lucky imagine getting a doable geo our geo was so hard anyway, AB=DE implies AE || BD which implies $\angle EAD =\angle EBD$ which by orthocenter reflection stuff blah blah implies BHD and AHE both equilateral so we're done

This is my solution provided during the exam

Let $ABC$ be an acute angled triangle with circumcircle $\Gamma$. The perpendicular from $A$ to $BC$ intersects $\Gamma$ at $D$, and the perpendicular from $B$ to $AC$ intersects $\Gamma$ at $E$. Prove that if $|AB| = |DE|$, then $\angle ACB = 60^o$.

jamboard solve!!! pretty easy??, nothing used beyond the fact that angles intercepting the same arc are equal + angles in a triangle sum to 180.

Notice $\triangle ABE\cong\triangle BED$ by SAS so $H=\overline{AD}\cap\overline{BE}$ is the center of $\Gamma.$ Hence, $$2\angle C=\angle AHB=180-\angle C$$and $\angle C=60.$ $\square$

Let $AD \cap BE = H$, $AC \cap BE = X$, $BC \cap AD = Y$. Using the fact that angles subtended by arcs of equal lengths are equal, $\angle DBE = \angle ADB$ so $HB = HD$. Since $AXYB$ is cyclic, $\angle HBY = \angle HAX$. By construction $ABDC$ is cyclic so $\angle HAX = \angle DBC$ so $BY$ is the perpendicular bisector of $\triangle HBD$ thus $BD = HB$. Hence $\triangle HBD$ is equilateral and $\angle ACB = \angle ADB = \angle HDB = 60^\circ$.

Let $X$ be the foot from $A$ to $BC$ and $Y$ be the foot from $B$ to $AC$. By reflecting the orthocenter, $XY=\frac{1}{2}ED=\frac{1}{2}AB$. Since $AYXB$ is cyclic, we have $\triangle CYX\sim \triangle CBA$ so $CX=\frac{1}{2}AC$ and hence $\cos C=\frac{1}{2}\implies C=60^{\circ}$.

Let $AD \cap BC = X$ and $BE \cap CA = Y$. The Orthocenter Reflection Lemma yields $$XY = \frac{DE}{2} = \frac{AB}{2}.$$Now, since $CXY \overset{-}{\sim} CAB$, we know $$| \cos ACB | = \frac{CX}{CA} = \frac{XY}{AB} = \frac{1}{2}$$so $\angle ACB = 60^{\circ}$ or $\angle ACB = 120^{\circ}$ must hold. $\blacksquare$ Remarks: I'm pretty sure $\angle ACB = 120^{\circ}$ is possible too. Also, what is the CJMO? Personally, I only know of one CJMO, and this problem isn't from there.

Canada Junior Math Olympiad?

ike.chen wrote: Also, what is the CJMO? Personally, I only know of one CJMO, and this problem isn't from there. Canadian Junior Math Olympiad

Notation : $(XY)$ stands for small arc $XY$ $(DE)= (EC)+(CD)=2\angle EAC+2<DBC=2(180^o-\angle AEB) + 2(180^o-\angle DBC) = 180^o +180^o-4(AB)-4(AB)=360^o-8(AB)=360^o- 4\angle ACB=360^o-4 \cdot 60^o=360^o -240^o=120^o =\angle (DCE) \Rightarrow AB= DE$ my solution might be modified to prove that the converse is also true, and so the problem could have been asked with iff condition