$ and therefore, $T \in \ell$ meaning that $MN=\ell$.

mathuz wrote: We have \[ KL\bot CI \]$I$ is incenter of the $ \triangle ABC $. Hence, $MKLP$ - cyclic. Why ? I cannot notice the perpendicularity

Solution: Let $PQ\cap MN=R $. Claim: $RC\perp AB $. Proof: Let $RC\cap AB=S $. Let $I_a $ and $I_b $ denote the centers of $\omega_a $ and $\omega_b $ respectively. Applying sine law in $\Delta RCQ $ and $\Delta RCN $ and using $\sin \angle APQ=\sin \angle CQR $ and $\sin \angle BMN=\sin \angle CNR $, we get $\frac {SM}{SP}=\frac{CN}{CQ}=\frac{I_bM}{I_aP}\implies RC$ bisects $\angle I_aSI_b $ externally. Now, let $I_aI_b\cap AB=J $. As $I_aA $ and $I_bA $ are the bisectors of $\angle BAC $, so, $(I_a,I_b;C,J)=-1\implies RC\perp AB $ (as $RC $ bisects $\angle I_aSI_b $ externally). Main problem: Consider an inversion around $R $ with radius $\sqrt{RC\cdot RS} $. The result now follows immediately.

monsterrr wrote: mavropnevma wrote: In a triangle $ABC$, the excircle $\omega_a$ opposite $A$ touches $AB$ at $P$ and $AC$ at $Q$, while the excircle $\omega_b$ opposite $B$ touches $BA$ at $M$ and $BC$ at $N$. Let $K$ be the projection of $C$ onto $MN$ and let $L$ be the projection of $C$ onto $PQ$. Show that the quadrilateral $MKLP$ is cyclic. (Bulgaria) It suffice to prove that triangles $CKL$ and $AIB$ is similar, where $I$ is the incenter of $ABC$. It is obvious that $ \angle KCL\ = 90^{\circ}+\angle\frac{C}{2} = \angle AIB\ $. Also, we can easily count that $ \frac{CK}{CL}=\frac{(p-a){\cos{\beta}}}{(p-b){\cos{\alpha}}}\ =\frac{AI}{BI}$, where $\angle A\ = 2\alpha$, $\angle B\ = 2\beta$, $p$ is the semi-perimetr of $ABC$. Seeing as there is no synthetic(no projective, trig, inversion etc.) solution, this is the nicest one by far.

Dear Mathlinkers, here Problem 3 Sincèrement Jean-Louis

Let $X,Y$ be centers of $A$-excircle and $B$-excircle and $I$ be incenter of $ABC$. First note that $\angle BMN = \angle 90 - \frac{\angle B}{2}$ so $BY \perp MN$ so $IY || CK$. with same approach we have $IX || CL$. Step$1 : XIY$ and $LCK$ are similar. Let $MN$ and $PQ$ meet at $S$. $KCLS$ is cyclic and $\angle MSP = \frac{\angle A}{2} + \frac{\angle B}{2}$ so $\angle CKL = \angle 180 - \frac{\angle A}{2} - \frac{\angle B}{2}$ and $\angle XIY = \angle AIB = \angle 180 - \frac{\angle A}{2} - \frac{\angle B}{2}$. $\frac{CK}{CL} = \frac{CN}{CQ} . \frac{\sin{KNC}}{\sin{LQC}} = \frac{p-a}{p-b} . \frac{\cos{B}}{\cos{A}} = \frac{AI}{BI} = \frac{IY}{IX}$ $\angle QLK = \angle 90 - \angle KLC = \angle 90 - \angle YXI = \angle 90 - \angle CBI = \angle MNB = \angle NMB$ so $MKLP$ is cyclic.

Let $\omega$ be the incircle, $I$ be the incenter, and $I_A,I_B$ be the centers of $\omega_a,\omega_b.$ Notice $\overline{AI}\parallel\overline{CL}$ and $\overline{BI}\parallel\overline{CK}.$ Claim: $\overline{I_AI_B}\parallel\overline{KL}.$ Proof. Let $D$ be the foot from $I$ to $\overline{AB},$ noting $AD=CN$ and $BD=CQ.$ We see $$\measuredangle LCK=\measuredangle LCI_A+\measuredangle I_BCK=\measuredangle II_AI_B+\measuredangle I_AI_BI=\measuredangle I_AII_B$$and $$\frac{CL}{CK}=\frac{CQ\cos\angle LCQ}{CN\cos\angle NCK}=\frac{BD\cos\angle I_AAC}{AD\cos\angle CBI_B}=\frac{IB}{IA}=\frac{II_A}{II_B}$$so $\triangle CLK\sim\triangle II_AI_B.$ The conclusion is evident as $\overline{CL}\parallel\overline{AI_A}.$ $\blacksquare$ Then, $$\measuredangle MKL=\measuredangle NKL=\measuredangle (\overline{MN},\overline{I_AI_B})=90-\measuredangle I_AI_BB=90-\measuredangle I_AAB=\measuredangle APL.$$$\square$

omarius wrote: mathuz wrote: We have \[ KL\bot CI \]$I$ is incenter of the $ \triangle ABC $. Hence, $MKLP$ - cyclic. Why ? I cannot notice the perpendicularity Because $I_aI_b // KL$

Similar to monster's solution: Firstly, draw the incircle of $\triangle ABC$ and denote the tangency points at $AB$, $AC$ and $BC$ by $D$, $E$, $F$, respectively. Note that $AD=AE=CX=CN$ where $X$ is the tangency point of $B-excircle$ with $AC$. Similarly, $BD=CQ$. Let $I$ be the incenter of $\triangle ABC$. Observe that $AI$ and $BI$ are parallel to $CL$ and $CK$, respectively. So, $\angle BAI=\angle LCQ$ and $\angle ABI=\angle KCN$. Using the sine rule for $\triangle ABI$, $\triangle CKN$ and $\triangle CLQ$ we get that $AI/BI$=$CK/CL$.As a result, $\triangle ABI$ is similar to $\triangle KLC$. Therefore, if $\angle CLK=x$, then as $\angle ABI=x$, we have $\angle BAK=90+x=180-\angle PLK$. We are done

<ABP=<ACP=a. and <PBC=<BCQ=C and <CBQ=<BCP=b and <BAP=x , <PAQ=z , <QAC=y. BP=CQ=m , BQ=CP=n. In ∆ABP and ∆ACP m/AP=sinx/sina AP/n=sina/sin(y+z) m/n=sinx/sin(y+z) (1) In ∆ABQ and ∆ACQ m/AQ=siny/sin(a+b+c) AQ/n=sin(a+b+c) m/n=siny/sin(x+y) (2) From (1) and (2) Sinx/sin(y+z)=siny/sin(x+z) Sin(y-x)=0 x=y as desired.