Nice problem.

Attachments:
Problem 14.pdf (186kb)

I only attempted problems after 12, this problem was easily my favourite. Here's two famous lemmas that I'll post without a proof. Lemma 1: In triangle $ABC$, let $T$, $X$, $Y$ denote the projection of $A$, inceter and $A$-excenter onto $BC$, respectively. $J$ be the intersection of the $A$-angle bisector and $BC$. Then $(T, J; X, Y)$ is harmonic. Lemma 2: In triangle $ABC$, if $X$ is the antipode of the $A$-intouch point in the incircle and $Y$ is the $A$-extouch point, then $A, X, Y$ are collinear. Let $M$ be the midpoints of segment $AC$, respectively. Note that $M$ is also the midpoint of segment $KL$. Let $DK$ and $DL$ be tangents to $(BKL)$, note that $D$ must be the midpoint of arc $AC$ since $A$ and $C$ are symmetric about the perpendicular bisector of $KL$ and hence $DA = DC$. Let $I$ be the incenter of $\triangle ABC$; $T$ the projection of $B$ onto $AC$; $J = BI \cap AC$. $B'$ be the point on $(ABC)$ such that $BB' \parallel AC$. Note that $B'$ is the reflection of $B$ about the perpendicular bisector of $AC$ and $BB'KL$ is an isosceles trapezium. [asy][asy] import olympiad; import geometry; size(10 cm); defaultpen(fontsize(10pt)); pair B = dir(165.5); pair A = dir(210), C = -1/A; pair I = incenter(A, B, C); pair D = - sqrt(A*C); pair K = foot(I, C, A), L = C + A - K, P = foot(I, D, 0); pair M = (C + A)/2; pair K1 = 2*I - K; pair O = foot(B, D, 0); pair B1 = 2*O - B; pair J = extension(A, C, B, I); circle w = circumcircle((point) B, (point) K, (point) L); pair T = foot(B, A, C); triangle t = triangle((point) A, (point) B, (point) C); draw(K--D--L, linewidth(1)); draw(circumcircle(P, K, L), linewidth(1.2)); draw(A--B--C--cycle); draw(circumcircle(A, B, C), dashed+linewidth(0.4)); draw(P--D, grey+dashed); draw(K--B--L, linewidth(0.4)); draw(K--K1^^I--P, grey); draw(w); draw(B--D, grey); draw(A--T, grey+dashed); draw(B--T, grey); draw(A--foot(D, A, B)); draw(line(foot(D, A, B), false, relpoint(line(A, foot(D, A, B)), 3), true), dashed); dot(A^^B^^C); dot(I^^P); dot(K^^L^^K1); dot(D); dot(M); dot(T); dot(J); dot(B1); label("$A$", A, dir(240)); label("$B$", B, B); label("$C$", C, E); label("$D$", D, D); label("$K$", K, 2*dir(270)); label("$K'$", K1, 2*dir(90)); label("$L$", L, 2*dir(270)); label("$P$", P, dir(225)); label("$I$", I, dir(45)); label("$M$", M, dir(225)); label("$T$", T, dir(225)); label("$J$", J, dir(60)); label("$B'$", B1, B1); [/asy][/asy] Claim 1: $P$ is the intersection of $BL$ and $B'K$. Proof. Let $K'$ be the reflection of $K$ in $I$. By Lemma 2, $K'$ lies on $BL$. $P$ is the intersection of the perpendicular bisector of $KK'$ and $KL$. Thus, $P$ is the circumcenter of $\triangle K'KL$. Since $\angle K'KL = 90^{\circ}$, $P$ is the midpoint of $K'L$. Now, $BL$ and $B'K$ intersect at the perpendicular bisector of $KL$ since they are symmetric about this line. Hence, $P$ is the intersection of $BL$ and $B'K$. $\square$ Claim 2: $BB'$ is the diameter of $(BB'KL)$. Proof. Let $\ell$ be the tangent to $(BKL)$ at $B$. Since $BD$ is the $B$-symmedian of $\triangle BKL$,, \[ (\ell \cap AC, J, K, L) \stackrel{B}{=} (\ell, BD; BK, BL) = -1 = (T, J; K, L).\]Thus, $\ell \equiv BT$. Therefore, the diameter of $(BKL)$ through $B$ is parallel to $AC$. $\square$ Claim 3: $D$ is the circumcenter of $PKL$. Proof. By the previous claim, $\angle BKP = \angle BKB' = 90^{\circ}$. Clearly, $DK = DL$. Now we prove that $DP = DK$, \begin{align*} \angle PKD &= \angle DKL + \angle PKL \\ &= \angle KBL + \angle KLB\\ &= 180^{\circ} - \angle BKL\\ &= \angle BKP + \angle IKL - \angle BKL \\ &= \angle IKP = \angle KPM\\ &= \angle KPD\tag*{$\square$} \end{align*}Since $B, I, D$ are collinear, there exists a homothety at $B$ sending the incircle to the $(PKL)$. Thus, the tangents from $B$ to the incircle are also the tangents to $(PKL)$. Therefore, $(PKL)$ is tangent to lines $AB$ and $BC$. $\qquad\blacksquare$

Nice problem!

Attachments:
S14_soln._Part_1.pdf (92kb)
S14_soln._Part_2.pdf (77kb)
S14_soln._Part_3.pdf (396kb)

I bash-fired this!! For some reason, I can't upload this. Anyway.

Personal favourite on this Sharygin Let $ABDC$ be a trapezoid and $M$ be midpoint of arc $\widehat{AC}$ opposite $B$. Since $\overline{BL}$ passes through antipode of $K$ in the incircle, we see that $(BK, BL; BI, BJ)=-1$ where $\overline{BJ}$ is the $B$-altitude in $ABC$. Thus, $\overline{BJ}$ is tangent to $\odot(BKL)$. Let $N$ be midpoint of $\overline{BD}$ and $S$ be midpoint of $\overline{AC}$. Then $N$ is circumcenter of $\triangle BKL$ as $\overline{MK}, \overline{ML}$ are tangent to $\odot(BKL)$. Now $P$ lies on $\overline{BL}$ and the reflection $Q$ of $P$ in $M$ lies on $\overline{BK}$. So $(PQ;SN)=(BL, BK, BS, BD)=(KL;\infty S)=-1$ proving $\odot(BKL)$ and $\odot(M, MP)$ are orthogonal. So $K, L$ lie on $\odot(M, MP)$. Since $MI=MA$ and $\angle IMP=\angle AMX$ where $\overline{MX} \perp \overline{BA}$, we get $\odot(M, MP)$ tangent to $\overline{BA}$; likewise to $BC$. Thus $\odot(PKL)$ tangent to $\overline{AB}, \overline{BC}$.

Very nice problem. Vrangr wrote: Let the side $AC$ of triangle $ABC$ touch the incircle and the corresponding excircle at points $K$ and $L$ respectively. Let $P$ be the projection of the incenter onto the perpendicular bisector of $AC$. It is known that the tangents to the circumcircle of triangle $BKL$ at $K$ and $L$ meet on the circumcircle of $ABC$. Prove that the lines $AB$ and $BC$ touch the circumcircle of triangle $PKL$. Solution. Let $M$ be the midpoint of $AC,$ $I$ be the incenter of $\triangle ABC,$ $(XYZ)$ denote the circle passing through $X,Y,Z,$ let $M_B$ be the intersection of tangents at $K,L$ to $(BKL).$ $X$ be the antipode of $K$ with respect to the incircle. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); import olympiad; real a = 2; real b = 1; pair M = (0,0); pair K = (-a,0); pair L = (a,0); pair I = (-a,b); pair K2 = (-a, 2b); pair P = (0,b); pair A = extension(P,K2,K,(P-K)*dir(90)+K); pair T = extension(A,I,P,M); pair D = tangent(A, I, b, 1); pair E = tangent(A, I, b, 2); pair B = extension(A,D,M,K); pair C = extension(A,E,K,M); pair X = extension(K,P,T,I); dot("$M$", M, dir(45)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$I$", I, dir(45)); dot("$X$", K2, NE); dot("$P$", P, dir(45)); dot("$B$", A, dir(A)); dot("$M_B$", T, dir(T)); dot("$A$", B, dir(B)); dot("$C$", C, dir(C)); draw(A--K); draw(A--L); draw(A--T); draw(K--T--L); draw(circle(I,b),cyan); draw(P--T,dashed); draw(A--B--C--A); draw(circumcircle(A,B,C),green); draw(K--K2,dashed); draw(I--P,dashed); draw(circumcircle(A,K,L),dotted+red); [/asy][/asy] Consider the following claims, Claim 1. $B,X,P,L$ are collinear. Proof. Since there exist a homothety centered at $B$ taking the incircle to the excircle, it is obvious that $B,X,L$ are collinear. Also note that $IKMP$ is a rectangle and $XK=2\cdot PM,$ hence considering a homothety of scale factor 2 at $L$ taking $PM$ to $XK,$ we get that $X,P,L$ are collinear. Hence the claim. $\square$ Claim 2. $M_B$ is the center of $(PKL).$ Proof. It is well known that $AK=LC$. Since, $M_BK=M_BL$ and $M_B\in(ABC)$ hence because of symmetry $M_B$ is actually the mid point of the arc $\widehat{AC}$ not containing $B,$ so, $M_B=BI\cap(ABC).$ Observe that $BM_B$ is the $B-$symmedian of $\triangle BKL.$ Therefore, $BI$ and $BM$ are isogonals with respect to $\angle KBL,$ moreover $BM$ is the median of $\triangle BKL$ and $BI$ is the median of $\triangle BKX.$ So $KX$ and $KL$ are antiparallels with respect to $\angle KBL.$ Thus, $\triangle BKX\sim\triangle BKL.$ So, \begin{align*} &\angle BKL = 90^{\circ}+\angle BKX\\ \implies & 180^{\circ}-\angle BLK-\angle KBL=90^{\circ}+\angle BLK\\ \implies & 90^{\circ}=2\angle BLK+\angle KBL\\ \implies & 90^{\circ}-\angle BLK=\angle BKL+\angle KBL. \end{align*}Note that $PK=PL$ and $M_BK=M_BL$ so \[\angle M_BPL=\angle MPL=90^{\circ}-\angle BLK=\angle BKL+\angle KBL=\angle BKL+\angle KLM_B=\angle PLM_B,\]hence $M_BL=M_BP$ and so $M_B$ is the center of $(PKL).\square$ Coming to the original problem, As $XI\parallel PM_B,$ there is a homothety at $B$ which takes $XI$ to $PM_B,$ this means under this homothety, the incircle will get mapped to the circle with center $M_B$ and radius $M_BP,$ therefore the incircle will get mapped to the circumcircle of $\triangle PKL,$ now since the incircle is tangent to $BC$ and $BA,$ we are done. $\square$

My first (and last!) time participating in Sharygin. This was one of two problems I solved... A bit of a bashy solution. [asy][asy] size(10cm); defaultpen(fontsize(10pt)); import olympiad; real a = 2; real b = 1; pair M = (0,0); pair K = (-a,0); pair L = (a,0); pair I = (-a,b); pair K2 = (-a, 2b); pair P = (0,b); pair A = extension(P,K2,K,(P-K)*dir(90)+K); pair T = extension(A,I,P,M); pair D = tangent(A, I, b, 1); pair E = tangent(A, I, b, 2); pair B = extension(A,D,M,K); pair C = extension(A,E,K,M); pair X = extension(K,P,T,I); dot("$M$", M, dir(45)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$I$", I, dir(45)); dot("$K_1$", K2, NE); dot("$P$", P, dir(45)); dot("$B$", A, dir(A)); dot("$T$", T, dir(T)); dot("$A$", B, dir(B)); dot("$C$", C, dir(C)); dot("$X$", X, dir(45)); draw(A--K--P); draw(A--L); draw(A--T, dashed); draw(K--T--L); draw(circle(I,b)); draw(P--T); draw(A--B--C--A); draw(circumcircle(A,B,C)); draw(K--K2,dashed); draw(I--P,dashed); draw(circumcircle(K,P,L),dotted); draw(B--conj((C-A)/dir(T-A))*dir(T-A)+A,Arrow); draw(circumcircle(A,K,L),dashed); [/asy][/asy] Let $I$ be the incenter, let $M$ be the midpoint of $AC$, let the intersection of the tangent lines to the circumcircle of $\triangle BKL$ at $K$ and $L$ meet at $T$, let $K_1$ be the point diametrically opposite $K$ across the incircle, and let $(P_1P_2P_3)$ be the circumcircle of $\triangle P_1P_2P_3$ for arbitrary points $P_1$, $P_2$, and $P_3$. Firstly, it is well known that if $K$ is the intouch point to $AC$ and $L$ is the $B$-extouch point to $AC$, then $AK=CL$, thus $KM=ML$. Also, observe that by symmetry, $T$ must be the midpoint of minor arc $AC$ of $(ABC)$. This implies that $B$, $I$, and $T$ are collinear along the angle bisector. Since $TK$ is tangent to $(BKL)$, we have that $\angle TKL = \angle KBL$. Let $\angle KBL = \alpha$. Now, it is well known that cevian $BL$ must contain the antipode of $K$ with respect to the incircle, i.e. $K_1$. Since quadrilateral $IKMP$ is a rectangle, $K_1I = IK$, and $KM = ML$, it is obvious that $P$ lies on segment $K_1L$, and is in fact the midpoint of this segment. Thus $B$, $K_1$, $P$, and $L$ are collinear. We now state, without proof, a well-known projective lemma: Lemma (Some useful projective ratio from EGMO) Suppose $A$, $B$, $C$, and $D$ lie on a line in this order, and $P$ is a point not on this line. Then any two of the three following statements implies the third: - $(A,C;B,D) = -1$, i.e. $A$, $C$, $B$, and $D$ form a harmonic bundle. - $\angle APB = \angle BPC$ - $\angle BPD$ is right. Let $X$ be the intersection between $IT$ and $KP$, so $B$, $I$, $X$, and $T$ are collinear in this order. Observe that $\angle IPT$ is right, and $\angle BPI = \angle IPX$. Thus $(B,X;I,T) = -1$. But then we have that the cross ratio is preserved unto the cross ratio of sines of angles around $K$, i.e.: $$\frac{\sin(\angle IKB)}{\sin(\angle IKX)} \div \frac{\sin(\angle TKB)}{\sin(\angle TKX)} = 1$$Let $\angle IK_1P = \angle IKX = \theta$. Then $\angle IKB = \theta - \angle KBP = \theta - \alpha$. Since $\angle IKM = 90^\circ$, and $\angle TKM = \alpha$, we have that $\angle TKB = \theta - \alpha + 90^\circ + \alpha = \theta + 90^\circ$. Lastly, we note that $\angle MKP = 90^\circ - \theta$, so $\angle TKX = \alpha + 90 - \theta$. Substituting and rearranging: $$\frac{\sin(\theta - \alpha)}{\sin(\theta)} = \frac{\sin(\theta + 90^\circ)}{\sin(\alpha + 90 - \theta)} = \frac{\cos(\theta)}{\cos(\theta - \alpha)}$$So $\sin(\theta - \alpha)\cos(\theta - \alpha) = \sin(\theta)\cos(\theta)$, or $\sin(2\theta - 2\alpha) = \sin(2\theta)$. We conclude that the arguments of the sines are either equal or supplementary. The equality case is absurd, so they are supplementary. Thus $4\theta - 2\alpha = 180^\circ$, or $2\theta - \alpha = 90^\circ$. So, $\angle TKP = \alpha + 90^\circ - \theta = (2\theta - 90^\circ) + 90^\circ - \theta = \theta = \angle KPT$, and so $\triangle TKP$ is isosceles, as well as $\triangle TPL$, and $TK = TP = TL$. We conclude that the circumcircle of $\triangle KPL$ is centered at $T$. Consider the homothety $\mathcal{H}$ centered at $B$ that maps $I$ to $T$. Clearly, $\mathcal{H}(IK_1) = \mathcal{H}(TP)$, so the corresponding radii of the incircle and $(KPL)$ are homothetic. Thus $(KPL)$ is homothetic with the incircle with respect to $B$. Since the incircle is tangent to $BA$ and $BC$, we conclude that $(KPL)$ is tangent to $BA$ and $BC$ as well.

Good Problem.

Attachments:
Sharygin- 14.pdf (40kb)