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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
an exponential inequality with two variables
teresafang   0
9 minutes ago
x and y are positive real numbers.prove that [(x^y)/y]^(1/2)+[(y^x)/x]^(1/2)>=2.
sorry.I’m not good at English.Also I don’t know how to use Letax.
0 replies
teresafang
9 minutes ago
0 replies
Arbitrary point on BC and its relation with orthocenter
falantrng   31
N 17 minutes ago by NZP_IMOCOMP4
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
31 replies
falantrng
Apr 27, 2025
NZP_IMOCOMP4
17 minutes ago
IMO Genre Predictions
ohiorizzler1434   23
N 24 minutes ago by ohiorizzler1434
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
23 replies
ohiorizzler1434
Yesterday at 6:51 AM
ohiorizzler1434
24 minutes ago
Number theory
gggzul   0
33 minutes ago
Is the number
$$10^{32}+10^{28}+...+10^4+1$$a perfect square?
0 replies
gggzul
33 minutes ago
0 replies
No more topics!
Incenter, Excenter and intersections
JuanDelPan   23
N Mar 27, 2025 by Ilikeminecraft
Source: Pan-American Girls’ Mathematical Olympiad 2021, P6
Let $ABC$ be a triangle with incenter $I$, and $A$-excenter $\Gamma$. Let $A_1,B_1,C_1$ be the points of tangency of $\Gamma$ with $BC,AC$ and $AB$, respectively. Suppose $IA_1, IB_1$ and $IC_1$ intersect $\Gamma$ for the second time at points $A_2,B_2,C_2$, respectively. $M$ is the midpoint of segment $AA_1$. If the intersection of $A_1B_1$ and $A_2B_2$ is $X$, and the intersection of $A_1C_1$ and $A_2C_2$ is $Y$, prove that $MX=MY$.
23 replies
JuanDelPan
Oct 6, 2021
Ilikeminecraft
Mar 27, 2025
Incenter, Excenter and intersections
G H J
G H BBookmark kLocked kLocked NReply
Source: Pan-American Girls’ Mathematical Olympiad 2021, P6
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JuanDelPan
122 posts
#1 • 1 Y
Y by FaThEr-SqUiRrEl
Let $ABC$ be a triangle with incenter $I$, and $A$-excenter $\Gamma$. Let $A_1,B_1,C_1$ be the points of tangency of $\Gamma$ with $BC,AC$ and $AB$, respectively. Suppose $IA_1, IB_1$ and $IC_1$ intersect $\Gamma$ for the second time at points $A_2,B_2,C_2$, respectively. $M$ is the midpoint of segment $AA_1$. If the intersection of $A_1B_1$ and $A_2B_2$ is $X$, and the intersection of $A_1C_1$ and $A_2C_2$ is $Y$, prove that $MX=MY$.
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FabrizioFelen
241 posts
#2 • 5 Y
Y by FaThEr-SqUiRrEl, mijail, kamatadu, lian_the_noob12, endless_abyss
From Brocard's theorem $X$ and $Y$ belongs to the polar of $I$ wrt $\Gamma$, then $X$ and $Y$ are midpoints of $A_1B_1$ and $A_1C_1$, respectively. Thus $$MY=\frac{AC_1}{2}=\frac{AB_1}{2}=MX$$done, this problem is too easy for P6.
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MathLuis
1522 posts
#3 • 1 Y
Y by kamatadu
By brokard theorem on $C_1C_2A_1A_2$ we have that $I$ lies on the polar of $Y$ w.r.t. $\Gamma$ and by brokard on $A_2A_1B_2B_1$ we have that $I$ lies on the polar of $X$ w.r.t. $\Gamma$. Also note that $Y \in \mathcal P_B$ and $X \in \mathcal P_C$ thus by La'Hire we have that $C \in \mathcal P_X$ and $B \in \mathcal P_Y$ thus we have that $\mathcal P_X=BI$ and $\mathcal P_C=CI$ meaning that $XI_A \perp CI$ and $YI_A \perp BI$ but since $\angle IBI_A=\angle ICI_A=90$ we have that $C,X,I_A$ and $B,Y,I_A$ are colinear meaning that $X$ is midpoint of $A_1B_1$ and $Y$ is midpoint of $A_1C_1$. Now by midbase we have $2MY=AC_1=AB_1=2MX$ meaning that $MY=MX$ are desired thus we are done :D.

Girls that knew projective geo should've killed this problem...
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daniel73b
14 posts
#4
Y by
No need for projective, elementary methods suffice.

Lemma: $X$ is the midpoint in $A_1B_1$. Similarly, $Y$ is the midpoint in $A_1C_1$.

Proof: Since $I$ is inside $ABC$ and lines $AB,AC$ are tangent to $\Gamma$, $A_1$ is inside segment $IA_2$ and $B_2$ is inside segment $IB_1$. By Menelaus' theorem applied to $A_2,X,B_2$ on the sides of triangle $IA_1B_1$, and using that $IA_1\cdot IA_2=IB_1\cdot IB_2$ is the power of $I$ with respect to $\Gamma$, the Lemma is equivalent to $IB_1\cdot B_1B_2=IA_1\cdot A_1A_2$. Note that the power of $I$ with respect to $\Gamma$ also equals $II_A^2-\rho^2$, where $\rho=I_AA_1$ is the radius of $\Gamma$. Or, denoting by $\alpha$ the angle between $IA_1$ and $BC$, we have $\angle IA_1I_A=90^\circ+\alpha$, and using the Cosine Law we obtain
$$IA_1\cdot A_1A_2=IA_1\cdot IA_2-IA_1^2=II_A^2-I_AA_1^2-IA_1^2=-2I_AA_1\cdot IA_1\cos\angle IA_1I_A=2\rho\cdot IA_1\sin\alpha=2\rho\cdot r.$$Similarly, $IB_1\cdot B_1B_2=2\rho\cdot r=IA_1\cdot A_1A_2$. The first part of the Lemma follows, and the second one is proved analogously, exchanging $B$ and $C$.

Since by the Lemma $X$ is the midpoint in $A_1B_1$, $Y$ is the midpoint in $A_1C_1$ and by definition $M$ is the midpoint of $A_1A$, triangle $XYM$ is similar to triangle $B_1C_1A$. Or since $AB_1C_1$ is isosceles at $A$, $MXY$ is isosceles at $M$. The conclusion follows.
This post has been edited 1 time. Last edited by daniel73b, Oct 7, 2021, 8:02 PM
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Supertinito
45 posts
#5
Y by
This problem took me ages to solve as I somehow forgot about proyective geo. Anyways here is my solution.

Note that $BI\parallel C_1A_1$ as $\angle ABI=\angle AC_1A_1=\angle \frac{B}{2}$. Then we use the tangents and parallel lines to get that
\[\angle BIC_2= \angle C_2C_1A_1=\angle BA_1C_2 \text{ and } \angle ABI=\angle AC_1A_1=\angle C_1A_1A_2 \]thus both $BIA_1C_2$ and $BIA_2C_1$ are cyclic. Then we get that $\angle C_2BA_1=\angle C_2IA_1=\angle C_1IA_2=\angle C_1BA_2$.Thus $\angle C_1BC_2=\angle A_2BA_1 (*)$.

Let $I_a$ be the $A$-excenter. $(*)$ together with the fact that $BI_a$ is the angle bisector of $\angle C_1BA_1$ imply that $BI_a$ is the angle bisector of $\angle C_2BA_2$. The perpendicular bisector of $C_2A_2$ also goes through $I_a$ as $C_2$ and $A_2$ both lie on the excircle with center $I_a$. Thus by the incenter excenter lemma, $BC_2A_2I_a$ is cyclic.

Clearly $C_1BA_1I_a$ is cyclic. Then we get that $BI_a$, $A_1C_1$ and $C_2A_2$ are the radical axii of $(BC_2A_2I_a),(A_1A_2B_1B_2),(C_1BA_1I_a)$, which implies that $BI_a$ goes through $X$. As $BA_1I_aC_1$ is a kite, $X$ is the midpoint of $A_1C_1$. Analogously $Y$ is the midpoint of $A_1B_1$. Thus $MXY$ is similar to $AC_1B_1$ which is clearly isosceles as desired.
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hydo2332
435 posts
#6
Y by
Very nice problem.

The main claim is that $X,Y$ are midpoints of segments $A_1B_1$ and $A_1C_1$.
In order to prove that, we see that $X$ and $Y$ belong to the polar of $I$ wrt $\Gamma$, so the polar of $I$ is $XY$, hence $II_A \perp XY$, and since $II_A \perp C_1B_1$, we have that $XY \parallel C_1B_1$ (fact I)).
Now, notice that $X$ belongs to the polar of $C$, and hence the polar of $X$ is $CI$, so $I_AX \perp CI$ hence $X$ belongs to the segment $CI_A$. So $X,Y$ are the midpoints of segments $A_1B_1$ and $A_1C_1$, as claimed.
Now, let $N$ be the midpoint of segment $A_1I_A$, $P$ be the midpoint of $XY$ and $L$ be the midpoint of $C_1B_1$. Take an homothety centered at $A_1$ and scaling factor $\frac12$. Due to fact I, $L$ goes to $P$; $I_A$ clearly goes to $N$ and $A$ goes to $M$. Since $A,L,I_A$ are collinear points, then $M,P,N$ must also be, and we're done.
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hydo2332
435 posts
#7
Y by
FabrizioFelen wrote:
From Brocard's theorem $X$ and $Y$ belongs to the polar of $I$ wrt $\Gamma$, then $X$ and $Y$ are midpoints of $A_1B_1$ and $A_1C_1$, respectively. Thus $$MY=\frac{AC_1}{2}=\frac{AB_1}{2}=MX$$done, this problem is too easy for P6.

Lol this sol is much easier than mine. Nicely done.
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EJGarcia
52 posts
#8 • 1 Y
Y by NoobMaster_M
Clearly, $\angle{ABI}=\angle{BC_1A_1}$, so $BI\parallel{A_1C_1}$. By property of angles between parallels, $\angle{BIC_1}=\angle{IC_1A_1}$ and by property of semi-inscribed angle, $\angle{BC_1I}=\angle{C_1A_1C_2}$. This implies that $\triangle{IBC_1}\sim{\triangle{A_1C_1C_2}}$. It then follows that $\angle{IC_2A_1}=180^\circ-\angle{C_1C_2A_1}=180^\circ-\angle{C_1BI}=\angle{ABI}=\angle{A_1BI}=\frac{\angle{ ABC}}{2}$. From which it follows that $IBC_2A_1$ is cyclic. Let $N$ denote the second intersection of $BC_2$ with $\tau$. By Reim's Theorem, $BI\parallel{A_2N}$. Note that $A_1NC_1C_2$ is a harmonic quadrilateral, so if we project from $A_2$ onto the line $A_1C_1$, also parallel to $A_2N$ (remember that $BI\parallel{A_1C_1}$), we have

$$-1=(N, C_2; C_1, A_1)\stackrel{A_2}{=}(P_\infty, Y; C_1, A_1).$$
This implies that $Y$ is the midpoint of $A_1C_1$. Similarly, $X$ is the midpoint of $A_1B_1$. By the Midsegment Theorem, $MY=\frac{AC_1}{2}$ and $MX=\frac{AB_1}{2}$. But $AC_1=AB_1$ for being common tangents, therefore $MY=MX$.
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levifb
61 posts
#9
Y by
Claim: $BIA_{2}C_1$ and $BIC_{2}A_{1}$ are cyclic.
This is just an angle chasing: $\angle  C_{1}BA_{1} = 180 - B \Rightarrow \angle BC_{1}A_{1} = \angle C_{1}A_{2}A_{1} = \frac{B}{2}$. Since $\angle ABI = \frac{B}{2}$, $BIA_{2}C_{1}$ is cyclic. Note that $\angle IC_{2}A_{1} = \angle IA_{2}C_{1} = \frac{B}{2} = \angle IBA_{1}$, then $BIC_{2}A_{1}$ is cyclic.

A known fact in a cyclic complete quadrilateral $ABCDPQ$ is that the intersection of the diagonals (let $R$) is the inverse of the Miquel Point M. This is because with an angle chasing we get $\angle AR'B = \angle APB$, so $R'$ lies on $(PAB)$. Similarly, we can achieve that R' lies on $(PCB), (QBC), (QDA)$, so $R' = M$.
Note that in this problem, $B$ is the Miquel Point of the cyclic quadrilateral $A_{1}A_{2}C_{1}C_{2}$, so $Y$ is the inverse of $B$ wrt $\Gamma$ $\Rightarrow$ Y is the midpoint of $A_{1}C_{1} \Rightarrow MY = \frac{AC_{1}}{2}$. Similarly, $X$ is the midpoint of $A_{1}B_{1} \Rightarrow MX = \frac{AB_{1}}{2}$. Since $AC_1 = AB_1$, $MX = MY$ and we are done.
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v_Enhance
6877 posts
#10 • 6 Y
Y by HamstPan38825, kamatadu, Mango247, Mango247, Mango247, busy-beaver
The main content of the problem is the following claim.
Claim: Point $Y$ coincides with the midpoint of $\overline{A_1C_1}$.
[asy]size(6cm);
pair A_1 = dir(40); pair C_1 = dir(140); pair B = 2*A_1*C_1/(A_1+C_1); pair Y = midpoint(A_1--C_1); pair C_2 = dir(80); pair I = extension(C_1, C_2, B+dir(0), B); pair A_2 = extension(C_2, Y, A_1, I);
filldraw(unitcircle, invisible, blue); draw(C_1--B--A_1, blue); draw(C_1--I--A_2--C_2, lightred); draw(B--I, deepgreen); draw(C_1--A_1, deepgreen); pair J = origin;
dot("$A_1$", A_1, dir(0)); dot("$C_1$", C_1, dir(C_1)); dot("$B$", B, dir(B)); dot("$Y$", Y, dir(Y)); dot("$C_2$", C_2, dir(C_2)); dot("$I$", I, dir(I)); dot("$A_2$", A_2, dir(A_2)); dot("$J$", J, dir(90));
/* TSQ Source:
A_1 = dir 40 R0 C_1 = dir 140 B = 2*A_1*C_1/(A_1+C_1) Y = midpoint A_1--C_1 C_2 = dir 80 I = extension C_1 C_2 B+dir(0) B A_2 = extension C_2 Y A_1 I
unitcircle 0.1 lightcyan / blue C_1--B--A_1 blue C_1--I--A_2--C_2 lightred B--I deepgreen C_1--A_1 deepgreen J = origin R90
*/
[/asy]
Proof. By Brokard theorem both $I$ and $B$ lie on the polar of $Y$, hence $\overline{BI}$ is exactly the polar of $Y$. In particular, if $J$ is the excenter, then $\overline{YJ} \perp \overline{BI}$. As $\overline{BI} \parallel \overline{A_1C_1}$, we're done. $\blacksquare$
Similarly $X$ is the midpoint of $\overline{A_1B_1}$. All that remains is the extraction $MX = \frac{1}{2} AC_1 = \frac{1}{2} AB_1 = MY$.
This post has been edited 2 times. Last edited by v_Enhance, Oct 30, 2022, 6:36 PM
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Mogmog8
1080 posts
#11 • 1 Y
Y by centslordm
Notice that $\angle BA_2C_1=\tfrac{1}{2}\angle ABC=\angle IBA_1$ so $\overline{IB}\parallel\overline{A_1C_1}.$ Also, by Brokard, $I$ lies on the polar of $Y$ and $B$ lies on the polar of $Y$ by La Hire. Hence, the line perpendicular to $\overline{IB}$ and therefore $\overline{A_1C_1}$ through $Y$ passes through the center of $\Gamma,$ so $Y$ is the midpoint of $\overline{A_1C_1}.$ Similarly, $X$ is the midpoint of $\overline{A_1B_1}$ so \[MX=\tfrac{1}{2}AB_1=\tfrac{1}{2}AC_1=MY,\]as desired. $\square$
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jbaca
225 posts
#12
Y by
Solution. An alternative way to show that $Y$ is the midpoint of $A_1B_1$ without projective geometry.
Since $BI \parallel C_1A_1$, we have $\angle ABI = \angle IA_1B = \angle BA_1C_1 = \angle IC_2A_1$ where the last equality holds by virtue of the semi-inscribed angle theorem. Thus $IBC_2A_1$ is cyclic. Therefore
$$\angle IC_2B = \angle IA_1B = \angle CA_1A_2 = \angle A_1CA_2 =\angle A_1C_2Y$$which means that $BC_2$ and $YC_2$ are isogonal conjugates of $\bigtriangleup C_1C_2A_1$. In particular, $BC_2$ is the $C_2$-symmedian of $\bigtriangleup C_1C_2A_1$, which forces $C_2Y$ to be a median and hence $Y$ is the midpoint of $A_1C_1$. $\square$
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IAmTheHazard
5001 posts
#13 • 1 Y
Y by centslordm
oops

By using Brocard on $A_1A_2B_1B_2$, $\overline{IP}$ is the polar of $X$, so $\overline{I_AX} \perp \overline{IP}$. Since $X$ clearly lies on the polar of $C$, $C$ lies on $\overline{IP}$ by La Hire's. Furthermore, since $\angle ICI_A=90^\circ$, it follows that $I_A,X,C$ are collinear, hence $X$ is the midpoint of $\overline{A_1B_1}$. Likewise, $Y$ is the midpoint of $\overline{A_1C_1}$, so a homothety of scale factor $2$ at $A_1$ sends $\triangle MXY$ to $\triangle AB_1C_1$. Since $AB_1=AC_1$, we are done. $\blacksquare$

Completely unnecessary
This post has been edited 1 time. Last edited by IAmTheHazard, Aug 10, 2023, 3:37 PM
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eibc
600 posts
#14
Y by
lol why does the statement say "$A$-excenter $\Gamma$"

Let $I_A$ be the $A$-excenter. By Brokard on $A_1B_2B_1A_2$, we find that $I$ lies on the polar of $X$, and by La Hire's we can see that $C$ lies on the polar of $X$, too (as line $A_1B_1$ is the polar of $C$). Therefore since $\overline{IC} \perp \overline{I_AC}$, $X$ must lie on $\overline{I_AC}$, which implies that $X$ is the midpoint of $\overline{A_1B_1}$. Similarly, $Y$ must be the midpoint of $\overline{A_1C_1}$. By taking a homothety at $A_1$, we find that $MX = \tfrac{1}{2}AB_1 = \tfrac{1}{2}AC_1 = MY$, so we are done.
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huashiliao2020
1292 posts
#15
Y by
Brocard's on $A_1A_2C_1C_2$ implies that I lies on the polar of Y. Since B also lie on the polar of Y (tangents construction), we get that BI is the polar of Y; in particular, $T_aY$ perp. IB ($T_a$ is the the center of circle T), but since we know $T_aB$ perp. IB we get that $T_aB$ passes through Y, which readily implies that Y is the midpoint of $A_1C_1$.

Doing analogous work on $A_1A_2B_1B_2$, ... gives that X is the midpoint of $A_1B_1$ as well. Then $MY=\frac{AC_1}{2}=\frac{AB_1}{2}=MX$, as desired.
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YaoAOPS
1538 posts
#16 • 1 Y
Y by Shifted777
By Brokard's on $A_1A_2C_1C_2$ and $A_1A_2B_1B_2$ it follows that $XY$ is the polar of $I$.

Claim: $XA_1 = XB_1$. Similarily, $YA_1 = YC_1$.
Proof. The polar of $X$ is $IC$.
As such, $IC \perp XI_A$. However, $CI \parallel A_1B_1$ follows since $\measuredangle ICA_1 = \measuredangle B_1A_1C$. Thus, $XI_A \perp A_1B_1$ so $XA_1 = XB_1$.
The proof of $YA_1 = YC_1$ is similar. $\blacksquare$
Thus, taking a homothety of size $2$ at $A_1$ means that $MX = MY$ is the same as $AC_1 = AB_1$, which holds.
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smileapple
1010 posts
#17
Y by
who else is here because of OTIS lol

Since $A_1C$ and $B_1C$ are both tangent to $\Gamma$, with $X$ lying on $A_1B_1$, we find that $C$ is on the polar of $X$. However, observe that since $A_1B_2B_1A_2$ is a cyclic quadrilateral inscribed in $\Gamma$, with intersection points $I=A_1A_2\cap B_1B_2$, $K=A_1B_2\cap A_2B_1$, and $X=A_1B_1\cap A_2B_2$, the line $KI$ must be the polar of $X$ due to Brokard's Theorem, forcing $K$, $C$, and $I$ to be collinear.

Angle chasing on the angle bisectors of $\angle A_1CA$, we find that
\begin{align*}
    \angle ICA_1&=\frac{\angle C}2\\
    &=\frac{180^\circ-\angle A_1CB_1}2\\
    &=\angle CA_1B_1,
\end{align*}implying that $CI\parallel A_1B_1$. Again using Brokard's Theorem on $A_1B_2B_1A_2$, we find that the orthocenter of $\triangle IKX$ is exactly $I_A$, so that $KI\perp XI_A$. Combined with the results that $CI\parallel A_1B_1$ and that $K$, $C$, and $I$ are collinear, as found previously, we find that $XI_A\perp A_1B_1$. Since $A_1$ and $B_1$ both lie on $\Gamma$, centered at $I_A$, it thus follows that $A_1I_A=B_1I_A$, and we therefore conclude that $X$ is the midpoint of $A_1B_1$.

Symmetrical logic yields that $Y$ is the midpoint of $A_1C_1$. Then, a homothety centered at $A_1$ with ratio $2$ thus maps $\triangle MXY$ to $\triangle AB_1C_1$, implying that $\frac{MX}{MY}=\frac{AB_1}{AC_1}$. Since $AB_1$ and $AC_1$ are both tangents to $\Gamma$ from $A$, respectively at $B_1$ and $C_1$, it follows that $AB_1=AC_1$, so that $MX=MY$, as desired. $\blacksquare$

- Jörg
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Shreyasharma
682 posts
#18
Y by
Brockards sobad.

We first claim that $XY$ is the polar of $I$ with respect to $\Gamma$. Indeed note that by Brokards on $A_1A_2B_1B_2$ we have that $I$ lies on the polar of $X$. Also we have that $B$ lies on the polar of $X$, so $BI$ is the polar of $X$ with respect to $\Gamma$. Similarly $CI$ is the polar of $Y$. Then by La Hire's we have $X$ and $Y$ lie on the polar of $I$, so $XY$ is the polar of $I$.

Now some simple angle chasing gives $IB \parallel A_1C_1$, hence $XI_A perp A_1B_1$ from which we can conclude that $X$ is the midpoint of $A_1B_1$. Now a homothety centered at $A_1$ with scale factor $2$ finishes.
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dolphinday
1324 posts
#19
Y by
By Brokard's we have $CI$ being the polar of $X$ wrt $\Gamma$. Note that since $\angle ICI_A = 90^{\circ}$, we have $X$ lying on $I_AC$ which implies that $X$ is the midpoint of $A_1B_1$, similarly for $Y$. Then taking a homothety centered at $A_1$ sending $\triangle MYX \to \triangle AC_1B_1$ with factor of $2$ finishes.
This post has been edited 1 time. Last edited by dolphinday, Feb 7, 2024, 7:19 PM
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SenorSloth
37 posts
#20
Y by
By definition, $Y$ lies on $A_1C_1$, which is the polar of $B$ with respect to $\Gamma$. By Brokard's Theorem on $A_1A_2C_1C_2$, $Y$ also lies on the polar of $I$ with respect to $\Gamma$. La Hire's Theorem then tells us that the polar of $Y$ with respect to $\Gamma$ must be the line $IB$. We note that $\angle IBA_1 = \angle BA_1C_1 = \frac12 \angle B$, so $IB \parallel A_1C_1$. Since $IB$ is the polar of $Y$, $IB\perp OY$, so we also have $OY \perp A_1C_1$, so $Y$ must be the midpoint of $A_1C_1$.

We can follow the same steps to show that $X$ is the midpoint of $A_1B_1$. Then we note that $\triangle MXY$ is the image of $\triangle AB_1C_1$ under a homothety centered at $A_1$ with scale factor $\frac12$. $AB_1 = AC_1$, so we also have $MX = MY$, as desired.
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joshualiu315
2534 posts
#21 • 1 Y
Y by dolphinday
Let $I_A$ be the center of $\Gamma$.

Note that $B$ lies on the polar of $Y$ because of the tangents construction. Also, Brokard on $A_1A_2C_1C_2$ gives that $I$ lies on the polar of $Y$ as well. Hence, the polar of $Y$ is simply $\overline{BI}$. Since $\overline{IB} \parallel \overline{A_1C_1}$, we must have that $Y$ is the midpoint of $\overline{A_1C_1}$ as $\overline{I_AY} \perp \overline{IB}$.

Similarly, we can determine that $X$ is the midpoint of $\overline{A_1C_1}$. Thus, a homothety centered at $A_1$ with scale factor $2$ maps $\triangle MXY$ to $\triangle AB_1C_1$; we consequently have

\[MX = \frac{1}{2} AB_1 = \frac{1}{2} AC_1 = MY. \ \square\]
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ohiorizzler1434
768 posts
#22
Y by
ok bluds lets use projective geometry

from brocard twice I is the pole of XY. now reverse reconstruct with X'Y' as midpoints to see that IB and IC are polars of X and Y, so I is the pole of X'Y' midpoints = XY. now dilate MXY to get the equal lengths.
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Ihatecombin
59 posts
#23
Y by
A diagram is presented below
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[/asy]
The main claim is that there exists a homothety centered at \(A_1\) of scale \(2\) which sends \(\triangle MXY\) to \(\triangle AB_{1}C_{1}\). This clearly finishes since \(AB_1 = AC_1\).
To do this it suffices to show that \(X\) is the midpoint of \(A_1B_1\), notice that \(BI \parallel A_{1}B_{1}\),
and thus it suffices to show \(I_{A}X \perp BI\), which is equivalent to showing that \(BI\) is the polar of \(X\) wrt the excircle.

Define \(Z_B = A_{1}B_{2} \cap A_{2}B_{1}\), notice that by Brokard on \(A_{1}B_{2}B_{1}A_{2}\) we obtain the fact that \(IZ_B\) is the polar of \(X\), thus \(I\) lies on the polar of \(X\). We shall show that \(B\) lies on the polar of \(X\). Notice that by La Hire's it suffices to show that \(X\) lies on the polar of \(B\), the polar of \(B\) is \(A_{1}B_{1}\) and so we are done.
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Ilikeminecraft
614 posts
#24
Y by
nice

Let $E, F$ be the intersections of $(M_A)$ and $\Gamma.$
With respect to $\Gamma:$ by Brokard’s on $C_1C_2A_1A_2$ gives $I$ lies on the polar of $Y$. Similarly, $I$ lies on the polar of $X$. Hence, $XY$ is the polar of $I.$ However, $EF$ is also the polar of $I,$ as $\angle IFI_A = \angle IEI_A = 90.$
Thus, $-1 = (C_1C_2;EF) = (A_1A_2;EF),$ and projection through $EF\cap A_1C_1$ implies that $C_2, EF\cap A_1C_1, A_2$ are collinear. Thus, define $Y$ to be the concurrency of $A_1C_1, C_2A_2, EF.$
Radax on $(BC_1I_AA_1), (IBFI_AE),(C_1FA_1EB_1A_2)$ implies $Y$ also lies on $BI_A.$
Thus, $Y$ is midpoint of $A_1C_1.$
Homothety about $A_1$ finishes.
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