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Find all possible values of BT/BM
va2010   54
N 32 minutes ago by lpieleanu
Source: 2015 ISL G4
Let $ABC$ be an acute triangle and let $M$ be the midpoint of $AC$. A circle $\omega$ passing through $B$ and $M$ meets the sides $AB$ and $BC$ at points $P$ and $Q$ respectively. Let $T$ be the point such that $BPTQ$ is a parallelogram. Suppose that $T$ lies on the circumcircle of $ABC$. Determine all possible values of $\frac{BT}{BM}$.
54 replies
va2010
Jul 7, 2016
lpieleanu
32 minutes ago
A Familiar Point
v4913   52
N 35 minutes ago by SimplisticFormulas
Source: EGMO 2023/6
Let $ABC$ be a triangle with circumcircle $\Omega$. Let $S_b$ and $S_c$ respectively denote the midpoints of the arcs $AC$ and $AB$ that do not contain the third vertex. Let $N_a$ denote the midpoint of arc $BAC$ (the arc $BC$ including $A$). Let $I$ be the incenter of $ABC$. Let $\omega_b$ be the circle that is tangent to $AB$ and internally tangent to $\Omega$ at $S_b$, and let $\omega_c$ be the circle that is tangent to $AC$ and internally tangent to $\Omega$ at $S_c$. Show that the line $IN_a$, and the lines through the intersections of $\omega_b$ and $\omega_c$, meet on $\Omega$.
52 replies
v4913
Apr 16, 2023
SimplisticFormulas
35 minutes ago
Tangential quadrilateral and 8 lengths
popcorn1   72
N 39 minutes ago by cj13609517288
Source: IMO 2021 P4
Let $\Gamma$ be a circle with centre $I$, and $A B C D$ a convex quadrilateral such that each of the segments $A B, B C, C D$ and $D A$ is tangent to $\Gamma$. Let $\Omega$ be the circumcircle of the triangle $A I C$. The extension of $B A$ beyond $A$ meets $\Omega$ at $X$, and the extension of $B C$ beyond $C$ meets $\Omega$ at $Z$. The extensions of $A D$ and $C D$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively. Prove that \[A D+D T+T X+X A=C D+D Y+Y Z+Z C.\]
Proposed by Dominik Burek, Poland and Tomasz Ciesla, Poland
72 replies
popcorn1
Jul 20, 2021
cj13609517288
39 minutes ago
An algorithm for discovering prime numbers?
Lukaluce   3
N an hour ago by TopGbulliedU
Source: 2025 Junior Macedonian Mathematical Olympiad P3
Is there an infinite sequence of prime numbers $p_1, p_2, ..., p_n, ...,$ such that for every $i \in \mathbb{N}, p_{i + 1} \in \{2p_i - 1, 2p_i + 1\}$ is satisfied? Explain the answer.
3 replies
Lukaluce
May 18, 2025
TopGbulliedU
an hour ago
Random concyclicity in a square config
Maths_VC   5
N an hour ago by Royal_mhyasd
Source: Serbia JBMO TST 2025, Problem 1
Let $M$ be a random point on the smaller arc $AB$ of the circumcircle of square $ABCD$, and let $N$ be the intersection point of segments $AC$ and $DM$. The feet of the tangents from point $D$ to the circumcircle of the triangle $OMN$ are $P$ and $Q$ , where $O$ is the center of the square. Prove that points $A$, $C$, $P$ and $Q$ lie on a single circle.
5 replies
Maths_VC
Tuesday at 7:38 PM
Royal_mhyasd
an hour ago
Basic ideas in junior diophantine equations
Maths_VC   3
N an hour ago by Royal_mhyasd
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
3 replies
Maths_VC
Tuesday at 7:54 PM
Royal_mhyasd
an hour ago
Prime number theory
giangtruong13   2
N 2 hours ago by RagvaloD
Find all prime numbers $p,q$ such that: $p^2-pq-q^3=1$
2 replies
giangtruong13
2 hours ago
RagvaloD
2 hours ago
Problem 2
delegat   147
N 2 hours ago by math-olympiad-clown
Source: 0
Let $n\ge 3$ be an integer, and let $a_2,a_3,\ldots ,a_n$ be positive real numbers such that $a_{2}a_{3}\cdots a_{n}=1$. Prove that
\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]

Proposed by Angelo Di Pasquale, Australia
147 replies
delegat
Jul 10, 2012
math-olympiad-clown
2 hours ago
Coloring points of a square, finding a monochromatic hexagon
goodar2006   6
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part 2-P1
Prove that for each coloring of the points inside or on the boundary of a square with $1391$ colors, there exists a monochromatic regular hexagon.
6 replies
goodar2006
Sep 15, 2012
quantam13
2 hours ago
Van der Warden Theorem!
goodar2006   7
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part 2-P2
Suppose $W(k,2)$ is the smallest number such that if $n\ge W(k,2)$, for each coloring of the set $\{1,2,...,n\}$ with two colors there exists a monochromatic arithmetic progression of length $k$. Prove that


$W(k,2)=\Omega (2^{\frac{k}{2}})$.
7 replies
goodar2006
Sep 15, 2012
quantam13
2 hours ago
Maxi-inequality
giangtruong13   0
2 hours ago
Let $a,b,c >0$ and $a+b+c=2abc$. Find max: $$P= \sum_{cyc} \frac{a+2}{\sqrt{6(a^2+2)}}$$
0 replies
giangtruong13
2 hours ago
0 replies
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
1850 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
1850 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.
1737 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
995 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
1748 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
5005 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
3449 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
7357 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
183 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
799 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
899 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
920 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 $IX \perp BC$
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cursed_tangent1434
650 posts
#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
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#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
69 posts
#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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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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