Let $k$ be a circle with radius $r$ and $AB$ a chord of $k$ such that $AB > r$. Furthermore, let $S$ be the point on the chord $AB$ satisfying $AS = r$. The perpendicular bisector of $BS$ intersects $k$ in the points $C$ and $D$. The line through $D$ and $S$ intersects $k$ for a second time in point $E$. Show that the triangle $CSE$ is equilateral. Proposed by Stefan Leopoldseder
Problem
Source: Austrian Mathematics Olympiad Regional Competition (Qualifying Round) 2018, Problem 2
Tags: geometry, Equilateral Triangle, perpendicular bisector, Chords, Austria, national olympiad, AUT
WizardMath
12.06.2018 09:34
Define $F$ in a similar manner to $E$, except that $D$ is replaced by $C$. The points where the circle with center $A$ and radius $r$ meets $k$ are $E', F'$, where $E', C$ are on the same side of $AB$. Claim : $E' \equiv E, F' \equiv F$. Proof : Redefine $C, D$ as the points where $FS, ES$ meet $k$ again. Now angle chasing gives $$\angle DBS = \angle SE'A = \angle ASE'$$and thus $DB = DS$. So $D$ is the same point as in the problem statement. Thus $E' \equiv E, F' \equiv F$. So we also have $\angle CSE = 60^\circ$. Now if the center of $k$ is $K$, we have $\angle EAF = \angle EKF$. This means that $\angle SCE = \angle SEC$, which gives the desired conclusion.
dikhendzab
09.05.2020 19:40
Let $O$ be the center of this circle, let sides $BS$ and $CD$ intersect each other in point $F$. Since $DF$ is perpendicular, we see that triangles $BSD$ and $BSC$ are isosceles meaning that $BD=SD, BC=SC, \angle SDC=\angle BDC$. Since this two angles are equal, it means that their arcs must be equal, so $CE=CB=CS$. $\angle EAS=\angle SDB$ (same arc $EB$), $\angle ASE=\angle BSD \implies$ triangles $BSD$ and $EAS$ are similar, so $EAS$ is also an isosceles triangle with $EA=ES=r$, also triangle $AOE$ is equilateral with $\angle AOE=60^\circ$, so $\angle ABE=\frac{\angle AOE}{2}=30^\circ$. Similarly, $\angle ECS=2\angle ABE=60^\circ$. Because $CS=CE$ and $\angle ECS=60^\circ$, triangle $CES$ is an equilateral triangle.
A-Thought-Of-God
10.09.2020 12:13
Unnecessarily long maybe, but anyway
Let $O$ be the center of the circle and $\overline{CS} \cap \odot O=F$.
Claim : $AS=AE=AO$ and $AS=AF=AO$
Proof : Notice that $$\angle ASE=\angle DSB=\angle DBS=\angle DBA=\angle DEA=\angle SEA$$Hence $AE=AS=AO$. Similarily we can show that $AF=AS=AO$ which proves the claim $\square$
Now observe that $AFOE$ forms a rhombus with $\angle EAF=120^\circ \implies 60^\circ=\angle FCE=\angle SCE$. Also $$\angle ABC=\angle AFC=\angle AFS=90^\circ-\tfrac{1}{2}(\angle FAS)$$and
\begin{align*}
\angle AES=\angle ASE &=90^\circ-\tfrac{1}{2}(\angle EAS) \\
&=90^\circ-\tfrac{1}{2}(120^\circ-\angle FAS) \\
&=30^\circ+\tfrac{1}{2}(\angle FAS) \\
\end{align*}Finally
\begin{align*}
\text{We have that}~& \angle ABC+\angle AEC=180^\circ \\
&\implies 90^\circ-\tfrac{1}{2}(\angle FAS) + \angle AES+\angle CES=180^\circ \\
&\implies 90^\circ-\tfrac{1}{2}(\angle FAS) + 30+\tfrac{1}{2}(\angle FAS) +\angle CES=180^\circ \\
&\implies \angle CES=60^\circ \\
\end{align*}which shows that $\triangle CES$ is equilateral, finishing the problem $\blacksquare$