IMO 2008, Question 1

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3647 posts

orl

#1 h Jul 16, 2008, 1:24 PM • 30 Y

Y by Davi-8191, Smita, microsoft_office_word, OlympusHero, stayhomedomath, Jc426, centslordm, Adventure10, jhu08, mathlearner2357, megarnie, HWenslawski, bjump, lian_the_noob12, Mango247, ItsBesi, Sedro, Tastymooncake2, Rounak_iitr, ehuseyinyigit, DEKT, AlexCenteno2007, cubres, and 7 other users

Let $H$ be the orthocenter of an acute-angled triangle $ABC$ . The circle $\Gamma_{A}$ centered at the midpoint of $BC$ and passing through $H$ intersects the sideline $BC$ at points $A_{1}$ and $A_{2}$ . Similarly, define the points $B_{1}$ , $B_{2}$ , $C_{1}$ and $C_{2}$ .

Prove that the six points $A_{1}$ , $A_{2}$ , $B_{1}$ , $B_{2}$ , $C_{1}$ and $C_{2}$ are concyclic.

Author: Andrey Gavrilyuk, Russia

This post has been edited 4 times. Last edited by orl, Jul 20, 2008, 8:14 AM

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Jan

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Jan

#2 h Jul 16, 2008, 1:47 PM • 24 Y

Y by kgo, microsoft_office_word, Williamgolly, OlympusHero, myh2910, centslordm, Adventure10, Adventure10, jhu08, EpicBird08, Mango247, Tastymooncake2, Sedro, Adi1005247, MS_asdfgzxcvb, cubres, and 8 other users

Call $M_1,M_2,M_3$ the midpoints of $BC,AC$ and $AB$ .

We know that $BP \perp AC$ , so $BP$ and $M_1M_3$ are also perpendicular, and $B$ lies on the radical axis of $\Gamma_A$ and $\Gamma_C$ . It follows that $BC_1\cdot BC_2 = BA_1 \cdot BA_2$ , so $A_1,A_2,C_1$ and $C_2$ all lie on a circle, whose center is clearly $O$ , the circumcenter of $\Delta ABC$ .

Now we can show that $B_1$ and $B_2$ also lie on the circle centered at $O$ , passing through $A_1$ and $A_2$ , from which the conclusion follows.

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TomciO

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TomciO

#3 h Jul 16, 2008, 1:52 PM • 28 Y

Y by mathmaster2012, AdBondEvent, Arthur., pad, Lifefunction, OlympusHero, MiraclesINmaths, myh2910, Adventure10, SPHS1234, jhu08, Numbertheorydog, rayfish, EpicBird08, Mango247, Tastymooncake2, Sedro, cubres, and 10 other users

The radical axis of $\Gamma_{A}$ and $\Gamma_{B}$ is perpendicular to the line connecting the centers of these circles, i.e. it is perpendicular to $AB$ . Since both circles pass trough $H$ their radical axis is a height from $C$ , so $C$ lies on a radical axis. It means that $CA_1CA_2 = CB_1CB_2$ so $A_1, A_2, B_2, B_1$ lie on one circle. From the same reasoning $B_1, B_2, C_2, C_1$ and $C_1, C_2, A_2, A_1$ lie on a one circle as well. Suppose that this three circles doesn't coincide. Then we obtain a contradiction since radical axis of these circles - the sides of the triangle $ABC$ - don't intersect in one point.

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Lepuslapis

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Lepuslapis

#4 h Jul 16, 2008, 1:59 PM • 4 Y

Y by jhu08, Adventure10, Mango247, cubres

Another possibility to prove that, let's say, $A_1A_2B_1B_2$ is cyclic consists in transforming the relation $CA_1\cdot CA_2 = CB_1\cdot CB_2$ into $CD^2 - DH^2 = CE^2 - EH^2$ (where $D$ and $E$ are the respective midpoints of $[BC]$ and $[AC]$ ). Adding $CH^2$ on both sides and applying the generalized version of Pythagoras' theorem (the one with cosines, I ignore the correct English name), we get a condition which is obviously true (simplifying cosines in right-angled triangles)...

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plane geometry

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plane geometry

#5 h Jul 16, 2008, 2:11 PM • 5 Y

Y by Mathmick51, jhu08, Adventure10, Mango247, cubres

A,B,C lie on the radix axis of each two circles
so conclusion apparently follows

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TTsphn

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TTsphn

#6 h Jul 16, 2008, 2:17 PM • 11 Y

Y by Mobashereh, jhu08, Adventure10, Mango247, Tastymooncake2, AlexCenteno2007, cubres, and 4 other users

I have a same solution with you .
Call $M,N,P$ is the midpoint of $BC,CA,AB$
From condition we have :
$BA_1.BA_2=BM^2-HM^2$
$BC_1.BC_2=BN^2-HN^2$
But from $MN||AC,BH\perp AC$ therefore $BH\perp MN$
It gives $BM^2-BN^2=HM^2-HN^2$
Therefore $A_1,B_1,A_2,B_2$ are cyclic on circle $O_c$
Similar for $A_1,A_2,C_1,C_2$ lie on $O_b$ and $B_1,B_2,C_1,C_2$ lie on circle $O_a$
Easy to check that three circle are coincide .
So problem claim.

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pohoatza

1145 posts

pohoatza

#7 h Jul 16, 2008, 2:41 PM • 12 Y

Y by iarnab_kundu, trumpeter, Jc426, Adventure10, jhu08, Tastymooncake2, cubres, and 5 other users

Indeed, the proof involving the three radical axis is the best. In any case, the circle belongs to Droz-Farny as Darij well mentioned (see http://www.pandd.demon.nl/drozf.htm; scroll down till figure 3).

By the way, it is obvious that the triangle doesn't need to be acute-angled. I guess the coordinators didn't want messy diagrams

.

In addition, here is a natural generalization:

Theorem. Let $P$ , $Q$ be two isogonal points with respect to a given triangle $ABC$ . Let $P_{A}$ , $P_{B}$ , $P_{C}$ be the orthogonal projections of $P$ on the sidelines $BC$ , $CA$ and $AB$ , respectively. Denote by $\Gamma_{A}$ the circle centered at $P_{A}$ which passes through $Q$ and let $A_{1}$ , $A_{2}$ be the intersections points of $\Gamma_{A}$ with the sideline $BC$ . Similarly, define $B_{1}$ , $B_{2}$ , $C_{1}$ , $C_{2}$ . Then, the points $A_{1}$ , $A_{2}$ , $B_{1}$ , $B_{2}$ , $C_{1}$ , $C_{2}$ are on a same circle centered at $P$ .

I guess it would have been a better IMO problem, at least to avoid the "well-known"-type commentaries.

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Ahiles

374 posts

Ahiles

#8 h Jul 16, 2008, 4:12 PM • 11 Y

Y by thunderz28, microsoft_office_word, Adventure10, Mango247, Tastymooncake2, cubres, and 5 other users

It is enough to prove that points $A_1,A_2,B_1,B_2$ are concyclic. Let $O$ be circumcenter of the triangle $ABC$ . We'll prove that $O$ is circumcenter of quadrilateral $A_1A_2B_1B_2$ . Then $O$ is situated on the perpendicular bisectors of the segments $A_1A_2$ and $B_1B_2$ $\Longrightarrow OA_1 = OA_2, OB_1 = OB_2$ . So it is enough to prove that $OA_1 = OB_1$ .
Call $r_1,r_2$ circumcenters of the $\Gamma_a$ and $\Gamma_b$ respectively and $A',B'$ midpoints of the segments $BC,CA$ . Then $\left\|\begin{array}{cc} OA_1 = r_1^2 + OA'^2 \\ OB_1 = r_2^2 + OB'^2 \end{array} \right\|$ .
By Stewart we have
$\boxed{\begin{array}{cc} 4r_1^2 = 2(CH^2 + BH^2) - a^2 = \\ = 8R^2(\cos^2{B} + \cos^2{C}) - 4R^2\sin^2{A} = \\ = 4R^2(2\cos^2{B} + 2\cos^2{C} - \sin^2{A})\end{array}}$ .

By analogy $4r_2^2 = 4R^2(2\cos^2{A} + 2\cos^2{C} - \sin^2{B})$ .

But $\left\| \begin{array}{cc} OA'^2 = \frac {1}{4}a^2\cos^2{A} = R^2\cot^2{A} \\ OB'^2 = \frac {1}{4}b^2\cos^2{B} = R^2\cot^2{B} \end{array} \right\|$ .

Then $OA_1^2 = 4R^2(2\cos^2{B} + 2\cos^2{C} - \sin^2{A} + \cot^2{A})$ and $OB_1^2 = 4R^2(2\cos^2{A} + 2\cos^2{C} - \sin^2{B} + \cot^2{B})$ .

So we have to prove that:
$\boxed{\begin{array}{cc} 2\cos^2{B} + 2\cos^2{C} - \sin^2{A} + \cot^2{A} = 2\cos^2{A} + 2\cos^2{C} - \sin^2{B} + \cot^2{B} \\ \cos^2{B} + (\cos^2{B} + \sin^2{B}) + \cot^2{A} = \cos^2{A} + (\cos^2{A} + \sin^2{A}) + \cot^2{B} \\ \cos^2{B} + \cot^2{A} = \cos^2{A} + \cot^2{B} \\ \cos^2{B} + \frac {\cos^2{A}}{\sin^2{A}} = \cos^2{A} + \frac {\cos^2{B}}{\sin^2{B}} \\ \cos^2{B}\sin^2{A} + \cos^2{A} = \cos^2{A}\sin^2{B} + \cos^2{B} \\ \cos^2{B}(\sin^2{A} - 1) = \cos^2{A}(\sin^2{B} - 1) \\ - \cos^2{B}\cos^2{A} = - \cos^2{B}\cos^2{A} \end{array} }$
and we are done !!!!

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cyshine

236 posts

cyshine

#9 h Jul 16, 2008, 4:29 PM • 3 Y

Y by Adventure10, Mango247, cubres

Fix $O$ as center of a coordinate vector system, so that $H = A + B + C$ . If $M = \frac{A + B}2$ we have to prove that $MC_1^2 + OM^2$ is symmetric in $A$ , $B$ and $C$ (it would be the square of the radius of the circle).

But

$MC_1^2 + OM^2 = HM^2 + OM^2 = \left(A + B + C - \frac{A+B}2\right)\cdot \left(A + B + C - \frac{A+B}2\right) + \left(\frac{A+B}2\right)\cdot \left(\frac{A+B}2\right) = \frac12A\cdot A + \frac12B\cdot B + C\cdot C + A\cdot B + B\cdot C + C\cdot A = 2R^2 + A\cdot B + B\cdot C + C\cdot A$

and we are done.

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msecco

154 posts

msecco

#10 h Jul 16, 2008, 5:26 PM • 3 Y

Y by Adventure10, Mango247, cubres

No no no!!
More one trivial Geometry.
Denote AB=2c, AC=2b and BC=2a.
Denote BA_1=x and AC_1=y.Denote too M, N and P the midpoints of BC, AB and AC respectively.
Then, MA_1=a-x=MA_2. We have to prove that BA_1.BA_2=BC_2.BC_1.
Applying the Cosines'Law in triangles BNH and BMH, we have:

a²-2ax+x²=a²+BH²-2a.BH.sen<C and c²-2cy+y²=c²+BH²-2c.BH.sen<A <=>
2ax-x²=2a.BH.sen<C-BH² and 2cy-y²=2c.BH.sen<A-BH².
But 2ax-x²=BA_1.BA_2 and 2cy-y²=BC_2.BC_1.
Then, we have to prove that: 2c.BH.sen<A-BH²=2a.BH.sen<C-BH² <=>2a.sen<C=2c.sen<A (obviously by Sines'law).
Hence, A_1,A_2,C_1,C_2 are concyclics and by the same form, the other 2 quadruples of points are concyclics.
Conclusion: The six points are concyclics!

Have fun!

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not_trig

2049 posts

not_trig

#11 h Jul 16, 2008, 5:34 PM • 6 Y

Y by Adventure10, Mango247, DEKT, cubres, and 2 other users

What is a sideline!? What is sideline $BC$ ?

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kaloi01

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kaloi01

#12 h Jul 16, 2008, 5:51 PM • 7 Y

Y by Adventure10, Tastymooncake2, cubres, and 4 other users

See point $A$ first. let $X$ is the midpoint of $AB$ and $Y$ is the midpoint $AC$ . It follows that $XY \perp AH$ . Thus, $AH$ is the radical axis of circle $B_1B_2H$ and $C_1C_2H$ . It follows that $AC_1*AC_2 = AB_1*AB_2 \Leftrightarrow B_1B_2C_1C_2$ cylcic, by similar argument on $B$ and $C$ we´re done.

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campos

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campos

#13 h Jul 16, 2008, 6:04 PM • 7 Y

Y by Adventure10, Tastymooncake2, cubres, and 4 other users

it's enough to prove that $O$ is at the same distance of any of the six points... obviously $OA_1 = OA_2$ , so it's enough to prove that the expression $OA_1$ is independent from $A$ . In fact, $OA_1^2 = OM^2 + MA_1^2 = R^2\cos^2A + MH^2$ ...

it can be proven that $MH^2 = R^2(4\cos^2B\cos^2C + \sin^2(B - C))$

so, $OA_1^2 = R^2(\cos^2A + 4\cos^2B\cos^2C + \sin^2(B - C))$ ...

finally, after some work (you can try $\cos A = \sin B\sin C - \cos B\cos C$ ) it follows that it equals $R^2(1-4\cos A\cos B\cos C)$ , and we're done.

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Virgil Nicula

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Virgil Nicula

#14 h Jul 16, 2008, 6:37 PM • 8 Y

Y by Adventure10, Mango247, Tastymooncake2, Rounak_iitr, cubres, and 3 other users

orl wrote:

Let $H$ be the orthocenter of an acute triangle $ABC$ . The circle centered at the midpoint of $BC$ and passing through $H$ intersects

the line $BC$ at $A_{1}$ , $A_{2}$ . Similarly define the pairs $B_{1}$ , $B_{2}$ and $C_{1}$ , $C_{2}$ . Prove that $A_{1}$ , $A_{2}$ , $B_{1}$ , $B_{2}$ , $C_{1}$ , $C_{2}$ are concyclically.

Proof. Denote the midpoints $D$ , $E$ , $F$ of the sides $[BC]$ , $[CA]$ , $[AB]$ respectively. Prove easily or is well-known that

$\boxed {\ HA^2 + a^2 = HB^2 + b^2 = HC^2 + c^2 = 4R^2\ }\ (*)$ . Thus, $4\cdot\overline {BA_1}\cdot\overline {BA_2} = 4\cdot\left(BD^2 - HD^2\right) =$ $a^2 - 4\cdot HD^2 =$

$a^2 - 2\cdot\left(HB^2 + HC^2\right) + a^2$ $\stackrel {(*)}{\ \ \implies\ \ }$ $\overline {BA_1}\cdot\overline {BA_2} = \frac 12\cdot \left(a^2 + b^2 + c^2\right) - 4R^2$ $\implies$ $\boxed {\ \overline {BA_1}\cdot\overline {BA_2} = 4R^2\prod\cos A\ }$

because prove easily or is well-known that $\boxed {\ \left(a^2 + b^2 + c^2\right) - 8R^2 = 8 R^2\prod\cos A\ } > 0$ (remark that $\{A_1,A_2\}\subset (BC)$ a.s.o.).

Thus, $\overline {BA_1}\cdot \overline {BA_2} = - \overline {A_1B}\cdot\overline {A_1C} = R^2 - OA_1^2\ \implies\ 4R^2\prod\cos A = R^2 - OA_1^2$ $\implies$ $OA_1^2 = R^2\left(1 - 4\prod\cos A\right)$

(symmetrically in $a$ , $b$ , $c$ ). In conclusion, $\boxed {\ \rho = OA_1 = OA_2 = OB_1 = OB_2 = OC_1 = OC_2 = R\sqrt {1 - 4\cos A\cos B\cos C}\ }$ .

Remark. Since $1 - 8\prod\cos A\ge 0$ obtain easily that $\rho\ge 2R\sqrt {\prod\cos A}$ .

Pohoatza wrote:

I guess it would have been a better IMO problem, at least to avoid the "well-known"-type commentaries.

All right !

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Akashnil

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Akashnil

#15 h Jul 16, 2008, 6:44 PM • 13 Y

Y by iarnab_kundu, AnonymousBunny, Adventure10, Adventure10, Mango247, Mango247, Mango247, Tastymooncake2, cubres, and 4 other users

Let $O$ circumcenter, $D$ midpoint of $BC$ , $N$ nine point center.
$OA_1^2=OD^2+DA_1^2=OD^2+DH^2=2(DN^2+ON^2)=const.$

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G0d_0f_D34th_h3r3

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G0d_0f_D34th_h3r3

#150 h Jun 16, 2024, 5:37 PM • 1 Y

Y by cubres

Let's prove this nice problem

.

If we prove that $B_1$ , $B_2$ , $C_1$ and $C_2$ are concyclic then rest of the cases follow similarly.
Let $\omega_2$ and $\omega_3$ to be $(HB_1B_2)$ and $HC_1C_2$ respectively.

Lemma: $A$ lies on the radical axis of $\omega_2$ and $\omega_3$ .
Proof We know that $\omega_2$ and $\omega_3$ have centers $M_2$ and $M_3$ respectively.

Since, $M_1$ and $M_2$ are midpoints of $AC$ and $AB$ , so
$M_2M_3 \parallel BC$ Also since line $AH \perp BC$ , so
$AH \perp M_2M_3$ Hence, $A$ lies on the radical axis of $\omega_2$ and $\omega_3$ .

Since, $B_1$ , $B_2$ and $C_1$ , $C_2$ lie on $\omega_2$ and $\omega_3$ and $B_1B_2$ and $C_1C_2$ intersect on their radical axis.
Therefore, $B_1$ , $B_2$ , $C_1$ and $C_2$ are concyclic.

Attachments:

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dudade

139 posts

dudade

#151 h Jun 24, 2024, 5:01 AM • 1 Y

Y by cubres

Let $M$ and $N$ be the midpoints of $BC$ and $CA$ , respectively. Since $NM \perp CH$ , then the second intersection of $\left(A_1HA_2\right)$ and $\left(B_1HB_2\right)$ , say $I$ , lies on $CH$ .

By Power of Point, on $\left(A_1HA_2\right)$ and $\left(B_1HB_2\right)$ yields:
$\begin{cases} A_1HA_2 &: A_1C \cdot CA_2 = CI \cdot CH \\ B_1HB_2 &: B_1C \cdot CB_2 = CI \cdot CH. \end{cases}$ Equating yields $A_1C \cdot CA_2 = B_1C \cdot CB_2$ which, by Power of Point, implies $A_1A_2B_1B_2$ is cyclic. Similarly, $B_1B_2C_1C_2$ and $C_1C_2A_1A_2$ are both cyclic, thus $A_1$ , $A_2$ , $B_1$ , $B_2$ , $C_1$ , and $C_2$ are concyclic, as desired.

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Jndd

1417 posts

Jndd

#152 h Jul 22, 2024, 9:20 PM • 1 Y

Y by cubres

okayy here's my hilariously bad writeup originally done on mathdash:

If $A_1, A_2, B_1, B_2, C_1, C_2$ were concyclic, then their center would have to be the circumcenter of $ABC$ , which we will call $O$ , since the perpendicular bisector of $A_1A_2$ is the same as the perpendicular bisector of $BC$ , and likewise for all the other sides of $ABC$ .

Let $D$ , $E$ , and $F$ be the midpoints of sides $BC$ , $AC$ , and $AB$ . In order to show that the distance from $O$ to all of $A_1, A_2, B_1, B_2, C_1, C_2$ are the same, we will show that $OA_1^2 = DA_1^2+DO^2=DH^2+DO^2$ and the same with the other points are equal. Let $r$ be the radius of the circumcircle of $ABC$ , so by Power of Point, we have $(r+DO)(r-DO)=BD^2$ , giving $DO^2=r^2-BD^2$ , so $DO^2+DH^2=r^2-(BD^2-DH^2)$ .

Let $X_D$ be the point on $\Gamma_A$ inside $ABC$ such that $BX_D$ is tangent to $\Gamma_A$ and let $X_F$ be the point on $\Gamma_C$ inside $ABC$ such that $BX_F$ is tangent to $\Gamma_C$ . Since $BD^2-DH^2=BX_D^2$ and $BF^2-FH^2=BX_F^2$ , we want to show $BX_D=BX_F$ . This is simply true because $B$ lies on the radical axis of $\Gamma_A$ and $\Gamma_C$ since $H$ definitely lies on the radical axis, and since that radical axis is perpendicular to the line connecting their centers, $B$ also lies on it since $BH\perp DF$ because $BH\perp AC$ .

Now that we know $BX_D=BX_F$ , we get $OA_1^2=DO^2+DH^2=r^2-(BD^2-DH^2) = r^2 - (BF^2-FH^2)=FO^2+FH^2=OC_2$ , meaning $OA_1=OC_2$ . Therefore, using symmetry, we now have that all of $A_1, A_2, B_1, B_2, C_1, C_2$ have the same distance from $O$ , as desired.

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RedFireTruck

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RedFireTruck

#153 h Jul 30, 2024, 6:43 AM • 1 Y

Y by cubres

https://mathdash.s3.us-east-2.amazonaws.com/user-uploads/65c15bae768c20d4b06de6f0/1722320783919-11

clearly the center of such a circle would have to be at the circumcenter (intersection of green lines) because the circles are centered at midpoints so the perp. bisector of $A_1A_2$ would be the same as the perp. bisector of $BC$ , for example

now it just suffices to show that the sum of squares of the purple lengths in the picture doesn't depend on which midpoint they are on (in the example they are on the midpoint of $BC$ )

let the circumcenter be $0$ and the vertices be $a,b,c$ s.t. $|a|=|b|=|c|=1$ and the orthocenter is $a+b+c$

then the sum of the squares of the purple lengths in the picture is $|\frac{b+c}{2}|^2+|(a+b+c)-\frac{b+c}{2}|^2=|\frac{b+c}{2}|^2+|a+\frac{b+c}{2}|^2$ $=\frac{b+c}{2}\frac{\frac1b+\frac1c}{2}+(a+\frac{b+c}{2})(\frac{1}{a}+\frac{\frac1b+\frac1c}{2})$ $=2+\frac{\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}}{2}$ which is symmetric in $a,b,c$ , as desired.

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ezpotd

1286 posts

ezpotd

#154 h Aug 30, 2024, 8:44 PM • 1 Y

Y by cubres

Claim: $A_2A_1B_1B_2$ and cyclic variants are all cyclic.

Proof: Let $X_C$ be the intersection of $(HA_1A_2), (HB_1B_2)$ , clearly $X_C$ is the reflection of $H$ over $M_AM_B$ , so $X_CH$ is just the $C$ altitude, and by radax we are done.

Now assume these three circles $A_1A_2B_1B_2$ and cyclic variants are all distinct, then their radaxes should concur but they happen to precisely be the sides of the triangle, so the three circles are the same and all six points are cyclic.

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MagicalToaster53

159 posts

MagicalToaster53

#155 h Sep 3, 2024, 11:16 PM • 1 Y

Y by cubres

Suppose $BH$ and $CH$ meet $\overline{CA}$ and $\overline{AB}$ at $E$ and $F$ , respectively. Then observe $\Gamma_A \cap \Gamma_B = X$ lies on $CH$ . Indeed, $XH$ is the radical axis of $\Gamma_A$ and $\Gamma_B$ , we find $XH \perp M_AM_B \implies XH \perp AB$ . Hence $C$ has equal power from both $\Gamma_A$ and $\Gamma_B$ so that $CA_2 \times CA_1 = CB_1 \times CB_2 \implies A_2, A_1, B_1, B_2$ are concyclic. Similarly $B_1, B_2, C_1, C_2$ and $C_1, C_2, A_1, A_2$ are separately concyclic, which follows from symmetry. Hence $A_1, A_2, B_1, B_2, C_1, C_2$ all lie on a circle (and further the circumcenter of such a circle is the circumcenter of $ABC$ ), as desired. $\blacksquare$

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megahertz13

3184 posts

megahertz13

#156 h Nov 26, 2024, 6:16 PM • 1 Y

Y by cubres

Let $X$ , $Y$ , and $Z$ be the midpoints of the sides opposite $A$ , $B$ , and $C$ , respectively. Also, let $\omega_1, \omega_2$ be the circles centered at $Y$ and $Z$ , respectively. Suppose that $\omega_1$ and $\omega_2$ intersect again at $P$ .

Since $PH\perp YZ$ and $AH\perp BC\implies AH\perp YZ$ , we have $P, A, H$ collinear.

Now, $A$ lies on the radical axis of $\omega_1,\omega_2$ , so $B_1, B_2, C_1,C_2$ are concyclic. Similarly, we can find that $A_1, A_2, B_1, B_2$ are concyclic, as are $C_1, C_2, A_1, A_2$ . Notice that the center of those circles is the circumcenter $O$ of $ABC$ . Since those three circles have the same radius of $OA_1=OA_2=OB_1=OB_2=OC_1=OC_2,$ they are the same circle and we are done.

This post has been edited 1 time. Last edited by megahertz13, Nov 26, 2024, 6:16 PM

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gladIasked

648 posts

gladIasked

#157 h Dec 1, 2024, 7:27 PM • 1 Y

Y by cubres

Let $M$ be the midpoint of $AC$ and $N$ be the midpoint of $AB$ . Note that $AH\perp BC$ and $BC\parallel MN$ by similar triangles. Thus, $AH\perp MN$ . Consider the circles $(C_1HC_2)$ and $(B_1HB_2)$ ; note that the centers of these two circles are $N$ and $M$ , respectively. However, because $H$ lies on both circles and $AH\perp MN$ , we know that $AH$ is the radical axis of the two circles. Therefore, by power of a point, $B_1B_2C_1C_2$ is cyclic. We can similarly conclude that $A_1A_2B_1B_2$ and $C_1C_2A_1A_2$ are cyclic.

Assume for the sake of contradiction that $(A_1A_2B_1B_2)\ne (B_1B_2C_1C_2)$ . Clearly, $AC$ is the radical axis of these two circles. However, the power of $B$ with respect to $(B_1B_2C_1C_2)$ is $BC_2\cdot BC_1 = BA_1\cdot BA_2$ by power of a point on circle $(C_1C_2A_1A_2)$ . Clearly, the power of $B$ with respect to $(A_1A_2B_1B_2)$ is $BA_1\cdot BA_2$ , so in fact $B$ must lie on the radical axis of $(A_1A_2B_1B_2)$ and $(B_1B_2C_1C_2)$ , or just $AC$ . This is a contradiction, so $A_1$ , $A_2$ , $B_1$ , $B_2$ , $C_1$ , $C_2$ are concyclic. $\blacksquare$

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Vedoral

89 posts

Vedoral

#158 h Dec 20, 2024, 9:55 PM • 1 Y

Y by cubres

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Maximilian113

575 posts

Maximilian113

#159 h Dec 30, 2024, 8:50 PM • 1 Y

Y by cubres

Let $D, E, F$ be the midpoints opposite $A, B, C.$ We begin by showing that $C_2, C_1, B_2, B_1$ are concyclic. Let $P$ be the second intersection of the circles, then clearly $PH \perp EF$ but since $EF \parallel BC$ by the Midpoint Theorem, it follows that $PH \perp BC,$ so $A$ lies on $PH.$ Therefore, $A$ lies on the radical axis of $\Gamma_B, \Gamma_C$ and by Power of a Point our claim is proven.

Similarly, $B_1, B_2, A_1, A_2$ are concyclic, and $C_1, C_2, A_1, A_2$ are too. But their centers are clearly $O,$ the circumcenter of $\triangle ABC,$ so we are done. QED

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clarkculus

244 posts

clarkculus

#160 h Feb 9, 2025, 10:07 PM • 2 Y

Y by centslordm, cubres

Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $AC$ . Because $AH\perp BC$ , $AH\perp MN$ , and because $H$ lies on the radical axis of $\Gamma_B$ and $\Gamma_C$ , $A$ also lies on the radical axis of $\Gamma_B$ and $\Gamma_C$ , implying $(AC_1)(AC_2)=(AB_1)(AB_2)$ . By the Converse of Power of a Point, this implies $C_1,C_2,B_1,B_2$ are concyclic. Similarly, $A_1,A_2,C_1,C_2$ are concyclic, which finishes.

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QueenArwen

110 posts

QueenArwen

#161 h Apr 2, 2025, 7:11 AM • 1 Y

Y by cubres

Let $M_1$ be the midpoint of $AB$ and $M_2$ be the midpoint of $AC$ . Since $M_1M_2$ is parallel to $BC$ , $AH$ is perpendicular to it, hence $A$ lies on the radical axis of $\Gamma_B$ and $\Gamma_C$ (since $H$ lies on both these circles). Since $C_1C_2$ and $B_1B_2$ intersect on the radical axis, $B_1B_2C_1C_2$ is cyclic by a well-known theorem. Similarly $A_1A_2B_1B_2$ and $A_1A_2C_1C_2$ are also cyclic. Since the perpendicular bisectors of $C_1C_2$ and $B_1B_2$ are the same as those of $AB$ and $AC$ respectively, the circumcenter of cyclic quadrilateral $B_1B_2C_1C_2$ and $\triangle{ABC}$ is the same, which we call $O$ . Similarly $O$ is also the circumcenter of $A_1A_2B_1B_2$ and $A_1A_2C_1C_2$ , so $OA_1 = OA_2 = OB_1=OB_2=OC_1=OC_2$ , hence $A_1,A_2,B_1,B_2,C_1,C_2$ are concyclic.

This post has been edited 2 times. Last edited by QueenArwen, Apr 3, 2025, 6:03 AM

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eg4334

637 posts

eg4334

#162 h Apr 8, 2025, 1:17 AM • 1 Y

Y by cubres

Clearly we need to only prove it true for two circles, lets say the $A$ and $B$ ones. Then, it suffices for $C$ to have the same power wrt to both, or $C, H, K$ to be collinear where $K$ is their second intersection. But $HK$ (because perpendicular to line joining centers) and $HC$ (orthocenter) are both perpendicular to $AB$ and the result follows.

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Ilikeminecraft

658 posts

Ilikeminecraft

#163 h Apr 25, 2025, 7:23 AM • 1 Y

Y by cubres

Let the circles through the midpoints of $AB, BC, AC$ be $\omega_C, \omega_A, \omega_B,$ respectively.

I will prove that $A_1, A_2, B_1, B_2$ is cyclic. Notice that the radical axis of $\omega_A, \omega_B$ must be perpendicular to $EF,$ and it also runs through $H.$ Thus, $C$ also lies on this radical axis. However, because $CH, AC, BC,$ are concurrent, we have that there must exist a circle running through $A_1, A_2, B_1, B_2.$

Similarly, there is a circle going through $A_1, A_2, C_1, C_2,$ and another one going through $B_1, B_2, C_1, C_2.$ However, in all circles, the center is clearly $O,$ the circumcenter. Thus, we are done.

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Siddharthmaybe

117 posts

Siddharthmaybe

#164 h May 11, 2025, 6:05 AM • 1 Y

Y by alexanderhamilton124

Let $M_A, M_B, M_C$ be the midpoints of $BC,CA,AB$ , then by midpoint theorem, $M_AM_B \perp HC$ , also $\Gamma_A \cap\Gamma_B = H$ (and some other point). Now these conditions are enough to deduce that $HC$ is the radax of $\Gamma_A, \Gamma_B$ . So evidently, $CA_1 \times CA_2 = CB_1 \times CB_2$ or $A_1A_2B_1B_2$ is cyclic and similarly for $A_1A_2C_1C_2, B_1B_2C_1C_2$ . Now assume FTSOC that they have distinct circumcircles, then the radax of the first two circles is the common chord $\overleftrightarrow{A_1A_2} \equiv \overleftrightarrow{BC}$ , similarly the lines $AB, CA$ are also the radical axes of the other two pairs, now by radical center they must be concurrent which means that $ABC$ is degenerate $\implies\impliedby$ .

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