Let $ABC$ be a triangle such that $\measuredangle BAC = 60^{\circ}$. Let $D$ and $E$ be the feet of the perpendicular from $A$ to the bisectors of the external angles of $B$ and $C$ in triangle $ABC$, respectively. Let $O$ be the circumcenter of the triangle $ABC$. Prove that circumcircle of the triangle $BOC$ has exactly one point in common with the circumcircle of $ADE$.
Problem
Source: 2nd Final Mathematical Cup Junior Division P4 and Senior Division P2 (2020)
Tags: geometry, tangent circles, perpendicular
01.10.2020 10:45
Let $I_a$ be $A$ - excenter of $\triangle ABC$ then $I_a$ lies on $(BOC)$. We have $\angle{I_aED} = \angle{I_aAD} =90^{\circ} - \angle{AI_aD} = \dfrac{\angle{ACB}}{2} = \angle{I_aCA},$ hence $DE$ $\parallel$ $BC$ or $(ADE)$ and $(BOC)$ tangent at $I_a$
01.10.2020 14:09
I wanted to say that I don't think that your solution is correct.... $I$ does not even lie on the circumcircle of $ADE$. The point of tangency is $EC$ intersection with $DB$ which is also the antipode of $I$ in circ(BOIC) aka the $A$-excenter (by the Incenter Excenter Lemma). I will post my solution soon... Edit: Above solution was corrected.
01.10.2020 18:48
Let $Z$ be the $A-$ excenter $(BD\cap CE)$, and $I$ the incenter of $\triangle ABC$, clearly $ADZE$ is cyclic. Notice that $IC\perp CZ$ and $IB\perp BZ$, so $IBZC$ is cyclic, but also easy angle chasing shows us $\angle BIC=120^{\circ}$ and since $O$ is the circumcenter we have $\angle BOC=120^{\circ}$. With this that $CIOB$ is cyclic and then $O, I, B, C, Z$ lie all in the same circle. Finally see $IC\|AE$ since they are both perpendicular to $CZ$ and $IB\|AD$ since they are both perpendicular to $BZ$, and with this $BC\|DE$ and the conclusion follows
01.10.2020 18:52
Ricochet wrote: Let $Z$ be the $A-$ excenter $(BD\cap CE)$, and $I$ the incenter of $\triangle ABC$, clearly $ADZE$ is cyclic. Notice that $IC\perp CZ$ and $IB\perp BZ$, so $IBZC$ is cyclic, but also easy angle chasing shows us $\angle BIC=120^{\circ}$ and since $O$ is the circumcenter we have $\angle BOC=120^{\circ}$. With this that $CIOB$ is cyclic and then $O, I, B, C, Z$ lie all in the same circle. Finally see $IC\|AE$ since they are both perpendicular to $CZ$ and $IB\|AD$ since they are both perpendicular to $BZ$, and with this $BC\|DE$ and the conclusion follows Nice solution... I'll send my solution (which is different) soon.
01.10.2020 21:24
Miku3D wrote: Let $ABC$ be a triangle such that $\measuredangle BAC = 60^{\circ}$. Let $D$ and $E$ be the feet of the perpendicular from $A$ to the bisectors of the external angles of $B$ and $C$ in triangle $ABC$, respectively. Let $O$ be the circumcenter of the triangle $ABC$. Prove that circumcircle of the triangle $BOC$ has exactly one point in common with the circumcircle of $ADE$. Here is my solution which uses the Incenter - Excenter Lemma (https://web.evanchen.cc/handouts/Fact5/Fact5.pdf) and the fact that the external angle bisectors of angles $B$ and $C$ intersect at the $A$-excenter. Let ($XYZ$) denote the circumcircle of triangle $XYZ$. Let $M$ be the midpoint of arc $BC$ not containing $A$, $I_A$ be the $A$-excenter of triangle $ABC$ (i.e, $BD$ $\cap$ $CE$), and $I$ be the incenter of triangle $ABC$. Firstly, I will prove the following lemma (which only holds when $\measuredangle BAC = 60^{\circ}$): Lemma: $O$ lies on ($BIC$) which has center $M$. Proof) Let $\alpha$ = $\measuredangle BAC$ By the Incenter - Excenter Lemma, $MI$ = $MB$ = $MC$ = $2R$ sin($\alpha$/$2$) = $2R$ sin $30^{\circ}$ = $R$ because, sin $30^{\circ}$ = $1$/$2$ and $\measuredangle BAC$ = $60^{\circ}$ This implies that $O$ lies on ($BIC$) as $M$ is equidistant to $B$, $O$, $I$, and $C$. $\square$ Furthermore, ($BIC$) contains $I_A$ by the Incenter Excenter Lemma. Moreover, $I_A$ = $BD$ $\cap$ $EC$. We know that $\angle ADB$ = $90^{\circ}$, $\angle AEC$ = $90^{\circ}$. $\implies$ $\angle ADI_A$ = $90^{\circ}$ and $\angle AEI_A$ = $90^{\circ}$ $\implies$ $ADI_AE$ cyclic. Moreover, $D$, $E$ lie on the circle with diameter $AI_A$ meaning that ($ADE$) has diameter $AI_A$. Let $O_1$ be the center of ($ADE$). Then, we now that $A$, $I$, $M$ $I_A$, and $O_1$ are collinear because $O_1$ is on $AI_A$. This suffices as the centers of ($ADE$) and ($BOC$) and one of the intersections of ($ADE$) and ($BOC$) are collinear (namely, $O_1$, $M$, and $I_A$, respectively), as $B$, $O$, $I$, $C$, and $I_A$ are concyclic by the lemma proven above and the Incenter Excenter Lemma. $\square$
20.01.2021 19:52
Posting for fun. Let be the $A$-excentre of $ABC$. We claim that $BOCI_a$ and $ADI_aE$ are cyclic and show that centres of those circles (let those be $X,Y$ respectively). $\angle BOC=2\cdot \angle{BAC}=120^{\circ}$. Also if $I$ is the incentre of $ABC$, then $BICI_a$ is cyclic, and since $\angle BIC=90^{\circ} +\dfrac{\angle{ACB}}{2}=120^{\circ}\implies \angle CI_aB=60^{\circ}$, thus $BOCI_a$ is cyclic. Note that centre of this circle, $X$, lies on the angle bisector of $\angle BAC$, since $II_a$ is the diameter of $BIOCI_a$. Since $AD\perp DI_a$ and $AE\perp EI_a$, we have $$\angle ADI_a+\angle AEI_a=90^{\circ}+90^{\circ}=180^{\circ},$$thus $ADEI_a$ is cyclic and since $AI_a$ is the diameter, centre of this circle, $Y$, lies on the angle bisector of $\angle BAC$. We conclude that $X,Y,I_a$ are collinear and since $I_a$ lies on the both circles, then indeed circumcircle of the triangle $BOC$ has exactly one point in common with the circumcircle of $ADE$.
30.10.2021 09:25
I just realized this was a JBMO question.