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Many Equal Sides
mathisreal   3
N 23 minutes ago by QueenArwen
Source: Brazil EGMO TST 2023 #1
Let $ABC$ be a triangle with $BA=BC$ and $\angle ABC=90^{\circ}$. Let $D$ and $E$ be the midpoints of $CA$ and $BA$ respectively. The point $F$ is inside of $\triangle ABC$ such that $\triangle DEF$ is equilateral. Let $X=BF\cap AC$ and $Y=AF\cap DB$. Prove that $DX=YD$.
3 replies
mathisreal
Nov 10, 2022
QueenArwen
23 minutes ago
Calculus
youochange   1
N an hour ago by youochange
Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$
1 reply
youochange
2 hours ago
youochange
an hour ago
Indian Geo
Adywastaken   0
an hour ago
Source: NMTC 2024/5
$\triangle ABC$ has $\angle A$ obtuse. Let $D$, $E$, $F$ be the feet of the altitudes from $A$, $B$, $C$ respectively. Let $A_1$, $B_1$, $C_1$ be arbitrary points on $BC$, $CA$, $AB$ respectively. The circles with diameter $AA_1$, $BB_1$, $CC_1$ are drawn. Show that the lengths of the tangents from the orthocentre of $ABC$ to these circles are equal.
0 replies
Adywastaken
an hour ago
0 replies
Angle ratio
Adywastaken   0
an hour ago
Source: NMTC Junior 2024/4
An acute angle triangle $\triangle PQR$ is inscribed in a circle. Given $\angle P=\frac{\pi}{3}$ and $\angle R>\angle Q$. Let $H$ and $I$ be the orthocentre and incentre of $\triangle PQR$ respectively. Find the ratio of $\angle PHI$ to $\angle PRQ$.
0 replies
Adywastaken
an hour ago
0 replies
concyclic , touchpoints of incircle related
parmenides51   2
N an hour ago by Blackbeam999
Source: All-Russian MO 1994 Regional (R4) 11.3
A circle with center $O$ is tangent to the sides $AB$, $BC$, $AC$ of a triangle $ABC$ at points $E,F,D$ respectively. The lines $AO$ and $CO$ meet $EF$ at points $N$ and $M$. Prove that the circumcircle of triangle $OMN$ and points $O$ and $D$ lie on a line.
2 replies
parmenides51
Aug 26, 2024
Blackbeam999
an hour ago
Geometry Parallel Proof Problem
CatalanThinker   5
N 2 hours ago by Tkn
Source: No source found, just yet, please share if you find it though :)
Let M be the midpoint of the side BC of triangle ABC. The bisector of the exterior angle of point A intersects the side BC in D. Let the circumcircle of triangle ADM intersect the lines AB and AC in E and F respectively. If the midpoint of EF is N, prove that MN || AD.
I have done some constructions, but still did not quite get to the answer, see diagram attached below
5 replies
CatalanThinker
Yesterday at 3:33 AM
Tkn
2 hours ago
find angle
TBazar   6
N 2 hours ago by sunken rock
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
6 replies
TBazar
May 8, 2025
sunken rock
2 hours ago
Iranian geometry configuration
Assassino9931   1
N 3 hours ago by sami1618
Source: Al-Khwarizmi Junior International Olympiad 2025 P7
Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$, such that $CD$ is not a diameter of its circumcircle. The lines $AD$ and $BC$ intersect at point $P$, so that $A$ lies between $D$ and $P$, and $B$ lies between $C$ and $P$. Suppose triangle $PCD$ is acute and let $H$ be its orthocenter. The points $E$ and $F$ on the lines $BC$ and $AD$, respectively, are such that $BD \parallel HE$ and $AC\parallel HF$. The line through $E$, perpendicular to $BC$, intersects $AD$ at $L$, and the line through $F$, perpendicular to $AD$, intersects $BC$ at $K$. Prove that the points $K$, $L$, $O$ are collinear.

Amir Parsa Hosseini Nayeri, Iran
1 reply
Assassino9931
Today at 9:39 AM
sami1618
3 hours ago
Hard geometry
Lukariman   0
4 hours ago
Given triangle ABC, a line d intersects the sides AB, AC and the line BC at D, E, F respectively.

(a) Prove that the circles circumscribing triangles ADE, BDF and CEF pass through a point P and P belongs to the circumcircle of triangle ABC.

(b) Prove that the centers of the circles circumscribing triangles ADE, BDF, CEF and ABC are all on the circle.

(c) Let $O_a$,$ O_b$, $O_c$ be the centers of the circles circumscribing triangles ADE, BDF, CEF. Prove that the orthocenter of triangle $O_a$$O_b$$O_c$ belongs to d.

(d) Prove that the orthocenters of triangles ADE, ABC, BDF, CEF are collinear.
0 replies
Lukariman
4 hours ago
0 replies
CGMO5: Carlos Shine's Fact 5
v_Enhance   61
N 4 hours ago by Adywastaken
Source: 2012 China Girl's Mathematical Olympiad
As shown in the figure below, the in-circle of $ABC$ is tangent to sides $AB$ and $AC$ at $D$ and $E$ respectively, and $O$ is the circumcenter of $BCI$. Prove that $\angle ODB = \angle OEC$.
IMAGE
61 replies
v_Enhance
Aug 13, 2012
Adywastaken
4 hours ago
Reflecting triangle sides across angle bisector
PEKKA   33
N Apr 1, 2025 by quantam13
Source: Canada MO 2024/1
Let $ABC$ be a triangle with incenter $I$. Suppose the reflection of $AB$ across $CI$ and the reflection of $AC$ across $BI$ intersect at a point $X$. Prove that $XI$ is perpendicular to $BC$.
33 replies
PEKKA
Mar 8, 2024
quantam13
Apr 1, 2025
Reflecting triangle sides across angle bisector
G H J
Source: Canada MO 2024/1
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PEKKA
1848 posts
#1 • 1 Y
Y by Rounak_iitr
Let $ABC$ be a triangle with incenter $I$. Suppose the reflection of $AB$ across $CI$ and the reflection of $AC$ across $BI$ intersect at a point $X$. Prove that $XI$ is perpendicular to $BC$.
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PEKKA
1848 posts
#2
Y by
Sketch: (Might write up when I get home)
Prove that X,I and D, the point of tangency of the incircle and BC are collinear, the end.
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math_comb01
662 posts
#3 • 1 Y
Y by bjump
Call $(AIB) \cap AC,BC = U,V$ and $(AIC) \cap AB,BC = T,S$ and let $D$ be the foot of perp from $I$ to $BC$. Clearly $DV=DS$ and $\measuredangle XVS = \measuredangle XSV=\measuredangle BAC$ by cyclic quads so done.
This post has been edited 1 time. Last edited by math_comb01, Mar 8, 2024, 4:28 PM
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awesomeming327.
1717 posts
#4 • 1 Y
Y by PRMOisTheHardestExam
My method: extend the reflections of the sides across angle bisectors to hit BC at R, S then prove that XI is perpendicular bisector of that
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IRANIAN
5 posts
#5
Y by
Let $C'$ and $A'$ be the reflections of $C$ and $A$ across $BI$ and let $B'$ and $A'_1$ be the reflections of $B$ and $A$ across $CI$. Note that $A'$ and $A'_1$ lie on $BC$ and $B'$ and $C'$ lie on $AC$ and $AB$ respectively. By the fact that $AA'_1 \perp BB'$ and $AA' \perp CC'$, we get that $\widehat{XA'A'_1}=\widehat{XA'_1A'}=\widehat{BAC}$ so $XA'=XA'_1$ and since $AB'IBA'_1$ and $AC'CA'I$ are cyclic, $I$ is the incenter of $XA'A'_1$ so $XI\perp BC$
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khina
994 posts
#6 • 8 Y
Y by PRMOisTheHardestExam, PEKKA, bookstuffthanks, LLL2019, Plasma_Vortex, CyclicISLscelesTrapezoid, Bluesoul, EpicBird08
mine :D hope you all enjoyed it. The shortest solution I know is:

Suppose the reflection of $AB$ across $CI$ intersects $BC$ at $E$, and define $F$ similarly. Then $\angle{XEC} = \angle{BAC} = \angle{XFB}$, so $XE = XF$. But also since $E$ and $A$ are reflections across $IC$, $IE = IA$ and hence $IA = IF$. So $XI$ is the perpendicular bisector of $EF$.
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BVKRB-
322 posts
#7
Y by
What!?

Let $A_B$ and $A_C$ respectively be the reflections of $A$ about $CI$ and $BI$
Notice that both of these lie on line $BC$ and $IA_B=IA=IA_C$
Obviously $\angle XA_BA_C=\angle BAC = \angle XA_CA_B$ which gives us $XA_B=XA_C$ which combined with the above fact gives us that $XI$ is the perpendicular bisector of $A_BA_C$ which implies it is perpendicular to $BC$

Sniped @above xD Nice problem! :D
This post has been edited 1 time. Last edited by BVKRB-, Mar 8, 2024, 5:09 PM
Reason: Khina orz
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rrc08
767 posts
#8
Y by
awesomeming327. wrote:
My method: extend the reflections of the sides across angle bisectors to hit BC at R, S then prove that XI is perpendicular bisector of that

I did this as well
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vsamc
3789 posts
#9 • 1 Y
Y by bookstuffthanks
Solution
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alexgsi
139 posts
#10 • 1 Y
Y by Fatemeh06
Solution
Diagram
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GrantStar
821 posts
#11 • 1 Y
Y by OronSH
hard for a problem 1, or is it just since im bad with lengths?? Really cute though

Let $B'$ and $C'$ be where the reflections hit $BC$. As $\angle XB'C'=\angle XC'B'=\angle BAC$, it suffices to show the midpoint of $B'C'$ is the intouch point. But as $CB'=CA$ and $BC'=BA$ be reflection, this is clearly true say by coordinates or anything to keep track of lengths really.
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eibc
600 posts
#12
Y by
Let $BI$ meet $AC$ at $E$ and $CI$ meet $AB$ at $F$. Also, let $XF$ and $XE$ meet $BC$ at $A_1$ and $A_2$, respectively. Note that
$$\measuredangle A_2A_1X = \measuredangle CA_1F = \measuredangle FAC = \measuredangle BAC,$$and similarly $\measuredangle XA_2A_1 = \measuredangle BAC$, so $\triangle XA_1A_2$ is isosceles. But from the reflections we find that $A_1I$ bisects $\angle XA_1A_2$ and $A_2I$ bisects $\angle XA_2A_1$, so $I$ is the incenter of $\triangle XA_1A_2$. Thus $XI$ bisects $\angle A_1XA_2$, which is enough to imply $\overline{XI} \perp \overline{BC}$.

Edit: I think in the obtuse case $I$ is actually the $X$-Excenter oops but it should be similar
This post has been edited 1 time. Last edited by eibc, Mar 8, 2024, 10:01 PM
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sixoneeight
1138 posts
#13 • 1 Y
Y by OronSH
sus problem
why are there config issues

Let the reflection of $AB$ over $CI$ intersect $BC$ at $A_1$ and let the reflection of $AC$ over $BI$ intersect $BC$ at $A_2$. Note that $\angle XA_1A_2 = \angle A = \angle XA_2A_1$. Therefore, $XA_1A_2$ is isosceles, so it suffices to show that the foot from $I$ to $BC$, which we call $D$, is the midpoint of $A_1A_2$.

By reflections, we have that $IA_1 = IA = IA_2$. Therefore, by the Pythagorean Theorem or congruent triangles or something, we are done.

P.S. I initially used length chase to show the midpoint, but that runs into SO MANY config issues
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OronSH
1745 posts
#14 • 3 Y
Y by sixoneeight, megarnie, alsk
Let $E,F=BI\cap AC,CI\cap AB.$ Let $B',C'$ be the reflections of $B,C$ over $CI,BI.$ Pappus on $BB'FCC'E$ gives $X$ lies on the line through $I$ and the orthocenter of $BIC$ which finishes.
This post has been edited 2 times. Last edited by OronSH, Mar 9, 2024, 3:02 AM
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LLL2019
834 posts
#15
Y by
Define $P$, $Y$ to be the intersection of the perpendicular line to $AC$ through $I$ with $AC$ and $AB$. Let $X_1$ be the intersection with the perpendicular line to $BC$ through $I$ with the reflection of $AB$ about $CI$. We can easily prove $X_1IF$ and $IYF$ are congruent. If $Q$ and $Z$ are defined similarly, we see $IY=IZ$, thus $IX_1=IX_2$, hence done.
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IAmTheHazard
5001 posts
#16 • 1 Y
Y by apotosaurus
khina wrote:
mine :D hope you all enjoyed it. The shortest solution I know is:

Suppose the reflection of $AB$ across $CI$ intersects $BC$ at $E$, and define $F$ similarly. Then $\angle{XEC} = \angle{BAC} = \angle{XFB}$, so $XE = XF$. But also since $E$ and $A$ are reflections across $IC$, $IE = IA$ and hence $IA = IF$. So $XI$ is the perpendicular bisector of $EF$.

Here is a shorter solution:

Let $DEF$ be the contact triangle. Use complex numbers wrt the incircle, so the reflection of $\overline{AB}$ is the tangent at $\tfrac{de}{f}$ and similarly the reflection of $\overline{AC}$ is the tangent at $\tfrac{df}{e}$. Thus $x=\tfrac{2}{\tfrac{f}{de}+\tfrac{e}{df}}$ and $\tfrac{x}{d}=\tfrac{ef}{e^2+f^2} \in \mathbb{R}$ so $I,D,X$ collinear; done.
This post has been edited 1 time. Last edited by IAmTheHazard, Mar 9, 2024, 2:20 AM
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CT17
1481 posts
#17
Y by
Let $X$ and $Y$ be the reflections of $A$ over $CI$ and $BI$. If $T$ is the desired intersection point, then $\angle TXY = \angle TYX = \angle A$, so the foot from $T$ to $BC$ has distance $\frac{BX+BY}{2} = \frac{BC-AC+AB}{2}$ from $B$ as desired.
This post has been edited 1 time. Last edited by CT17, Mar 9, 2024, 2:28 AM
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ihatemath123
3446 posts
#18 • 2 Y
Y by OronSH, alsk
Very simple and beautiful too :D

Let $E$ be the intersection of the angle bisector of $\angle ABC$ with $\overline{AC}$; let $F$ be the intersection of the angle bisector of $\angle ACB$ with $\overline{AB}$; let $L$ be the reflection of $A$ over the angle bisector of $\angle ACB$; let $K$ be the reflection of $A$ over the angle bisector of $\angle ABC$.

Because of how angle bisectors work, $K$ lies on line $\overline{BC}$ such that $AB = BK$. Furthermore, line $BE$ is the perpendicular bisector of $\overline{AK}$. So, by symmetry, $\angle EKB = \angle EAB = \angle A$. Similarly, $\angle FLC = \angle A$, so $\triangle XLK$ is isosceles. Lastly, since $I$ lies on the perpendicular bisectors of $AL$ and $AK$, it also lies on the perpendicular bisector of $LK$, so it follows that $XI \perp BC$ as desired.
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DottedCaculator
7352 posts
#19 • 1 Y
Y by mrtheory
If $A_1C=AC$ and $A_2B=AB$ with $A_1,A_2$ on $BC$, then $XA_1=XA_2$ and $IA_1=IA_2$, so $XI\perp BC$.
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lelouchvigeo
182 posts
#20
Y by
Nice geo. But still it took time
Let $A_B$ and $A_C$ respectively be the reflections of $A$ about $CI$ and $BI$. Let $IP \perp BC$
Observe that $AA_BA_C$ is isosceles. Now by standard length calculations we can find that $ IP$ is bisects $A_BA_C$. We are done
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CyclicISLscelesTrapezoid
372 posts
#21 • 1 Y
Y by MS_asdfgzxcvb
Let $E=\overline{BI} \cap \overline{AC}$ and $F=\overline{CI} \cap \overline{AB}$. Let $B'$ and $C'$ be the reflections of $B$ and $C$ over $\overline{CI}$ and $\overline{BI}$, respectively. Notice that $H=\overline{BB'} \cap \overline{CC'}$ is the orthocenter of $IBC$. Pappus on $BB'FCC'E$ gives $I$, $H$, and $X$ collinear, which finishes.
This post has been edited 1 time. Last edited by CyclicISLscelesTrapezoid, Mar 9, 2024, 5:26 AM
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blackbluecar
303 posts
#22 • 1 Y
Y by OronSH
Let $\omega$ denote the incircle and $D$ be it's tangent at $BC$. By reflection properties, we have $\angle BFI = \angle IFX$. Thus, if we let $FX$ intersect $AC$ at $R$ then $\omega$ is the excircle of $AFR$ which implies $FX$ is tangent to $\omega$. Similarly, $EX$ is tangent to $\omega$. Thus, by Brianchon's on $BFXECD$ we have $BE$, $CF$, and $XD$ concur. But, this point of concurrence is $I$. Thus, $D, I, X$ are collinear which implies the result.
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shendrew7
795 posts
#23 • 1 Y
Y by MS_asdfgzxcvb
Denote the intersection of $BC$ and the two reflections as $K$ and $L$. Note
\[\angle XKC = \angle XLB = \angle A, \quad d(I, XK) = d(I, AB) = d(I, AC) = d(I, XL),\]
so $\triangle XKL$ is isosceles and $I$ lies on the $X$-altitude, as desired. $\blacksquare$
This post has been edited 1 time. Last edited by shendrew7, Mar 9, 2024, 6:46 AM
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Mogmog8
1080 posts
#24 • 1 Y
Y by centslordm
Let $A_1$, $A_2$ be the reflections of $A$ over $\overline{CI}$, $\overline{BI}$, respectively. Let $D=\overline{AB}\cap\overline{CI}$ and $E=\overline{AC}\cap\overline{BI}$.

Claim: $I$ is the incenter of $\triangle XA_1A_2$.
Proof. Notice \[\measuredangle IA_2B=\measuredangle BAI=\measuredangle IAD=\measuredangle DA_2I\]and similarly $\overline{IB}$ bisects $\angle A_2A_1X$. $\blacksquare$

Notice $\measuredangle DA_2B=\measuredangle BAD=\measuredangle EAC=\measuredangle CA_1E$ so $\triangle XA_1A_2$ is isosceles and we finish. $\square$
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Bluesoul
898 posts
#25
Y by
Let $A,B$ are reflected to $A', B'$ and $A,C$ are reflected to $A'', C'$.

Denote the tangency points of the incircle and $AB, BC, CA$ are $D,E,F$ respectively. The condition implies $AC=A'C, AB=A''B, AB=B'C, BC'=BC$ where $A', A''\in BC; B'\in AC; C'\in AB$. Let $AF=AD=a, CD=CE=b, BE=BD=c, AC=a+c=A'C=A'E+EC=A'E+c, A'E=a$. Similar reason yields $A''E=a, A'E=A''E$. Now, the question is equivalent to prove $\angle{B'A'A"}=\angle{C'A"A'}$.

Let $\angle{ACI}=\angle{A'CI}=\alpha, \angle{ABI}=\angle{A"BI"}=\beta$. Since $A'B'$ is the reflection of $AB$ about $CI$, $A'B'$ must intersect $AB$ on $CI$, denote the intersection as $K$. We have $\angle{KB'B}=\angle{KBB'}=\angle{KAA'}=\angle{KA'A}$.

Since $AA'||BB', \angle{CA'A}=\angle{CBB'}=90-\alpha, \angle{B'BA}=90-\alpha-2\beta=\angle{B'A'A}, \angle{B'A'C}=180-2\alpha-2\beta$

By the same reason, we get $\angle{C'A''A}=90-\beta-2\alpha, \angle{C'A"A'}=180-2\alpha-2\beta=\angle{B'A'A"}$

Thus, we have $\angle{XA'A"}=\angle{XA"A'}, AE=A"E$, we know an isosceles triangle's height, angle bisector, and median to the base are the same line, we get $X,I,E$ are collinear which yields $XI\bot BC$ as desired.

(I probably add too many details since I wrote basically nothing in my scratch work but i want to make sure the sol is right <<
This post has been edited 1 time. Last edited by Bluesoul, Mar 9, 2024, 7:40 PM
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apotosaurus
79 posts
#26
Y by
Hmm maybe I should return to trying to bash every problem I do...
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P2nisic
406 posts
#27
Y by
PEKKA wrote:
Let $ABC$ be a triangle with incenter $I$. Suppose the reflection of $AB$ across $CI$ and the reflection of $AC$ across $BI$ intersect at a point $X$. Prove that $XI$ is perpendicular to $BC$.

Let $D,E,F$ be the tengency points of the incercly.$M=EF\cap CI,N=EF\cap BI$.
By Iran lemma we have that $F,M,I,D,B$ lie on the same circle the symmetry of thiw circle with respect to $CI$ is the circle $Z,M,I,E,B'$.
Now we have that $\measuredangle MZE=\measuredangle EIC=90-\frac{\measuredangle C}{2}=\measuredangle DFE=\measuredangle DZE$ we get that $Z,M,D$ are collinear.SIlilary we have $D,Y,N$.
$\angle ZDI=\angle MDI=\angle MBI=\angle MBN=\angle MCN=\angle ICN=\angle IDN=\angle IDY$ means that $DI$ is the angel bisector of $ZDY$ but also$I$ is the circumcenter of $ZDY$ so $DY=DZ$ witch givesw that $YZ//BC$ it is enougt to prove that $IX$ perpendicular to $YZ$.
This is true because $XY,XZ$ is tengent to the incircle.
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sami1618
908 posts
#28
Y by
Let the reflections be $l_C$ and $l_B$ respectively. Let $l_C$ and $l_B$ intersect $BC$ at $C'$ and $B'$ respectively. Let the incirle touch $BC$ at $D$. Using angle chasing $\measuredangle(l_B,BC)=\measuredangle(l_C,BC)$. Use the fact that $AC=CC'$ and $AB=BB'$ to show that $DB'=DC'$. Now it follows that $D$, $X$, and $I$ are collinear proving the claim.
This post has been edited 1 time. Last edited by sami1618, Mar 16, 2024, 9:31 PM
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NTguy
23 posts
#29
Y by
Let the reflection of $AB$ across $CI$ be $A_1B'$, and the reflection of $AC$ across $BI$ be $A_2C'$. Since we have reflected, $\angle ACI = \angle A_1CI = \angle BCI$ and similarly $\angle ABI = \angle A_2BI = \angle CBI$. So, $A_1$ and $A_2$ both lie on $BC$. Also, $\angle XA_1A_2 = \angle B'A_1C = \angle BAC$ and $\angle XA_2A_1 = \angle C'A_1B = \angle CAB$ so $\triangle XA_1A_2$ is isosceles, implying that the incentre and orthocentre are collinear. Since $A_1$ is the reflection of $A$ across $CI$, $\angle IA_1A_2 = \angle IA_1C = \angle IAC = \angle BAC/2 = \angle XA_1A_2/2$ and similarly $\angle IA_2A_1 = \angle XA_2A_1/2$, so $A_1I$ and $A_2I$ bisect $\angle XA_1A_2$ and $\angle XA_2A_1$ respectively, which means $I$ is the incentre of $\triangle XA_1A_2$. So, $XI \perp A_1A_2 \implies XI \perp BC$ and we are done.
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kiemsibongtoi
25 posts
#30
Y by
PEKKA wrote:
Let $ABC$ be a triangle with incenter $I$. Suppose the reflection of $AB$ across $CI$ and the reflection of $AC$ across $BI$ intersect at a point $X$. Prove that $XI$ is perpendicular to $BC$.

Let $D$, $E$, $F$ be points of tangency of $(I)$ with $BC$, $CA$, $AB$, respectively
$\hspace{0.4cm}$$E'$, $F'$ be refelections of $E$, $F$ across $BI$, $CI$ respectively
We see that $E'$, $F'$ are points of tangency of $(I)$ with tangents from $X$ to $(I)$
So $IX \perp E'F'$
Next, cuz $BI$, $CI$ are perpendicular bisecter of segments $DF$, $DE$ respectively, so $DE' = EF = DF'$
Which means $\triangle DE'F'$ is an isosceles triangle at $D$, $ID \perp E'F'$
Thus,$E'F' \| BC$ and $IX \perp BC$
Attachments:
canada-2024.pdf (67kb)
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cursed_tangent1434
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#31
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Very very easy problem. Took me around 5 minutes. Let $A_B$ and $A_C$ denote the reflections of $A$ across lines $BI$ and $CI$ respectively. Further let $B'$ and $C'$ be the reflections of $B$ across $CI$ and $C$ across $BI$. We can first locate all of these points.

Claim : Points $A_B$ and $A_C$ lie on $\overline{BC}$ and points $B'$ and $C'$ lie on $\overline{AC}$ and $\overline{AB}$ respectively.
Proof : The proof of all of these are entirely similar so we only show one of them. Let $D$ be the intersection of line $CI$ with side $AB$. Then, let $A_C'$ denote the intersection of the reflection of side $AB$ across $\overline{CI}$. Note that, $\measuredangle CDA_B' = \measuredangle ADC$ and $\measuredangle A_B'CD = \measuredangle DCA$ which since $DC$ is a common side implies that $\triangle A_B' CD \cong \triangle DCA$. Thus, $DA_B' = DA$ which implies that $A_B'$ is the reflection of $A$ across $\overline{CI}$ and $A_B'\equiv A_B$ as desired.

Now, note that $IA_B = IA=IA_C$ due to reflections so $I$ lies on the perpendicular bisector of segment $A_BA_C$. Further,
\[\measuredangle XA_BA_C = \measuredangle DA_BA_C = \measuredangle CAB = \measuredangle EAB = \measuredangle BA_CE = \measuredangle BA_CX\]so $\triangle XA_BA_C$ is isosceles and in particular, $X$ lies on the perpendicular bisector of segment $A_BA_C$. Thus, $\overline{XI}$ is the perpendicular bisector of segment $A_BA_C$ which implies $XI \perp BC$ as we wished to show.
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maths_enthusiast_0001
133 posts
#32
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Nice and easy problem.
Walkthrough
$1$. Let $A_{1}$ and $A_{2}$ be the reflections of $A$ across $CI$ and $BI$ respectively. It is evident that, $B-A_{1}-A_{2}-C$ because $CI$ and $BI$ are angle bisectors of $\angle{C}$ and $\angle{B}$ respectively.
$2$. After some trivial angle chasing we get that the points $(A,I,A_{1},B)$ and $(A,I,A_{2},B)$ are concyclic. Also since, $A_{1}$ and $A_{2}$ are the reflections of $A$ across $CI$ and $BI$ respectively, we have $IA=IA_{1}=IA_{2}$ i.e, $I$ is the center of $(AA_{1}A_{2})$.
$3$. Again angle chase to show that, $XA_{1}=XA_{2},\angle{XA_{1}A_{2}}=\angle{XA_{2}A_{1}}=180^{\circ}-\angle{A}$ and, $\angle{IA_{1}A_{2}}=\angle{IA_{2}A_{1}}=\frac{\angle{A}}{2}$.
$4$. Let $IA_{1}=IA_{2}=x$,$XA_{1}=XA_{2}=y$ and replace $\angle{B}+\angle{C}$ with $\theta$. Then, $\angle{XA_{1}A_{2}}=\angle{XA_{2}A_{1}}=\theta$ and $\angle{IA_{1}A_{2}}=\angle{IA_{2}A_{1}}=90^{\circ}-\dfrac{\theta}{2}$. Now, $A_{1}A_{2}=2x\sin\left(\frac{\theta}{2}\right)=2y\cos(\theta) \implies \dfrac{x}{y}=\dfrac{\cos(\theta)}{\sin\left(\frac{\theta}{2}\right)} \implies \boxed{\dfrac{A_{1}I}{A_{1}X}=\dfrac{\sin(90^{\circ}-\theta)}{\sin\left(\frac{\theta}{2}\right)}}$.
But by Sine Rule, $\dfrac{A_{1}I}{A_{1}X}=\dfrac{\sin(\angle{A_{1}XI})}{\sin(\angle{A_{1}IX})}$ and also, $\angle{A_{1}XI}+\angle{A_{1}IX}=180^{\circ}-\angle{IA_{1}X}=90^{\circ}-\dfrac{\theta}{2}=(90^{\circ}-\theta)+\left(\frac{\theta}{2}\right)$.
Thus we get, $\boxed{\dfrac{\sin(\angle{A_{1}XI})}{\sin(\angle{A_{1}IX})}=\dfrac{\sin(90^{\circ}-\theta)}{\sin\left(\frac{\theta}{2}\right)}}$ with $\boxed{\angle{A_{1}XI}+\angle{A_{1}IX}=(90^{\circ}-\theta)+\left(\frac{\theta}{2}\right)}$.
$5$. By "$\alpha+\gamma=\beta+\delta$ Lemma" conclude that, $\angle{A_{1}XI}=90^{\circ}-\theta$ and $\angle{A_{1}IX}=\dfrac{\theta}{2}$. We had $\angle{XA_{1}A_{2}}=\angle{XA_{2}A_{1}}=\theta$ thus, $XI \perp BC$ as desired. $\blacksquare$ ($\mathcal{QED}$)
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Tony_stark0094
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#33
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claim $XA'A_1'$ is isosceles
CI is the perpendicular bisector of AA'
so we can deduce $\angle B'A'B=\angle BAB' =\angle BAC $
claim $AA_1'CC'$ is cyclic
proof : BI is perpendicular bisector of CC' so we can deduce $\angle AC'A_1'=\angle ACA_1'$

so $\angle B'A'B=\angle BAC= \angle=BAA1'+\angle A1'AC=\angle A1'CC'+\angle A1'C'C=\angle C'A1'B$
SO$ XA'=XA_1'$
similarly we can get $A_1'X=A'A_1'$
also we can see I is the incentre of the new trianlge and the required result immidiately follows
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This post has been edited 3 times. Last edited by Tony_stark0094, Mar 23, 2025, 3:46 AM
Reason: error
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quantam13
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#34
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Complex bash with respect to the incircle finishes :)
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