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Source: Iran 2005

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1275 posts

#1 h Aug 27, 2005, 1:02 PM • 2 Y

Y by Adventure10, Mango247

Suppose $H$ and $O$ are orthocenter and circumcenter of triangle $ABC$ . $\omega$ is circumcircle of $ABC$ . $AO$ intersects with $\omega$ at $A_1$ . $A_1H$ intersects with $\omega$ at $A'$ and $A''$ is the intersection point of $\omega$ and $AH$ . We define points $B',\ B'',\ C'$ and $C''$ similiarly. Prove that $A'A'',B'B''$ and $C'C''$ are concurrent in a point on the Euler line of triangle $ABC$ .

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#2 h Aug 28, 2005, 5:26 PM • 2 Y

Y by Adventure10, Mango247

I think the statement obscures what's really going on: if you replace the circumcircle with a circumconic, and $O,H$ with any two points $U,V$ , you still have a valid problem

. So let's restate it (it won't be the exact analog, but something equivalent):

Problem: Let $BB_1B''B',CC_1C''C'$ be two quadrilaterals inscirbed in a conic $\mathcal T$ having the same intersection $V$ of the diagonals (i.e. $V=B_1B'\cap BB''=C_1C'\cap CC''$ ). Put $U=CC_1\cap BB_1,T=C'C''\cap B'B''$ . Then $U,V,T$ are collinear.

If we define $A_1,A',A''$ in a similar manner and prove the analogous statements for the pairs of configurations corresponding to $(A,B),(A,C)$ , we will have shown that each two among the lines $A'A'',B'B'',C'C''$ intersect on $UV$ , which, except maybe for some uninteresting limit cases, is different from all of them, meaning that all three of them actually concur on $UV$ , as desired.

Now for a proof of the problem stated above:

Fix the quadrilateral $BB_1B''B'$ and move $C$ on $\mathcal T$ . This gives rise to a self-map $C'\mapsto C''$ of $\mathcal T$ which is clearly projective. When $C$ reaches the position that $C_1$ initially had, $C'$ and $C''$ are interchanged, so this map is an involution, meaning that $C'C''$ passes through a fixed point. When $C=B_1$ , the line $C'C''$ coincides with $B''B'$ , so this fixed point is actually the intersection $T$ between $C'C''$ and $B'B''$ . Finally, when $C=B''$ , the collinearity of $U,V,T$ follows from a direct application of Pascal's Theorem applied to the hexagon $CC_1C'C''B_1B'$ (in this configuration $C=B'',C''=B$ ).

There's probably a slicker and more direct proof of the problem based on a couple of applications of Pascal's Theorem as well, but I didn't want to go point-chasing

.

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#3 h Aug 29, 2005, 1:50 AM • 2 Y

Y by Adventure10, Mango247

what is point chasing?

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#4 h Aug 29, 2005, 11:09 AM • 1 Y

Y by Adventure10

shobber wrote:

what is point chasing?

I believe it is not mathematically defined. Anyway, grobber's meaning of point chasing is understandable here: Applying Pascal Theorem to one orientation of six of $B,B_1,B'',B',C,C_1,C'',C'$ . I've tried it last night(GMT+8), and applying two times is enough.
Anyway, I prefer using trigo rather than Pascal to solve it-they are just the same

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#5 h Jun 27, 2006, 9:28 AM • 2 Y

Y by Adventure10 and 1 other user

Just wanted to note:

The points H, O, H lie on the Euler line of triangle ABC.
The triangle ABC is the circumcevian triangle of the point H with respect to triangle A''B''C''.
The triangle $A_1B_1C_1$ is the circumcevian triangle of the point O with respect to triangle ABC.
The triangle A'B'C' is the circumcevian triangle of the point H with respect to triangle $A_1B_1C_1$ .

Thus, the statement of the problem, namely that the lines A''A', B''B', C''C' concur at one point on the Euler line of triangle ABC, follows from the "circumcevian pingpong theorem".

Darij

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jensen

#6 h Jun 29, 2006, 2:35 PM • 1 Y

Y by Adventure10

here is my solution:
if we use Pascal theorem about inscribed hexagons we will have:
in $A_1AA'C_1CC'$ $\rightarrow$
$[(A_1A , C_1C)=O]$ & $[(AA',CC')=H]$ & $[(A_1C' , C_1A')=X]$ are on one line.(2)
and in $C_1A'A''A_1C'C''$ $\rightarrow$
$[(C_1A' , C'A_1)=X]$ &
$[(A'A'' , C'C'')=Q]$ & $[(C_1C'' , A_1A'')=H]$ are on one line.(1)
and we have $[(C_1A' , C'A_1)=(A_1C' , C_1A')]$ AND
$[(C_1C'' , A_1A'')=(AA' , CC')=H]$ (3)
and from (1),(2) and (3) we get that $H,O,Q,X$ are on one line.
and it means that $A'A'',C'C'',OH$ are concurrent.
Similary we get that $B'B'',C'C'',OH$ are concurrent and we R done

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#7 h Jun 13, 2018, 7:40 PM • 2 Y

Y by Adventure10, Mango247

PRETTY EASILY USING COMPLEX

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#8 h Jun 13, 2018, 8:25 PM • 2 Y

Y by Adventure10, Mango247

Trivial cause using a composition of three involutions in $A'$ to $A$ to $A_1$ to $A''$ with focus $H,O,H$ resp, makes $A',A"$ an involution with focus at $OH$

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Yaghi

#9 h Jun 14, 2018, 4:17 AM • 2 Y

Y by Adventure10, Mango247

Another way to do this is poles and polar:
Let $X_a$ be the pole of $A'A"$ .Define $X_b,X_c$ similarly.Since $(A',A";B,C)=-1 \implies X_a \in BC$ .Note that
$X_aO^2 -X_aH^2=X_aA'^2+R^2-X_aH^2=R^2$ So by Carnot's theorem, $X_a,X_b,X_c$ are collinear and the line passing through them is perpendicular to $OH$ ,so their polar are concurrent on $OH$ and we are done.

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#10 h Jun 14, 2018, 4:29 AM • 2 Y

Y by Adventure10, Mango247

It can also be proven very easily using inversion .

This post has been edited 2 times. Last edited by enhanced, Jul 7, 2019, 2:21 AM

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#11 h Aug 17, 2023, 5:11 PM

Y by

Perform an inversion centered at $H$ with radius $\sqrt{-HA\cdot HA''}$ . So this inversion fixes the circumcircle.

After this inversion, it is sufficient to show circumferences $(A''H)$ , $(B''H)$ , $(C'C''H)$ are coaxial for the concurrency of $A'A''$ , $B'B''$ , $C'C''$ . Notice that $\angle A'A''H=\angle A'A_1O=\angle A_1A'O$ So $OA'$ is tangent to $(A'A''H)$ , similarity for the other two. Hence $O$ has equal power wrt these three circles, so $OH$ is the common radical axis of these three circles. So these circles are coaxial. Done for the first part.

Now suppose $A'A''$ , $B'B''$ , $C'C''$ concur at $X$ . By power of point notice that $X$ lies on their common radical axis which is $OH$ . Done.

This post has been edited 1 time. Last edited by Tafi_ak, Aug 17, 2023, 5:13 PM

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