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Guessing Point is Hard

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Source: IMO Shortlist 2023 G5

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#1 h Jul 17, 2024, 12:01 PM • 9 Y

Y by OronSH, peace09, Rounak_iitr, buratinogigle, crazyeyemoody907, Funcshun840, ehuseyinyigit, GeoKing, wizixez

Let $ABC$ be an acute-angled triangle with circumcircle $\omega$ and circumcentre $O$ . Points $D\neq B$ and $E\neq C$ lie on $\omega$ such that $BD\perp AC$ and $CE\perp AB$ . Let $CO$ meet $AB$ at $X$ , and $BO$ meet $AC$ at $Y$ .

Prove that the circumcircles of triangles $BXD$ and $CYE$ have an intersection lie on line $AO$ .

Ivan Chan Kai Chin, Malaysia

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#2 h Jul 17, 2024, 12:01 PM • 5 Y

Y by BorivojeGuzic123, kinnikuma, Funcshun840, wizixez, Sadigly

Just bash it :yup:

:yup:

Set $(ABC)$ as the unit circle then we find:
$d=-\frac{ac}b ~,~x=\frac{c^2(a+b)}{c^2+ab}$ Now define the point $T$ such that:

$t=\frac{a^2+bc}{b+c}~,~$ clearly $T\in AO$

So if we prove that $B,D,X,T$ are concyclic we will be done by symmetry
and this is indeed true because:
$\frac{x-b}{t-b}\div\frac{x-d}{t-d}=\frac{a(c^2-b^2)}{c(a^2-b^2)}\in\mathbb{R}~~\blacksquare$ remark

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#3 h Jul 17, 2024, 12:04 PM • 8 Y

Y by peace09, OronSH, Math-48, ehuseyinyigit, Rounak_iitr, Funcshun840, Sedro, Kingsbane2139

Synthetic is also equally short

$[asy] size(7cm); defaultpen(fontsize(10pt)); import olympiad; import geometry; pair A = (1,3.2); pair B = (0,0); pair C = (4,0); pair O = circumcenter(A,B,C); pair D = 2*foot(O,B,foot(B,A,C)) - B; pair X = extension(C,O,A,B); pair T = 2*foot(circumcenter(B,O,C),A,O)-O; pair P = extension(X,X+B-D,A,O); draw(A--B--C--cycle, linewidth(1)); draw(circle(B,X,D), linewidth(0.7)+dashed); draw(circle(B,O,C), black); draw(A--T, linewidth(0.7)); draw(C--O, linewidth(0.7)); draw(O--X, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm)); draw(O--P, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm)); draw(circle(A,B,C), black); draw(O--B, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm)); draw(O--D, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm)); dot("$A$", A, dir(115)); dot("$B$", B, dir(179)); dot("$C$", C, dir(0)); dot("$O$", O, 1.5*dir(-108)); dot("$D$", D, dir(70)); dot("$X$", X, dir(154)); dot("$T$", T, dir(-56)); dot("$P$", P, dir(165)); [/asy]$
Let $AO$ intersect $\odot(BOC)$ again at $T$ . We claim that $T$ is the concurrency point.

Note that since angle bisector of $\angle AOC$ is perpendicular to $AC$ , lines $AO$ , $OX$ , and $BD$ form an isosceles triangle. Combining with $OB=OD$ , it follows that there exists point $P$ on $AO$ such that $BXPD$ is isosceles trapezoid. Clearly, $P\in\odot(BXD)$ . Now, notice that
$\measuredangle BTP = \measuredangle BTO = \measuredangle BCO = \measuredangle OBC = \measuredangle ABD = \measuredangle XBD = \measuredangle BDP,$ so $T\in\odot(BXDP)$ . Analogously, $T\in\odot(CYE)$ , so we are done.

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#4 h Jul 17, 2024, 12:04 PM

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Let $AO\cap (BOC)=\{O,K\}$ and $H$ be the orthocenter. We will show that $K$ lies on both $(BXD)$ and $(CYE)$ .
Take the inversion centered at $A$ with radius $\sqrt{AB.AC}$ and reflect according to the angle bisector of $A$ . We have $K\leftrightarrow H, \ B\leftrightarrow C,\ Y\leftrightarrow Y^*, \ E\leftrightarrow E^*$
Since $O^*$ is the reflection of $A$ with respect to $BC$ , we get $Y^*C\parallel BO$ and $Y^\in AB$ . Also $\angle E^*AC=\angle BAE=90-\angle B$
Let's show that $Y^*,B,H,E^*$ are cyclic and the other is similar.
Since $ACE^*\sim AHB,$ we have
$\frac{CE^*}{HB}=\frac{AC}{AH}=\frac{\sin B}{\cos A}=\frac{Y^*C}{Y^*B}\implies \frac{Y^*B}{BH}=\frac{Y^*C}{CE^*}$ Also $\angle HBY^*=90+\angle A=\angle E^*CY$ hence $Y^*BH\sim Y^*CE$ . Thus, $\angle Y^*HB=\angle Y^*E^*C=\angle Y^*E^*B$ as desired. $\blacksquare$

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#5 h Jul 17, 2024, 12:33 PM

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Let $N = AO \cap (BOC)$ . We show that $N$ is the desired intersection point.
$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair O = (-2.74,1.36); pair A = (-0.22,4.82); pair B = (-6.49312,-0.69816); pair C = (0.94790,-0.81287); pair X = (-3.58643,1.85871); pair D = (1.52203,0.96367); pair N = (-6.90682,-4.36111); pair P = (-6.42790,3.53287); pair Q = (-2.16161,2.15412); import graph; size(10cm); draw(circle(O, 4.28042), linewidth(1)); draw(A--B, linewidth(1)); draw(B--C, linewidth(1)); draw(C--A, linewidth(1)); draw(circle((-2.80673,-2.96935), 4.32986), linewidth(1)); draw(A--N, linewidth(1)); draw(P--C, linewidth(1)); draw(D--B, linewidth(1)); draw(circle((-1.81924,-3.08088), 5.24619), linewidth(1) + linetype("4 4")); dot("$O$", O, dir((3, -10))); dot("$A$", A, dir((8.000, 20.000))); dot("$B$", B, dir((-10, 0))); dot("$C$", C, dir((10, 0))); dot("$X$", X, dir((0, 20))); dot("$D$", D, dir((15, 15))); dot("$N$", N, dir((-49.318, -3.889))); dot("$P$", P, dir((-10, 10))); dot("$Q$", Q, dir((2, -5)));[/asy]$
Let $P = CO \cap \omega$ , and $\ell$ be the perpendicular bisector of $BD$ . Notice $A, P$ are symmetric wrt $\ell$ . Let $Q$ be a point symmetric with $X$ wrt $\ell$ , which clearly lies on $AO$ . Since
$\measuredangle QDB = \measuredangle DBX = \measuredangle CBO = \measuredangle ONB,$ we get that $B, X, Q, D, N$ are concyclic. From this, we conclude $N = (BXD) \cap (CYE)$ . $\ \ \blacksquare$

This post has been edited 2 times. Last edited by pokmui9909, Jul 17, 2024, 12:44 PM

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#6 h Jul 17, 2024, 12:33 PM • 1 Y

Y by Funcshun840

To solve the problem, we need to solve this subproblem:
“Let $ABC$ be a triangle.
$AO$ meets $(BOC)$ again at $T$ and $OC$ meets $AB$ at $X$ .
$(XBT)$ meets $(ABC)$ again at $D$ .
Prove that $AC \perp BD$ cyclic.”

https://i.imgur.com/zH9EHA5.png

Let $I$ be the circumcenter of $(TBXD)$ .
We must show $OI \parallel AC$

Claim 1: $XOIT$ is cyclic.
Proof:
By angle-chase and incenter-excenter lemma,
$\angle XIT = 2\angle XDT = 2(180 - \angle XBT) = 2\angle ABC = \angle AOC = \angle XOT$
as desired. $\square$

Claim 2: $OI \parallel AC$
Proof:
By angle-chase,
$\angle COI = \angle XTI = 90 - \angle XIC/2 = 90 - \angle XDT = 90 - \angle ABC = \angle OCA$ ,
as desired. $\square$ .

Now, $BO=OD$ and $BI=ID \implies OI \perp BD$
By claim 2 and since $OI \perp BD$ , we have $AC \perp BD$ and we are done. $\square$

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#7 h Jul 17, 2024, 1:27 PM

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I claim that the intersection point is the second intersection, $T$ , of the line $AO$ with the circle $(BOC)$ (other than $O$ ). We will now employ complex numbers with $(ABC)$ as the unit circle.

We know that $T \in AO$ , so $t = \lambda a$ , where $\lambda \in \mathbb R$ . Moreover, from $T \in (BOC)$ , we know that $\frac{t-b}{t-c} \div \frac{o-b}{o-c} \in \mathbb R \Longleftrightarrow \frac{c(\lambda a - b)}{b(\lambda a - c)} \in \mathbb R \Longleftrightarrow \frac{c(\lambda a - b)}{b(\lambda a - c)} = \frac{\lambda b - a}{\lambda c-a}$ Now, we solve for $\lambda$ , which is given by $\begin{align*} c(\lambda a-b)(\lambda c - a) &= b(\lambda a-c)(\lambda b - a) \\ ac^2\lambda^2 - (a^2c+bc^2)\lambda + abc &= ab^2\lambda^2 - (a^2b+b^2c) \lambda + abc \\ a(c-b)(c+b)\lambda &= a^2(c-b) + bc(c-b) \\ \lambda &= \frac{a^2+bc}{a(b+c)} \end{align*}$ Thus, we see that $t = \frac{a^2+bc}{b+c}$ . Moreover, we know that $e = -ab\bar c$ . We compute for the point $Y$ , it is the intersection of $AC$ and the line passing through $B$ and its antipodal point. So we see that $y = \frac{ac(b+(-b))+b^2(a+c)}{ac+b^2} = \frac{b^2(a+b)}{ac+b^2}$
Finally, we show that $T, Y, E, C$ are concyclic: $\frac{t-e}{t-c} \div \frac{y-e}{y-c}\in \mathbb R$ After some tedious computation you'll check that this is true.

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#8 h Jul 17, 2024, 1:31 PM • 12 Y

Y by khina, peace09, Seicchi28, CyclicISLscelesTrapezoid, AlephG_64, Assassino9931, sami1618, Kosiu, SerdarBozdag, pingupignu, Sedro, Funcshun840

My proposal

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#9 h Jul 19, 2024, 5:28 AM • 1 Y

Y by Funcshun840

Here is a solution with lengths that sheds some light on how one might guess the correct point.

Part I: Setup. First note that $D$ and $E$ are reflections over $\overline{AO}$ . Let $S$ and $R$ be the reflections of $B$ and $C$ over $\overline{AO}$ ; we claim that the desired point of intersection is $P = \overline{BR}\cap \overline{CS}$ , which clearly lies on $\overline{AO}$ . We will show that $P\in (CYE)$ , and by symmetry we will also have $P\in (BXD)$ .

Observe that $\triangle BAD\sim \triangle BYC$ and $\triangle CAX\sim \triangle CEB$ , so we can write
$\frac{CY}{AD} = \frac{BC}{BD}, \quad \frac{BX}{AE} = \frac{BC}{CE} \implies \frac{CY}{CE} = \frac{BX}{BD}.$ Since $\angle XBD = \angle YCD = 90^\circ - \angle A$ , this implies $\triangle BXD \sim \triangle CYE$ .
$[asy] defaultpen(fontsize(10pt)); size(300); pair A, B, C, H, O, D, E, X, Y, R, S, P; A = dir(115); B = dir(210); C = dir(330); H = orthocenter(A, B, C); O = (0,0); D = 2*foot(H, A, C) - H; E = 2*foot(H, A, B) - H; X = extension(C, O, A, B); Y = extension(B, O, A, C); R = 2*foot(C, A, O) - C; S = 2*foot(B, A, O) - B; P = extension(B, R, C, S); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(B--Y^^C--X^^A--P, grey+dotted); draw(B--X--D--cycle, lightblue+dotted); draw(B--P--S, magenta+dashed); draw(C--Y--E--cycle, lightblue+linewidth(0.9)); draw(B--X--D--cycle, lightblue+linewidth(0.9)); draw(circumcircle(C, Y, E), heavycyan+dashed); draw(circumcircle(B, X, D), heavycyan+dashed); draw(B--P--S, magenta+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(200)); dot("$C$", C, dir(0)); dot("$D$", D, dir(70)); dot("$E$", E, dir(150)); dot("$X$", X, dir(150)); dot("$Y$", Y, dir(60)); dot("$R$", R, dir(R)); dot("$S$", S, dir(S)); dot("$P$", P, dir(P)); dot("$O$", O, dir(250)); [/asy]$

Remark: [Motivation for $P$ ] Here's how one might guess the correct construction of $P$ after showing $\triangle BXD\sim \triangle CYD$ . The line $AO$ is the perpendicular bisector of $\overline{DE}$ , so we want the intersection $P = (CYE)\cap (BXD)$ to satisfy $PD = PE$ . The similarity gives a nice way to compute $PD$ and $PE$ : we have
$\frac{PD}{PE} = \frac{R_{(BXD)}\sin \angle DBP}{R_{(CYE)}\sin \angle ECP} = \frac{BD}{CE} \cdot \frac{\sin \angle DBP}{\sin \angle ECP},$ so we want $P$ to satisfy $\sin \angle DBP / \sin \angle ECP = CE/BD$ . This motivates the construction of $R$ and $S$ since $CE = DR$ and $BD = ES$ give this ratio automatically.

Part II: Length Calculation. The key claim is the following:

Claim: We have the similarity $\triangle XAD\sim \triangle PCE$ .

Proof. Using the reflections over $\overline{AO}$ , we have
$\angle PCE = 180^\circ - \angle SCE = 180^\circ - \angle BCD = \angle XAD,$ so it suffices to prove that
$\frac{PC}{AX} = \frac{CE}{AD} \iff PC\cdot AD = AX\cdot CE \iff PC\cdot AD = AC\cdot BE,$ where we've used $\triangle CAX\sim \triangle CEB$ in the last step. But we have $AD = AE$ and $\triangle AEB\sim \triangle ACP$ (using $\angle AEB = \angle ACP = 180^\circ - \angle C$ and $\angle BAE = \angle PAC = 90^\circ - \angle B$ ), which gives
$\frac{AE}{AC} = \frac{BE}{PC} \implies PC\cdot AE = AC\cdot BE \implies PC\cdot AD = AC\cdot BE,$ as needed. $\blacksquare$

Now $\angle CPE + \angle CYE = \angle AXD + \angle BXD = 180^\circ$ , so $P\in (CYE)$ . This completes the proof.

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#10 h Jul 20, 2024, 7:49 AM

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Let $\Psi$ be the inversion centered at $A$ with radius $\sqrt{\frac{AB \cdot AC}{2}}$ and reflect according to the angle bisector of $A$ .
And Let $M,N$ is midpoint of $AB, AC$ respectively and $P$ is foot of the altitude from $A$ , $H$ is Orthocenter of $\Delta ABC$
Then $\Psi (B)=N, \Psi(C)=M, \Psi(Y)=AB \cap \odot(ANP)$ and $\Psi(E)$ is on $MN$ and $\measuredangle \Psi(E)AP=\measuredangle CAB$
And let $T$ is midpoint of $AH$ , $X = MN\cap \odot (ANP)$ , $F = X\Psi(Y) \cap AH$ , $K=AP \cap MN$ , $\Psi(\odot (CEY))=\odot (M\Psi(E)\Psi(Y))$
Since nine-point circle and Reim, we have $AX \parallel NP$
By angle chasing, we have $X,F,A,\Psi(E)$ is cyclic
Also, we have $M,\Psi(Y),F,\Psi(E)$ is cyclic
And by Reim, we have $F,\Psi(E),T,M$ is cyclic
So, $\Psi(E), \Psi(Y), M, T$ is cyclic
Similarly $\Psi(D), \Psi(X), N, T$ is cyclic
Since $\Psi(AO)=AH$ , we have $\Psi(T)(=\odot (BXD) \cap \odot (CYE)$ is on $AO$

This post has been edited 2 times. Last edited by foolish07, Jul 20, 2024, 7:53 AM

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#11 h Jul 21, 2024, 12:32 AM

Y by

Solution

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#12 h Jul 21, 2024, 3:08 PM • 1 Y

Y by Funcshun840

Very Beautiful! I think same point as ISL 2022 G6

:

Let the reflections of $BC$ over $AB$ and $AC$ meet at $Z$ ( $A$ is the $Z$ -excenter). Let $I$ be the incenter of $ZBC$ thus $Z$ , $I$ , $O$ , and $A$ are collinear. We show that $Z$ is the desired intersection point. It is sufficient to show that $Z$ lies on $(BXD)$ .

Let the parallel line to $CI$ through $Z$ meet $(CIZ)$ again at $P$ . As $CIZP$ and $CIBD$ are both isosceles trapezoid, $ZPDB$ is also an isosceles trapezoid (thus cyclic). Notice we also have that $P$ , $C$ , and $X$ are collinear as $\angle PCI+\angle ICB+\angle BCO=\angle CIZ+180^{\circ}-\angle A-\angle C=180^{\circ}$ Also $PZBX$ is cyclic as $\angle XPZ=\angle XCI=\angle B=180^{\circ}-\angle XBZ$

Attachments:

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#13 h Jul 22, 2024, 1:39 AM

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Let $AO$ intersect $(BOC)$ again at $S$ . We will show that both $(BXD)$ and $(CYE)$ passes through $S$ .

Let $S'$ be the reflection of $S$ over $\ell$ . Then $BSS'D$ is an isosceles trapezoid, and also $S'$ lies on line $OX$ as $\ell$ is the external angle bisector of $\angle AOC$ . By symmetry, we see that $\triangle BOS \cong \triangle DOS'$ . Therefore,
$\measuredangle XBD = \measuredangle OBC = \measuredangle BCO = \measuredangle BSO = \measuredangle OS'D = \measuredangle XS'D,$ which shows that $XBS'D$ is cyclic. Therefore, combining the two, we see that $BXDS$ is cyclic. Similarly, $CYES$ is also cyclic, so we are done.

Some remarks: Personally, this is the most enjoyable problem in the G shortlist this year, barring G8 which I couldn't solve last year. Somehow it reminds me a lot of 2013 ISL G2.This is also the fastest I've ever solved a G5 (<10 mins) because I managed to guess the point $S$ right away.

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#14 h Jul 23, 2024, 4:14 PM • 2 Y

Y by ehuseyinyigit, Amy_Chen

Similarity :w00tb:

:w00tb:

:wow:

:w00t:

Let $Z = AO \cap (OBC)$ , $BD \cap AZ = P$ and $CE \cap AZ = Q$ .

$(1)$ $BZP \sim XCB \sim ACE$ because $\angle ACE = \angle OCB = \angle OZB$ and $\angle AEC = \angle ABC = \angle OBC + \angle OBA = \angle PBA + \angle BAP = \angle BPZ$ .

$(2)$ $ABC \sim APD$ because $\angle OAD = 90 -\angle ABD = \angle BAC$ and $\angle ADB = \angle ACB$ .

$(3)$ $AE = AD$ because $E$ and $D$ are reflections of $H$ over $AB$ and $AC$ .

Claim: $ZBX \sim ZPD$
Proof. $\angle ZBX = \angle ABC + \angle ZBC = \angle B + \angle ZOC = 180 - \angle B = 180 - \angle BPZ = \angle ZPD$ . Lastly,
$\frac{BX}{PD} \overset{3} = \frac{BX}{EA} \cdot \frac{AD}{PD} \overset{1,2} = \frac{CX}{CA} \cdot \frac{AC}{BC} = \frac{CX}{BC} \overset{1} = \frac{ZB}{ZP}$ which is sufficient. $\square$

The claim gives $\angle BXZ = \angle BDZ$ which shows $Z \in (BXD)$ . By symmetry, $Z \in (CYE)$ so we are done.

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#16 h Jul 24, 2024, 5:35 PM • 1 Y

Y by sami1618

Angle Chase:

Let $T = AO \cap (BOC)$ . We will show that $T$ is our desired point. It suffices to prove that $BXDT$ is cyclic as the other part can be proved similarly.

First, let $C^\prime$ be the antipode of $C$ in $\omega$ , and $Q = DC^\prime \cap AO$ .

Claim: $AQXC^\prime$ is cyclic.
Proof: $\angle XC^\prime Q = \angle CC^\prime D = \angle CBD = \angle ABO = \angle BAO = \angle XAQ$ . $\blacksquare$

Claim: $BXQD$ is cyclic.
Proof: Since, $AQXC^\prime$ is cyclic

$\begin{align*} \angle AXQ &= AC^\prime Q = \angle AC^\prime D = \angle ABD = \angle CBO\\ &= \angle OCB = \angle C^\prime CB = \angle C^\prime DB\\ &= \angle QDB. \blacksquare \end{align*}$
To finish the problem, observe that: $\angle BTQ = \angle BTO = \angle BCO = \angle BCC^\prime = \angle BDC^\prime = \angle BDQ$ . Therefore, making $BXQDT$ cyclic. Hence, we are done.

Nice Problem

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ddami

#17 h Jul 28, 2024, 12:34 AM

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I could not figure out a nice way to construct the point we wanted, so I came up with something rather different.

Let $H$ be the orthocenter of $ABC$ and let $AH$ meet $\omega$ at $P$ . Let $O_1$ and $O_2$ be the circumcenters of $CEY$ and $BDX$ respectively. Note that $O_1E = O_1Y$ , $OE = OA$ and
$\angle EO_1Y = 2\angle ECY = 2\angle EOA.$ Thus $E$ is the center of spiral similarity that sends $O_1Y$ to $OA$ . Hence it also sends $O_1O$ to $YA$ . Similarly $D$ sends $O_2O$ to $XA$ . Now notice that
$\frac{O_1O}{YA} = \frac{EO}{EA} = \frac{DO}{DA} = \frac{OO_2}{AX}.$ Furthermore $\angle O_1OO_2 = \angle YAX$ , thus $OO_1O_2 \sim AYX$ . Moreover, note that line $AO$ divides $\angle O_1OO_2$ in the same way line $AP$ divides $\angle YAX$ . Thus we might consider $T$ in $AO$ , with $O$ between $A$ and $T$ , such that quadrilaterals $OO_1TO_2$ and $AYPX$ are similar. Now note that
$\frac{XP}{O_2T} = \frac{AX}{OO_2} = \frac{XD}{OO_2}.$ Since $XP = XD$ we thus obtain $O_2T = OO_2$ which means $T \in (BDX)$ . Similarly we get that $T \in (CEY)$ .

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#18 h Jul 28, 2024, 6:38 AM

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We claim that the desired point of concurrency is the second intersection of $AO$ with $(BOC)$ . Call this point $P$ . We shall show that $PYCE$ is cyclic, from which the result follows by symmetry.

Since there are many circles passing through $C$ , let us invert with radius $\sqrt{CA\cdot CB}$ and reflect about the $\angle C$ -angle bisector. Then we have: $A$ and $B$ swap places, $E$ is sent to the intersection of $CO$ with $AB$ , $O$ is sent to the reflection of $C$ across $AB$ , and $Y$ is the intersection of $(B^*O^*C)$ with $A^*C^*$ . The task is therefore to show that $B^* O^*$ , $Y^* E^*$ and $(CA^*O^*)$ are concurrent. Relabelling points for convenience, we have the following equivalent problem:

ISL 2023/G5, inverted wrote:

Let $\triangle ABC$ have circumcenter $O$ , and suppose $A'$ is the reflection of $A$ across $BC$ . Let $AB$ intersect $(ACA')$ at $D$ and $A'C$ intersect $(ABA')$ at $E$ . Show that $DE$ , $BC$ and $AO$ are concurrent.

$[asy] size(8cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair AA = 2*foot(A,B,C)-A; path ABAA = circumcircle(A,B,AA); path ACAA = circumcircle(A,C,AA); pair D = intersectionpoints(A--B+10*(B-A),ACAA)[1]; pair E = intersectionpoints(AA--C+10*(C-AA),ABAA)[1]; pair O = circumcenter(A,B,C); draw(A--B--C--A--cycle); draw(B--AA--C); draw(B--D--E--C); draw(ABAA); draw(ACAA); pair P = extension(A,O,B,C); draw(A--P); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$A'$",AA,dir(AA)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$P$",P,dir(P)); dot("$O$",O,dir(45)); [/asy]$
It would be great if someone could provide a synthetic solution, it doesn't seem too hard to find but I can't seem to do it.

Anyway, here's a bary bash. Set $\triangle ABC$ to be the reference triangle, and denote $S_A = \frac{-a^2+b^2+c^2}{2} = bc\cos A$ and likewise $S_B$ and $S_C$ for convenience. Note that $S_A+S_B=c^2$ , as well as cyclic variants. Let $AO$ intersect $BC$ at $P$ .

First, we have $O=(a^2 S_A:b^2 S_B: c^2 S_C)$ so $P = (0:b^2 S_B : c^2 S_C)$ . Next, the foot from $A$ to $BC$ is $(0:b\cos C:c\cos B) = (0:S_C : S_B) = (0, \frac{S_C}{a^2},\frac{S_B}{a^2})$ . Therefore,
$A' = 2\left(0,\frac{S_C}{a^2},\frac{S_B}{a^2}\right)-(1,0,0) = \left(-1,\frac{2S_C}{a^2},\frac{2S_B}{a^2}\right) = (-a^2: 2S_C : 2S_B).$ Now we find the equation for $(ABA')$ . The generic equation is $-a^2 yz - b^2 zx - c^2 xy + (x+y+z)(ux+vy+wz)=0$ . Substituting in $A=(1,0,0)$ , we have $u=0$ . Similarly, $B=(0,1,0)$ gives $v=0$ . To find $w$ , we substitute $A'$ to get
$-a^2\cdot 4S_B S_C + b^2 \cdot 2a^2 S_B + c^2 \cdot 2a^2 S_C + a^2 \cdot w \cdot 2S_B = 0.$ Cancelling $a^2$ and simplifying, we have
$w\cdot S_B = 2S_B S_C - b^2 S_B - c^2 S_C = 2S_B S_C - (S_A+S_C) S_B - (S_A+S_B)S_C = -S_A(S_B+S_C) = -a^2 S_A.$ Therefore,
$w = -\frac{a^2 S_A}{S_B}.$ Similarly, the equation for $(ACA')$ is
$-a^2 yz - b^2 zx - c^2 xy + (x+y+z)\left(\frac{a^2 S_A}{S_C}\right)y = 0.$
Let $D = (k,1-k,0)$ . Then
$-c^2 k(1-k) - (1-k)\left(\frac{a^2 S_A}{S_C}\right) = 0 \quad\Rightarrow\quad k = -\frac{a^2 S_A}{c^2 S_C}.$ Therefore, $D = (-a^2 S_A: a^2S_A + c^2 S_C : 0)$ .

The strategy now is to intersect $DP$ and $A'C$ and show that this intersection lies on $(ABA')$ .

The line $DP$ is given by
$0 =\begin{vmatrix}x & y & z \\ -a^2 S_A & a^2 S_A + c^2 S_C & 0 \\ 0 & b^2 S_B & c^2 S_C \end{vmatrix} = c^2 S_C(a^2 S_A+c^2 S_C)x + a^2 c^2 S_A S_C y - a^2 b^2 S_A S_B z.$ The line $A'C$ is given by
$0 = \begin{vmatrix}x& y & z \\ -a^2 & 2S_C & 2S_B \\ 0& 0 & 1 \end{vmatrix} = 2S_C x + a^2 y.$ The intersection of these two lines is
$\begin{align*} &\left(a^4 b^2 S_A S_B : -2a^2 b^2 S_A S_B S_C : a^2 c^2 (a^2 S_A S_C + c^2 S_C^2 - 2S_A S_C^2)\right) \\&= \left(a^2 b^2 S_A S_B : -2b^2 S_A S_B S_C : c^2 S_C(a^2 S_A + c^2 S_C - 2S_A S_C)\right) \\&= \left(a^2 b^2 S_A : -2b^2 S_A S_C : c^2 S_C(S_A+S_C)\right) \\&= \left(a^2 S_A : -2S_A S_C: c^2 S_C\right). \end{align*}$ Now we verify that this point lies on $(ABA')$ , which has equation $-a^2 yz - b^2 zx - c^2 xy - \frac{a^2 S_A}{S_B}(x+y+z)z = 0$ .
$\begin{align*} &2a^2 c^2 S_A S_C^2 - a^2 b^2 c^2 S_A S_C + 2a^2 c^2 S_A^2 S_C - \frac{a^2 S_A}{S_B}\cdot S_B(S_A+S_C)\cdot c^2 S_C \\&= 2a^2 c^2 S_A S_C(S_A+S_C) - a^2 c^2 S_A S_C (S_A+S_C) - a^2 c^2 S_A S_C (S_A+S_C) = 0 \end{align*}$ by using the fact that $S_A+S_C=b^2$ .

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Kosiu

#19 h Aug 4, 2024, 11:15 PM

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I'll show that the desired point of concurrency is the second intersection of $AO$ with circle $BOC$ . Let's call this point $P'$ . I'll show that $P'CYE$ is cyclic.

It's equivalent to $\angle P'YE = \angle CYO$ which is the same as $\angle CYP' = \angle BYE$ .

Now let's define a bunch of useful points:

$R$ is the second intersection of $BO$ with circle $ABC$
$G$ is orthogonal projection of $P'$ onto $AC$
$T$ is orthogonal projection of $A$ onto $BC$
$S$ is the second intersection of $AP'$ with circle $GYP'$
$K$ is point on segment $AB$ such that $KY||AR$

See that $\triangle AGP'$ ~ $\triangle ATB$ ~ $\triangle REB$

Now I will show that point $S$ in $\triangle AGP'$ is the same point as $Y$ in $\triangle REB$ , by proving: $RY/YB = AS/SP'$ .

See that $RY/YB = AK/KB$ (because $KY||AR$ )

It's left to show that $KS||BP'$ , but fortunately it's easy by simple angle chasing, because $\angle YKA = 90$ and $AYSK$ is cyclic.

$\angle AP'B = \angle OCB = 90 - \angle BAC$ and $\angle ASK = \angle AYK = 90 - \angle BAC$

Then $\angle GSP' = \angle BYE$ but $\angle GSP' = \angle GYP'$

That finishes the proof.

I enjoyed this problem:)

Attachments:

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crazyeyemoody907

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crazyeyemoody907

#20 h Aug 14, 2024, 11:39 PM • 1 Y

Y by GeoKing

$[asy] //23slg5 init //setup defaultpen(fontsize(10pt)); size(6cm); pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); blu1=RGB(233,242,255); blu2=RGB(212,227,255); lightpurple=RGB(234,218,255);// blu1 lighter //define int i=0; pair inverse(pair P) {return P/(P.x*P.x+P.y*P.y); } pair O,A,B,C; O=(0,0); A=(-.6,.8); B=(-.8,-.6); C=(.8,-.6); pair X,Y,Z,X1,Y1,Z1; X=extension(A,O,B,C); Y=extension(B,O,A,C); Z=extension(C,O,A,B); X1=inverse(X); Y1=inverse(Y); Z1=inverse(Z); //draw filldraw(A--B--C--cycle,blu1,blu); draw(A--X1^^B--Y,blu); draw(circle(O,1),blu); draw(circumcircle(B,O,C),purple); draw(circumcircle(C,X1,Y),magenta); draw(C--2*foot(C,A,B)-A-B-C,magenta+dashed); clip((1.2,1.2)--(1.2,-1.2)--(-.4,-1.2)--(-1.2,-.4)--(-1.2,1.2)--cycle); //label void pt(string s,pair P,pair v, pen a){filldraw(circle(P,.015),a,linewidth(.1)); label(s,P,v);} string labels[]={"$O$","$A$","$B$","$C$","","$Y$",""}; //7 pair points[]={O,A,B,C,X1,Y,2*foot(C,A,B)-A-B-C}; real dirs[]={-100,120,-160,-20,0,60,0}; pen colors[]={blu,blu,blu,blu,purple,magenta,magenta}; for (i=0; i<7; ++i) {pt(labels[i],points[i],dir(dirs[i]),colors[i]);} [/asy]$

Desired intersection is stated in earlier posts. Do refactoring of points as follows:

refactoring wrote:

Triangle $ABC$ has circumcircle $\Omega$ with center $O$ . Define $X=\overline{AO}\cap\overline{BC}$ and $Y,Z$ similarly, and $X'=\overline{AO}\cap(BOC)$ to be the inverse of $X$ in $\Omega$ (and $Y',Z'$ similarly). Then $A,Y,Z'$ , and the intersection of the $A$ -altitude with $(ABC)$ are concyclic.

$[asy] //23slg5 final //setup defaultpen(fontsize(10pt)); size(9cm); pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); blu1=RGB(233,242,255); blu2=RGB(212,227,255); lightpurple=RGB(234,218,255);// blu1 lighter //define int i=0; pair inverse(pair P) {return P/(P.x*P.x+P.y*P.y); } pair O,A,B,C; O=(0,0); A=dir(110); B=dir(-160); C=dir(-20); pair X,Y,Z,X1,Y1,Z1; X=extension(A,O,B,C); Y=extension(B,O,A,C); Z=extension(C,O,A,B); X1=inverse(X); Y1=inverse(Y); Z1=inverse(Z); pair K,T; K=extension(Y,Z1,Z,Y1); T=2*foot(A,B,C)-A-B-C; pair[] xook=intersectionpoints(circumcircle(Y,Z,Y1),circle(O,1)); //draw filldraw(A--B--C--cycle,blu1,blu); draw(circle(O,1)^^circumcircle(A,O,B)^^circumcircle(A,O,C),blu); draw(A--X1^^B--Y1^^C--Z1,blu); draw(Y--Z1^^Z--Y1,purple); draw(circumcircle(Y,Z,Y1),purple); draw(A--T,magenta); draw(circumcircle(A,Y,Z1),magenta+dotted); draw(1.2*xook[1]-.2*xook[0]--1.2*xook[0]-.2*xook[1],red); clip( box((-1.5,-2.6),(2.2,1.5)) ); //label void pt(string s,pair P,pair v, pen a){filldraw(circle(P,.015),a,linewidth(.1)); label(s,P,v);} string labels[]={"$O$","$A$","$B$","$C$","$X$","$Y$","$Z$","$X'$","$Y'$","$Z'$","$K$","$T$"}; //12 pair points[]={O,A,B,C,X,Y,Z,X1,Y1,Z1,K,T}; real dirs[]={-110,100,-130,-50, -110,-100,-80,0,-40,100,130,-90}; pen colors[]={blu,blu,blu,blu,blu,blu,blu,blu,blu,blu,purple,magenta}; for (i=0; i<12; ++i) {pt(labels[i],points[i],dir(dirs[i]),colors[i]);} [/asy]$
Define $K=\overline{Y'Z}\cap\overline{YZ'}$ ; observe that there is a spiral similarity at $A$ sending $Z',B\to C,Y'$ so $Y',Z'$ are $\sqrt{bc}$ inverses. As a consequence $\overline{AY'}$ and $\overline{AZ'}$ are isogonal in $\angle BAC$ .

Claim 1: $K$ lies on the $A$ -altitude.
Proof. Apply DDIT to $A$ and $YY'ZZ'$ , to obtain the involution $A(BC;Y'Z';OK)$ , which, based on the first two pairs, is isogonality at $\angle BAC$ , so $\overline{AK}$ and $\overline{AO}$ are isogonal as well. $\qquad\qquad\square$

Claim 2: $\overline{Y'Z}\cap\overline{YZ'}$ lies on the radical axis of $\Omega$ and $(YY'ZZ')$ .
Proof. Said radical axis is also the polar of $O$ wrt $(YY'ZZ')$ because $OA^2=OY\cdot OY'=OZ\cdot OZ'$ by design.
The claim then reduces to Brocard on that quadrilateral. $\qquad\qquad\square$
By PoP at the concurrency point in claim 1, if $T$ is the intersection of the $A$ -altitude with $(ABC)$ $KY\cdot KZ'=\text{Pow}(K,\Omega)=KA\cdot KT$ , and we win.

remark: guessing the concurrency point

This post has been edited 3 times. Last edited by crazyeyemoody907, Aug 16, 2024, 5:50 AM

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#21 h Aug 28, 2024, 8:05 AM • 1 Y

Y by Amy_Chen

I will give a second solution using Ptolemy-Sinus Lemma. Solved with my bros from ENKA.

Let $Z = AO \cap (OBC)$ . I will prove that $Z \in (BXD)$ which is sufficient from symmetry.

$Z \in (BXD) \iff$ $BD \sin B = BX \sin (270 - 2B - C)+BZ \sin (90-A)$ $\iff$ $BD \sin B = BX \cos (B-A)+BZ \cos A$ $\overset{\text{Sinus Theorem in } \triangle BCX} \iff$ $BD \sin B = BC \cos A+BZ \cos A$ $\overset{\text{Sinus Theorem in } \triangle BOZ \text{ and } \triangle ABC} \iff$ $2R\cos (C-A)\sin B = R\sin 2A + R\sin 2C$ $\iff$ $2 \sin(C+A) \cos(C-A) = \sin 2A + \sin 2C$ which is true.

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#22 h Sep 1, 2024, 7:42 PM

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Let $K$ be the second intersection of $(BOC)$ with $AO$ . It now suffices to show that $K \in (BXD) \cap (CYE)$ . We will show $K \in (CYE)$ ; the proof for $K \in (BXD)$ is similar. Let $P$ be the intersection of $AO$ with the line through $Y$ parallel to $CE$ .

Claim: $EPYC$ is an isosceles trapezoid, and thus cyclic.
Proof: Let $Q$ be the point diametrically opposite $A$ on $\omega$ . Then $PY \parallel CE \parallel BQ$ . Since $OB = OQ$ and $\triangle OBQ \sim \triangle OYP$ , $OY = OP$ , so a line through $O$ perpendicular to $PY$ is the perpendicular bisector of both $PY$ and $CE$ . $\square$

Claim: $PYCK$ is cyclic.
Proof: We have $\angle PYC = 180^\circ - \angle YCE = \angle BAC + 90^\circ$ , and $\angle PKC = \angle OKC = \angle OBC = 90^\circ - \angle BAC$ , giving $\angle PYC + \angle PKC = 180^\circ$ .

Combining the two claims gives $EPYCK$ cyclic, so $K \in (CYE)$ .

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1180 posts

#23 h Sep 17, 2024, 3:04 PM

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MarkBcc168 wrote:

Synthetic is also equally short

$[asy] size(7cm); defaultpen(fontsize(10pt)); import olympiad; import geometry; pair A = (1,3.2); pair B = (0,0); pair C = (4,0); pair O = circumcenter(A,B,C); pair D = 2*foot(O,B,foot(B,A,C)) - B; pair X = extension(C,O,A,B); pair T = 2*foot(circumcenter(B,O,C),A,O)-O; pair P = extension(X,X+B-D,A,O); draw(A--B--C--cycle, linewidth(1)); draw(circle(B,X,D), linewidth(0.7)+dashed); draw(circle(B,O,C), black); draw(A--T, linewidth(0.7)); draw(C--O, linewidth(0.7)); draw(O--X, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm)); draw(O--P, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm)); draw(circle(A,B,C), black); draw(O--B, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm)); draw(O--D, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm)); dot("$A$", A, dir(115)); dot("$B$", B, dir(179)); dot("$C$", C, dir(0)); dot("$O$", O, 1.5*dir(-108)); dot("$D$", D, dir(70)); dot("$X$", X, dir(154)); dot("$T$", T, dir(-56)); dot("$P$", P, dir(165)); [/asy]$
Let $AO$ intersect $\odot(BOC)$ again at $T$ . We claim that $T$ is the concurrency point.

Note that since angle bisector of $\angle AOC$ is perpendicular to $AC$ , lines $AO$ , $OX$ , and $BD$ form an isosceles triangle. Combining with $OB=OD$ , it follows that there exists point $P$ on $AO$ such that $BXPD$ is isosceles trapezoid. Clearly, $P\in\odot(BXD)$ . Now, notice that
$\measuredangle BTP = \measuredangle BTO = \measuredangle BCO = \measuredangle OBC = \measuredangle ABD = \measuredangle XBD = \measuredangle BDP,$ so $T\in\odot(BXDP)$ . Analogously, $T\in\odot(CYE)$ , so we are done.

Did you mean perpendicular bisector?

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481 posts

#24 h Oct 1, 2024, 3:20 PM • 4 Y

Y by SilverBlaze_SY, S.Ragnork1729, HACK_IN_MATHS, GeoKing

Solved with SilverBlaze_SY, HACK_IN_MATHS and S.Ragnork1729.

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (10.69888,16.57077); pair B = (1.51632,-12.50733); pair C = (34.50552,-13.10250); pair O = (18.20964,-1.78998); pair D = (32.29102,12.18297); pair X = (7.29373,5.78779); pair T = (13.17850,10.50910); pair P = (30.63286,-32.15967); import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw(C--A, linewidth(0.6)); draw(circle(O, 19.83755), linewidth(0.6)); draw(circle((24.58069,-9.73103), 23.23086), linewidth(0.6) + blue); draw(circle((17.88747,-19.64757), 17.86050), linewidth(0.6) + linetype("4 4") + red); draw(X--T, linewidth(0.6)); draw(A--P, linewidth(0.6)); draw(B--D, linewidth(0.6)); draw(C--X, linewidth(0.6)); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$O$", O, NE); dot("$D$", D, NE); dot("$X$", X, NW); dot("$T$", T, W); dot("$P$", P, SE); [/asy]$

Let the line through $X$ parallel to $BD$ intersect $AO$ at $T$ .

Claim: $T$ lies on $\odot(BXD)$ .
Proof. Firstly, we have, $\begin{align*} \measuredangle OTX &= \measuredangle ATX\\ &= \measuredangle (AO,BD)\\ &= \measuredangle OAB+\measuredangle ABD\\ &= (90^{\circ}-\measuredangle BCA) + (90^{\circ}-\measuredangle CAB)\\ &= \measuredangle ACB + \measuredangle BAC\\ &= \measuredangle ABC .\end{align*}$ Also, $\begin{align*} \measuredangle OXT &= \measuredangle (OX,BD)\\ &= \measuredangle DBX + \measuredangle BXO\\ &= (90^{\circ}-\measuredangle CAB)+ (\measuredangle ABC+\measuredangle BCO)\\ &= (90^{\circ}-\measuredangle BCO)+\measuredangle ABC+ \measuredangle BAC\\ &= \measuredangle CAB + \measuredangle ABC + \measuredangle BAC\\ &= \measuredangle ABC .\end{align*}$ This gives us that $\measuredangle OTX=\measuredangle ABC= \measuredangle OXT$ . This further implies that $OX=OT$ . We also have that $OB=OD$ . Combining these with the fact that $XT\parallel BD$ , we get that $BXTD$ is an isosceles trapezium and thus is also cyclic. $\blacksquare$

Let $P=AO \cap \odot(BXD)$ . Then note that, $\measuredangle OPB=\measuredangle TPB=\measuredangle TDB =\measuredangle DBX =90^{\circ}-\measuredangle BAC =\measuredangle OCB .$
So $OBPC$ is cyclic. Similarly, if we define $Q=AO\cap \odot(CYE)$ , then $OBQC$ is also cyclic. Now note that $P=AO\cap \odot(OBC)=Q$ which gives $P\equiv Q$ . We are done.

This post has been edited 2 times. Last edited by kamatadu, Oct 1, 2024, 4:34 PM

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872 posts

#25 h Oct 17, 2024, 3:22 PM

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Bashing is nice!
take $\odot (ABC)$ as unit circle, then $d=-ac/b,$ $x=tc$ where $t\in\mathbb R.$ By $x+ab\overline x=a+b,$ $t=\frac{(a+b)c}{ab+c^2},$ $x=\frac{(a+b)c^2}{ab+c^2}.$ Some observation makes me guess that the intersection is on $\odot (BOC),$ say $\lambda a$ where $\lambda\in\mathbb R.$ We want $\frac{\lambda a-b}{\lambda a-c}/\frac bc\in\mathbb R.$ Here $\lambda =\frac{a^2+bc}{a(b+c)},$ $\lambda a=\frac{a^2+bc}{b+c}.$ Now we only need $b,-ac/b,\frac{(a+b)c^2}{ab+c^2},\frac{a^2+bc}{b+c}$ are concyclic. $\iff \dfrac{\frac{(a+b)c^2}{ab+c^2}-b}{\frac{(a+b)c^2}{ab+c^2}+\frac {ac}b}\div\dfrac{\frac{a^2+bc}{b+c}-b}{\frac{a^2+bc}{b+c}+\frac {ac}b}=\frac{a(c+b)(c-b)}{c(a+b)(a-b)}\in\mathbb R$ which is obvious. $\Box$

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1360 posts

#26 h Nov 1, 2024, 3:21 AM

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Heres my solution:

Define point $T = AO \cap (BOC)$ with $T \neq O$ .

Note that: $\angle BTO = \angle BCO = 90^\circ - A$ . The following claim finishes the problem: Let $P = (BXD) \cap AO$ .

Claim : $BXPD$ is a cyclic trapeziod.
Proof: Notice that: $BD$ and the angle bisector of $\angle AOC$ are parallel which implies that $BD$ and the angle bisector of $\angle XOA$ are perpendicular. Let $P'$ be a point on $AO$ such that $XP' \parallel BD$ . Therefore, $XP'$ and $BD$ have same perpendicular bisector which implies $BCP'D$ is an isosceles trapezoid (and thus cyclic) and therefore $P=P'$ .

Due to the above claim: $\angle PDB =\angle XBD = 90^\circ-A = \angle BTP$ and therefore $BXDT$ is cyclic.

Remark

This post has been edited 2 times. Last edited by Saucepan_man02, Nov 1, 2024, 3:26 AM
Reason: EDIT

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520 posts

#27 h Dec 10, 2024, 5:22 PM

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Sol:- Let $(BOC)$ meet $AO$ again at $V$ .Let $l$ be the line through $O$ parallel to $AB$ . Since $AO=OB$ ,the reflection $F$ of $Y$ across $l$ lies on $AO$ .Since $l$ is also the perpendicular bisector of $CE$ , $CYFE$ is a cyclic isosceles trapezoid.
$\measuredangle CVF=\measuredangle CVO=\measuredangle CBO=\measuredangle OCB=\measuredangle YCE=\measuredangle CEF \implies V \in (CYEF)$ . Similarly $V\in (BXD)$ .

https://cdn.discordapp.com/attachments/1247512024687181896/1316092227696594944/image.png?ex=6759c9e6&is=67587866&hm=abc2f48fb0ccc7d6b6d8baa84f813f2805ba6a447199a6e0f60e37f2de7a73d4&

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168 posts

#28 h Jan 11, 2025, 4:05 PM

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Easy G5
Solution With Complex Numbers

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#29 h Feb 16, 2025, 2:36 PM

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MarkBcc168 wrote:

Let $ABC$ be an acute-angled triangle with circumcircle $\omega$ and circumcentre $O$ . Points $D\neq B$ and $E\neq C$ lie on $\omega$ such that $BD\perp AC$ and $CE\perp AB$ . Let $CO$ meet $AB$ at $X$ , and $BO$ meet $AC$ at $Y$ .

Prove that the circumcircles of triangles $BXD$ and $CYE$ have an intersection lie on line $AO$ .

Ivan Chan Kai Chin, Malaysia

Define X': intersection of line through X and parallel to BD with AO and we define Y' similarly
Define X": second intersection of AO and BXD define Y" similarly
Claim 1: BXX'D is cyclic
Proof:
notice XX' and BD are parallel thus it's enough to show angels XBD and X'DB are equal with is equivalence to showing tan(A)=cot(pi/2-A)=cot(X'DB) now by similarity we have XB/AX =X'T/AX' and by law of sin in triangles CXB and CXA we have X'T/AX'=XB/AX=(sin(A)*cos(A))/((sin(B)*cos(B)) thus AX'/X'T=cos(B)*sin(B)/sin(A)*cos(A)
now by law of sin in triangles DX'A and DX'T we have (sin(C-X'DB)/sin(X'DB))*(sin(B)/(Sin(A))=AX'/X'T
thus we have sin(C)*cot(X'DB)-Cos(C)=Cos(B)/Cos(A)
thus it's enough to prove sin(C)*tan(A)-Cos(C)=Cos(B)/Cos(A) ifoif sin(A)*sin(C)-cos(A)*cos(C)=cos(B) ifoif
-Cos(A+C)=Cos(B) with is obvious
Claim 2: AY"=AX"= (sin(B)*sin(C)*2R)/Cos(A)
proof:by power of point we have AX*AB=AX'*AX" thus we have AX"=(AX/AX')*AB
by law of sin and famous identity sin(x)=sin (pi-x) we have AX/AX'=sin(B)/Cos(A) thus we have AX"= sin(B)*sin(C)*2R/cos(A) we could have similar argument for AY" thus claim is proved now for notice X", and Y" are at same side of ray AO because triangle is acute thus X"=Y" and problem is solved

This post has been edited 1 time. Last edited by Autistic_Turk, Feb 16, 2025, 2:38 PM
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#30 h Feb 22, 2025, 4:54 AM • 1 Y

Y by babarazamtruefan152-0

How is this a G5 bro

Let $T=\overline{AO} \cap (BOC)$ and we will prove this is the concurrency point.

We will do a sketch with complex numbers with $(ABC)$ as unit circle. See that $d=-\frac{ac}b, \text{ } t=\frac{a^2+bc}{b+c},\text{ }x=\frac{c^2(a+b)}{c^2+ab}$ And done just check $\frac{x-d}{x-b}:\frac{t-d}{t-b}= \frac{c(a^2b+abc+ac^2+b^2c}{ab(c-b)(c+b)}:\frac{a^2b+abc+ac^2+b^2c}{b(a-b)(a+b)}= \frac{c(a-b)(a+b)}{a(c-b)(c+b)} \in \mathbb{R}$ And done.

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#31 h Mar 12, 2025, 5:28 AM

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triv???
Let $T$ be the second intersection of $(OBC)$ and $AO.$ I claim $T$ is the desired concurrency point.
Let $P$ be the intersection inside of $ABC$ of $AO$ and $(BXD).$
Claim: $OX = OE$ and $BXED$ is isosceles trapezoid
Proof: Observe that the angle bisector of $EOC$ is perpendicular to the angle bisector of $AOX.$
Also note that we have $\angle(BH, OC) = 180 - \angle HBC - \angle OCB = 180 -(90 - C) - (90 - A) = 180 - \angle OAB - \angle HBA = \angle(BH, AO)$ . Thus, the angle bisector of $AOX$ is perpendicular to $EX$ and $BD,$ which finishes.

Note that $\angle XET = \angle OEX = \angle(BH, OC) = B = 90 - C + 90 - A = \angle BAO + \angle OTB = \angle XBT,$ which finishes.

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#32 h Apr 2, 2025, 11:15 PM

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Let the intersections of $BXD$ and $CYE$ be points $T$ and $S$ with $T$ being on the opposite side of $BC$ wrt $A$ .
Let $L$ be the intersection of the $A$ altitude with $(ABC)$ .

Claim 1: $H$ lies on $ST$ .
Proof: Since $BH\cdot HD=CH\cdot HE$ it means that $H$ has equal powers to circles $BXD$ and $CYE$ so it lies on $ST$ .

Claim 2: $ASLT$ is cyclic
Proof: $SH\cdot HT=BH\cdot HD=AH\cdot HL$ which implies that $ASLT$ is cyclic.

Let $OH$ intersect $(AOL)$ again at $J$ .

Claim 3: $JSOT$ is cyclic
Proof: $JH\cdot HO=AH\cdot HL=SH\cdot HT$ so $JSOT$ is cyclic.

Claim 4: $JSHL$ is cyclic
Proof: $\angle JSH+\angle JLH=\angle JOT+\angle JOA=180^\circ$ so $JSHL$ is cyclic.

Now by all the cyclics we found, we can chase angles

$\angle STO=\angle SJO=\angle SLO=\angle STA$ which implies that $A$ , $O$ , and $T$ are collinear.

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#33 h Apr 20, 2025, 9:27 AM

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cute!

Let $H$ be the orthocenter of $ABC$ . Perform a $\sqrt{bc}$ inversion with the standard reflection over the $A$ -angle bisector. Noting that $\angle CAD = 90 - \angle C$ , it follows that $D$ maps to a point $D'$ on ray $CB$ with $\angle D'AB = 90 - \angle C$ ; similarly, $E$ maps to $E'$ on ray $BC$ with $\angle E'AC = 90 - \angle B$ . As $H$ and $O$ are isogonal conjugates, and as $(ABC)$ maps to $BC$ , we note that $O$ maps to the reflection of $A$ over $BC$ , point $O'$ .

Thus point $Y$ maps to $(ACO') \cap AB$ . We claim that $Y'BHE'$ is cyclic. This will, by symmetry, give us that $X'CHD'$ is cyclic, and we will be done, as $(Y'BHE')$ would be the image of $(CYE)$ .

Observe that $\angle Y'CE' = \angle Y'CO' + \angle E'CO' = 90 - \angle B + 180 - \angle C = 90 + \angle A$ , and that $\angle Y'BH = 180 - \angle ABH = 90 + \angle A$ . Next, we'll show that $\frac{Y'B}{Y'C} = \frac{BH}{CE'}$ , which will yield similar triangles $Y'BH$ and $Y'CE'$ , from which we easily obtain $Y'BHE'$ cyclic.

Note that by the Law of Sines, $\frac{Y'B}{Y'C} = \frac{\cos \angle A}{\sin \angle B}$ . Moreover, we have $BH = 2 R \cos \angle B$ and $CE' = \frac{b \cos \angle B}{\cos \angle A}$ , again by the Law of Sines. Putting this together, we see that $\frac{BH}{CE'} = \frac{2R \cos \angle A}{b}$ , so as $\frac{b}{\sin \angle B} = 2R$ , we are done.

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