Here is the TST problem (notations changed): Problem 3 of 3rd German TST 2006. Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$. Note. A matter of notation: The $A$-excenter of a triangle $ABC$ is defined as the center of the excircle of triangle $ABC$ which touches the interior of the side $BC$ and the extensions of the sides $CA$ and $AB$. [See the attached graphic - click it to enlarge it.] Solution. We WLOG assume that the point X lies on the extension of the segment BC beyound C, and that the point Y lies inside the segment CD. The other case can be treated analogously. Let K' and L' be the Y-excenter and X-excenter of triangle XCY, respectively. Then, both of these points K' and L' lie on the exterior angle bisector of the angle XCY, what yields $\measuredangle XCL^{\prime}=90^{\circ}+\frac{\measuredangle XCY}{2}$ and $\measuredangle K^{\prime}CX=\measuredangle YCL^{\prime}=90^{\circ}-\frac{\measuredangle XCY}{2}$; also, the point K' lies on the exterior angle bisector of the angle CXY, what gives $\measuredangle K^{\prime}XC=90^{\circ}-\frac{\measuredangle CXY}{2}$, and the point L' lies on the exterior angle bisector of the angle CYX, leading to $\measuredangle CYL^{\prime}=90^{\circ}-\frac{\measuredangle CYX}{2}$. Thus, by the sum of angles in triangle CK'X, we get $\measuredangle CK^{\prime}X=180^{\circ}-\measuredangle K^{\prime}XC-\measuredangle K^{\prime}CX=180^{\circ}-\left(90^{\circ}-\frac{\measuredangle CXY}{2}\right)-\left(90^{\circ}-\frac{\measuredangle XCY}{2}\right)$ $=\frac{\measuredangle CXY+\measuredangle XCY}{2}=\frac{180^{\circ}-\measuredangle CYX}{2}$ (since < CXY + < XCY = 180° - < CYX by the sum of angles in triangle XCY) $=90^{\circ}-\frac{\measuredangle CYX}{2}$. Thus, < CK'X = < CYL'. Combining with < K'CX = < YCL', we see that triangles K'CX and YCL' are similar. Since ABCD is a parallelogram, we have AB || CD, so that Thales yields $\frac{XC}{XB}=\frac{XY}{XA}$. Now, the homothety with center X and ratio $\frac{XC}{XB}=\frac{XY}{XA}$ obviously maps the point B to the point C and the point A to the point Y, and leaves the point X fixed. Thus, this homothety must map the A-excenter of triangle XBA to the Y-excenter of triangle XCY; in other words, it maps the point K to the point K'. Since the center of this homothety is X and the ratio is $\frac{XY}{XA}$, it thus follows that the point K' lies on the line XK and satisfies $\frac{XK^{\prime}}{XK}=\frac{XY}{XA}$. This transforms into $\frac{KK^{\prime}}{XK}=\frac{XK^{\prime}-XK}{XK}=\frac{XK^{\prime}}{XK}-1=\frac{XY}{XA}-1=\frac{XY-XA}{XA}=\frac{AY}{XA}$. Similarly, the point L' lies on the line YL and we have $\frac{LL^{\prime}}{YL}=\frac{AX}{YA}$, so that $\frac{YL}{LL^{\prime}}=\frac{YA}{AX}$, what rewrites as $\frac{LY}{L^{\prime}L}=\frac{AY}{XA}$. Comparison with $\frac{KK^{\prime}}{XK}=\frac{AY}{XA}$ yields $\frac{KK^{\prime}}{XK}=\frac{LY}{L^{\prime}L}$. Thus, the points K and L are corresponding points in the similar triangles K'CX and YCL' (since they lie on the respective sidelines K'X and YL' and divide them in the same ratio $\frac{KK^{\prime}}{XK}=\frac{LY}{L^{\prime}L}$). Since corresponding points in similar triangles form equal angles, this implies < KCX = < LCL'. Thus, $\measuredangle KCL=\measuredangle KCX+\measuredangle XCL=\measuredangle LCL^{\prime}+\measuredangle XCL=\measuredangle XCL^{\prime}$ $=90^{\circ}+\frac{\measuredangle XCY}{2}=90^{\circ}+\frac{180^{\circ}-\measuredangle BCD}{2}=180^{\circ}-\frac{\measuredangle BCD}{2}$. Since this angle is independent of the line g, this solves the problem. Darij

madatmath wrote: Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$. Another solution 1) A bit of angle chasing first $\angle BAK = \frac{\angle BAX}{2} = \frac{\angle AYD}{2} = \angle ALD$. This last equality is because $Y$, $L$, $D$ and the incenter, say $I$ of triangle $AYD$ are concyclic so $\angle IYD = \angle ILD$. 2) $\angle ADL = \angle ABK \Rightarrow \triangle ADL$ and $\triangle ABK$ similar 3) Hence $BK \cdot DL = AB \cdot AD = BC \cdot CD$ Hence $\frac{BK}{BC} = \frac{CD}{DL}$ but 4) $\angle CBK = \angle LDC$ hence $\triangle CBK$ is similar to the $\triangle CLD$. Hence the angle between lines $CK$ and $CL$ is the angle between lines $BC$ and $DL$. This last clearly do not depend on the location of point $X$ hence result. Daniel

Changed form of this problem is Iran's TST,which is here http://www.mathlinks.ro/Forum/viewtopic.php?p=498848#

Assume wlog that $ X$ lies between $ A$ and $ Y$. Let $ S$ be the point of intersection of the lines $ BC$ and $ AL$ and let $ T$ be the point of intersection of $ DC$ and $ AK$. Moreover, let $ \angle DAL = \angle LAX = \alpha$ and $ \angle XAK = KAB = \beta$. First, we claim that $ SLCKT$ is a cyclic quadrilateral. We have $ \alpha + \beta = \angle SBK = \angle SAK$ and $ \alpha + \beta = \angle LDT = \angle LAT$. Thus, quadrilaterals $ ATLD$ and $ ABKS$ are cyclic. A simple angle chase shows that $ \angle BKA = \alpha$ and $ \angle ALD = \beta$. Therefore, $ \angle KSL = \angle KSB + \angle BSA = \angle KAB + \angle BKA = \beta + \alpha$ and $ \angle KTL = \angle ATD + \angle DTL = \angle ALD + \angle DAL = \beta + \alpha$. Thus, quadrilateral $ SLKT$ is cyclic. Moreover, $ \angle SKT = 180 - \angle AKS = 180 - \angle ABS = 2(\alpha + \beta)$ and $ \angle SCT = \angle BCD = 2(\alpha + \beta)$. Thus, $ SLCKT$ is cylic. It follows that $ \angle KCL = 180 - \angle KTL = 180 - ( \alpha + \beta) = 180 - \angle DAB/2$ and that is clearly constant. edit: Hmm, my solution is exactly the same as orznorz' solution…^^ it would be better to read the posts above before posting^^ orznorz even chose the same names for the angles $ \angle DAL$ and $ \angle KAB$...

Seems too easy to be right... By angle chasing, ABK~LDA, so AB/BK=LD/DA, also KB/BC=CD/DL and <KBC=<CDL, so KBC~CDL. Then by angle chasing again <KCL=180-<BAD/2, which is independent of g.

If you intersect CD with AK and BC with AL at S and T respectively, angle chasing yields that TKBA, SLDA, and KCTLS are all cyclic. This gives $\angle KCL=\angle KTL=\pi - \dfrac{\angle BAD}{2}$ as desired. you also get a bunch of similar triangles here, such as KBA ~ KCL ~ ADL.

Finally I came up with a solution with a geometric approach:(withuot any bash) Note that $\angle{ADL}=\angle{ABK}=90+\frac{B}{2}$ and $\angle{DAL}=\angle{AKB} \Rightarrow \frac{BK}{BC}=\frac{BK}{AD}=\frac{AB}{LD}=\frac{DC}{DL}$ .Also note that $\angle{LAC}=\angle{CBK} \Rightarrow \triangle{CBK} \sim \triangle{LDC}$. Hence $\angle{KCL}=360-\angle{KCB}-\angle{DCL}-\angle{DCB}=360-(\angle{KCB}+\angle{CKB})-\angle{A}=360-90+\frac{B}{2}-180+B=90+3\frac{B}{2}$ which is fixed,so we are done!!!

ISL 2005 G3 wrote: Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$. Proposed by Vyacheslev Yasinskiy, Ukraine Let $\ell_b$ and $\ell_d$ denote the external bisectors of $\measuredangle ADC$ and $\measuredangle ABC$, respectively. Consider the pencil of lines $\ell$ through $A$ and observe that $K, L$ move on ranges $\ell_b, \ell_d$ respectively, such that $\measuredangle KAL=\tfrac{1}{2}\measuredangle DAB$. Let $\Gamma : \ell_b \mapsto \ell_d$ be the composition of rotation about $A$ with angle $\tfrac{1}{2}\measuredangle DAB$ followed by projection through $A$. Let $\Phi: \ell_b \mapsto \ell_d$ be the composition of rotation about $C$ with angle $180^{\circ}-\tfrac{1}{2}\measuredangle DAB$ followed by projection through $C$. Note that both $\Gamma$ and $\Phi$ provide a one-one projective map from $\ell_b$ to $\ell_d$. Thus, in order to establish $\Gamma=\Phi$, it suffices to check this for three possibilities of line $\ell$. However, the result is immediate when line $\ell$ coincides with $\overline{AB}, \overline{AC},$ and $\overline{AD}$; so we are done! $\blacksquare$

All what you have to do is \(\Delta DLY \sim \Delta BKX\) and \(\Delta DLC \sim \Delta BCK\).

madatmath wrote: Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$. Let $\ell_B,\ell_D$ be the external angle bisectors of $\angle B,\angle D$ respectively. We consider the general version of this problem where $L,K$ are defined as the intersections of the angle bisectors of $\measuredangle DAY, \measuredangle YAB$ with $\ell_D,\ell_B$ respectively. Thus, they are the excenters when $Y \in \vec{DC},$ and might correspond to incenters if $Y$ is outside the segment $CD.$ We claim that $\theta=\measuredangle KCL=\pi-\tfrac{1}{2}\angle C$ works. Fix $ABCD$ and move $L$ on $\ell_D.$ Clearly, $L \mapsto K$ is projective since $\angle KAL$ is fixed. Also, let the rotation at $C$ by angle $\theta$ counterclockwise send $L \in \ell_D$ to $K^\ast \in \ell_B.$ It suffices to check $K=K^\ast,$ for which we just need three positions of $L.$ Pick $L=D, \infty_{AD}$ and the $A$ excenter of $\triangle ACD.$ It is easy to check these cases so done. $\square$

Denote the external angle bisectors of $B$, $C$ as $\ell_B$, $\ell_D$, and define $L'$ to be the intersection of the clockwise rotation of $CK$ about $C$ by $180-\frac{\angle C}{2}$ with $\ell_B$. We wish to show that $L=L'$, which we will do with moving points. If $K$ varies with degree $1$, note that $\angle KAL$ is fixed, so $AL$ and $L$ also vary with degree $1$. As $CL'$ is just a rotation of $CK$, $\deg CL'=1\implies \deg L'=1\implies \deg(L=L')=2$. The cases $g=AB, AC, AD$ are easily verified to work, so we are done. @below Thanks

Through angle chasing, we see that $AK\perp YL$ and $AL\perp XK$. Thus let $K_1 = AK\cap YL$, $L_1=AL\cap XK$ and $P = XK\cap YL$, the $C$-excenter of $\triangle CXY$. Notice that $P$ is the orthocenter of $\triangle AKL$. The following claim will imply the problem. Claim: $KCLP$ is cyclic. Proof: WLOG $P$ lie on the different side of $L$ w.r.t. $AK$. Let $T = PC\cap KK_1$. Notice that $$\angle XPT = 90^{\circ} - \angle XYP = \angle KAY$$or $A,X,T,P$ are concyclic. Moreover, notice that $$\angle XCT = 90^{\circ} - \tfrac{\angle A}{2} = \angle ALP = 180^{\circ}-\angle XKT$$thus $X,C,K,T$ are concyclic. Hence $$\angle KCP = \angle KXT = \angle KAP = \angle KLP$$as desired. $\blacksquare$ The claim implies the problem as $\angle KCL = 180^{\circ} - \angle KPL = 180^{\circ} - \angle KAL$.

madatmath wrote: Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$. Proposed by Vyacheslev Yasinskiy, Ukraine Solution from Twitch Solves ISL: Let $\alpha = \angle BAK = \angle KAX$ and $\beta = \angle XAL = \angle LAD$. Since $\angle ABK = \angle ADL = 180^{\circ}-(\alpha+\beta)$, it follows that \[ \triangle ABK \sim \triangle LDA \implies \frac{AB}{LD} = \frac{BK}{DA} \implies \frac{CD}{LD} = \frac{BK}{BC} \implies \triangle LDC \sim \triangle CBK. \]Then, $\angle BCK + \angle LCD = 180^{\circ}-(\alpha+\beta)$, which is fixed. On the other hand, $\angle BCD = 2(\alpha+\beta)$ is fixed too. Therefore $\angle KCL$ is fixed as desired.

We use the method of moving points. Let $K,L$ move along the exterior angle bisectors of $\angle ABC,\angle ADC$ respectively and note $\angle KAL$ is fixed. Thus the map $K\mapsto AK\mapsto AL\mapsto L$. Similarly the map $K\mapsto CK\mapsto CL'\mapsto L'=CL'\cap DL$ is projective where $\angle KCL'$ is fixed at $180^\circ-\angle BCD/2$. To show these maps are the same, it suffices to check three cases for $X$. At $X=B$, we have $K=B$ and $L$ is the point at infinity along the angle bisector of $\angle BCD$, so the maps must be the same. Similarly the result trivially holds when $Y=D$. For $X=C$, we get $X=Y=C$ so the result is obvious by excenter angles. Hence we are done.

We claim that $\angle KCL = 180^{\circ} - \frac{\angle BCD}{2},$ which is independent of $g.$(You can easily find this by considering the case $C=X=Y$) Let $\angle KCA=\theta$. It is easy to then se that $\angle BAC=\angle CAD=2\theta$ and $\angle BAD=\angle BCD=4\theta$. By incenter excenter duality, we know that $\angle I_1CK=90^{\circ}$. It then follows from simple angle chasing that $\angle KCL$(reflex) is equal to $180^{\circ}+2\theta,$ so the other part is just $180^{\circ}-2\theta=180^{\circ}-\frac{\angle BCD}{2},$ as desired. The key claim in this problem was really just the incenter-excenter duality.

Interesting how much G3 is easier than G2... Let AL meet BC at P and AK meet DC at Q. Step1 : QLDA and BKPA are cyclic. ∠LDQ = (180 - ∠ADC)/2 = ∠DAB/2 = ∠LAQ ---> QLDA is cyclic. ∠PBK = (180 - ∠ABC)/2 = ∠DAB/2 = ∠PAK ---> BKPA is cyclic. Step2 : QKCPL is cyclic. ∠CQL = ∠DAL = ∠CPA ---> CQLP is cyclic. ∠KPC = ∠KAB = ∠KQC ---> KQPC is cyclic. Now we have ∠KCL = ∠KPL = 180 - (∠KPA) = 180 - (∠KPC + ∠CPA) = 180 - ∠BAC/2 so ∠KCL is independent of line g. we're Done.

As always, I have an overcomplicated solution, so I won't be posting it here. However, I do have a diagram which I think will help with understanding this configuration .

Since $L,D,Y,$ and the incenter of $ADY$ are concyclic, we have $$\angle ALD = \frac{1}{2}\angle AYD = \frac{1}{2}\angle BAX = \angle BAK.$$Similarly $\angle BKA = \angle DAL$, so $\triangle BAK \sim \triangle DLA$. Therefore $\frac{BA}{BK} = \frac{DL}{DA}$. Since $AD=BC$ and $AB=CD$ we have $$\frac{DA}{BK}=\frac{DL}{BA}\Rightarrow \frac{BC}{BK}=\frac{DL}{CD}.$$However, note that $\angle CBK = \angle CDL = \frac{1}{2}(180^{\circ}-\angle ABC)$, so $\triangle BKC \sim \triangle DCL$. Therefore $\angle BCK + \angle DCL = \angle BCK + \angle BKC = 180^{\circ} - \angle CBK = 180^{\circ}-\frac{1}{2}(180^{\circ}-\angle ABC)$ and is thus fixed. Now $$\angle KCL = 360^{\circ}-\angle BCD - (\angle BCK +\angle DCL)$$which is fixed as $\angle BCD$ and $\angle BCK + \angle DCL$ are.

We have $\angle KAL=\frac{1}{2}\angle ABC=\angle KBC=\angle LDC$ so let $AK$ and $DC$ intersect at $E$ and $AL$ and $BC$ intersect at $F.$ $ADLE$ and $ABKF$ are cyclic. Now, $\angle AEL=\angle AFK=\angle KAL$ so $KFLE$ is cyclic. In particular, $\angle KEL$ and $\angle KFL$ are fixed. $\angle LEC=\angle LAD=\angle CFA$ so $ECFL$ is cyclic. Thus, $\angle KCL=\angle KFL$ is fixed.

Let $I$ and $L_1$ denote the incenter and $X$-excenter of $ABX$ respectively. It's easy to see $ABX$ and $YDA$ are homothetic. Now, if we let $C_1$ denote the image of $C$ under the homothety taking $YDA$ to $ABX$, $$\frac{BC_1}{C_1A} = \frac{DC}{CY} = \frac{AB}{CY} = \frac{BX}{XC}$$which means $AC \parallel C_1X$. Properties of internal and external bisectors imply $BIAL_1$ and $BIXK$ are cyclic with diameters $BA$ and $BX$ respectively. Because $I = AK \cap XL_1$, the Spiral Center Lemma yields $BAL_1 \overset{+}{\sim} BKX$. Now, combining this similarity with $AC \parallel C_1X$ gives $$BK \cdot BL_1 = BA \cdot BX = BC \cdot BC_1$$so $\frac{BK}{BC_1} = \frac{BC}{BL_1}$. In addition, we have $$\angle KBC = \angle KBX = \angle ABL_1 = \angle C_1BL_1.$$Thus, SAS Similarity implies $BKC \sim BC_1L_1$. Now, because the aforementioned homothety yields $BC_1L_1 \sim DCL$, we know $$\angle KCB + \angle BCD + \angle DCL = \angle BCD + \angle KCB + BC_1L_1$$$$= \angle BCD + \angle KCB + \angle BKC = \angle BCD + (180^{\circ} - \angle CBK)$$$$= \angle BCD + 180^{\circ} - \left( \frac{180^{\circ} - \angle ABC}{2} \right) = 180^{\circ} + \frac{\angle BCD}{2}$$so $\angle KCL = 180^{\circ} - \frac{\angle BCD}{2}$, which finishes. $\blacksquare$

at this point the questions are getting ridiculous. i find the needed points to add in and then i dont add them in because im convincing myself that they dont work. jeez i need to hone my intuition more!1 Let $C'$ be the point on $BK$ such that \[\frac{C'B}{C'K}=\frac{CD}{CY}=\frac{XA}{XY}=\frac{XB}{XC}\implies C'X\parallel CK.\] Also notice that $\triangle BKX\sim \triangle DYL$. Now notice that \[\measuredangle DCL+\measuredangle KCB=\measuredangle BC'X+\measuredangle KCB=\measuredangle BC'X+\measuredangle C'XB=\measuredangle KBX\]which is constant. Also notice that $\measuredangle BCD$ is constant, so $\measuredangle KCL$ is constant (since it is the last part of a $360^{\circ}$ angle).

Here is a sketch Observe BLC is similar to DCK.

We are done

Fix parallelogram $ABCD$ and animate $X$ on $\overline{BC}.$ Then $Y,L,K$ move as a function of $X.$ Observe the projective map \[L \xmapsto{A} \text{incenter(ADY)} \mapsto{Y} \xmapsto{A} X \mapsto \text{incenter(ABX)} \xmapsto{A} K.\]Thus, it suffices to check three positions of $X$ and show that in these cases, $\angle KCL$ are all equal. When $X=B,$ we have $X=K=B,$ and $L$ the point at infinity in the direction of the external angle bisector of $\angle BCD.$ Thus, in this case $\angle KCL = 180^\circ-\angle B/2.$ When $X=\infty_{BC},$ we have $Y=L=D$ and $K$ the point at infinity in the direction of the external angle bisector of $\angle BCD.$ Thus, in this case, $\angle KCL=180^\circ - \angle B/2$ as well. When $X=C,$ we have $X=Y=C,$ from which using the excentral angles, we also obtain $\angle KCL = 180^\circ - \angle B/2.$ We have checked three cases, and may conclude.