The point F lies on the line OI in triangle ABC

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The point F lies on the line OI in triangle ABC

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Source: All-Russian Olympiad 2012 Grade 10 Day 2

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WakeUp

#1 h May 31, 2012, 6:33 PM • 2 Y

Y by Amir Hossein, Adventure10

The point $E$ is the midpoint of the segment connecting the orthocentre of the scalene triangle $ABC$ and the point $A$ . The incircle of triangle $ABC$ incircle is tangent to $AB$ and $AC$ at points $C'$ and $B'$ respectively. Prove that point $F$ , the point symmetric to point $E$ with respect to line $B'C'$ , lies on the line that passes through both the circumcentre and the incentre of triangle $ABC$ .

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RSM

#2 h Jun 1, 2012, 4:24 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

This problem is just a different version of a well-known result.
We know that, if $A'B'C'$ is the intouch triangle of $ABC$ , then the Feuerbach point( $F$ ) of $ABC$ is the anti-steiner point of $I$ wrt $A'B'C'$ . So if $A_1B_1C_1$ is the circumcevian triangle of the orthocenter of $A'B'C'$ wrt $A'B'C'$ , then $F$ is the center of homothety of $A_1B_1C_1$ and $A_2B_2C_2$ where $A_2,B_2,C_2$ are the midpoints of $AH,BH,CH$ ( $H$ is the orthocenter of $ABC$ , $A_2\equiv E$ ). So $F$ lies on $A_1A_2$ .Reflection of $OI$ on $B'C'$ passes through $A_1$ and $F$ . So it passes through $A_2$ . So done.

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#3 h Jun 1, 2012, 5:58 PM • 5 Y

Y by khanhha1999, noway, Adventure10, Mango247, and 1 other user

Let $P$ be the reflection of $A$ in line $B'C'$ , $O$ be the circumcentre of $ABC$ , $H$ be the orthocentre of $ABC$ and $I$ be the incentre of $ABC$ . Now let $AH$ and $AO$ intersect $B'C'$ at $Q$ and $R$ , respectively. Since $\angle{HAC}=90^\circ - \angle{C} = \angle{OAB}$ , it follows that $AI$ is the bisector of angle $\angle{OAH}$ . Therefore, since $B'C'$ is perpendicular to $AI$ , it follows that $\angle{AQR}=\angle{ARQ}$ . Since $P$ and $F$ are the reflections of $E$ and $A$ in $B'C'$ , it follows that $P$ , $F$ and $Q$ are collinear and that $\angle{PQR}=\angle{AQR}=\angle{ARQ}$ . This implies that $PF$ is parallel to $AO$ . Now let $X$ and $Y$ be the intersections of $BH$ and $CH$ with $AC$ and $AB$ , respectively. Since $E$ is the midpoint of $AH$ and $\angle{AXH}=\angle{AYH}=90^\circ$ , it follows that $E$ is the circumcentre of triangle $AXY$ . Further, the fact that $BCXY$ is cyclic implies that $\angle{C}=\angle{AYX}$ which implies that $AXY$ and $ABC$ are similar. Therefore $AE/AO = AX/AB = \cos{(\angle{A})}$ . Now let $B'I$ and $C'I$ intersect $C'P$ and $B'P$ at $Z$ and $W$ , respectively. Note that $AB'PC'$ is a rhombus and that $IZPW$ and $AB'IC'$ are similar kites satisfying that $\angle{ZPW}=\angle{B'AC'}$ and $\angle{AB'I}=\angle{AC'I}=\angle{IZP}=\angle{IWP}=90^\circ$ . Therefore $PI/IA = PZ/AC' = PZ/PB' = \cos{(\angle{A})}$ . Therefore $PF/AO=AE/AO = PI/IA$ which implies that triangles $AOI$ and $PFI$ are similar since $PF$ is parallel to $AO$ . Therefore $\angle{AIO}=\angle{PIF}$ which implies that $O$ , $I$ and $F$ are collinear.

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#4 h Jun 3, 2012, 9:28 AM • 7 Y

Y by BuiBaAnh, ArseneLupin, Vietnamisalwaysinmyheart, Radmandookheh, Adventure10, Mango247, Mogmog8

Let $I'$ be the reflection of $I$ across $B'C'$ , we know that the line obtained by reflecting $OI$ across $B'C'$ is parallel to the line obtained by reflecting the same line across the perpendicular to $B'C'$ . Let $O '$ be the reflection of $O$ across $AI$ , to show $F$ is on $OI$ , it suffices to show that $EI'$ is parallel to $IO'$ . But since $I'$ is the orthocentre of $\Delta AB'C'$ and $O'$ is on $AH$ , $AI' : AE = AI \cos(\angle A) : R \cos(\angle A) = AI : AO'$ , done .

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#5 h Jun 14, 2012, 7:53 PM • 1 Y

Y by Adventure10

yeah it is a known result but:
Generalization:for any point $P$ on angle bisector of $A$ ,let we reflect $E$ wrt $MN$ (where $M,N$ are the projections of $P$ on $AB,AC$ )then we have that the point $F$ (reflection of $E$ )is on $OP$ .and it can be easily proved by a little computation.

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#6 h Feb 27, 2016, 11:58 PM • 2 Y

Y by Tafi_ak, Adventure10

I found two solutions to this. Here is the first one. (Will post the second later)

Let $F$ be the Feuerbach point of triangle $ABC$ . It is well known that $F$ is the anti-Steiner point of the Euler line of triangle $DEF$ .

Now, let $X,Y,Z$ be the midpoints of the segments $AH,BH,CH$ respectively and let the feet of altitudes from $D,E,F$ to $EF,FD,DE$ meet the Incircle of $ABC$ again at $U,V,W$ respectively. Now, it is clear by chasing few angles that $\triangle XYZ$ and $\triangle UVW$ are both similar to $\triangle ABC$ within the same orientation. Also, their corresponding sides are parallel. Thus, $\triangle UVW,\triangle XYZ$ are homothetic. Now clearly, the centre of Homothety is actually the Feuerbach point which is just the exsimilicentre of the Incircle and the nine point circle which are tangent at it. Thus, points $U,X,F$ are collinear and cyclically. Now, reflecting over $EF$ gives the result due to our first one line paragraph. $\blacksquare$

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#7 h May 24, 2016, 12:56 AM • 1 Y

Y by Adventure10

Still looking for the second solution

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pi37

#8 h May 24, 2016, 1:00 AM • 1 Y

Y by Adventure10

Let the intouch triangle be $A'B'C'$ , and let $K$ be its orthocenter. It's well known that $K$ lies on $OI$ (which follows from an inversion about $(I)$ ), so let $\ell$ be this line. Let $K_A$ be the reflection of $K$ across $B'C'$ , and let $\ell_A$ be the reflection of $\ell$ across $B'C'$ . It's also well-known that the Feurbach point $Fe$ is the anti-steiner point of $\ell$ with respect to $A'B'C'$ , so $Fe$ lies on $\ell_A$ . Noting that $E$ lies on the nine-point circle and $K_A$ lies on the incircle, the collinearity of $Fe$ , $K_A$ , and $E$ is equivalent to $K_A$ and $E$ being corresponding points on the circles through the homothety centered at $Fe$ .

Now because $A'K_A\perp B'C'$ , the tangent to $(I)$ at $K_A$ is the reflection of $BC$ over the bisector of $AB$ and $AC$ . But a homothety about $H$ maps $E$ and the nine-point circle to $A$ and the circumcircle, which implies the tangent to the nine-point center at $E$ is also antiparallel to $BC$ . Thus the two tangents are parallel , and $Fe,K_A,E$ are collinear.

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#9 h Jun 18, 2017, 7:39 AM • 2 Y

Y by myang, Adventure10

A solution without the Feuerbach point:

My solution:

Let $DEF$ be the intouch triangle of $ABC$ , $K$ be the miquel point of $EFBC$ , $H, H_2$ the orthocenter of $ABC, DEF$ , $H_1$ be the midpoint of $AH$ , $H_3, I_1$ be the reflections of $H_2$ and $I$ , the incenter of $ABC$ , in $EF$ , $IH_3 \cap EF = Y, H_2H_3 \cap EF = X$ . $O$ is the circumcenter of $ABC$ and the line through $I_1$ and perpendicular to $BC$ meets $EF$ at $Z$ .

Since $OI$ is the Euler line of the intouch triangle, upon reflecting in $EF$ , we want that $H_1$ is on $H_3I_1$ . By inversion around the incircle, since the nine point circle of the intouch triangle is sent to the circumcircle of $ABC$ and $EF$ is sent to $AEF$ , $K$ is the inverse of $X$ under this inversion so $K,X,I$ are collinear. $I$ is the antipode of $A$ in $(AEF)$ , so $I_1$ is the orthocenter of the isosceles $AEF$ .
Since the reflection of the Steiner line of $EFBC$ in $EF$ passes through $K$ , and $I$ is the reflection of $I_1$ in $EF$ , so the reflection of the Steiner line of $EFBC$ in $EF$ is precisely $KXI$ . So reflecting back, $I_1, X, H$ are collinear and $DX$ bisects $\angle HXI$ .
$\angle IZE = \angle I_1ZE = 90^\circ +(C-B)/2 = \mathrm{angle \ between \ EF\ and \ IH_1}$ , so $H_3, Z,I$ are collinear.
Cross ratio of 4 concurrent lines is a function of the angles between them, so reflection of a harmonic bundle is a harmonic bundle, and thus $I_1(HAH_3Z)=I(XAH_2Z)=I(XAH_2H_3)=-1$
So $I_1H_3$ bisects $AH$ and thus we are done.

This post has been edited 1 time. Last edited by WizardMath, Jun 18, 2017, 7:40 AM
Reason: Spacing

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DSD

#10 h Jun 18, 2017, 7:51 AM • 2 Y

Y by Adventure10, Mango247

Pedram-Safaei wrote:

yeah it is a known result but:
Generalization:for any point $P$ on angle bisector of $A$ ,let we reflect $E$ wrt $MN$ (where $M,N$ are the projections of $P$ on $AB,AC$ )then we have that the point $F$ (reflection of $E$ )is on $OP$ .and it can be easily proved by a little computation.

This is also a known result and very easy to prove (see here)

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#11 h Oct 28, 2018, 3:50 PM • 4 Y

Y by char2539, Rg230403, Adventure10, Mango247

WakeUp wrote:

The point $P$ is the midpoint of the segment connecting the orthocentre $H$ of the scalene $\triangle ABC$ and the point $A$ . The incircle of $\triangle ABC$ is tangent to $AB$ and $AC$ at points $F$ and $E$ respectively. Prove that point $P'$ , the point symmetric to point $P$ with respect to line $EF$ , lies on the line that passes through both the circumcentre $O$ and the incentre $I$ of triangle $ABC$ .

Here's my solution (No Feuerbach point

): Seeing the floating point $P'$ , we get the idea of reflecting the figure about $EF$ . So let $A',I',O'$ be the reflections of $A,I,O$ about $EF$ . Also let $R_1$ and $R_2$ be the circumradii of $\triangle ABC$ and $\triangle AEF$ . Then it suffices to show that $P,I',O'$ are collinear. Note that, as $AI \perp EF$ , $I'$ must be the orthocenter of $\triangle AEF$ . Also, $A'$ must lie on $AI$ , with $I'A'=AI$ . As $AI=2R_2$ , we get that $\frac{I'A}{I'A'}=\frac{AI'}{AI}=\frac{2R_2 \cos \angle EAF}{2R_2}=\cos A$ Now, $AA'OO'$ is a cyclic trapezoid, which gives $\frac{PA}{O'A'}=\frac{AP}{AO}=\frac{R_1 \cos \angle BAC}{R_1}=\cos A=\frac{I'A}{I'A'}$ But, as $AH$ and $AO$ are isogonal in $\angle BAC$ , and $O'A'$ are isogonal wrt $EF$ , and cause $EF$ is perpendicular to the internal angle bisector of $\angle BAC$ , we get that $A'O'$ must be parallel to $AH$ . But this gives that $\angle PAI'=\angle O'A'I'$ , which together with the previous equalities, implies that $\triangle PAI' \sim \triangle O'A'I'$ . Thus, $\angle PI'A=\angle O'I'A'$ . However, as $A,I',A'$ are collinear, we have that $O',I',P$ are also collinear. Hence, done. $\blacksquare$

This post has been edited 3 times. Last edited by math_pi_rate, Feb 4, 2020, 2:53 PM
Reason: Fixed typos

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#12 h Sep 6, 2021, 1:40 PM • 3 Y

Y by hakN, starchan, Mango247

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Let $I'$ be the reflection of $I$ with resepct to $C'B'$ . Suppose $J,K,L$ are the projection of $F,I,O$ on $C'B'$ . Then $I'$ is the orthocenter of $\triangle C'A'B'$ . Moroever, $AI$ is the diameter of $(AC'B')$ . Therefore,
$\frac{AE}{AI'}=\frac{AO\sin A}{AI\sin A}=\frac{AO}{AI}$ Notice that $AH,AO$ are isogonals w.r.t. $AI$ , hence $\triangle AEI'\sim\triangle AOI$ .
Therefore,
$\angle FIK=\angle EI'K=180^{\circ}-\angle AIO$ as desired.

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#13 h Apr 25, 2024, 6:00 PM • 2 Y

Y by Titibuuu, Mutse

Lets take reflection of $A$ wrt $B'C'$ be $S$ , $AH$ intersect $B'C'$ at $K$ . Then because $AC'B'$ is isosceles and $AH$ and $AO$ are isogonal $AKSO$ will become a rhombus. Thus if we prove that $\Delta AIO$ and $\Delta SIH$ are similar $F$ , $I$ , $O$ will become collinear. Thus we have to prove that $AI : IS = AO : FS$ which comes from simple calculations leading to it being equal to $\cos(\angle A)$ .

This post has been edited 1 time. Last edited by Akacool, Apr 25, 2024, 6:00 PM

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#14 h Apr 20, 2025, 2:34 PM

Y by

Wow, so isogonalaty of $AH$ and $AO$ allows us to bash it smoothly. That's cool solution i would say. Anyway here i will prove lemma that $pi37$ used in their solution.

Let $A'B'C'$ be the intouch triangle and $K$ be it's orthocenter. Then prove that $I-O-K$ are collinear.
Proof: Let $R$ be the $9 point center$ of $\triangle A'B'C'$ . Then it suffices to prove that $I-O-R$ are collinear, since $I-K-R$ are clearly collinear. Let $G$ be the antipode of $A$ in $(ABC)$ . Let $M$ be the midpoint of minor arc $BC$ . Let $P$ be the intersection of $MA'$ and $(ABC)$ . Then we claim that $P-I-G$ are collinear, in order to prove this let $G'=MG \cap BC$ . Then $IA'MG'$ are definitely concyclic, by the inversion at $(BIC)$ we get the desired collinearity.

By angle chase $PAB'IC'$ is concyclic. Let $X=AP \cap B'C'$ . Let $D$ be the $A'$ altitude in $\triangle A'B'C'$ . Then by proving $\frac{C'D}{B'D}=\frac{BA'}{CA'}$ , we can easily conclude that $D$ lies on $PG$ . Let $N$ be the midpoint of $B'C'$ . It's clear that $PAND$ is cyclic.

So $X$ is the radical center of circles $(ABC)$ and $(A'B'C')$ and $9 point circle of \triangle A'B'C'$ . But also we know that there is two more points (by symmetry on the other vertices) which is radical center of these three circles. Which means they have one radical line $\implies$ their centers collinear. And we're done.

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