Oi! These lines concur

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Oi! These lines concur

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: LMAO 2021 P5, LMAOSL G3(simplified)

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

222 posts

#1 h May 10, 2021, 6:39 PM • 4 Y

Y by A-Thought-Of-God, samrocksnature, Ya_pank, ohhh

Let $I, O$ and $\Gamma$ respectively be the incentre, circumcentre and circumcircle of triangle $ABC$ . Points $A_1, A_2$ are chosen on $\Gamma$ , such that $AA_1 = AI = AA_2$ , and point $A'$ is the foot of the altitude from $I$ to $A_1A_2$ . If $B', C'$ are similarly defined, prove that lines $AA', BB'$ and $CC'$ concurr on $OI$ .
Original Version from SL
Proposed by Mahavir Gandhi

This post has been edited 4 times. Last edited by Rg230403, May 13, 2021, 11:41 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

hellomath010118

373 posts

hellomath010118

#2 h May 10, 2021, 7:03 PM • 4 Y

Y by samrocksnature, Ya_pank, math_comb01, Exposter

Note that $A'B'C'$ is the incircle of $\triangle ABC$ because of tangents from the midpoint of arc $BC$ not containing $A$ and poncelet.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

71 posts

PUjnk

#3 h May 10, 2021, 7:03 PM • 3 Y

Y by samrocksnature, Mango247, Mango247

Very nice configurations

\input{fig1.tex}

%% write the problem proof here:

Let w denote the incircle of $\triangle{ABC},D,E,F$ are the intouch points. Let $P_D$ be the foot of perpendicular from D onto EF. $P_E,P_F$ are defined analogously.
{\textbf{Claim 1}}: YZ is tangent to $w$ .

\begin{proof}
Let $M_D = DP_D \cap w$ and define $M_E,M_F$ analogously. Let the tangent to $w$ at $M_D$ meet AB, AC at $Z_1$ and $Y_1$ respectively. Now by Newton's Theorem on quadrilateral $BCY_1Z_1$ , we have $BY_1 \cap CZ_1$ = $DM_D \cap EF = P_D$

$\Rightarrow P_D \in\ BY_1$ and $P_D \in CZ_1$
$\because Y = Y_1$ and $Z$ = $Z_1$ .
This proves the claim.
\end{proof}

{\textbf{Claim 2}}:
Let M be any point on $\circledcirc{ABC}$ . Let the tangents from X to
$w$ intersect $w$ at $Y_1$ , $Y_2$ and $BC$ extended at $X_1$ , $X_2$ and let the point of tangency between the A-Mixtilinear incircle and $\circledcirc{ABC}$ be $U$ . Then $\circledcirc{MX_1X_2}$ passes through U.

%\begin{figure}
\input{fig2.tex}
%\end{figure}

\begin{proof}
Let $N$ = $AM\cap BC$
By Dual of Desargues Involution Theorem on complete quadrilateral $ABDC$ athrough $M$ , giving the involutive pairing
( $MA,MD); (MB, MC); (MY_1,MY_2$ ). Now projecting this onto line $BC$ ,
we get the pairs :
( $MN,MD); (MB,MC); (MX_1,MX_2$ ). Now we know that every involution is an inversion about some center. Let this center be $K$ .
$KB\times KC = KD\times KN = KX_1\times KX_2$ . So
$\circledcirc{AMBC}$ , $\circledcirc{MDN} , \circledcirc{MX_1X_2}$ are
co- axial circles. So it suffices to prove $U \in\circledcirc{MDN}$ .
Now let $DU \cap \circledcirc{ABC}$ = $A_1$ . By properties of mixtilinear incircles, $AA_1 \Vert BC$ . $\because \angle ANB$ = $\angle A_1AM$ = $\angle DUM$ . So $UDNM$ is cyclic as required.
\end{proof}

{\textbf{Claim 3}}:
Let $XY \cap\circledcirc{ABC}$ = $J$ . Then the tangents from $X$ and $J$ to $w$ intersect at $\circledcirc{ABC}$ .

\begin{proof}
Let the tangents from $X$ and $J$ to $w$ meet $BC$ at
$H$ , $L$ and let $T$ = $YZ \cap BC$ . Now by claim 1, $TX$ and $TH$ are tangent to $w$ . So by claim 2, $\circledcirc{TXH}$ and $\circledcirc{TJL}$
pass through $U$ .
$\Longrightarrow$ by Miquel's Theorem, $U$ is the miquel point of quadrilateral $XHLJ$ .
$\because$ $XH \cap JL$ lies on $\circledcirc{XUJ}$ = $\circledcirc{ABC}$ .
\end{proof}

{\textbf{Claim4}}: $M_D=A',M_E=B',M_F=C'$ .

\begin{proof}:
We shall prove $M_d=A'$ . The others can be proved analogously.
By Claim 3, we know that the tangents from $X$ and $J$ to $w$ intersect at $\circledcirc{ABC}$ at a point say K. Now w is the inscribed circle in $\triangle{XJK}$ . So I is the incenter of $\triangle{XJK}$ . Now we know $BCYZ$ is cyclic. So YZ is antiparallel to BC wrt $\angle{BAC}$ . Now since, O is the circumcenter of $\triangle{ABC}, AO \perp XJ$ . So $AX=AJ$ .
Thus A is the midpoint of XJ in $\circledcirc{XJK}$ . Now since I is the incenter of $\triangle{XJK}$ , by incircle-excircle lemma, we have that $AX=AJ=AI$ .
Thus $X=D$ and $J=E$ . So $M_D$ is the foot of perpendicular from I onto DE which is exactly the definition of $A'$ . This proves the claim.
\end{proof}

{\textbf{Claim 5}}: $\triangle{M_DM_EM_F}$ is similar to $\triangle{ABC}$

\begin{proof}:
Note that $\angle{M_DDF}=\angle{M_EDF}=\angle{M_EEF}=\frac{C}{2} \Longrightarrow \angle{M_DM_FM_E}=\angle{M_DDM_E}=C$ .
Similarly it can be shown that $\angle{M_EM_DM_F}=A,\angle{M_DM_EM_F}=B$ . This proves the claim.
\end{proof}

Back to the main problem, combining Claim 4 and Claim 5, we see that $\triangle{A'B'C'}$ is similar to $\triangle{ABC}$ .
Thus there exists a centre of homothety T, swapping these two triangles.
Now since I is the circumcenter of $\triangle{A'B'C'}$ and O is the circumcenter of $\triangle{ABC}$ , by properties of homothety, we have that $T \in IO$ . We also notice that $AA',BB',CC'$ concur at T.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

168 posts

#4 h May 10, 2021, 7:05 PM • 4 Y

Y by DebayuRMO, A-Thought-Of-God, samrocksnature, trigocalc

Hopefully correct.

Solution. Let $\odot(I)$ denote the incircle, $\odot(AI)$ the circle centered at $A$ with radius $AI$ , and $(ABC)$ the circumcircle; $D, E, F$ the respective intouch points opposite to $A, B, C$ respectively, and $M_A$ the midpoint of arc $BC$ opposite to $A$ .

Claim. $AI$ is the external bisector of $\angle A'ID$ .

Proof. $\measuredangle DIM_A = \measuredangle AM_AO = \measuredangle OAM_A = \measuredangle AIA'.$ As both $ID, OM_A$ are perpendicular to $BC$ , and $IA', OA$ are perpendicular to $A_1A_2$ (the radical axis). $\quad\square$

Claim. $A' \in \odot(I)$ .

Proof. Let the tangent from $M_A$ to $\odot(I)$ intersect $(ABC)$ at $A_1'$ and $A_2'$ . So, by Poncelet's Porism $A_1'A_2'$ is tangent to $\odot(I)$ . Now, by Fact 5 we get $AA_1' = AI = AA_2'$ , so $A_1' \equiv A_1$ and $A_2' \equiv A_2$ (as $\odot(AI)$ and $(ABC)$ already intersect at $A_1, A_2$ ). So, $A_1A_2$ is tangent to $\odot(I)$ , and that at $A'$ (using the right angle there). $\quad\square$

So, $ID=IA'$ , and thus, the internal bisector of $\angle A'ID$ is perpendicular to $A'D$ . This along with the first claim yields $DA' \parallel AI$ . Hence, $DA' \perp EF$ (as $AI \perp EF$ ). So, $A'$ is the intersection of the perpendicular from $D$ to $EF$ with $\odot(I).$

Suppose $AA'$ meets $(ABC)$ at $T_A$ .

Claim. $T_A$ is the $A$ -mixtilinear intouch point.

Further let $\omega_A$ denote the $A$ -mixtilinear incircle, and $E_1, F_1$ be the intouch points of $\omega_A$ on $AC, AB$ respectively.

Proof. Note that $AE \cdot AE_1=AF \cdot AF_1=AI^2$ , so $\odot(I)$ and $\omega_A$ are inverses w.r.t $\odot(AI)$ .

Let $E_1F_1$ intersect $BC$ at $Z$ , as $ZI \perp AI$ , so $ZI$ is tangent to both $\odot(AI)$ and $\odot (BIC)$ , yielding $Z$ to lie on the radical axis of $(ABC)$ and $\odot(AI)$ , and thus, $A'Z$ is the radical axis of $(ABC)$ and $\odot(AI)$ (since $A' \in A_1A_2$ ). In other words, $A'Z$ and $(ABC)$ are inverses w.r.t $\odot(AI)$ .

On inverting w.r.t $\odot(AI)$ , $A'$ goes to $T_A$ . But, as $A' \in \odot(I)$ , so $T_A \in \omega_A$ and we get the desired. $\quad\square$

Likewise define $T_B, T_C$ , and get them as the $B, C$ -mixtilinear intouch points; and further let $\omega_B$ and $\omega_C$ to be the respective mixtilinear incircles.

By Monge's theorem applied to $\omega_A, (ABC), \odot(I)$ , we get the exsimilicenter of $(ABC)$ and $\odot(I)$ to lie on $AT_A$ . Analogous holds for the lines $BT_B$ and $CT_C$ . Whence, $OI, AT_A, BT_B, CT_C$ concur at $K$ , the isogonal point of the Nagel point of $\triangle ABC$ (appealing to the well known fact that respective mixtilinear cevian acts as the isogonal of the Nagel line generating from the respective vertex; in other words $AT_A$ and $AQ_A$ are isogonals, where $Q_A$ is the $A$ -extouch point on $BC$ ). Since $A' \in AT_A$ , etc, we're done. $\quad \blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1349 posts

i3435

#5 h May 10, 2021, 7:06 PM • 2 Y

Y by Aryan-23, samrocksnature

If I'm not mistaken SORY P6 was very similar.

Invert around $(A,AI)$ . This takes $\overline{A_1A_2}$ to $(ABC)$ and takes the incircle to the $A$ -mixtilinear incircle. Thus the incircle is tangent to $\overline{A_1A_2}$ . Since $\overline{A_1A_2}\perp\overline{AO}$ , $\overline{IA'}||\overline{AO}$ . Thus the positive homothety taking the circumcircle to the incircle takes $A$ to $A'$ , so $\overline{AA'}$ goes through the exsimillicenter of the incircle and circumcircle.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

222 posts

#6 h May 10, 2021, 7:23 PM • 1 Y

Y by samrocksnature

Yes, that works. Can you please share what SORY P6 was? I have not seen that problem.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

141 posts

#7 h May 10, 2021, 7:24 PM • 4 Y

Y by samrocksnature, Muaaz.SY, TheorM, MatBoy-123

The official solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

hellomath010118

373 posts

hellomath010118

#8 h May 10, 2021, 7:25 PM • 3 Y

Y by samrocksnature, math_comb01, Exposter

For @above

Rg230403 wrote:

Yes, that works. Can you please share what SORY P6 was? I have not seen that problem.

Attachments:

SORY_solutions.pdf (273kb)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

222 posts

#9 h May 10, 2021, 7:40 PM • 1 Y

Y by samrocksnature

Oh I see, I think configurations of this sort have been explored. We did not know how much of it has appeared before, but I think it still serves well as an easy problem. The test-solvers and contestants had not seen the results beforehand on the basis of the response we have received.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

95 posts

#10 h May 10, 2021, 9:42 PM • 1 Y

Y by samrocksnature

Very cool problem! Here is the solution I submitted (cleaned it up a bit

) .

Let $A_1A_2 \cap B_1B_2=X$ , and define $Y,Z$ similarly. The key claim of the problem is that $I$ is incenter of $\triangle XYZ$ .

Since $B_1,B_2,A_1,A_2$ are concyclic, we deduce that $X$ lies on the radical axis of $(A)$ and $(B)$ $\Rightarrow AB \perp XI$ and similarly.

Let $A_1A_2 \cap AB=V,B_1B_2 \cap AB=U$ . Note that $\measuredangle BB_1U=\measuredangle BB_1B_2=\measuredangle B_1B_2B=\measuredangle B_1AB \Rightarrow \triangle BUB_1 \sim \triangle BB_1A$ and similarly $\triangle AA_1V \sim \triangle ABA_1$ . But $\measuredangle BB_1A=\measuredangle BA_1A$ so we have $\measuredangle XUV=\measuredangle UVX$ . Since $\triangle VXU$ is isoceles and $XI \perp \overline{UV}\equiv \overline{AB}$ , we deduce $\measuredangle VXI= \measuredangle IXU$ . Since similar results hold, we deduce that $I$ is the incenter of $\triangle XYZ$ .

Now simply note that $AA_1=AA_2 \Rightarrow A_1A_2 \perp AO$ , and similarly. Thus, we have $\triangle A'IB'$ and $\triangle AOB$ are isoceles with two sides parallel, so $AA',BB',OI$ , and similarly $CC'$ as well, concur at the center of homothety of the two circles and we are done.

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

924 posts

#11 h May 11, 2021, 5:02 AM • 2 Y

Y by samrocksnature, Muaaz.SY

Let $M_A$ be the midpoint of arc $\widehat{BC}$ not containing $A$ . Let $A_1',A_2'$ be points on $\odot(ABC)$ such that $\overline{M_AA_1'}$ and $\overline{M_AA_2'}$ are tangent to incircle of $\triangle{ABC}$

Claim 1: $\overline{AA_1'} = \overline{AA_2'}$
Proof: Obviously $\overline{M_AI}$ is the angle bisector of $\angle A_1'M_AA_2'$ . Hence the result. $\qquad \square$

Claim 2: $\overline{A_1'A_2'}$ is tangent to incircle of $\triangle{ABC}$
Proof: Poncelets porism. $\qquad \square$

Claim 3: $\overline{AI} = \overline{AA_1'} = \overline{AA_2'}$ . Thus $\{A_1',A_2'\} = \{A_1,A_2\}$ .
Proof: From previous results we get that $I$ is the incenter of $\triangle{M_AA_1'A_2'}$ and the result follows from fact 5 $\qquad \square$

Now clearly $A'$ is the tangency point of $\overline{A_1A_2}$ with incircle of $\triangle{ABC}$ and let $D,E,F$ be the tangency points of incircle with $\overline{BC},\overline{CA},\overline{AB}$ respectively.

Claim 4: $\overline{DA'} \perp \overline{EF}$
Proof: Now notice that clearly $\overline{IA'} \parallel \overline{AO}$ . Let $T$ be the midpoint of $\widehat{EF}$ not containing $D$ in incircle of $\triangle{ABC}$ and let $D'$ be the $D$ antipode in the incircle. Now we have that $\angle A'IT = \angle IAO = \angle OM_AA = \angle DIM_A = \angle D'IT$ but this would clearly imply $\overline{DA'},\overline{DD'}$ are isogonal w.r.t $\angle{EDF}$ and hence done $\qquad \square$

Claim 5: $\overline{A'B'} \parallel {AB}$ similiarly for others.
Proof: $\overline{IF} \perp \overline{AB}$ and also $\angle FA'B'=\angle FEB' = 90^\circ -\angle EFD = \angle A'DF = \angle A'B'F$ so $\overline{IF} \perp \overline{A'B'}$ $\qquad \square$

Now just apply homothety on $\triangle{A'B'C'}$ and $\triangle{ABC}$ to finish $\qquad \blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1184 posts

L567

#12 h May 11, 2021, 5:11 AM • 2 Y

Y by p_square, samrocksnature

Here's another way to get that $A_1A_2$ is tangent to the incircle.

Let $A_1A_2$ meet $AB,AC$ at $X,Y$ . Let $\angle AA_2A_1 = \angle AA_1A_2 = x$

Then, we can easily get that $\triangle AXY \sim \triangle ACB$ .

Since $\angle AA_1Y = \angle YCA_1$ , $AA_1$ is tangent to $(A_1YC)$ and so $AA_1^2= AY.AC$ .

Since $AI = AA_1$ , $AI^2 = AY.AC$ and so $AI$ is tangent to $(IYC)$ and so $\angle AIY = \angle ICY$ and now its easy enough to prove by angel chasing that $I$ is the A-excenter in $\triangle AXY$ . So because $AX,AY$ are already tangent to the incircle, it must be the excircle. So, $A_1A_2$ is tangent to the incircle

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

993 posts

khina

#13 h May 13, 2021, 11:44 PM

Y by

why are the solutions above so ridiculously complicated >.<

Note that $AO$ is perpendicular to $A_1A_2$ . Thus, it suffices to prove that the ratio between the distance from $I$ to $A_1A_2$ , and $AO$ , is constant (when replaced with $B$ and $C$ instead). Since $AO$ is just the circumradius of $ABC$ it suffices to prove $I$ is equidistant from $A_1A_2$ , $B_1B_2$ , and $C_1C_2$ .

We in fact claim all three lines are tangent to the incircle of $ABC$ , which finishes. Indeed, let $AI \cap (ABC) = A, M$ , and let the tangents from $M$ to the incircle of $ABC$ meet the circumcircle of $ABC$ again at $X$ and $Y$ . Note by Poncelet's Porism, $XY$ is tangent to the incircle of $ABC$ as well. Now by fact five $AX = AI = AY$ , so $\{ X, Y \}$ is some permutation of $\{A_1, A_2 \}$ , and so we are done!

This post has been edited 2 times. Last edited by khina, May 13, 2021, 11:47 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

173 posts

#14 h May 14, 2021, 3:29 PM • 1 Y

Y by Mango247

Here is a way to sneakily avoid Poncelet's Porism using (one of many) Euler's Formula.

Let $\triangle DEF$ and $\triangle D'E'F'$ be the intouch triangle and circumcevian triangle of $I$ . By the incenter lemma, it follows that $I$ is the incenter of $\triangle D'A_1A_2$ , and by Euler's formula, the inradius of $\triangle ABC$ is the same as that of $\triangle A_1A_2D'$ , so it follows that they share the same incircle. Now let $\overline{A_1A_2}$ cut $\overline{AB}$ and $\overline{AC}$ at $X$ and $Y$ respectively. Then as $AO\perp A_1A_2$ ,
$\measuredangle AYX=90^\circ-\measuredangle OAC=\measuredangle CBA$ and thus quadrilateral $XYCB$ is bicentric. It is then well-known that $DA'\perp EF$ . (This can be easily proven via angle chasing.) Similarly, we can deduce that $EB'\perp DF$ , and $FC'\perp DE$ . Since $\triangle DEF$ and $\triangle D'E'F'$ are homothetic, $\triangle ABC$ and $\triangle A'B'C'$ must also be homothetic as well. Therefore, $\overline{AA'}$ , $\overline{BB'}$ and $\overline{CC'}$ are concurrent at the homothetic center of two triangles, which lies on $\overline{OI}$ .

This post has been edited 2 times. Last edited by KST2003, May 14, 2021, 3:37 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

466 posts

#15 h Oct 31, 2021, 5:58 AM

Y by

L567 wrote:

Here's another way to get that $A_1A_2$ is tangent to the incircle.

Inverting at $A$ with radius $AI$ maps the incircle to the $A-$ mixtilinear incircle (can be easily proved).
Also $A_1A_2$ goes to the circumcircle of $\triangle ABC$ .
‐

Simple homothety: $OA || IA'$ , the ratios $\frac{OA}{IA'}=\frac{R}{r}$ are constant and $O$ and $I$ are the circumcenters of $ABC$ and $A'B'C'$ ....

This post has been edited 1 time. Last edited by SPHS1234, Oct 31, 2021, 5:58 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

659 posts

#16 h Dec 25, 2023, 3:39 PM

Y by

Funny Problem.
We make use of following 3 well-known claims
Claim 1: $A_1A_2$ is tangent to incircle at $P$ s.t. $DP \perp EF$
Claim 2 IF $T_a$ is the mixti touch point then $A-P-T_a$
Claim 3: $AT_a,BT_b,CT_c,OI$ concurr
Hence we're done

This post has been edited 1 time. Last edited by math_comb01, Dec 25, 2023, 3:39 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1247 posts

ezpotd

#17 h Aug 2, 2024, 3:33 AM

Y by

woah...

Let $M_a$ be the arc midpoint of $BC$ and cyclic variants. By Poncelet's Porism, there exist two unique points $X,Y$ such that $M_aXY$ has both the same circumcircle and incircle as $ABC$ . Since $M_aI$ is the bisector of $\angle M_aY$ , we can in fact conclude the arc midpoint of $XY$ is $A$ , thus the center of $(XYI)$ is $A$ , clearly forcing $A_1$ , $A_2$ = $X$ , $Y$ . Thus the foot from $I$ to $A_1A_2$ lies on the incircle.

Let $AA'$ meet $OI$ at $K$ . We prove $\frac{KI}{KO}$ is symmetric in $AB$ , $BC$ , $AC$ . Since $IA'$ is parallel to $AO$ , we just want $\frac{IA'}{AO} = \frac rR$ , so we are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

249 posts

L13832

#18 h Wednesday at 1:14 PM

Y by

SORY P6

solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

586 posts

#19 h Yesterday at 6:23 PM • 1 Y

Y by ohhh

I've done tooooooo much config geo in life.

Invert at $A$ with radius $AI$ .Note that the incircle and the mixtillinear incircle gets swapped and $(ABC)$ gets swapped to $A_1A_2$ . So the incircle is tangent to $A_1A_2$ as well.Also note that $A,A',T_A$ become collinear since $A'$ becomes the tangency point of the incircle with $A_1A_2$ and on inversion it swaps with $T_A$ (the $A-$ mixtillinear intouch point).Now it's well known that $AT_A,BT_B,CT_C,OI$ are concurrent and we are done $\blacksquare$

Click to reveal hidden text

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a