Let $ABC$ be a scalene triangle with $\angle A = 60^{\circ}$. Let $E$ and $F$ be the feet of the angle bisectors of $\angle ABC$ and $\angle ACB$, respectively, and let $I$ be the incenter of $\triangle ABC$. Let $P,Q$ be distinct points such that $\triangle PEF$ and $\triangle QEF$ are equilateral. If $O$ is the circumcenter of of $\triangle APQ$, show that $\overline{OI}\perp \overline{BC}$. Proposed by Vincent Huang
Problem
Source: 2017 ELMO Shortlist G2
Tags: geometry
anantmudgal09
03.07.2017 05:45
Nice problem! Congratulations on making it tastymath75025 wrote: Let $ABC$ be a scalene triangle with $\angle A = 60^{\circ}$. Let $E$ and $F$ be the feet of the angle bisectors of $\angle ABC$ and $\angle ACB$, respectively, and let $I$ be the incenter of $\triangle ABC$. Let $P,Q$ be distinct points such that $\triangle PEF$ and $\triangle QEF$ are equilateral. If $O$ is the circumcenter of of $\triangle APQ$, show that $\overline{OI}\perp \overline{BC}$. Proposed by Vincent Huang Firstly, note that $\overline{EF}$ is the perpendicular bisector of $\overline{PQ}$ hence $O$ lies on $\overline{EF}$. Let $L=\overline{AI} \cap \overline{EF}$ and $N$ be the circumcenter of $\triangle AEF$. Claim: $LION$ is a rhombus. (Proof) Firstly, we see that $I$ lies on $(AEF)$ since $\measuredangle EAF+\measuredangle EIF=60^{\circ}+\left(90^{\circ}+\tfrac{1}{2} \cdot 60^{\circ} \right)=180^{\circ}.$ Also, $IE=IF$ by Fact 5, so $I, N$ are reflections in line $\overline{EF}$. Note also that $P \in (AEF)$ is the midpoint of arc $EAF$ so $\overline{ON}$ is the perpendicular bisector of $\overline{AP}$; hence $\overline{ON} \parallel \overline{LI}$. It follows that $LION$ is indeed a rhombus just as we claimed. $\blacksquare$ Evidently, $\overline{IO} \parallel \overline{LN}$; however, we see from Brokard's Theorem that $\overline{LN} \perp \overline{BC}$ so we are done! $\blacksquare$
Einstein314
03.07.2017 06:32
Let $O'$ be the center of the circumcircle of triangle $FAE$. Call $M$ the midpoint of $FE$ and $K$ the intersection of $AI$ and $FE$.
Clearly, $P$ lies on the circumcircle of $FAE$ because $\angle{FPE} = \angle{FAE} = 60$. Also, $I$ and $O$ are reflections across $FE$ and $IO'$ is perpendicular to $FE$ by symmetry.
Brokard's on quadrilateral $FAEI$ implies that $O'K$ is perpendicular to $BC$. However, $IA$ is perpendicular to $PA$ because $PI$ is the diameter of $(PFE)$ and by radical axis, $OO'$ is perpendicular to $PA$ as well. Thus, $OO'$ is parallel to $IK$. Notice that this implies by symmetry that $OI$ is parallel to $O'K$, so $OI$ is perpendicular to $BC$ as desired.
Tumon2001
03.07.2017 09:32
Nice problem! Solution: Let $A $ and $P $ lie on the same side of $EF $. It is easy to observe that $APEIF $ is cyclic with $PI $ being a diameter. Let $O'$ be its center. Let $AI\cap FE = X $. Claim: $Q $ lies on $BC $. Proof of the claim: Let $M $ be the Miquel point of $AFIE$ , which lies on $BC $. Let $\odot (EFM) $ intersect $BC$ again at $Q'$. As $BI $ and $CI $ are bisectors, so, $I $ is the center of $\odot (EFMQ) $. This gives, $\angle FEQ' = \angle FMB = 60$° and $\angle EQ'F = \angle EMF = 60$°. So, $\Delta EQ'F $ is equilateral $\implies$ $Q = Q'$. Main problem: It is now obvious that $I$ and $O'$ are reflections of each other in $EF$. Also, $AI ||OO'$, as both are $\perp $ to $AP $. So, $IOO'X $ is a rhombus. Applying Brokard's Theorem, we find that $O'X\perp BC $. As, $OI || O'X $, the result follows immediately.
RC.
14.07.2017 12:33
tastymath75025 wrote: Let $ABC$ be a scalene triangle with $\angle A = 60^{\circ}$. Let $E$ and $F$ be the feet of the angle bisectors of $\angle ABC$ and $\angle ACB$, respectively, and let $I$ be the incenter of $\triangle ABC$. Let $P,Q$ be distinct points such that $\triangle PEF$ and $\triangle QEF$ are equilateral. If $O$ is the circumcenter of of $\triangle APQ$, show that $\overline{OI}\perp \overline{BC}$. Proposed by Vincent Huang It can also be done using elementary geo only/.
TheDarkPrince
23.12.2018 18:04
tastymath75025 wrote: Let $ABC$ be a scalene triangle with $\angle A = 60^{\circ}$. Let $E$ and $F$ be the feet of the angle bisectors of $\angle ABC$ and $\angle ACB$, respectively, and let $I$ be the incenter of $\triangle ABC$. Let $P,Q$ be distinct points such that $\triangle PEF$ and $\triangle QEF$ are equilateral. If $O$ is the circumcenter of of $\triangle APQ$, show that $\overline{OI}\perp \overline{BC}$. Proposed by Vincent Huang Nice problem. Solution. Firstly assume that $P$ lies on the same side of $EF$ as $A$. We have \[\angle FAE = 60^{\circ} = \angle FPE \implies F,A,P,E,I \text{ are concyclic points}.\]Also the point where the tangents at $E$ and $F$ to $\odot(AEF)$ meet is $Q$. Therefore $AP$ and $AI$ form external and internal angle bisectors of $\angle EAF$. So $PI$ is the diameter of $\odot(AEF)$. Let $O'$ be the center of $\odot(AEF)$ and $M$ be the midpoint of $EF$. Also let $X$ be the point where $AI$ meet $EF$ and let $Y$ be the reflection of $X$ over $M$. $O'$ lies on $PI$ and $\angle FO'E = 2\angle EAF = 120^{\circ} = \angle EIF$, so $O'M = MI$, therefore $XO'YI$ is a parallelogram. Therefore $XI||YO'$ and as $\angle IAP = 90^{\circ}\implies O'Y\perp AP$, so $AY = PY$. Also as $EF$ is the perpendicular bisector of $PQ$, $AY = PY = QY$, therefore $Y$ is the center of $\odot(PAQ)$ which gives $Y\equiv O$. Now by Brokard's theorem on $AEIF$, $O'X\perp BC$. As $O'XIO$ is parallelogram, $OI\perp BC$ and we are done.$\qquad\square$
BobaFett101
19.03.2019 08:38
Let $Q, A$ lie on the same side of $EF$. We know that $\angle A = 60 \implies \angle BIC = \angle FIE = 120$, so $I$ lies on $(AQEF)$. In addition, $AI$ bisects $\angle A$, so $I$ is the midpoint of arc $EF$ and $IF = IE$. Thus, $I$ lies on the perpendicular bisector of $EF$, which is $PQ$, so $P, I, Q$ collinear.
Notice that $EF$ is also the perpendicular bisector of $PQ$, so $O$ lies on $EF$. Because $QI \perp EF$, to prove $OI \perp BC$ it suffices to show that $\angle QIO = \angle (EF, BC)$.
Assume WLOG $AB < AC$. Then, $\angle (EF, BC) = \angle B - \angle AFE$. We'll get back to this later, chasing $\angle QIO$ in the meantime.
Let $N$ be the circumcenter of $(AQEFI)$. $N, I$ are reflections over $EF$, so $\angle QIO = \angle ONI$. We know $NO$ is the perpendicular bisector of $QA$, so if $M$ is the midpoint of $AQ$, we see $N, O, M$ are collinear and $\angle NMQ = 90$. Thus,
$$\angle ONI = \angle MNQ = 90 - \angle AQN = 90 - \angle AQI = 90 - \angle AEI.$$However, $90 - \angle AEI = 90 - (30 + \angle AEF) = 60 - \angle AEF$.
We have now reduced the problem to showing $\angle B - \angle AFE = 60 - \angle AEF$. Because $\angle AFE = 120 - \angle AEF$, we see $\angle B - \angle AFE = \angle B + \angle AEF - 120$, so we want to show $\angle B + 2\angle AEF = 180$. But
$$\angle AEF = 180 - \angle BEF - \angle BEC = 180 - 30 - (180 - \frac{1}{2}\angle B - (120 - \angle B)) = 90 - \frac{1}{2}\angle B,$$and the conclusion follows.
amar_04
07.02.2020 20:53
I've never seen any ELMOSL Problem which can be solved by Elementary methods too! . Anyways Nice Problem. 2017 ELMOSL G2 wrote: Let $ABC$ be a scalene triangle with $\angle A = 60^{\circ}$. Let $E$ and $F$ be the feet of the angle bisectors of $\angle ABC$ and $\angle ACB$, respectively, and let $I$ be the incenter of $\triangle ABC$. Let $P,Q$ be distinct points such that $\triangle PEF$ and $\triangle QEF$ are equilateral. If $O$ is the circumcenter of of $\triangle APQ$, show that $\overline{OI}\perp \overline{BC}$. Proposed by Vincent Huang We first deal with this Problem. Problem wrote: $ABC$ is a triangle with $\angle A =60^\circ$ and let $E,F$ be the feet of $B,C-$ bisectors respectively and let the perpendicular from $I$ to $BC$ intersects $EF$ at $X$ and if $K$ is the circumcenter of $\odot(AEF)$ then $KX\|AI$. Proof:- Notice that $A,F,I,E$ are concyclic and $\angle FKI=60^\circ\implies FI=FK$ and $EI=EO\implies K$ is the reflection of $I$ over $EF$. So we just need to prove that $KI$ bisects $\angle KIX$ which again suffices to prove that $\angle AIF=\angle EIX$. Let $\angle ABE=\theta$. So, $\angle AIF=\angle AEF=120^\circ-\angle AFE=120^\circ-(30^\circ+\theta)-90^\circ-\theta$ and $\angle CIE=\angle DIB=90^\circ-\theta=\angle AIF$. Hence, $\angle KIA=\angle KIX=\angle IKX\implies KX\|AI$. Now returning back to our Main Problem. Note that $\angle FPE=\angle FAE=60^\circ\implies P\in\odot(AIFE)$ and $Q$ is just the reflection of $P$ accross $EF$ also $I\in$ Perpendicular bisector of $EF\implies\overline{P-I-Q}$. So, circumcenter of $\odot(AEF)$ let it be $T$ is the midpoint of $PI$. Let the perpenicular from $I$ to $BC$ intersect $BC,EF$ at $D,O$ respectively. From our Problem we get that $ TO\|AI\implies TO\perp AP$. So $O$ must lie on the perpendicular bisector of $AP$. Also $EF$ is the perpendicular of $PQ$ as $Q$ is just the reflection of $P$ accross $EF$. Hence, $O$ is the circumcenter of $\odot(APQ)$ and also by our construction $OI\perp BC$. $\blacksquare$