Let $ABC$ be a triangle with circumcircle $\omega$. The internal angle bisectors of $\angle ABC$ and $\angle ACB$ intersect $\omega$ at $X\neq B$ and $Y\neq C$, respectively. Let $K$ be a point on $CX$ such that $\angle KAC = 90^\circ$. Similarly, let $L$ be a point on $BY$ such that $\angle LAB = 90^\circ$. Let $S$ be the midpoint of arc $CAB$ of $\omega$. Prove that $SK=SL$.
Problem
Source: Baltic Way 2020, Problem 12
Tags: geometry, geometry proposed
Gryphos
14.11.2020 19:31
Let $I$ be the incenter of $\triangle ABC$. It is well-known that $AY=BY=IY$ and $AX=CX=IX$. By Thales, this implies that $Y$ is the midpoint of $BL$ and $X$ is the midpoint of $CK$. In particular, $IY=LY$ and $IX=KX$.
Next, observe that $\measuredangle SYL = 180^\circ - \measuredangle BYS = \measuredangle SCB = \measuredangle CBS = \measuredangle CYS = \measuredangle IYS$, which means that $SY$ is the angle bisector of $\angle IYL$. Since $IY=LY$, this implies that $IS=LS$. Analogously, we can show that $IS=KS$, which completes the proof.
zuss77
14.11.2020 20:18
$XK=XC$ as radii in $(CAK)$. Similarly $YL=YB$.
$\angle BXS=(\angle B+\angle C)/2=\angle XBY$, so $BXSY$ - isoscales trapezoid, the same as $CYSX$.
Hence $XS=BY=YL$ and $XK=XC=YS$. So $\triangle SXK \cong \triangle LYS$ by SAS.
Kimchiks926
14.11.2020 23:45
Here is solution, which I found during the contest. After some synthetic observations complex bash is vanilla Observe that $YB = YA$.Since $\triangle LAB$ is right angled we must have $YB = YL$. Similarly $CX =XK$. Now we use complex numbers with $\odot(ABC)$ as unit circle. So we have that $A = a^2$, $B =b^2$, $C = c^2$, $Y = -ab$ and $X =-ac$. Since $S$ is antipode of $M_A$, which is midpoint of arc $BC$ not containing $A$, we get that $S = bc$. Since $\frac{B+L}{2} = Y$ we can compute $L = -2ab - b^2$ and similarly $K =-2ac - c^2$. The problem is equivalent to check that: $$ (S-L)( \overline{S} - \overline{L}) = (S-K)(\overline{S} -\overline{K}) $$Or equivalently: \begin{align*} (bc +2ab +b^2)(\frac{1}{bc}+\frac{2}{ab} + \frac{1}{b^2}) = (bc +c^2 +2ac)(\frac{1}{bc} + \frac{1}{c^2}+\frac{2}{ac}) \\ c(bc + 2ab +b^2) (ab + 2bc +ac) =b (bc+c^2 +2ac)(ac+ab+2bc) \\ bc(b+c) +2abc = bc(b+c) + 2abc \\ 0 = 0 \end{align*}as desired.
MeineMeinung
26.06.2021 06:33
Let $O$ and $Z$ be the circumcenter of triangle $ABC$ and the midpoint of arc $BC$ not containing $A$, respectively. Notice that $\angle BZC = 180^{\circ} - \angle A$ and $\angle YOX = 2\angle YZX = 2\left( \frac{1}{2} \angle B + \frac{1}{2} \angle C\right) = \angle B + \angle C = 180^{\circ} - \angle A$. So, the isosceles triangle $\triangle OXY$ is similar to $\triangle ZCB$. As previously mentioned, we know that $X$ and $Y$ are the midpoints of $BL$ ad $CK$. But, $O$ is also the midpoint of $SZ$, therefore we can apply MGT on $BYL, ZOS, CXK$ to derive $\triangle ZCB \sim \triangle OXY \sim \triangle SKL$. Hence, $SK = SL$.
cursed_tangent1434
08.09.2024 11:09
Let $I$ denote the incenter of $\triangle ABC$. We start off by locating $L$ and $K$. Claim : Points $L$ and $K$ are respectively the $B-$antipode and $C-$antipode in circles $(AIB)$ and $(AIC)$. Proof : It is well known that $Y$ is the center of $(AIB)$, and thus $L$ lies on the $BY-$diameter. Further, since $\measuredangle BAL = 90^\circ$, it follows that it is in fact the $B-$antipode as we wished to show. Similarly, $K$ must also be the $C-$antipode in $(AIC)$. Next, note that \[2\measuredangle SYI = 2\measuredangle SYC = 2\measuredangle SBC = \measuredangle BSC = \measuredangle BAC = \measuredangle BYC\]Thus, $SY$ is the internal $\angle LYI-$bisector. Thus, in triangles $\triangle LYS$ and $\triangle SYI$, $LY=YI$ and $\measuredangle LYS = \measuredangle SYI$ with a common side $SY$. It then follows that $\triangle LYS \cong \triangle SYI$, so $LS = SI$. Repeating this argument with $K$ yields, $\triangle KSX \cong \triangle ISX$ and $KS=SI$. Thus, \[LS = LI = LK\]which was the desired result.
onyqz
08.09.2024 19:39
We first make the following synthetic observations: $BY = LY$ and $CX=KX$ by chasing angles and the fact that $\triangle ACK$ and $\triangle ALB$ are right-angled. Moreover, note that $X$ are the midpoints of the arcs of $AC$ and $AB$ (not containing the third point) respectively.
Now we will complex bash the rest: Let $\omega$ be the unit circle, $a=u^2$, $b=v^2$, $c=w^2$ ($|u|=|v|=|w|=1$). Then $x=-uw$, $y=-uv$ and $s=vw$ (since $S$ is the antipode of the midpoint of arc $BC$ not containing $A$, which is located at $-vw$).
Now since $CX=KX$ we have $k=2x-c=-2uw-w^2$ and since $BY=LY$ we have $l=2y-b=-2uv-v^2$. It now remains to prove that $|k-s|=|l-s|$. Now $|k-s|=|w(2u-w-v)|=|v|\cdot|w(2u-v-w)|=|vw(2u-v-w)|=|w|\cdot|v(2u-v-w)|=|v(2u-v-w)|=|l-s|$, which finishes our proof. $\square$
reni_wee
21.09.2024 20:58
WLOG let $AB < AC$. Let $\angle SCA = \angle SBA =\gamma$, $\angle ACY = \angle YCB =\alpha$ $$\angle YAB=\angle YCB=\angle YCA=\angle YBA \rightarrow YA=YB\rightarrow \text{$Y$ is the circumcenter of $\triangle LAB$.}$$$\therefore YL = YB$, Similarly, $ KX = XC $. As $S$ is the midpoint of arc $CAB$, $\rightarrow SB=SC$. Hence, $\angle SBX = \alpha$, $\angle XBC = \alpha + \gamma$ $$\angle YSB = \angle YCB = \angle SBX =\alpha \rightarrow SY //XB$$Hence, $SX=YB$, Similarly, $SY=XC \rightarrow \triangle SYB \equiv \triangle SXC $ As $X,Y$ are the midpoints of $KC \text{ and } LB$ respectively, $$LY = SX\text{, }SY = LX \rightarrow \triangle LYS \equiv \triangle SXL$$Which implies that, $SK = SL$