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Source: USA December TST for IMO 2013, Problem 3

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6876 posts

#1 h Jul 30, 2013, 5:19 AM • 9 Y

Y by anantmudgal09, Math-Ninja, srijonrick, HamstPan38825, megarnie, son7, Adventure10, A21, Rounak_iitr

Let $ABC$ be a scalene triangle with $\angle BCA = 90^{\circ}$ , and let $D$ be the foot of the altitude from $C$ . Let $X$ be a point in the interior of the segment $CD$ . Let $K$ be the point on the segment $AX$ such that $BK = BC$ . Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$ . The circumcircle of triangle $DKL$ intersects segment $AB$ at a second point $T$ (other than $D$ ). Prove that $\angle ACT = \angle BCT$ .

This post has been edited 1 time. Last edited by v_Enhance, May 7, 2015, 1:33 AM
Reason: 90\dg should be 90^{\circ}

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#2 h Jul 30, 2013, 11:52 AM • 2 Y

Y by Adventure10, Mango247

this a little long solution but nice :

let the line perpendicular to $BK$ intersects $CD$ at point $S$ and let $BK \cap AL=M$ we will prove that $XM \cap AB=T$
let $XM \cap AB=T'$
we can easily see that $X$ is orthocenter of $ASB$ and $\angle SKB=\angle SLA=90$

$\frac{AT'}{BT'}=\frac{sin(\angle AXT')}{sin(\angle BXT'}.\frac{AX}{XB}=\frac{sin(\angle KXM)}{sin(\angle LXM}.\frac{cos(SAB)}{cos(SBA)} =\frac{sin(\angle XKM)}{sin(\angle XLM}.\frac{\frac{AD}{SA}}{\frac{BD}{SB}}=\frac{sin(\angle KSB)}{sin(\angle LSA}.\frac{AD}{BD}.\frac{SB}{SA} =\frac{\frac{KB}{SB}}{\frac{LA}{SA}}.\frac{AD}{BD}.\frac{SB}{SA}=\frac{BC}{AC}.\frac{AD}{BD}=\frac{BC}{AC}.\frac{AD.AB}{BD.AB} =\frac{BC}{AC}.\frac{AC^2}{BC^2}=\frac{AC}{BC} \Longrightarrow T' \equiv T$

so by harmonic division we have $(B,A,T,N)=-1$ where $(KL \cap BC=N)$

so we have $CT,CN$ are the internal and external bisectors of $C$

we want to prove that $NL.NK=ND.NT$

$\angle NCT=\angle CDT=90 \Longrightarrow NC^2=ND.NT$

so we need to prove that $NC$ is tangent to the circumcircle of $CLK$

and this is not hard

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#3 h Jul 30, 2013, 2:31 PM • 1 Y

Y by Adventure10

for the last part we can do like this :

we want to prove $NC$ is tangent to circumcircle of triangle $CLK$

so we need to prove that $\angle CKL=\angle NCL$

but we had $\angle NCB=45$ so we need to prove that $\angle CKL-\angle BCL=45$

$\angle CKL-\angle BCL=\angle CAL+\angle KCA+\angle KLA-\angle LCB$

let $J=CB \cap SL$ the we have $JCLA$ is cyclic , $AC=AL, JC=JL \Longrightarrow \angle BCL=\frac{\angle CAL}{2}$

so we need to prove that $\angle LSK+\angle LAC+\angle KBC=90$
$\Longleftrightarrow \angle LSB+\angle CBS=\angle LAC$ and this obvious ,so we are done

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Lyub4o

#4 h Jul 30, 2013, 5:17 PM • 2 Y

Y by Adventure10, Mango247

sco0orpiOn wrote:

so by harmonic division we have $(B,A,T,N)=-1$ where $(KL \cap BC=N)$

so we have $CT,CN$ are the internal and external bisectors of $C$

Doesn't it follow that $CA$ and $CB$ are bisectors of $\angle NCT$ ?

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#5 h Jul 30, 2013, 5:44 PM • 1 Y

Y by Adventure10

Lyub4o wrote:

Doesn't it follow that $CA$ and $CB$ are bisectors of $\angle NCT$ ?

yes ,and i used it ( $\angle NCB=45$ )

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#6 h Jul 30, 2013, 11:04 PM • 2 Y

Y by Adventure10, Mango247

We know $X$ is on the radical axis $CD$ of the orthogonal circles $\omega_A = (A,AC)$ and $\omega_B = (B,BC)$ , so letting $\{K,K'\} = AX\cap \omega_B$ and $\{L,L'\} = BX\cap \omega_A$ , we see that $K,K',L,L'$ lie on a circle $\omega$ centered at $O$ . Since $OB\perp XA$ and $OA\perp XB$ , $O$ must be the orthocenter of $\triangle{XAB}$ ; in particular, $O\in CXD$ . Furthermore, $AL^2 = AL'^2 = AC^2 = AK\cdot AK'$ , so $BLXL'$ is the polar of $A$ (with respect to $\omega$ ) and $AKXK'$ is the polar of $B$ . Hence $\triangle{XAB}$ is self-polar.

Since $D$ lies on the polar of $X$ and $O,X,D$ are collinear, $D$ is the inverse of $X = KK'\cap LL'$ . It follows from the standard Yufei configuration that $D$ is the spiral center taking $LK$ to $K'L'$ (and $LK'$ to $KL'$ ), and $T = LK'\cap KL'$ (note that $DK'L'T$ is also cyclic, and that $LK'\cap KL'$ lies on the polar of $X$ ).

To complete the diagram, let $T' = LK\cap K'L'$ so $\triangle{XTT'}$ is self-polar by Brokard's theorem (by duality, we should have $CT'$ the external bisector of $\angle{BCA}$ ). But $-1 = K(K,K';L,L') = K(B,A;T',T)$ , so it suffices to show $\angle{TCT'} = 90^\circ$ . However, $O$ has fixed power $OX\cdot OD$ with respect to any polar triangle $\triangle{XYZ}$ through $X$ , and thus lies on the radical axis of the circles with diameters $AB,TT'$ . Since line $ABTT'$ is perpendicular to $OCD$ , we conclude that $\angle{TCT'} = 90^\circ$ , as desired. (Since $L\in (BX)$ , $K\in (AX)$ , we have $T'\notin (BA)\implies T\in (BA)$ .)

Comment. After the first two paragraphs, it's straightforward to complex bash with unit circle $\omega$ ( $(k+k')(l+l') = 2(kk'+ll')$ or $(k+k'-l-l')^2 = (k-k')^2 + (l-l')^2$ from $KLK'L'$ harmonic), with the observation that $c,2d-c$ are inverses with respect to $\omega$ : we get $1-\frac{c}{d} = \pm\frac{(k-k')(l-l')}{2(kk'-ll')}$ .

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#7 h Aug 3, 2013, 5:27 AM • 5 Y

Y by Stranger8, JasperL, myh2910, Adventure10, and 1 other user

An easy exercise if you've seen IMO 2012 #5 before..

From that problem it follows that $K, L$ are pairs of antihomologous points w.r.t. the two orthogonal circles centered at $A, B$ with radii $AC, BC$ ; hence the line $KL$ goes through the exsimilicenter $Se$ of the two circles, and our goal is to prove that $T$ is the insimilicenter $Si$ .. For this we consider the inversion centered at $Se$ that swaps the two circles, and it remains to show that the points $Si$ and $D$ are inverses.. Indeed, $CSi$ is tangent to the circle of inversion since it is perpendicular to $CSe$ .. hence the two points are indeed inverses and the problem is finished.

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#8 h Aug 13, 2013, 1:20 PM • 6 Y

Y by mathuz, R8450932, Hamel, HolyMath, Adventure10, Mango247

v_Enhance wrote:

Let $ABC$ be a scalene triangle with $\angle BCA = 90\dg$, and let $D$ be the foot of the altitude from $C$ . Let $X$ be a point in the interior of the segment $CD$ . Let $K$ be the point on the segment $AX$ such that $BK = BC$ . Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$ . The circumcircle of triangle $DKL$ intersects segment $AB$ at a second point $T$ (other than $D$ ). Prove that $\angle ACT = \angle BCT$ .

As shown in the figure below.

$E, F$ are the intersection points of $AX, BX$ with the circumcircle of $Rt\triangle ABC$ .

$G$ is the intersection of $AF, BE$ .

It's obvious $X$ is the orthocenter of $\triangle ABG$ and $G$ is on $CD$ .

we have

$\angle DKB = \angle DAE = \angle DGE$

so $B,G,K,D$ are concyclic.

$\angle GKB=\angle GDB=90$ , $GK^2= GE\times GB$ .

Also

$\angle DLA = \angle DBF = \angle DGA$

so $A,D,L,G$ are concyclic.

$\angle ALG=\angle ADG=90$ , $GL^2= GF\times GA=GE\times GB =GK^2$ .

So
$GK=GL$ ,
Thus
$SK=SL$ , $KN=ML$

As it's quite obvious that
$TM//BX$ , $TN//AX$ ,
we have
$\frac{AM}{ML} =\frac{AT}{TB}=\frac{KN}{NB}$ ,
so
$ML\times KN=AM\times NB=(AL-ML)(BK-KN)$
$AL\times BK= AL\times KN + BK\times ML=(AC+BC)\times ML$
$ML=KN=\frac{AC\times BC}{AC+BC}$
$AM=AL-ML=\frac{AC^2}{AC+BC}$

Consequently
$\frac{AT}{TB}=\frac{AM}{ML}=\frac{AC}{BC}$
so
$CT$ is the bisector of $\angle ACB$ , namely

$\angle ACT = \angle BCT$ .

Attachments:

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mathuz

#9 h Oct 7, 2013, 12:13 PM • 1 Y

Y by Adventure10

My solution same with @Andrew 's solution

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#10 h Apr 16, 2017, 10:12 AM • 1 Y

Y by Adventure10

As it's quite obvious that
$TM//BX$ , $TN//AX$ ,

Can you show me why we have that?

This post has been edited 1 time. Last edited by Trafalgar0246, Apr 16, 2017, 10:17 AM

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#11 h Nov 9, 2017, 10:44 AM • 2 Y

Y by Adventure10, Mango247

We proved that A,D,L,G are concyclic so $\angle ALD = \angle AGD = \angle ABF$ . And $\angle ATM = \angle ALD$ since MTDL is an cyclic quadrilateral. Therefore, $\angle ATM = \angle ABF$ . So $TM//BX$

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#12 h Apr 9, 2018, 12:43 PM • 4 Y

Y by srijonrick, Aritra12, Adventure10, Mango247

v_Enhance wrote:

Let $ABC$ be a scalene triangle with $\angle BCA = 90^{\circ}$ , and let $D$ be the foot of the altitude from $C$ . Let $X$ be a point in the interior of the segment $CD$ . Let $K$ be the point on the segment $AX$ such that $BK = BC$ . Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$ . The circumcircle of triangle $DKL$ intersects segment $AB$ at a second point $T$ (other than $D$ ). Prove that $\angle ACT = \angle BCT$ .

WLOG $CA<CB$ . Let $\omega_A, \omega_B$ be circles centered at $A, B$ respectively, passing through point $C$ . Let $\overline{BL}$ meet $\omega_A$ again at $L'$ and $\overline{AK}$ meet $\omega_B$ again at $K'$ . Observe that $X$ lies on the radical axis of $\omega_A, \omega_B$ hence $XL \cdot XL'=XK \cdot XK'$ implying $KLK'L'$ is cyclic with circumcircle $\omega$ .

Now let $T=\overline{KL'} \cap \overline{K'L}$ and $S=\overline{KL} \cap \overline{K'L'}$ . Let $Y$ be the orthocenter of $\triangle XAB$ . Then $\overline{AY}$ is the perpendicular bisector of $\overline{LL'}$ and $\overline{BY}$ is the perpendicular bisector of $\overline{KK'}$ . Thus, $Y$ is the center of $\omega$ .

Observe that $A=\overline{LL} \cap \overline{LL'}$ while $B=\overline{KK} \cap \overline{K'K'}$ proving $A,B,T,S$ are collinear and $(AB, ST)=-1$ . Now $\angle ACB=90^{\circ}$ so $\overline{CA}, \overline{CB}$ are bisectors of angle $TCS$ . Now by Brokard's Theorem, $Y$ is the orthocenter of $\triangle TXS$ . Thus, $DT \cdot DS=DX \cdot DY=DA \cdot DB=DC^2$ proving $\angle TCS=90^{\circ}$ . Hence $\angle ACT=\angle BCT$ .

Finally, we show $T$ lies on $\odot(DKL)$ ; or equivalently $D$ is the spiral center of $\overline{KL'} \mapsto \overline{LK'}$ . Let $N=\overline{AY} \cap \overline{BX}$ then $OL^2=ON \cdot OA=OX \cdot OD$ proving $X, D$ are inverses in $\omega$ . Hence $D$ is the desired spiral center. Thus, our proof is complete! $\blacksquare$

Remark: This is an example of "completing the picture". By appending $Y$ we are able to see things from $\triangle AYB$ and $X$ suddenly becomes more natural. This gives us a foothold to start with and subsequent work is more motivated.

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Kayak

#13 h Nov 2, 2018, 9:57 AM • 1 Y

Y by Adventure10

Eh sorry if I'm reasoning circularly, but isn't it exceedingly trivial after $\sqrt{ab}$ inversion at $C$ :huh:

:huh:

?

Let the ray $CD$ hit the circumcircle of $\omega_{ABC}$ again at $H$ . Perform a $\sqrt{CA \cdot CB}$ inversion at $C$ followed by the reflection at $C$ -angel bisector. Now the problem reads:

USA 2013 TST P3, inverted wrote:

Let $\Delta CAB$ be a right angled traingle with $\angle CAB = 90$ . Let $H$ be the circumcentre of $\omega_{ABC}$ , $D$ be the $C$ -antipode and $E$ be the midpoint of arc $AB$ not containing $C$ .

Define $\ell_A, \ell_B$ to the lines through $H$ passing through the midpoint of $\bar{AC}, \bar{AB}$ respectively. For a point $X \in \vec{CH}$ outside $\Delta ABC$ , define $K_X \in \{ \ell_B \cap \omega_{ACH} \} , L_X \in \{ \ell_A \cap \omega_{BCH} \}$ such that $K_X, A$ are in the same side of $\vec{CH}$ (similar for $L_X$ ). Then $K_X, L_X, D, E$ are concyclic.

Claim $HK_X = HL_X$
Proof Define $K_Z$ to be the other intersection of $\ell_B$ with $\omega_{ACH}$ . $\ell_B$ is clearly perpendicular to $BC$ , so the circumcentre of $\omega_{ACH}$ lies on $\ell_A$ (and so does $H$ ). So $HK_X = HK_Z$ by symmetry.

By PoP $HA \cdot HX = HK_X \cdot HK_Z = HK_X^2 \Rightarrow HK_X = \sqrt{HA \cdot HX}$ . Similary $HL_X = \sqrt{HA \cdot HX}$ . The conclusion follows. $\blacksquare$

Claim $K_X, L_X, D, E$ is an isosceles trapezium.
Proof (Angles are NOT directed) Assume $AC < AB$ , and let $G$ be the antipode of $I$ . Then $\angle KHE = \angle GH \ell_B = \angle GHB = \angle \ell_B HB = 90 - \frac{\angle CHB}{2} = 90 - \angle CAB = \angle CBA = \angle CH\ell_A = \angle LHD$ . Coupled with $HI = HD$ and $HK = HL$ , we get $\Delta KHE \sim \Delta KHD$ , from which the conclusion follows. $\blacksquare$

Since any isosceles trapezium is a cyclic quad, we're done (?)

This post has been edited 1 time. Last edited by Kayak, Nov 2, 2018, 9:57 AM

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yayups

#14 h Oct 18, 2019, 6:45 AM • 2 Y

Y by srijonrick, Adventure10

$[asy] unitsize(0.9inches); pair B=dir(180); pair A=dir(0); pair C=dir(117); pair CC=2*foot(C,A,B)-C; pair D=(C+CC)/2; pair X=0.77*D+(1-0.77)*C; pair K=intersectionpoints(A--((X-A)*10+A),circle(B,abs(B-C)))[0]; pair KK=intersectionpoints(A--((X-A)*10+A),circle(B,abs(B-C)))[1]; pair L=intersectionpoints(B--((X-B)*1000+B),circle(A,abs(A-C)))[0]; pair LL=intersectionpoints(B--((X-B)*1000+B),circle(A,abs(A-C)))[1]; pair T=extension(KK,L,A,B); pair S=extension(K,L,A,B); pair BB=2*foot(0,B,X)-B; pair AA=2*foot(0,A,X)-A; pair J=extension(B,AA,A,BB); draw(A--B); draw(circumcircle(A,B,C)); draw(circle(B,abs(C-B))); draw(circle(A,abs(C-A))); draw(B--LL); draw(A--KK); draw(S--B); draw(J--CC); draw(J--B); draw(J--A); draw(LL--T); draw(KK--T); draw(LL--S); draw(K--S); dot("$A$",A,dir(A)); dot("$B$",B,dir(180+45)); dot("$C$",C,2*dir(C)); dot("$C'$",CC,dir(CC)); dot("$D$",D,dir(270-45)); dot("$X$",X,1.5*dir(60)); dot("$K$",K,1.5*dir(70)); dot("$K'$",KK,dir(130)); dot("$L$",L,dir(L)); dot("$L'$",LL,dir(LL)); dot("$S$",S,dir(S)); dot("$T$",T,dir(-90)); dot("$A'$",AA,1.6*dir(120)); dot("$B'$",BB,dir(65)); dot("$J$",J,dir(90)); [/asy]$
Let $\omega_A$ be the circle with center $A$ passing through $C$ and let $\omega_B$ be the circle with center $B$ passing through $C$ . Let $C'$ be the reflection of $C$ over $AB$ . Let $\{K,K'\}=AX\cap\omega_B$ and $\{L,L'\}=BX\cap\omega_A$ with $K$ and $L$ inside $(AB)$ . Also, let $A'$ and $B'$ be the second intersections of lines $AX$ and $BX$ with $(AB)$ respectively.

Note that $\angle BA'X=\angle BDX=\pi/2$ so $(A'XDB)$ is cyclic, and similarly $(B'XDA)$ is also cyclic. By radical axis along with $(AB)$ , we see that $A'B$ , $B'A$ , and $CC'$ concur at some point $J$ with
$JC\cdot JC'=JX\cdot JD=JA'\cdot JB=JB'\cdot JA.$ \
Let $T=K'L\cap KL'$ and $S=KL\cap K'L'$ . We'll show that $(DLK)\cap AB=T$ and that $\angle BCT=\angle ACT$ . Note that
$XK\cdot XK'=XC\cdot XC'=XL\cdot XL'$ so $\omega\equiv(KLK'L')$ is cyclic. We see that $BA'$ is the perpendicular bisector of $KK'$ and $AB'$ is the perpendicular bisector of $LL'$ , so $J$ is the center of $\omega$ .

Since $JC\cdot JC'=JX\cdot JD$ and $D$ is the midpoint of $CC'$ , we have that $(CC';JX)=-1$ . Therefore, the polar of $J$ with respect to $\omega_B$ passes through $X$ , but is also perpendicular to $BJ$ . Thus, the polar of $J$ with respect to $\omega_B$ is $KK'$ , so
$JK^2=JC\cdot JC'.$ Therefore, inversion in $\omega$ swaps the pairs $(C,C')$ , $(X,D)$ , $(B,A')$ , and $(A,B')$ .

By Brokard's theorem on $KLK'L'$ , we see that $ST$ is the polar of $X$ with respect to $\omega$ . Since $JX\cdot JD=JK^2$ and $DA\perp JX$ , we see that $AB$ is the polar of $X$ with respect to $\omega$ , so $S$ and $T$ lie on $AB$ . Furthermore,
$(ST;AB)\stackrel{L}{=}(KK';XA)=-1$ since $CC'$ is the polar of $A$ with respect to $\omega_B$ (fundamentally this is because $\omega_A$ and $\omega_B$ are orthogonal). Thus, $S$ and $T$ are on $AB$ and are in fact harmonic conjugates with respect to $AB$ .

We see that by the same application of Brokard's theorem that $S$ is on the polar of $T$ with respect to $\omega$ and vice versa, so by a well known lemma, we see that $(ST)\perp\omega$ . This means that if $\{C_1,C_1'\}\equiv(ST)\cap CC'$ , then $(C_1,C_1')$ are swapped under inversion about $\omega$ , since $(ST)$ should be fixed under the inversion. Therefore,
$JC_1\cdot JC_1'=JK^2=JC\cdot JC',$ so in fact $\{C_1,C_1'\}=\{C,C'\}$ . Thus, $\angle SCT=\pi/2$ . Therefore $CS$ and $CT$ are harmonic conjugates with respect to lines $CB$ and $CA$ , and furthermore $CS\perp CT$ . This can only happen if $CT$ and $CS$ are the internal and external angle bisectors of $CB$ and $CA$ , so we have $\angle BCT=\angle ACT$ , as desired.

All that's left to show is that $DTKL$ is cyclic. We see that $X$ and $D$ are inverses with respect to $\omega$ , and since $X=KK'\cap LL'$ , this implies that $D$ is the Miquel point of $KLK'L'$ . Thus, $D\in(TLK)$ , as desired. This completes the proof.

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#15 h Feb 13, 2020, 2:18 PM • 1 Y

Y by Adventure10

Nice blend of rich geometry configuration and projective geometry.
Take $H$ as the orthocenter of $\triangle AXB$ .
$\textbf{Claim 01.}$ $(H, X; C, C') = -1$ .
$\textit{Proof.}$ Notice that $HX \cdot HD = HA' \cdot HB = HC \cdot HC'$
$\textbf{Claim 02.} LKL'K'$ is cyclic.
$\textit{Proof.}$ We have $XK' \cdot XL = XC \cdot XC' = XL \cdot XL'$
$\textbf{Claim 03.}$ $H$ is the center of $KLK'L'$ .
$\textit{Proof.}$ Notice that the center of $KLK'L'$ must lie on both the perpendicular bisector of $KK'$ and $LL'$ which are $BA'$ and $AB'$ . This then follows by the fact that $AB' \cap A'B = H$ .

$\textbf{Claim 04.}$ $HK$ is tangent to $(K'CKC')$ .
$\textit{Proof.}$ Notice that $(H, X; C, C') = -1$ and the fact that $K'C'CK$ is harmonic (This follows by the fact that $AC$ tangent to $(K'CKC')$ .) Hence
$-1 = K(K', K; C, C') = (X, KK \cap CB; C, C')$ It follows that $HK$ is tangent to $BC$ .

Redefine $T = K'L \cap KL'$ . and $L'K' \cap LK = Y$ .
By Brocard Theorem on $K'L'KL$ , we have that $H$ is the orthocenter of $\triangle TXY$ .
Furthermore this gives us $HX \perp TY$ , or $TY \parallel AB$ .

$\textbf{Claim 05.}$ $A,B,T,Y$ are collinear. In fact, $(A,B;T,Y) = -1$ .
$\textit{Proof.}$ Pascal on $LLL'K'KK$ gives us $A,B, L'K' \cap LK = Y$ are collinear. Since we have $TY \parallel AB$ , then $A,B,T,Y$ are collinear. Now, notice that
$-1 = X(Y, TX \cap K'L' ; K', L') = (Y, T; A, B)$
$\textbf{Claim 06.}$ $\angle YCD = 90^{\circ}$ .
$\textit{Proof.}$ To prove this, notice that
$DY \cdot DT = DX \cdot DH = DA \cdot DB = DC^2$ Since $\angle YCT = 90^{\circ}$ and $(A,B;T,Y) = -1$ . Then, we must have $CT$ bisects $\angle ACB$ , which is what we wanted.

It suffices to prove that $D,T,K,L$ is cyclic.
$\textbf{Claim 07.}$ $D$ is the Miquel Point of $K'L'KL$ .
$\textit{Proof.}$ Notice that $(HL')^2 = HC \cdot HC' = HX \cdot HD$ . Since $X = KK' \cap LL'$ and $H$ is the center of $(KK'LL')$ , it follows that $D$ is the Miquel Point of $K'L'KL$ .
Therefore, this gives us $DTKL$ being cyclic.

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#16 h Oct 18, 2020, 10:46 PM

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Let $\overline{AL}$ and $\overline{BK}$ intersect at a point $M$ . Let $\omega_A$ and $\omega_B$ be the circles centered at $A$ and $B$ with radius $AC$ and $BC$ , respectively. Suppose that lines $AK$ and $BL$ intersect $\omega_B$ and $\omega_A$ at points $P$ and $Q$ , respectively. Since $KX \cdot XP = CX \cdot FX = LX \cdot QX$ , we find that $KLPQ$ is cyclic. Note that $\omega_A$ and $\omega_B$ are orthogonal, so $AK \cdot AP = AC^2 = AL^2$ , so $\overline{AML}$ is tangent to $(KLPQ)$ . Similarly, $\overline{BMK}$ is tangent to $(KLPQ)$ .

Note that $D$ is the miquel point of cyclic quadrilateral $KLPQ$ , so $T$ lies on line $PL$ and $QK$ . Furthermore, lines $KL$ and $PQ$ intersect a point $S$ on line $AB$ by Brokard's theorem.

Finally, $-1 = (PK;XA) \stackrel{Q}{=} (ST;AB)$ . Now consider the inversion around $T$ with radius $TC=TF$ . Note that this inversion swaps $\omega_A$ and $\omega_B$ , so it swaps $L$ and $P$ . Now $TD \cdot TS = TL \cdot TP = TC^2$ , so $\angle TCS = 90^\circ$ . This implies $\overline{CT}$ bisects $\angle ACB$ , as desired.

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#17 h Nov 22, 2020, 5:14 PM

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Let $C'$ be the reflection of $C$ over $AB$ . Then, let $E$ and $M$ be where $AX$ hits the circle centered at $B$ with radius $BC$ , and let $F$ and $N$ be where $BX$ hits the circle centered at $A$ with radius $AC$ . Let $AE>AM$ and $BE>BM$ . Note that $EX \cdot XM = C'X \cdot XC = FX \cdot XN$ and so $FENM$ is cyclic. Now, we see that the circle through $C$ with center $B$ has two tangents $AC$ and $AC'$ , meaning that $(AX; ME) = -1$ and $(BX; NF) = -1$ too. So, $EF$ , $MN$ and $AB$ concur at a point because such a point is needed to map $(AX; ME)$ to $(BX; NF)$ . Now, with one of those, let's say $(BX; NF) = -1$ , taking perspectivity with respect to $E$ gets that since $EM$ and $FN$ hit $AB$ at $T$ , $(PT;AB) = -1$ . Because symmetry holds meaning $C'$ is on the circumcircle of $PCT$ , then $\angle PCT = \angle PC'T = 90^{\circ}$ . Finally, $(PT;AB) = -1$ and $\angle PCT = 90^{\circ}$ gets that $\angle ACT = \angle BCT$ and we are done.

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#18 h Dec 7, 2020, 11:08 AM • 1 Y

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v_Enhance wrote:

Let $ABC$ be a scalene triangle with $\angle BCA = 90^{\circ}$ , and let $D$ be the foot of the altitude from $C$ . Let $X$ be a point in the interior of the segment $CD$ . Let $K$ be the point on the segment $AX$ such that $BK = BC$ . Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$ . The circumcircle of triangle $DKL$ intersects segment $AB$ at a second point $T$ (other than $D$ ). Prove that $\angle ACT = \angle BCT$ .

Solution.

Claim 1. If $\overline{AKX}\cap(\omega_B) = K' \ne K$ and $\overline{BLX}\cap(\omega_A) = L' \ne L$ , where $\omega_A$ and $\omega_B$ denotes the circles with radii $AC = AL$ and $BC = BK$ respectively. Then, we have $K,L,K',L'$ concyclic.

Proof: Since, $\{X\} \in \overline{CD}$ which's perpendicular to $\overline{AB}$ , so, $X$ lies on the radical axis of $\omega_A$ and $\omega_B$ . Hence, $|\text{pow}_{(\omega_A)}X| = |\text{pow}_{(\omega_B)}X| \implies XL \cdot XL' = XK \cdot XK'.\ \square$ Let $\Omega$ denote the circle here.

We re-define $T$ as: $\overline{K'L} \cap \overline{L'K} = \{T\} - (1)$ , we wish to show $\odot(DKL) \cap \overline{AB} =T \ne D.$ Next, we let $\overline{KL} \cap \overline{L'K'} = \{R\} - (2)$ and $J$ to be the orthocenter of $\triangle XAB.\ (\spadesuit)$

Claim 2. $J$ is the center of $\Omega.$

Proof. Since, $\overline{LL'}$ and $\overline{KK'}$ are the radical axes of $\left\{\omega_A, \Omega\right\}$ and $\left\{\omega_B, \Omega\right\}$ respectively, hence, $\overline{AJ}$ and $\overline{BJ}$ are perpendicular bisectors of $\overline{LL'}$ and $\overline{KK'}$ respectively, and we're through. $\square$

From $(1)$ and $(2)$ , by Brokard's, we get $\overline{RT}$ as the polar of $X.$ Now, we observe that $\overline{K'K'} \cap \overline{KK}=\{B\}$ and $\overline{L'L'} \cap \overline{LL}=\{A\}$ , hence, $\overline{AB}$ is the polar of $X$ . Thus, $R, B, T, A$ are collinear and $(A,B;T,R) = -1\ (\diamondsuit)$ . Since, $\angle BCA = 90^{\circ}$ , ergo, $\overline{CA}, \overline{CB}$ are bisectors of $\angle TCS$ .

Again by Brokard's, $J$ is the orthocenter of $\triangle TRX.$ Hence, by PoP WRT foot of altitude $D$ , with $J$ as the orthocenter of $\triangle TRX\ (\heartsuit)$ and $\triangle XAB\ (\spadesuit)$ gives: $DR \cdot DT \overset{(\heartsuit)}{=} DX \cdot DJ \overset{(\spadesuit)}{=} DA \cdot DB.\qquad(\clubsuit).$ At last, since $\angle BCA = 90^{\circ}$ , whence $DA \cdot DB = DC^2$ . So, from $(\clubsuit)$ , we have $DT \cdot DR=DC^2$ and thus, $\angle TCR = 90^{\circ}$ ; this along with $(\diamondsuit)$ , gives $\angle ACT = \angle BCT$ , and we're good.

Lemma.[Well known] Let $ABCDEF$ be a complete quadrilateral, where $\overline{AB}\cap\overline{CD} = \{E\}, \overline{AD}\cap\overline{BC} = \{F\}$ and $\overline{AC}\cap\overline{BD}=\{K\}.$ Then, $M$ , the Miquel point of the quadrilateral is the inverse of $K$ WRT $\odot(ABCD)$ . (Proof. Brokard suffices.)

Hence, we just need to show $T, K, L, D$ are concyclic. But before that, we note that, due to Orthocentric system surrounding $(\spadesuit)$ , we have $A, D, X$ and $\overline{AJ} \cap \overline{BX}$ concyclic, since $\overline{AJ} \perp \overline{LL'} \equiv \overline{BX}$ ; say $\overline{AJ} \cap \overline{BX} = \{Q\}$ . So, $JX \cdot JD = JQ \cdot JA = JL^2,$ and thus, $X, D$ are inverses of each other WRT $\Omega$ . Since, $\overline{LL'} \cap \overline{KK'}=\{X\}$ thus, by our lemma, we get $D$ to be the Miquel point of $KL'K'L$ ; hence, it belongs to $\odot(TLD)$ , and we're done. $\blacksquare$

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#19 h Dec 7, 2020, 11:32 AM

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anantmudgal09 wrote:

Let $N=\overline{AY} \cap \overline{BX}$ then $OL^2=ON \cdot OA=OX \cdot OD$ proving $X, D$ are inverses in $\omega$ . Hence $D$ is the desired spiral center.

What's $O$ here?? (By usual notations, it denotes the circumcenter of $\triangle ABC$ , but then how does that Power of Point hold?) It will be the center of $KL'KL'$ (in other words, the orthocenter of $XAB$ ), I guess :maybe:

:maybe:

; unless it's PoP WRT something else. Anyways, nice sol.

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GeronimoStilton

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GeronimoStilton

#20 h Feb 13, 2021, 4:28 AM

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Indeed, the solution relies on IMO 2012/5.

By inversion at $C$ , we get the following problem: $ACBD$ is a rectangle, $X$ lies on the extension of line $CD$ past $D$ , $(XAC)$ intersects the perpendicular bisector of $BC$ again at $K$ and $(XBC)$ intersects the perpendicular bisector of $AC$ again at $L$ , show $(DKL)$ passes through the arc midpoint of arc $AB$ not passing through $C$ . Let the perpendicular bisectors of $AC$ and of $BC$ intersect again at $O$ , which is the center of $(ABC)$ . Remark that the statement of IMO 2012/5 after this inversion would imply that a circle passing through $K$ and $L$ is tangent to both perpendicular bisectors, so we can eschew the information about $X$ and instead write $OK=OL$ . If we define $T$ as the arc midpoint of arc $AB$ not passing through $C$ , it suffices to show $DTKL$ is an isosceles trapezoid. As $OT=OD,OK=OL$ , it is sufficient to show $DT\parallel KL$ . Indeed, note that by a rotation by $90^\circ$ about $O$ , we have
$\measuredangle KOT=\measuredangle LOA=\measuredangle COL=\measuredangle DOL,$ hence we are done.

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#21 h Sep 21, 2021, 3:43 PM

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Let $K^*$ and $L^*$ be the inverses of $K$ and $L$ with respect to the circles they do not lie on. Observe that $K^*$ lies on the circle at $B$ and $L^*$ lies on the circle at $A$ because the two circles are orthogonal. In fact, radical axis implies $KL^*K^*L$ is cyclic. Next, we cliam that the center of this circle is $\overline{LL} \cap \overline{KK}$ . First we will show that this point also lies on $\overline{CD}$ .

Let $O$ be the polar of $\overline{BX}$ , which obviously lies on $\overline{CD}$ . Now $O$ lies on the polar of $L$ as well, implying $\overline{OL}$ is tangent to the circle at $A$ . This implies that $X$ is the orthocenter of $\triangle OAB$ , because $\overline{OX}$ is the polar of $B$ with respect to the circle with center $A$ . Hence, $\overline{AX} \perp \overline{BO}$ , and because $B$ lies on the polar of $A$ , $\overline{BO}$ is the polar of $A$ , and thus $\overline{AX}$ is the polar of $O$ . Hence $\overline{OK}$ is tangent to the circle at $B$ as well, proving the concurrence.

Now $\overline{AX}$ is the polar of $O$ with respect to the circle with center $B$ , implying that $OK = OK^*$ . Similarly, $OL = OL^*$ and thus $O$ is the center of $(K^*LKL^*)$ as required.

Next, denote $S = \overline{LK} \cap \overline{L^*K^*}$ . Since $\measuredangle ODB= \measuredangle OKB = 90^{\circ}$ , $BDKOK^*$ is cyclic, and so is $ADLLOL^*$ . It follows that $D$ is the Miquel Point of complete quadrilateral $KLK^*L^*$ . Henceforth both $\overline{KL^*} \cap \overline{LK^*}$ and $\overline{KL} \cap \overline{K^*L^*}$ lie on $\overline{AB}$ . Observe that the first point msut be $T$ because it lies on $(DLK)$ ; call the second point $S$ . The main claim is now the following.

Claim. $(ST;BA)=-1$ .

Project through $X$ . Then we need $(S, XT \cap \overline{L^*K^*}; L^*K^*) = -1$ which is obvious. $\blacksquare$

Hence it suffices to show $\angle SCT = 90^{\circ}$ by the bisectors lemma. Notice that $O$ lies on the radical axis of $(ABC)$ and $(ST)$ , and thus $\overline{OD}$ is the radical axis of these two circles, which passes through $C$ . Thus $(ST)$ passes through $C$ as well, and we are done.

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jrsbr

#23 h Mar 29, 2023, 8:53 PM

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Nice projective geometry.

First, let $R=\underbrace{\Gamma(A,AC)}_{\omega_a}\cap BX\ne L$ and $S=\underbrace{\Gamma(B,BC)}_{\omega_b}\cap AX\ne K$ . Since $\angle ACB=90^{\circ},$ we must have $\omega_a$ and $(ABC)$ orthogonal circles, and the same for $\omega_b$ and $(ABC)$ . This implies that
$(V,SL\cap AB,A,B)\overset{S}{=}(R,L,X,B)\overset{C}{=}(R,L,C,E)=-1,$ where $V=RS\cap AB$ . If we do the same for $(V,RK\cap AB,A,B)$ we can easily conclude that $SL$ , $RK$ and $AB$ concur. Furthermore, we have
$(LK\cap AB,T',A,B)\overset{K}{=}(L,R,X,B)=-1,$ from which we get $LK$ , $AB$ and $RS$ concur at $V$ .
Also, if we let $CD\cap (ABC)=E\ne C$ , we have $XR\cdot XL=XC\cdot XE=XD\cdot XK$ , concluding that $RSKL$ is a cyclic quadrilateral.
Since $X$ lies on $AD$ , we have $XD\perp VT'$ and therefore $D$ is $RSLK$ 's Miquel Point, and thus
$KLDT'\text{ is cyclic}\implies T=T'.$ Finally, see that $XAB$ is an auto polar triangle w.r.t (RSLK) and, by Brokard's theorem, $XVT$ is also an auto polar triangle. If we let $O$ to be the center of $(RSLK)$ , we conclude that $X$ is the orthocenter of $OVT$ and $OAB$ . This implies that $O$ lies on the radical axis of the circunferences with diameter $AB$ and $TV$ , which implies that $CEVT$ is cyclic.
Since $E$ is the reflection of $C$ w.r.t to $AB$ , we must have $\angle VCT=90^{\circ}$ , and since $(V,T,A,B)=-1$ it follows that $CT$ bisects $\angle BCA.$ $\blacksquare$

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#24 h Jul 5, 2023, 10:41 PM

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WLOG suppose that $CD = 1$ , and invert about the circle with diameter $CD$ , denoting the inverse of a point $P$ with $P'$ .
Then, let $C$ be the origin and $D = (\cos\theta, \sin\theta)$ in the Cartesian plane. We then deduce that $\begin{align*} A' &= (\cos\theta,0)\\ B' &= (0,\sin\theta)\\ X' &= (r\cos\theta, r\sin\theta),\quad r > 1. \end{align*}$
Claim: The equation of the inverse of the circle centered at $A$ with radius $AC$ is $x = \frac{1}{2}\cos\theta,$ and similarly, the inverse of the circle centered at $B$ with radius $BC$ has equation $y = \frac{1}{2}\sin\theta.$ Proof. Let $C^\ast$ be the reflection of $C$ over $A$ . Then the inverse of $C^\ast$ is the midpoint of $CA'$ , and the image of the circle goes through this point and is perpendicular to line $AC$ , which implies the result. The other half follows analogously. $\blacksquare$
Now, we will compute the circumcenter and circumcircle of $(B'CX')$ . The perpendicular bisector of $CX'$ has equation $-\cot\theta(x - \frac{1}{2} r\cos\theta) = y - \frac{1}{2} r\sin\theta,$ and the perpendicular bisector of $B'C$ is $y = \frac{1}{2}\sin\theta$ , so the $x$ -coordinate of their intersection is given by $-\cot\theta(x - \frac{1}{2} r\cos\theta) = \frac{1}{2}(1 - r)\sin\theta,$ which, upon solving, yields the point $\left(\frac{(r - 1)\sin^2\theta}{2\cos\theta}{2\cos\theta} + \frac{r\cos\theta}{2}, \frac{\sin\theta}{2}\right).$ We can clean up the $x$ -coordinate a little bit by rewriting it as $\frac{r\sin^2\theta - \sin^2\theta + r\cos^2\theta}{2\cos\theta} = \frac{r - 1 + \cos^2\theta}{2\cos\theta} = \frac{r - 1}{2\cos\theta} + \frac{\cos\theta}{2}.$ Then, the equation of the corresponding circle is $\left(x - \left(\frac{r - 1}{2\cos\theta} + \frac{\cos\theta}{2} \right)\right)^2 + \left(y - \frac{\sin\theta}{2}\right)^2 = \left(\frac{r - 1}{2\cos\theta} + \frac{\cos\theta}{2}\right)^2 + \frac{\sin^2\theta}{4}.$ To compute $L'$ , we want to intersect this circle with $x = \frac{1}{2}\cos\theta$ , which gives us the following: $\begin{align*} \left(\frac{\cos\theta}{2} - \left(\frac{r - 1}{2\cos\theta} + \frac{\cos\theta}{2} \right)\right)^2 + \left(y - \frac{\sin\theta}{2}\right)^2 &= \left(\frac{r - 1}{2\cos\theta} + \frac{\cos\theta}{2}\right)^2 + \frac{\sin^2\theta}{4}\\ \left(\frac{1 - r}{2\cos\theta}\right)^2 + \left(y - \frac{\sin\theta}{2}\right)^2 &= \left(\frac{r - 1}{2\cos\theta}\right)^2 + \frac{r - 1}{2} + \frac{\cos^2\theta}{4} + \frac{\sin^2\theta}{4}\\ \left(y - \frac{\sin\theta}{2}\right)^2 &= \frac{r - 1}{2} + \frac14 = \frac{2r - 1}{4}\\ y &= \frac{\pm\sqrt{2r-1}}{2} + \frac{\sin\theta}{2}. \end{align*}$ There is now a positive root and a negative root; $L'$ corresponds to the positive root because it's above line $CD$ , not below. Thus, $L'$ has coordinates $\left(\frac{\cos\theta}{2}, \frac{\sqrt{2r-1}}{2} + \frac{\sin\theta}{2}\right).$ Similarly, we can find that $K'$ has coordinates $\left(\frac{\sqrt{2r-1}}{2} + \frac{\cos\theta}{2}, \frac{\sin\theta}{2}\right).$ Now, let $\widehat{T}$ in the original picture be the intersection of the bisector of $\angle BCA$ with $AB$ (so we want to show that $D$ , $K$ , $L$ , $\widehat T$ are concyclic). Then $\widehat T'$ is the midpoint of arc $\widehat{A'DB'}$ on the circle with diameter $CD$ ; since $A'B'$ is a diameter of this circle too, this corresponds to a counterclockwise $90^{\circ}$ rotation of $A'$ about the center. This means $\widehat T' = \left(\frac{\sin\theta + \cos\theta}{2}, \frac{\sin\theta + \cos\theta}{2}\right).$ It now suffices to prove that $\widehat T'$ , $D$ , $K'$ , and $L'$ are concyclic. We will show that $D\widehat T'$ and $K'L'$ share a perpendicular bisector, implying that they are the bases of an isosceles trapezoid. To do this, observe that the midpoints of the aforementioned segments are $\begin{align*} M &= \left(\frac{\sin\theta + \cos\theta}{4} + \frac{\cos\theta}{2}, \frac{\sin\theta + \cos\theta}{4} + \frac{\sin\theta}{2}\right),\\ N &= \left(\frac{\sqrt{2r-1}}{4} + \frac{\cos\theta}{2}, \frac{\sqrt{2r-1}}{4} + \frac{\sin\theta}{2}\right) \end{align*}$ respectively; then vector $\overrightarrow{MN}$ , which is $\overrightarrow{NM} = \left(\frac{\sqrt{2r - 1}}{4} - \frac{\sin\theta + \cos\theta}{4}, \frac{\sqrt{2r - 1}}{4} - \frac{\sin\theta + \cos\theta}{4}\right),$ has a slope of 1, while segments $D\widehat{T}'$ and $K'L'$ both have slopes of $-1$ , so we are done!

This post has been edited 1 time. Last edited by ricegang67, Jul 6, 2023, 1:41 AM
Reason: typo

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AdventuringHobbit

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AdventuringHobbit

#25 h Jul 12, 2023, 7:40 PM

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Wait inversion at $C$ trivializes this. What?

Perform a $\sqrt{AC\cdot BC}$ inversion centered at $C$ followed by a reflection over the $C$ internal angle bisector. This map swaps $A$ and $B$ , it swaps $AB$ and $ABC$ . It swaps the point $D$ and the $C$ antipode $C'$ , it swaps $X$ with a point $X'$ on $CO$ , it swaps the circle centered at $B$ with radius $BC$ that is tangent to $AC$ at $C$ to a line $\ell_a$ parallel to $BC$ passing through the center $O$ of $(ABC)$ , and it swaps the circle centered at $A$ with radius $AC$ to a similarly defined line $\ell_b$ . Let $K_1$ and $K_2$ be the intersections of $(BXC)$ with $K_1$ being the point outside of triangle $ABC$ that $K$ inverts to. Define $L_1$ and $L_2$ similarly. The picture now looks like a rectangle. Since the center of $BXC$ must lie on the bisector of $BC$ which is $\ell_a$ , we must have that $OK_1=OK_2$ . Now PoP gives $CO\cdot XO = K_1O \cdot K_2O=K_1O^2$ . We can similarly get that $CO\cdot XO = L_1O^2$ . Thus $K_1O=L_1O$ , and $K_1$ and $L_1$ are on $\ell_b$ and $\ell_a$ , which are parallel to the sides $AC$ and $AB$ . We now apply Cartesian coordinates with $\ell_a$ and $\ell_b$ as the $x$ and $y$ axes. Now set $A=(x,y)$ . Since we have a rectangle, we compute $C=(-x,y)$ , $B=(-x,-y)$ , $C'=(x,-y)$ . Since $M$ is a $90$ degree rotation of $A$ about the origin, it has coordinates $(y,-x)$ . Since $OK_1=OL_1$ , we can set $L_1=(a,0)$ and $K_1=(0,-a)$ . Now we can easily check that $K_1L_1$ and $C'M$ have the same slopes so are parallel, and we can also check that $K_1M=L_1C'$ and $K_1C'=L_1M$ with the distance formula. These imply that $K_1L_1C'M$ is a cyclic isosceles trapezoid. Inverting back, we get that $LKDT'$ is cyclic, where we define $T$ as the intersection of the $C$ bisector and $AB$ . Thus, we are done.

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eibc

#26 h Dec 15, 2023, 2:32 AM

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Let $\omega_A$ be the circle with center $A$ and radius $AC$ , and $\omega_B$ be the circle with center $B$ and radius $BC$ . Denote by $K^{\ast}$ the second intersection of $KX$ with $\omega_B$ , and $L^{\ast}$ the second intersection of $LX$ with $\omega_A$ . Then because $\omega_A$ and $\omega_B$ are both symmetric about line $AB$ , we find that $CD$ is their radical axis, so $KK^{\ast}LL^{\ast}$ is cyclic.

Moreover, since $\omega_A$ and $\omega_B$ are orthogonal, we find that $\overline{BK}$ and $\overline{BK^{\ast}}$ are tangent to $\omega = (KK^{\ast}LL^{\ast})$ , so $KK^{\ast}$ is the polar of $B$ . Similarly, we see that $LL^{\ast}$ is the polar of $A$ , and since $X = \overline{LL^{\ast}} \cap \overline{KK^{\ast}}$ , by La Hire's we find that $AB$ is the polar of $X$ . Thus, $D$ is the inverse of $X$ wrt $\omega$ , so it must be the Miquel point of $KLK^{\ast}L^{\ast}$ , and $T = \overline{KL^{\ast}} \cap \overline{LK^{\ast}}$ . We also find that $S = \overline{KL} \cap \overline{K^{\ast}L^{\ast}}$ lies on $AB$ .

Now, by Brokard, we find that $S$ lies on the polar of $T$ wrt $\omega$ , so if $O$ is the center of $\omega$ , then $\text{Pow}_{(ST)}(O) = R_{\omega}^2$ . We also find that $A$ lies on the polar of $B$ , so $\text{Pow}_{(AB)}(O) = R_{\omega}^2$ , and hence line $OCD$ is the radical axis of $(ST)$ and $(AB)$ . Therefore $C$ lies on $(ST)$ , so $\angle SCT = 90^{\circ}$ . However, since
$-1 = (K, K^{\ast}; L, L^{\ast}) \overset{K}{=} (B, A; S, T),$ by the right angles and bisectors lemma we indeed have $\angle ACT = \angle BCT$ .

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#27 h Jan 12, 2024, 4:26 AM

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Let $K^*$ and $L^*$ be the inverses of $K$ and $L$ wrt $\omega_A$ and $\omega_B$ , the circles through $C$ centered at $A$ and $B$ , respectively.

$\omega_A$ and $\omega_B$ are orthogonal, so $K^*$ lies on $\omega_B$ and $L^*$ lies on $\omega_A$ .
$X$ lies on the radax of $\omega_A$ , and $\omega_B$ , so $KLK^*L^*$ is cyclic. Let its circumcircle be $\gamma$ .
$KLK^*L^*$ is also harmonic, with tangents from $K$ and $K^*$ meeting at $B$ and tangents from $L$ and $L^*$ meeting at $A$ , as
$AK \cdot AK^* = AC^2 = AL^2.$

Thus, Brocard on $KLK^*L^*$ tells us the polar of $X$ is $AB$ . Let $O$ be the center of $\gamma$ and $P = K^*L^* \cap KL$ . Now we note

$O$ is the orthocenter of $\triangle XPT$ , so $O$ lies on $CD$ .
Master Miquel says $D$ is the Miquel point of $KLK^*L^*$ , so $T = KL^* \cap K^*L$ .
Self-polar triangles tell us $\gamma$ is orthogonal to both $(PT)$ and $(AB)$ , so the radical axis of $(PT)$ and $(AB)$ is the altitude from $O$ to $AB$ .

Therefore $(ST)$ passes through $C$ . We finish using Apollonius, as $(PT;AB)=-1$ and $\angle PCT=90$ . $\blacksquare$

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gvole

#28 h Jan 28, 2024, 12:42 PM

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math154 wrote:

We know $X$ is on the radical axis $CD$ of the orthogonal circles $\omega_A = (A,AC)$ and $\omega_B = (B,BC)$ , so letting $\{K,K'\} = AX\cap \omega_B$ and $\{L,L'\} = BX\cap \omega_A$ , we see that $K,K',L,L'$ lie on a circle $\omega$ centered at $O$ . Since $OB\perp XA$ and $OA\perp XB$ , $O$ must be the orthocenter of $\triangle{XAB}$ ; in particular, $O\in CXD$ . Furthermore, $AL^2 = AL'^2 = AC^2 = AK\cdot AK'$ , so $BLXL'$ is the polar of $A$ (with respect to $\omega$ ) and $AKXK'$ is the polar of $B$ . Hence $\triangle{XAB}$ is self-polar.

Since $D$ lies on the polar of $X$ and $O,X,D$ are collinear, $D$ is the inverse of $X = KK'\cap LL'$ . It follows from the standard Yufei configuration that $D$ is the spiral center taking $LK$ to $K'L'$ (and $LK'$ to $KL'$ ), and $T = LK'\cap KL'$ (note that $DK'L'T$ is also cyclic, and that $LK'\cap KL'$ lies on the polar of $X$ ).

To complete the diagram, let $T' = LK\cap K'L'$ so $\triangle{XTT'}$ is self-polar by Brokard's theorem (by duality, we should have $CT'$ the external bisector of $\angle{BCA}$ ). But $-1 = K(K,K';L,L') = K(B,A;T',T)$ , so it suffices to show $\angle{TCT'} = 90^\circ$ . However, $O$ has fixed power $OX\cdot OD$ with respect to any polar triangle $\triangle{XYZ}$ through $X$ , and thus lies on the radical axis of the circles with diameters $AB,TT'$ . Since line $ABTT'$ is perpendicular to $OCD$ , we conclude that $\angle{TCT'} = 90^\circ$ , as desired. (Since $L\in (BX)$ , $K\in (AX)$ , we have $T'\notin (BA)\implies T\in (BA)$ .)

Comment. After the first two paragraphs, it's straightforward to complex bash with unit circle $\omega$ ( $(k+k')(l+l') = 2(kk'+ll')$ or $(k+k'-l-l')^2 = (k-k')^2 + (l-l')^2$ from $KLK'L'$ harmonic), with the observation that $c,2d-c$ are inverses with respect to $\omega$ : we get $1-\frac{c}{d} = \pm\frac{(k-k')(l-l')}{2(kk'-ll')}$ .

Another finish is to look at inversions at $A$ , $B$ , $T'$ and $T$ fixing $\omega$ . The fourth is the composition of the first three, so it fixes $C$ . This gives $\angle TCT'=90^{\circ}$ .

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#29 h Apr 24, 2025, 2:00 AM

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Let the image of $K$ after inversion centered at $A$ with radius $AC$ be $K^*.$ Since the circle of radius $BC$ centered at $B$ is orthogonal to this circle, $K^*$ lies on the circle centered at $B$ with radius $BC.$
Define $L^*$ similarly.

We have $AL^2 = AK\cdot AKK^* = A{L^*}^2$ by inversion, which implies $(KK^*;LL^*) = -1,$ and $KK^*LL^*$ cyclic.
Furthermore, $A$ is the pole of $LL^*$ while $B$ is the pole of $KK^*,$ implying that $X$ is the pole of $AB.$
Since $XD\perp AB,$ it follows that $D$ is the Miquel point of quadrilateral $KLK^*L^*.$
Since $T$ is the intersection of $X$ ’s polar with $(DKL),$ it follows that by Miquel Point properties, we have $TLK^*$ and $TKL^*$ are both collinear.
Let $AB, KL, K^*L^*$ concur at a point $V.$ They concur by radax on $(TDKL), (KLK^*L^*), (TDK^*L^*).$
Observe that $-1 = (KK^*;LL^*) \stackrel K= (BA;VT).$
By apollonian circles configuration, since $\angle VCT = 90,$ we have $CT$ bisects $\angle BCT$ as desired.

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