Circles with same radical axis

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Circles with same radical axis

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Source: DGO 2021, Individual stage, Day2 P3

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#1 h Dec 26, 2021, 7:19 PM

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Let $O$ be the circumcenter of triangle $ABC$ . The altitudes from $A, B, C$ of triangle $ABC$ intersects the circumcircle of the triangle $ABC$ at $A_1, B_1, C_1$ respectively. $AO, BO, CO$ meets $BC, CA, AB$ at $A_2, B_2, C_2$ respectively. Prove that the circumcircles of triangles $AA_1A_2, BB_1B_2, CC_1C_2$ share two common points.

Proporsed by wassupevery1

This post has been edited 1 time. Last edited by Jalil_Huseynov, Dec 28, 2021, 12:33 PM

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Davsch

#2 h Dec 27, 2021, 11:41 AM • 1 Y

Y by PRMOisTheHardestExam

We use complex numbers with $(ABC)$ as the unit circle. Then $a_1=-\frac{bc}a,a_2=\frac{a^2(b+c)}{a^2+bc}$ . The condition $Z\in (AA_1A_2)$ is equivalent to $\frac{(z-a)(a^3b+a^3c+a^2bc+b^2c^2)}{a(az+bc)}=\frac{(a\bar z-1)(a^3+bc(a+b+c))}{bc\bar z+a}$ , or, after diving by $a^2+bc$ , $z\bar z(b^2c^2-a^4)+za(a+b)(a+c)-\bar zabc(a+b)(a+c)-a(b+c)(a^2-bc)=0.$ Analogous equations hold for $Z\in (BB_1B_2),(CC_1C_2)$ . We will now show that these equations are linearly dependent. We compute that (to evaluate the $3\times 3$ -determinants, we always subtract row $1$ from rows $2,3$ )
$\det\begin{pmatrix}a(a+b)(a+c)&-abc(a+b)(a+c)\\b(b+a)(b+c)&-abc(b+a)(b+c)\end{pmatrix}=-abc(a+b)^2(b+c)(c+a)(a-b)\neq 0,$ $\det\begin{pmatrix}a(a+b)(a+c)&-abc(a+b)(a+c)&b^2c^2-a^4\\b(b+a)(b+c)&-abc(b+a)(b+c)&c^2a^2-b^4\\c(c+a)(c+b)&-abc(c+a)(c+b)&a^2b^2-c^4\end{pmatrix}$ $=-abc\det\begin{pmatrix}a(a+b)(a+c)&(a+b)(a+c)&b^2c^2-a^4\\(b-a)(b+a)(a+b+c)&(b-a)(b+a)&(a-b)(a+b)(a^2+b^2+c^2)\\(c-a)(c+a)(a+b+c)&(c-a)(c+a)&(a-c)(a+b)(a^2+b^2+c^2)\end{pmatrix}=0,$ $\det\begin{pmatrix}a(a+b)(a+c)&-abc(a+b)(a+c)&a(b+c)(a^2-bc)\\b(b+a)(b+c)&-abc(b+a)(b+c)&b(c+a)(b^2-ca)\\c(c+a)(c+b)&-abc(c+a)(c+b)&c(a+b)(c^2-ab)\end{pmatrix}$ $=-abc\det\begin{pmatrix}a(a+b)(a+c)&(a+b)(a+c)&b^2c^2-a^4\\(b-a)(b+a)(a+b+c)&(b-a)(b+a)&(b-a)(b+a)(ab+bc+ca)\\(c-a)(c+a)(a+b+c)&(c-a)(c+a)&(c-a)(c+a)(ab+bc+ca)\end{pmatrix}=0,$ as desired.

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BelieverofMaths

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BelieverofMaths

#3 h Jan 18, 2022, 5:04 PM • 2 Y

Y by Noob_at_math_69_level, vuanhnshn

is there any geomerical proof of this one ?

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#4 h Sep 25, 2024, 5:30 PM • 1 Y

Y by ehuseyinyigit

Nice problem!

Replace $A_2,B_2,C_2$ with $D,E,F$ . Let $H$ be the orthocenter of $\triangle ABC$ . Set $(BB_1E)\cap (CC_1F)=P,Q$ .
Claim: $H$ lies on $PQ$ .
Proof:
$Pow(H,(BB_1EPQ))=HB.HB_1=HC.HC_1=Pow(H,(CC_1FPQ))$ Thus, $H$ lies on the radical axis of $(BB_1E)$ and $(CC_1F)$ which is $PQ$ . $\square$
Claim: $A,A_1,P,Q$ are concyclic.
Proof:
$HA.HA_1=HB.HB_1=HP.HQ$ Which gives the desired result. $\square$
Take $\sqrt{bc}$ inversion and reflect over the angle bisector of $\angle BAC$ .
New Problem Statement: $ABC$ is a triangle whose circumcenter is $O$ . $BC$ intersects the reflection of $AO$ with respect to $AB,AC$ at $B_1,C_1$ respectively. Parallel lines to $AB_1,AC_1$ through $C,B$ intersect $AB,AC$ at $E,F$ respectively. $D$ is on $(ABC)$ which holds $AD\perp BC$ . Prove that $D$ lies on the radical axis of $(B_1EC)$ and $(C_1FB)$ .
Lemma: $ABC$ is a triangle whose circumcenter is $O$ . $AO\cap (BOC)=G,BO\cap AC=E$ . $C_1$ is the intersection of the altitude from $C$ to $AB$ and $(ABC)$ . Then, $C_1,E,C,G$ are concyclic.
Proof: Let $(GCE)\cap AG=S$ .
$\measuredangle ESG=180-\measuredangle GCA=\measuredangle C=90-\measuredangle GAB$ Hence $AS\perp AB\perp CC_1$ . Also $\measuredangle OES=90-\measuredangle ABS=\measuredangle C=\measuredangle ESO$ which implies $OE=OS$ . Combining this with $OC_1=OC,$ we conclude that $CESC_1$ is an isosceles trapezoid. Thus, $G,C,E,C_1,S$ are concyclic. $\square$
Let $B_1F\cap C_1E=T$ . Let $(B_1EC)\cap EC_1=L,(C_1BF)\cap B_1F=K$ . By inverting the configuration of the lemma with $\sqrt{bc}$ , we get that $H,B,C_1,E$ are concyclic.
$\measuredangle ECB_1=90-\measuredangle A=\measuredangle EC_1H=\measuredangle EC_1B+\measuredangle BC_1H=\measuredangle EC_1B+\measuredangle DC_1B_1=\measuredangle EC_1B+\measuredangle C_1B_1D=\measuredangle(EC_1,B_1D)$ Hence $B_1,D,L$ are collinear. Similarily, $C_1,D,K$ are collinear.
Claim: $TB_1=TC_1$ .
Proof:
$\frac{BB_1}{BF}=\frac{BC}{BF}.\frac{AB_1}{CE}=\frac{BC}{CE}.\frac{AB_1}{BF}=\frac{CC_1}{CE}$ And $\measuredangle FBB_1=90+\measuredangle A=\measuredangle C_1CE$ subsequently $FBB_1\sim ECC_1$ . So $\measuredangle BB_1F=\measuredangle EC_1C$ which yields $TB_1=TC_1$ . $\square$
Claim: $TK=TL$ .
Proof: Let $M=(B_1EC)\cap B_1F,N=(C_1FB)\cap C_1E$ .
$\measuredangle EMB_1=\measuredangle ECB_1=90-\measuredangle A=\measuredangle C_1BF=\measuredangle C_1NF$ Thus, $M,N,E,F$ are concyclic. By using this,
$TM.TF.TB_1.TK=(TK.TF)(TM.TB_1)=(TN.TC_1)(TE.TL)=(TN.TE).TC_1.TL=TM.TF.TB_1.TL$ Hence $TK=TL$ . $\square$
Since $TK=TL$ and $TB_1=TC_1,$ we see that $B_1C_1LK$ is an isosceles trapezoid whose diagonals intersect at $D$ .
$Pow(D,(B_1EC))=DB_1.DL=DC_1.DK=Pow(D,(C_1FB))$ Which completes our proof as desired. $\blacksquare$

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soryn

#5 h Sep 25, 2024, 5:41 PM

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Very nice problem and very nices solutions!

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#6 h Sep 26, 2024, 8:32 PM

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To continue, prove that the same is true when A1, B1 and C1 are the feet of the altitudes and A2, B2 and C2 are the antipodes of A, B and C on the circumcircle, and that the radical axis then is the Euler's line.
On the figure below, the red circles and the green circles corespond to the initial problem and the continuation, respectively.

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#7 h Apr 17, 2025, 8:06 AM

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Two beautiful proof for this beautiful problem. Here i provides two lemmas that technically solves the problem. How to prove them is your choice, it can be shorter may be. But i have nicer proofs here.

Lemma 1: (just for the sake of proving lemma 2). $O$ is the circumcircle of $\triangle ABC$ . $AO \cap (BC)=A_2$ , and $S$ be the intersection of $A$ altitude with $(ABC)$ . Let's assume $B_2$ and $C_2$ defined similarly as $A_2$ . Let $B'$ and $C'$ be the antipodes of $B$ and $C$ respectively. Let $D=SA_2 \cap (ABC)$ , $D'=DO \cap (ABC)$ . And let $B'C'$ and tangent of $(ABC)$ at $A$ intersect at $F$ . Then $F$ lies on $B_2C_2$ .

Lemma 2: Let's assume all the points defined same as Lemma 1. Let $X$ be the projection of $A_2$ to line $B_2C_2$ . Then $X$ lies on $(AA_1A_2)$ , which is actually follows from $F$ lies on D'S.

Proof for the lemma 1:
For the convenience let's assume that $B_2C_2 \ AA=F$ , then let's prove that $F$ lies on $B'C'$ . By the converse of Menelaus theorem it suffices to prove that $\frac{FC_2}{FB_2} \cdot \frac{C'O}{C'C_2} \cdot \frac{B'B_2}{B'O}=\frac{FC_2}{FB_2} \cdot \frac{B'B_2}{C'C_2}=1$ , which can turn to mess if you don't use trigonometry instantly. In $\triangle AC_2B_2$ :
by extended sines theorem, $\frac{FC_2}{FB_2}=\frac{AC_2}{AB_2} \cdot \frac{sin \gamma}{sin \beta}$ . Now we can express everything in terms of $\triangle ABC$ , so it's not hard for you.

Proof for the lemma 2:
I wont use any analytic technique here. Once you realize $FAA_2S$ is cyclic with diameter $A_2F$ , we just need to prove that $F-D'-S$ are collinear. Let's forget about $B_2, C_2$ and use lemma 1. Then new definition of $F$ will be $F=C'B' \cap AA$ . Since there is too many antipodes let's use inversion which swaps them. (which is the combination of inversion with $(ABC)$ , and reflection with point $O$ ).

Let $O$ be the circumcenter of $\triangle ABC$ . Let $S$ be the intersection of $(ABC)$ and $A$ altitude. Let $A'$ and $S'$ be the antipodes of $A$ and $S$ . Let $A_2=AO \cap BC$ and $D=SA_2 \cap (ABC)$ . Let $F'$ be the intersection of $(OBC)$ and circle with diameter $A'O$ . Prove that $OS'DF'$ are concyclic.

Let $F=OF' \cap BC$ . By regular inversion at $(ABC)$ (we're not using that, just motivational thing), we know that $FA'$ is tangent to $(ABC)$ . By some angle chase $DA_2A'F$ is cyclic, which proves the collinearity $F-D-S'$ . And we're done.

This post has been edited 1 time. Last edited by Nari_Tom, Apr 17, 2025, 9:12 AM

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#8 h Apr 17, 2025, 9:57 AM

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Actually they should mention that $\triangle ABC$ is not isosceles, at least not equilateral for this one.

Let $H$ and $H_2$ be the orthocenters of $\triangle ABC, \triangle A_2B_2C_2$ , respectively.
Since $Pow(H, (AA_1A_2))=Pow(H, (BB_1B_2))=Pow(H, (CC_1C_2))=HA \cdot HA_1=HB \cdot HB_1= HC \cdot HC_1$ , thus $H$ is radical center of these circles. By using the lemma 2 and, easy power of a point we can deduce that $H_2$ is also a radical center of these circles. Which means circles $(AA_1A_2)$ , $(BB_1B_2)$ , $(CC_1C_2)$ have two different radical centers $\implies$ these circles have common radical axis.

This post has been edited 1 time. Last edited by Nari_Tom, Apr 17, 2025, 9:58 AM

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#9 h Apr 17, 2025, 10:19 AM

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Another solution using very nice lemma.
Lemma: Let $ABCD$ be a convex quadrilateral. Let $E=AB \cap CD$ , $F=AD \cap BC$ . let $H_1, H_2, H_3, H_4$ be the four orthocenters of triangles formed by lines $AB,AD, CB,CD$ . Circles with diameter $AC, BD, EF$ have common radical axis, which passes through these orthocenters.
(Proof consists some power of a power argument, which is not hard. So you will do it yourselves.)

Let's solve the main problem. Let $X, Y, Z$ be the antipodes of $A_2, B_2, C_2$ in circles $(AA_1A_2), (BB_1B_2), (CC_1C_2)$ , respectively. We proved $X, Y, Z$ lies on the lines $B_2C_2, A_2C_2, A_2B_2$ , respectively. So let's just prove that $X-Y-Z$ are collinear. While proving the lemma 1, we've proved that $\frac{XC_2}{XB_2}=\frac{AC_2}{AB_2} \cdot \frac{AB}{AC}$ . So just by the Menelaus theorem on the $\triangle A_2B_2C_2$ and points $X, Y, Z$ , we're done.

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#10 h Apr 17, 2025, 10:22 AM

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For the second solution some should prove these circles should intersect (not tangent and have intersections with each other), For the first solution some should prove $H$ and $H_2$ are different. May be i will comeback for these later

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