The triangle $ABC$ is isosceles with $AB=AC$, and $\angle{BAC}<60^{\circ}$. The points $D$ and $E$ are chosen on the side $AC$ such that, $EB=ED$, and $\angle{ABD}\equiv\angle{CBE}$. Denote by $O$ the intersection point between the internal bisectors of the angles $\angle{BDC}$ and $\angle{ACB}$. Compute $\angle{COD}$.
Problem
Source: JBMO 2006
Tags: geometry, incenter
silouan
30.06.2006 16:41
Let D,E inside AC Let $\widehat{ABD}=\widehat{CBE}=x$ . By Angle chasing we find that $\widehat{DOC}=180-B+x$ .(1) But $O$ is the incenter in the triangle $DBC$ so $\widehat{DOC}=90+\frac{\hat{B}}{2}-\frac{x}{2}$ (2) (1)=(2) so $B-x=60$ and $\widehat{DOC}=120$ . When $D\equiv C$ then $\widehat{DOC}=90$
Umut Varolgunes
30.06.2006 18:10
what a bad question havent they got any questions its like a pre-middle school olympiad problem
Jessen1111
23.09.2014 06:54
can you explain the angle chasing ??
Jessen1111
23.09.2014 07:07
JBMO is hard
oolite
21.04.2022 20:43
Found this today (apologies for the necromancy) and, just like Jessen1111, I was a little confused: Jessen1111 wrote: can you explain the angle chasing ??
Perversely writing $\measuredangle XY$ to mean the angle mod $180^\circ$ through which line $XY$ must be rotated anticlockwise to become parallel with $BC$, we're told that\begin{align*}\measuredangle BC&=0&&\text{by choice of angle baseline} \\
\measuredangle AB + \measuredangle AC &= 0&&\text{because }\triangle BAC\text{ is isosceles}\\
\measuredangle AB-\measuredangle BD &= \measuredangle BE-\measuredangle BC&&\text{because }\angle ABD = \angle CBE\\
\measuredangle AC-\measuredangle BD&=\measuredangle BD-\measuredangle BE&&\text{because }\triangle DEB\text{ is isosceles}\end{align*}Substituting the first two into the sum of the second two gives $3\measuredangle BD = 0\pmod{180^\circ}$, so $\measuredangle BD=60^\circ$.
We're asked to find $\angle COD$. Mod $180^\circ$, that's equal to $\measuredangle DO-\measuredangle CO$. We're given one angle bisector ($2\measuredangle CO=\measuredangle AC+\measuredangle BC=\measuredangle AC$) and from the existence of the incentre we know another ($2\measuredangle DO=\measuredangle BD+\measuredangle AC$), so subtracting we get $2\angle COD=\measuredangle BD=60^\circ\pmod{180^\circ}$, so $\angle COD=120^\circ$.
huashiliao2020
17.03.2023 07:18
Got it! (most of the time I don't post because the solutions are already clear but this time I will clarify this one) silouan wrote: Let D,E inside AC Let $\widehat{ABD}=\widehat{CBE}=x$ . By Angle chasing we find that $\widehat{DOC}=180-B+x$ .(1) But $O$ is the incenter in the triangle $DBC$ so $\widehat{DOC}=90+\frac{\hat{B}}{2}-\frac{x}{2}$ (2) (1)=(2) so $B-x=60$ and $\widehat{DOC}=120$ . When $D\equiv C$ then $\widehat{DOC}=90$ We have <ODC=1/2BDE=1/2DBE=1/2(B-2x), OCD=1/2B, so <DOC=180-1/2B-1/2B+x=180-B+x. On the other hand, B-2x=<DBE=<BDC=180-(B-x)-B => 180=3B-3x => 60=B-x. Substituting, we have <DOC=180-60=$\boxed{120}$.