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Source: Pan-American Girls’ Mathematical Olympiad 2021, P6

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122 posts

#1 h Oct 6, 2021, 10:34 PM • 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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#2 h Oct 6, 2021, 11:46 PM • 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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1475 posts

#3 h Oct 7, 2021, 1:02 AM • 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

.

Girls that knew projective geo should've killed this problem...

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#4 h Oct 7, 2021, 8:01 PM

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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#5 h Oct 7, 2021, 10:25 PM

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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435 posts

#6 h Nov 1, 2021, 2:44 PM

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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435 posts

#7 h Nov 1, 2021, 2:45 PM

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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#8 h Aug 5, 2022, 11:24 AM • 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

#9 h Aug 19, 2022, 1:50 PM

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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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6872 posts

#10 h Oct 26, 2022, 1:28 AM • 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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#11 h Nov 28, 2022, 4:14 AM • 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

#12 h Jul 21, 2023, 2:58 PM

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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5001 posts

#13 h Aug 10, 2023, 3:35 PM • 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

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eibc

#14 h Aug 10, 2023, 6:16 PM

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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#15 h Aug 10, 2023, 7:18 PM

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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#16 h Aug 30, 2023, 2:27 AM • 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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#17 h Sep 19, 2023, 3:50 PM

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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#18 h Dec 21, 2023, 5:40 AM

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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#19 h Feb 7, 2024, 7:18 PM

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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#20 h Jul 2, 2024, 11:29 PM

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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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#21 h Aug 21, 2024, 5:22 AM • 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

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ohiorizzler1434

#22 h Oct 13, 2024, 11:25 PM

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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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#23 h Feb 28, 2025, 2:28 PM

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A diagram is presented below
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -21.273661694116957, xmax = 31.630837108620977, ymin = -15.926946739290985, ymax = 12.835128267700076; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen cxvqqq = rgb(0.7803921568627451,0.3137254901960784,0); pen zzttff = rgb(0.6,0.2,1); pen ccqqqq = rgb(0.8,0,0); /* draw figures */ draw(circle((5,5), 5), linewidth(0.4) + qqwuqq); draw((0.7001449911989663,2.4482854973001658)--(9.299855008801034,2.4482854973001658), linewidth(0.4)); draw((3.4419996538801927,9.751066714064386)--(0.7001449911989663,2.4482854973001658), linewidth(0.4)); draw((3.4419996538801927,9.751066714064386)--(9.299855008801034,2.4482854973001658), linewidth(0.4)); draw(circle((5,0), 4.948015255938652), linewidth(0.4) + fuqqzz); draw((5.7806790337593545,-4.886040853211339)--(5.7806790337593545,2.4482854973001658), linewidth(0.4)); draw(circle((5.7806790337593545,-4.886040853211339), 7.334326350511505), linewidth(0.4) + cxvqqq); draw((-1.0856404758509592,-2.308057259828283)--(4.219320966240646,4.886040853211343), linewidth(0.4) + zzttff); draw((4.219320966240646,4.886040853211343)--(11.501847581651049,-0.29686081071766207), linewidth(0.4) + ccqqqq); draw((-4.256209886061545,-0.9850127331741488)--(12.44281412818235,-7.953335915552896), linewidth(0.4) + zzttff); draw((-4.256209886061545,-0.9850127331741488)--(4.219320966240646,4.886040853211343), linewidth(0.4) + zzttff); draw((-4.256209886061545,-0.9850127331741488)--(5.7806790337593545,2.4482854973001658), linewidth(0.4) + zzttff); draw((4.219320966240646,4.886040853211343)--(12.44281412818235,-7.953335915552896), linewidth(0.4) + blue); draw((12.44281412818235,-7.953335915552896)--(1.288620736994243,0.911696228984708), linewidth(0.4) + zzttff); draw((-1.0856404758509592,-2.308057259828283)--(5.7806790337593545,2.4482854973001658), linewidth(0.4) + zzttff); draw((4.219320966240646,4.886040853211343)--(11.281319824727936,1.4975337796439174), linewidth(0.4) + ccqqqq); draw((11.281319824727936,1.4975337796439174)--(5.7806790337593545,2.4482854973001658), linewidth(0.4) + ccqqqq); draw((11.281319824727936,1.4975337796439174)--(12.44281412818235,-7.953335915552896), linewidth(0.4) + ccqqqq); draw((12.44281412818235,-7.953335915552896)--(8.242517487884227,2.022771956642427), linewidth(0.4) + ccqqqq); draw((11.501847581651049,-0.29686081071766207)--(5.7806790337593545,2.4482854973001658), linewidth(0.4) + ccqqqq); draw((-4.256209886061545,-0.9850127331741488)--(11.281319824727936,1.4975337796439174), linewidth(0.4)); draw((5.7806790337593545,-4.886040853211339)--(3.4419996538801927,9.751066714064386), linewidth(0.4)); draw((0.7001449911989663,2.4482854973001658)--(5.7806790337593545,-4.886040853211339), linewidth(0.4)); draw((3.4419996538801927,9.751066714064386)--(5.7806790337593545,2.4482854973001658), linewidth(0.4)); draw((-1.0856404758509592,-2.308057259828283)--(0.7001449911989663,2.4482854973001658), linewidth(0.4)); draw((9.299855008801034,2.4482854973001658)--(11.501847581651049,-0.29686081071766207), linewidth(0.4)); /* dots and labels */ dot((5,5),linewidth(3pt) + dotstyle); label("$O$", (5.1153048139473,5.177898177830069), NE * labelscalefactor); dot((3.4419996538801927,9.751066714064386),linewidth(3pt) + dotstyle); label("$A$", (3.5648739279818944,9.955756622335734), NE * labelscalefactor); dot((0.7001449911989663,2.4482854973001658),linewidth(3pt) + dotstyle); label("$B$", (0.8120680692269899,2.6465824456416365), NE * labelscalefactor); dot((9.299855008801034,2.4482854973001658),linewidth(3pt) + dotstyle); label("$C$", (9.418541558667611,2.6465824456416365), NE * labelscalefactor); dot((5.7806790337593545,-4.886040853211339),linewidth(3pt) + dotstyle); label("$I_{A}$", (5.90634098025618,-4.694233177704817), NE * labelscalefactor); dot((5.7806790337593545,2.4482854973001658),linewidth(3pt) + dotstyle); label("$A_{1}$", (5.90634098025618,2.6465824456416365), NE * labelscalefactor); dot((-1.0856404758509592,-2.308057259828283),linewidth(3pt) + dotstyle); label("$B_{1}$", (-0.9598529433049025,-2.1312759988640293), NE * labelscalefactor); dot((11.501847581651049,-0.29686081071766207),linewidth(3pt) + dotstyle); label("$C_{1}$", (11.633442824332477,-0.10622341311328333), NE * labelscalefactor); dot((4.219320966240646,4.886040853211343),linewidth(3pt) + dotstyle); label("$I$", (4.355910094290775,5.082973837873003), NE * labelscalefactor); dot((8.242517487884227,2.022771956642427),linewidth(3pt) + dotstyle); label("$C_{2}$", (8.374373819139889,2.203602192508661), NE * labelscalefactor); dot((1.288620736994243,0.911696228984708),linewidth(3pt) + dotstyle); label("$B_{2}$", (1.4132555556217392,1.096151559676222), NE * labelscalefactor); dot((12.44281412818235,-7.953335915552896),linewidth(3pt) + dotstyle); label("$A_{2}$", (12.582686223903133,-7.763453502983291), NE * labelscalefactor); dot((-4.256209886061545,-0.9850127331741488),linewidth(3pt) + dotstyle); 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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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#24 h Mar 27, 2025, 5:42 PM

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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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