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Source: China TSTST Test 2 Day 1 Q3

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615 posts

#1 h Mar 13, 2017, 1:07 AM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

Let $ABCD$ be a quadrilateral and let $l$ be a line. Let $l$ intersect the lines $AB,CD,BC,DA,AC,BD$ at points $X,X',Y,Y',Z,Z'$ respectively. Given that these six points on $l$ are in the order $X,Y,Z,X',Y',Z'$ , show that the circles with diameter $XX',YY',ZZ'$ are coaxal.

This post has been edited 1 time. Last edited by fattypiggy123, Mar 13, 2017, 1:34 AM
Reason: Typo

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ABCDE

#2 h Mar 13, 2017, 1:47 AM • 7 Y

Y by rafayaashary1, WizardMath, kapilpavase, RC., mathleticguyyy, Adventure10, ohiorizzler1434

Note that by Desargues' Involution Theorem, $(XX';YY;ZZ')$ are pairs of an involution on $l$ . Let $P$ and $Q$ be the fixed points of the involution. Note that the midpoint of $PQ$ has the same power $\frac{PQ^2}{4}$ to all three circles by harmonics, so the perpendicular bisector of $PQ$ is the common radical axis of all three circles.

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#3 h Mar 13, 2017, 11:05 AM • 2 Y

Y by RC., Adventure10

Nice solution @ABCDE, my solution coincides with your solution.

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math90

#4 h Mar 13, 2017, 12:43 PM • 1 Y

Y by Adventure10

What is Desargues' Involution Theorem?

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#5 h Mar 13, 2017, 1:15 PM • 4 Y

Y by k12byda5h, Adventure10, Mango247, Kingsbane2139

I wonder why this problem got in China TST...(and even number 6!!) This is so easy if you know involution...

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math90

#6 h Mar 13, 2017, 6:10 PM • 1 Y

Y by Adventure10

Can someone explain/provide a link?

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#7 h Mar 14, 2017, 1:53 AM • 3 Y

Y by math90, Adventure10, Mango247

Dear math90:http://www2.washjeff.edu/users/mwoltermann/Dorrie/63.pdf

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#8 h Mar 14, 2017, 4:18 PM • 2 Y

Y by Adventure10, Mango247

toto1234567890 wrote:

I wonder why this problem got in China TST...(and even number 6!!) This is so easy if you know involution...

Geometry is usually the easiest part in our country.

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#9 h Mar 15, 2017, 12:19 AM • 2 Y

Y by Adventure10, Mango247

Just look at the problem today... Surprise me!

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#10 h Jul 7, 2017, 7:33 AM • 2 Y

Y by Adventure10, Mango247

ABCDE wrote:

Note that by Desargues' Involution Theorem, $(XX';YY;ZZ')$ are pairs of an involution on $l$ . Let $P$ and $Q$ be the fixed points of the involution. Note that the midpoint of $PQ$ has the same power $\frac{PQ^2}{4}$ to all three circles by harmonics, so the perpendicular bisector of $PQ$ is the common radical axis of all three circles.

This solution is slightly incorrect: not all involutions on a (real) line are guaranteed to have fixed points. Fortunately this can be fixed with a pretty small amount of work. What is true (you can verify this algebraically using mobius transforms; CantonMathGuy and I just did this) is that any involution on a line is an inversion of some nonzero (possibly negative) power (there are no fixed points when it is negative). It is clear then that the center of this inversion has equal power to all three circles, so it serves the same purpose as the midpoint of $PQ$ .

In fact, the center of this inversion is the midpoint of $PQ$ , at least when they exist. In the complex projective line, where $P,Q$ must exist, the center of inversion is still the midpoint of these two points, so this is another way to patch the original solution. One still must take a certain amount of care to work out the details, though, when working in the complex projective line.

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#11 h Jul 28, 2017, 4:44 AM • 12 Y

Y by anantmudgal09, tastymath75025, Pluto1708, Crystal., Pure_IQ, MarkBcc168, Kagebaka, Gaussian_cyber, Imayormaynotknowcalculus, guptaamitu1, Adventure10, Mango247

Claim. Let $\omega_a$ , $\omega_b$ , $\omega_c$ , $\omega_d$ be four circles. Define $P_{xy}$ to be the exsimilicenter of $(\omega_x, \omega_y)$ for $\{x, y\} \subset \{a, b, c, d\}$ . Then the circles with diameters $P_{ab}P_{cd}, P_{ac}P_{bd}, P_{ad}P_{bc}$ are coaxal.

Proof. By Monge, $P_{ab}, P_{ac}, P_{ad}$ belong to a line $\ell_a$ ; define $\ell_b, \ell_c, \ell_d$ similarly. If these lines are pairwise distinct, the claim follows from Gauss-Bodenmiller Theorem in the complete quadrilateral $\{\ell_a, \ell_b, \ell_c, \ell_d\}$ ; otherwise, these lines must be identical, and the claim follows by continuity. $\square$

Select circles $\omega_a, \omega_b, \omega_c, \omega_d$ centered at $A, B, C, D$ , so that the exsimilicenters of $(\omega_c, \omega_d), (\omega_a, \omega_d), (\omega_b, \omega_d)$ are the points $X', Y', Z'$ , respectively. (The circles may have negative radius.) By Monge, the points $X, Y, Z$ are the exsimilicenters of $(\omega_a, \omega_b), (\omega_b, \omega_c), (\omega_a, \omega_c)$ . The problem now follows from the claim.

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#12 h Aug 27, 2017, 10:12 AM • 2 Y

Y by Adventure10, Mango247

adamov1 wrote:

This solution is slightly incorrect: not all involutions on a (real) line are guaranteed to have fixed points. Fortunately this can be fixed with a pretty small amount of work. What is true (you can verify this algebraically using mobius transforms; CantonMathGuy and I just did this) is that any involution on a line is an inversion of some nonzero (possibly negative) power (there are no fixed points when it is negative). It is clear then that the center of this inversion has equal power to all three circles, so it serves the same purpose as the midpoint of $PQ$ .

Can you show the proof this?

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#13 h Aug 27, 2017, 10:16 AM • 2 Y

Y by Adventure10, Mango247

Can somebody explain what coaxal means?

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#14 h Aug 27, 2017, 10:54 AM • 2 Y

Y by Adventure10, Mango247

Three circles are coaxal if and only if they have common radical axis.

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#16 h Mar 15, 2019, 6:38 PM • 2 Y

Y by Adventure10, Mango247

adamov1 wrote:

ABCDE wrote:

Note that by Desargues' Involution Theorem, $(XX';YY;ZZ')$ are pairs of an involution on $l$ . Let $P$ and $Q$ be the fixed points of the involution. Note that the midpoint of $PQ$ has the same power $\frac{PQ^2}{4}$ to all three circles by harmonics, so the perpendicular bisector of $PQ$ is the common radical axis of all three circles.

This solution is slightly incorrect: not all involutions on a (real) line are guaranteed to have fixed points. Fortunately this can be fixed with a pretty small amount of work. What is true (you can verify this algebraically using mobius transforms; CantonMathGuy and I just did this) is that any involution on a line is an inversion of some nonzero (possibly negative) power (there are no fixed points when it is negative). It is clear then that the center of this inversion has equal power to all three circles, so it serves the same purpose as the midpoint of $PQ$ .

In fact, the center of this inversion is the midpoint of $PQ$ , at least when they exist. In the complex projective line, where $P,Q$ must exist, the center of inversion is still the midpoint of these two points, so this is another way to patch the original solution. One still must take a certain amount of care to work out the details, though, when working in the complex projective line.

Sorry but how does this complete the proof?

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#17 h Dec 26, 2019, 2:11 AM • 2 Y

Y by Adventure10, Mango247

Does this work? It doesn't seem to be the same as ABCDE's.

By Desargues' Involution Theorem, $(X,X'),(Y,Y'),(Z,Z')$ must be swapped under some inversion with center $K$ on $l.$ Since $KX\cdot KX'=KY\cdot KY'=KZ\cdot KZ',$ $K$ must then be the radical center of $(XX'),(YY'),(ZZ').$ However, we know that the centers of the circles are collinear, so $K$ must either be the point at infinity along a line perpendicular to $l$ or $(XX'),(YY'),(ZZ')$ must be coaxial; clearly, the former is impossible, so we're done. $\blacksquare$

oh wait this is basically the correct finish that adamov1 pointed out .-.

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#18 h Apr 10, 2020, 4:59 AM

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By Desargues Involution Theorem, $(X', X), (Y, Y'), (Z, Z')$ are pairs swapped under some involution. If this is a reflection, the problem is done. Otherwise, suppose that the involution is an inversion, say with center $P$ on line $l.$ Now observe that $PX \cdot PX' = PY \cdot PY' = PZ \cdot PZ'$ meaning that $P$ is the radical center of the circles with diameters $XX', YY',$ and $ZZ'.$ Now if you draw the line perpendicular to $l$ passing through $P$ this must be the radical axis of the $3$ circles, as desired. Note: $P$ may be the point at infinity, but this is clearly not true since then the line must be the line at infinity.

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#19 h May 31, 2020, 5:52 PM • 4 Y

Y by Kagebaka, MarkBcc168, Inconsistent, ohiorizzler1434

Who needs DIT when you can just kill with coordinates :love:

:love:

The condition is equivalent to the existence of a point $P$ such that $PX\cdot PX' = PY\cdot PY' = PZ \cdot PZ'$ (lengths are signed).

Claim: For any four points $X,X',Y,Y' \in l$ there exists a unique $P\in l$ such that $PX\cdot PX' = PY\cdot PY'$ .

Proof: Note that $P$ has equal power with respect to the circles with diameter $XX', YY'$ respectively, thus $P$ must lie on the radical axis of the two circles which is perpendicular to $l$ , so $P$ is the intersection of $l$ and the radical axis of the two circles. It is evident that this intersection satisfies the problem statement, as claimed. $\blacksquare$

For the main part of the proof, set up coordinates with $P$ as the origin and $l$ the $x$ -axis. Consider conics $\Xi :=AB \cup CD, \Psi := BC \cup DA$ and any other conic through $A,B,C,D$ (which can be expressed as $\lambda \Xi + \mu \Psi = 0$ for some reals $\lambda, \mu$ ). By Vieta it follows that the ratio (constant/leading coeff) of $\lambda \Xi + \mu \Psi$ is the same as that of $\Xi$ and $\Psi$ , hence for any conic through $A,B,C,D$ (including $AC\cup BD$ ) intersecting $l$ at points $Z,Z'$ we have (in signed distances) $PZ\cdot PZ' = PX\cdot PX' = PY\cdot PY'$ , as desired.

Note that this offers a proof of DIT using cartesian coordinates.

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#20 h Sep 13, 2020, 10:34 PM • 1 Y

Y by Frestho

This is pretty much just DIT using $\ell$ : the desired involution is a negative inversion about some point $P$ on $\ell$ which can be easily seen from the orientation of the points. $P$ has equal power to all three circles, and since the circle centers are collinear, we are done.

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#21 h Jan 12, 2021, 5:38 AM

Y by

This is kind of equivalent to DIT, we are obviously done if the involution is a reflection, if the involution is an inversion, there exists point $P$ on $l$ such that it's power to all three circles is constant, and hence the line $d$ which is perpendicular to $l$ and passes through $P$ would be our radical axis, so those three circles would be coaxial.

The other side is also the same, simply consider the intersection of the radical axis of the three circles with $l$ , then that point is the center of an inversion which maps X to X', Y to Y' and Z to Z'.

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#22 h Mar 18, 2021, 10:15 PM

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Using DIT, reflection does not exist, since in that case, circles are concentric and radical axis does not exist.

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#23 h Aug 16, 2022, 5:06 AM • 1 Y

Y by centslordm

By DIT, $(X,X'),(Y,Y'),(Z,Z')$ are swapped by an involution. If it is a reflection, we are done. If it is an inversion at $P,$ then $PX\cdot PX'=PY\cdot PY'=PZ\cdot PZ'$ so $P$ is the radical center of our circles. If the circles are not coaxal, then $P$ is the point at infinity along the line perpendicular to $\ell,$ which is absurd. $\square$

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#24 h Jul 28, 2023, 2:09 PM • 1 Y

Y by HoripodoKrishno

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings. */ pair A = (-112.56662,71.45286); pair B = (-140,-80); pair C = (98.17049,-82.53832); pair D = (14.16640,154.49649); pair X = (-180.85092,-305.52787); pair Y = (-72.06909,-80.72397); pair Zp = (118.67646,313.46266); pair Z = (-28.24650,9.83792); pair Xp = (25.82900,121.58806); pair Yp = (54.56188,180.96621); import graph; size(12.1cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--D, linewidth(0.5)); draw(D--A, linewidth(0.5)); draw(B--Zp, linewidth(0.5)); draw(Zp--X, linewidth(0.5) + red); draw(X--B, linewidth(0.5)); draw(D--Yp, linewidth(0.5)); draw(A--C, linewidth(0.5)); draw(circle((-77.51096,-91.96990), 237.24703), linewidth(0.5) + blue); draw(circle((-8.75360,50.12111), 145.35917), linewidth(0.5) + blue); draw(circle((45.21497,161.65029), 168.65226), linewidth(0.5) + blue); dot("$A$", A, NW); dot("$B$", B, W); dot("$C$", C, E); dot("$D$", D, NW); dot("$X$", X, SW); dot("$Y$", Y, 2*dir(270)); dot("$Z'$", Zp, NE); dot("$Z$", Z, 2*dir(270)); dot("$X'$", Xp, NE); dot("$Y'$", Yp, 2*E); [/asy]$

Note that from the definition of the points, using DIT we get that there exists an Involution on $\ell$ swapping $(X,X')$ , $(Y, Y')$ and $(Z,Z')$ . Moreover, as this is an Involution on a line, so we get that it is either a reflection w.r.t. some point on $\ell$ , or an Inversion w.r.t. some point on $\ell$ as its center. If this were a reflection, then the segments $XX'$ and $YY'$ share the same midpoint and thus the order of the points would be $\overline{X-Y-Y'-X'}$ or $\overline{Y-X-X'-Y'}$ . This however contradicts the order of $Y'$ with $X'$ as given in the problem statement.

Thus this Involution must be an Inversion. Note that as this Inversion swaps $(X,X')$ we get that this Inversion fixes the circle with diameter $\odot(XX')$ . Similarly, it fixes the circles $\odot(YY')$ and $\odot(ZZ')$ too. Now consider the radical axis of $\odot(XX')$ and $\odot(YY')$ . Note that this passes through the intersection points of $\odot(XX')$ and $\odot(YY')$ (due to the ordering of the points in the problem statement). Now further notice that as both the circles are fixed, we get that their intersection points are also fixed. Thus the intersection points lie on the circle with respect to which we are Inverting (say $\Gamma$ ). Thus $\left\{\Gamma,\odot(XX'),\odot(YY')\right\}$ share a common radical axis. Similarly, $\left\{\Gamma,\odot(XX'),\odot(ZZ')\right\}$ also share the common radical axis which means that the circles $\left\{\odot(XX'),\odot(YY'),\odot(ZZ')\right\}$ are coaxial and we are done. :stretcher:

:stretcher:

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#25 h Dec 13, 2023, 2:16 PM

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By DIT, there exists an involution swapping $(X,X');(Y,Y');(Z,Z')$ but this is just inversion for some point $O$ , so $OX \cdot OX' = OY \cdot OY' = OZ \cdot OZ'$ and hence the circles are coaxial.

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#26 h Feb 9, 2024, 3:45 AM

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Wow.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.699762129321672, xmax = 6.559203545586037, ymin = -7.052415271250787, ymax = 6.240149911824218; /* image dimensions */ pen qqwwzz = rgb(0,0.4,0.6); pen ccqqqq = rgb(0.8,0,0); /* draw figures */ draw((-4.757561676220963,-2.701927853701796)--(-2.9394930360882117,4.271331028839409), linewidth(0.7) + blue); draw((-2.9394930360882117,4.271331028839409)--(-1.2281172214791287,-0.413771068210891), linewidth(0.7) + blue); draw((-1.2281172214791287,-0.413771068210891)--(2.5671300893294786,0.03546708777647148), linewidth(0.7) + blue); draw((2.5671300893294786,0.03546708777647148)--(-4.757561676220963,-2.701927853701796), linewidth(0.7) + blue); draw(shift((-1.1277580349207321,0.9301856222504602))*xscale(2.627452391198322)*yscale(2.627452391198322)*arc((0,0),1,-23.479199827160983,156.52080017283902), linewidth(0.7) + qqwwzz); draw(shift((-1.1277580349207321,0.9301856222504602))*xscale(2.6274523911983216)*yscale(2.6274523911983216)*arc((0,0),1,156.52080017283905,336.520800172839), linewidth(0.7) + qqwwzz); draw(shift((-0.07117818042485069,0.4712276449589007))*xscale(1.918189669306153)*yscale(1.918189669306153)*arc((0,0),1,-23.479199827160983,156.52080017283902), linewidth(0.7) + qqwwzz); draw(shift((-0.07117818042485069,0.4712276449589007))*xscale(1.918189669306153)*yscale(1.918189669306153)*arc((0,0),1,156.52080017283905,336.520800172839), linewidth(0.7) + qqwwzz); draw(shift((2.3708766145451667,-0.5895540123530806))*xscale(2.526602380171547)*yscale(2.526602380171547)*arc((0,0),1,-23.479199827160983,156.52080017283902), linewidth(0.7) + qqwwzz); draw(shift((2.3708766145451667,-0.5895540123530806))*xscale(2.5266023801715476)*yscale(2.5266023801715476)*arc((0,0),1,156.52080017283902,336.520800172839), linewidth(0.7) + qqwwzz); draw((-2.9394930360882117,4.271331028839409)--(4.688288375657454,-1.5961931318725715), linewidth(0.7) + blue); draw((-4.757561676220963,-2.701927853701796)--(0.05346485343287957,0.4170851071664104), linewidth(0.7) + blue); draw((-3.5376699074458338,1.9770050119621412)--(4.688288375657454,-1.5961931318725715), linewidth(0.7) + ccqqqq); draw((-0.0061926112720241955,-1.4458608984408903)--(1.3742965084867471,1.7322012890723566), linewidth(0.7) + dotted); /* dots and labels */ dot((-4.757561676220963,-2.701927853701796),dotstyle); label("$A$", (-5.019368325982243,-3.3321114106731107), NE * labelscalefactor); dot((-2.9394930360882117,4.271331028839409),dotstyle); label("$B$", (-3.0318654377917826,4.507922624592221), NE * labelscalefactor); dot((-1.2281172214791287,-0.413771068210891),dotstyle); label("$C$", (-1.1537663783090537,-0.9799290943265048), NE * labelscalefactor); dot((2.5671300893294786,0.03546708777647148),dotstyle); label("$D$", (2.6388996835589813,0.22293933512359834), NE * labelscalefactor); dot((-3.5376699074458338,1.9770050119621412),linewidth(4pt) + dotstyle); label("$X$", (-4.2617979818075026,1.9551666223555948), NE * labelscalefactor); dot((-1.8305508995277082,1.2354653322718627),linewidth(4pt) + dotstyle); label("$Y$", (-2.5113367224837943,0.770532164350039), NE * labelscalefactor); dot((0.05346485343287957,0.4170851071664104),linewidth(4pt) + dotstyle); label("$Z$", (-0.35089794848817917,0.6240867069036397), NE * labelscalefactor); dot((1.2821538376043697,-0.11663376746122077),linewidth(4pt) + dotstyle); label("$X'$", (0.7890121101994477,0.3317694779008617), NE * labelscalefactor); dot((1.6881945386780068,-0.29301004235406136),linewidth(4pt) + dotstyle); label("$Y'$", (1.5454350824458382,-0.9799290943265048), NE * labelscalefactor); dot((4.688288375657454,-1.5961931318725715),linewidth(4pt) + dotstyle); label("$Z'$", (4.936380086421348,-1.6004578093311348), NE * labelscalefactor); dot((0.6840519486073616,0.14317019531573355),linewidth(4pt) + dotstyle); label("$P$", (0.21435516650176835,0.423513021013619), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $\omega_1$ , $\omega_2$ and $\omega_3$ be the circles $(XX')$ , $(YY')$ and $(ZZ')$ . Note that $(X, X')$ , $(Y, Y')$ and $(Z, Z')$ are reciprocal pairs for a unique involution a unique involution along $\ell$ . Now it is well-known that any involution along a line is an inversion with fixed center, say $P$ . However this value of $P$ satisfies,
$\begin{align*} \text{Pow}(\text{Inversion}) = PX \cdot PX' = PY \cdot PY' = PZ \cdot PZ' \end{align*}$ Then $P$ lies on the pairwise radical axes of $\omega_i$ and $\omega_{i+1}$ , where indices are taken modulo $3$ . However also note that all three radical axes are perpendicular to $\ell$ and hence each radical axis is simply the line perpendicular to $\ell$ through $P$ . Hence the three circles are coaxial. $\square$

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#27 h Mar 18, 2024, 3:50 AM

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Rizztastic problem! By the amazing DDIT theorem, $XX'$ , $YY'$ and $ZZ'$ are pairs of points swapped by an involution. Also, the involution can be defined as inversion from a point $P$ on the line $XX'YY'ZZ'$ . But now, $PX \cdot PX' = PY \cdot PY' = PZ \cdot PZ'$ by inversion. Now this means that $P$ is on all the radical axes. But because the centers of the circles are collinear, then they have to be coaxial, or otherwise $P$ is at infinity (but lies on the line).

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Eka01

#28 h Aug 27, 2024, 4:33 PM

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By Desargues' Involution Theorem; $X$ and $X'$ ; $Y$ and $Y'$ ; $Z$ and $Z'$ are swapped by some involution. Since involutions are inversions/reflections, the center $O$ of this inversion lies on $l$ and satisfies $OX.OX'=OY.OY'=OZ.OZ'$ so $O$ lies on the radical axes of $(XX'),(YY'),(ZZ')$ . Note that $l$ joins the centers of all these circles so their pairwise radical axes must be perpendicular to $l$ . But by the above assertion, all of these lines must pass through $O$ , so it is easy to see that all these radical axes must coincide so we are done.

This post has been edited 1 time. Last edited by Eka01, Sep 4, 2024, 9:50 AM

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cursed_tangent1434

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cursed_tangent1434

#29 h Dec 2, 2024, 12:59 AM

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Denote by $\Gamma_X , \Gamma_Y$ and $\Gamma_Z$ the circles with diameters $XX'$ , $YY'$ and $ZZ'$ respectively. By Desargue's Involution Theorem, there exists an involution swapping pairs $(X,X')$ , $(Y,Y')$ and $(Z,Z')$ . Since $X,Y,X'$ and $Y'$ are on the line $\ell$ in this order, this involution cannot be a reflection across a point on $\ell$ . Thus, it must be an inversion about a point $P \in \ell$ . Thus, there exists a point $P \in \ell$ such that,
$PX\cdot PX' = PY \cdot PY' = PZ \cdot PZ'$ Thus, the pairwise radical axes of $\Gamma_X$ , $\Gamma_Y$ and $\Gamma_Z$ have a common point $P \in \ell$ . Further, it is well known that the radical axis of two circles must be perpendicular to the line joining their centers. Thus, the pairwise radical axes of $\Gamma_X$ , $\Gamma_Y$ and $\Gamma_Z$ must all also be perpendicular to $\ell$ . But then, they must be the same line - the line through $P$ which is perpendicular to $\ell$ . Thus, the pairwise radical axes of $\Gamma_X$ , $\Gamma_Y$ and $\Gamma_Z$ are all the same line, which since these circles clearly intersect ( $X,Y,Z,X',Y',Z'$ lie on $\ell$ in this order) implies that the circles $\Gamma_X , \Gamma_Y$ and $\Gamma_Z$ are coaxial.

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#30 h Jan 30, 2025, 4:48 AM

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I just built my new blog to collect bashing.

Nice solution from my teacher: Consider $\ell :y=c$ where $c$ is moving. $X(u,c),Y(v,c),Z(w,c),X'(u',c),Y'(v',c),Z'(w',c).$
The Analytical formula of $\odot (XX')$ is $(x-(u+u')/2)^2+(y-c)^2=(u-u')^2/4\iff x^2-(u+u')x+uu'+(y-c)^2=0.$ To prove the three functions are linearly dependent, we only need
$\frac{u+u'-v-v'}{uu'-vv'}=\frac{u+u'-w-w'}{uu'-ww'}.$ However we may write $u=P(c)$ where $\deg P=1,$ thus the above equality is $Q(c)$ with degree 3. Therefore we only need to plug in 4 special $c,$ and using $A,B,C,D$ we are done! $\Box$

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#31 h Apr 17, 2025, 2:32 PM

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For more information about DIT, see here:
[url] https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZS9jLzYzYWY3MmU0MmJjNjhiZmYwMDVhMjEzOWQzYmZjMGVmODVlOTZjLnBkZg==&rn=ZGVzYXJndWVzLWludm9sdXRpb24tdGhlb3JlbS5wZGY=[/url]

Very sad that this makes the problem seem too easy...

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#32 h May 14, 2025, 1:30 PM

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Take DIT on $ABCD$ with line $l.$ It follows that by desargues assistance theorem, it must be either an inversion/negative inversion or it is a reflection. However, it is obviously impossible for it to be a reflection, as we are given the order of the points. If it is an inversion about point $K,$ it follows that the radius of inversion squared is $= KX\cdot KX' = KY \cdot KY' = KZ \cdot KZ'.$ Thus, $K$ must be the radical center. However, we know that $K$ lies on the point $l,$ which implies that they are coaxial.

This post has been edited 1 time. Last edited by Ilikeminecraft, May 14, 2025, 1:30 PM

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