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#1 h Aug 19, 2011, 6:11 AM • 1 Y

Y by Adventure10

The vertices A,B,C of an acute-angled triangle ABC lie on the sides B1C1, C1A1, A1B1 respectively of a triangle A1B1C1 similar to the triangle ABC (∠A = ∠A1, etc.). Prove that the orthocenters of triangles ABC and A1B1C1 are equidistant from the circumcenter of △ABC.

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#2 h Aug 19, 2011, 1:12 PM • 1 Y

Y by Adventure10

Let $H$ be the orthocenter of $\triangle ABC$ . Then $\angle CHA=180-\angle CB_1A$ , so $B_1CHA$ is cyclic. Hence $\angle HB_1A=\angle HCA=90-\angle B_1A_1C_1$ . Thus $H$ is the circumcenter of $\triangle A_1B_1C_1$ . Let $K,N$ be the orthocenter and nine-point center of $\triangle A_1B_1C_1$ . Then $N$ is the midpoint of $HK$ . Let $A_2B_2C_2$ be the medial triangle of $\triangle A_1B_1C_1$ . Let $O$ be the circumcenter of $\triangle ABC$ . It is easy to see that triangles $ABC$ and $A_2B_2C_2$ are homothetic about $H$ . Thus $\triangle HON\sim\triangle HAA_2$ . So $\angle HNO=90$ and finally $HO=OK$ .

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#3 h Aug 19, 2011, 2:00 PM • 1 Y

Y by Adventure10

oneplusone wrote:

It is easy to see that triangles $ABC$ and $A_2B_2C_2$ are homothetic about $H$ .

My solution is same as yours. But I think u made a tying mistake in your proof. $ABC$ and $A_2B_2C_2$ are not homothetic about $H$ . There exists a spiral similarity centered at $H$ takes $ABC$ to $A_2B_2C_2$ .

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#4 h Aug 19, 2011, 4:28 PM • 2 Y

Y by Adventure10, Mango247

The same solution as yours and oneplusone's ^_^ , maybe it's just another saying of spiral similar because he mentioned $\triangle HON \sim \triangle HAA_2$

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#5 h Aug 19, 2011, 8:31 PM • 2 Y

Y by Adventure10, Mango247

My solution starts from the construction of such triangle $\triangle ABC$ : take $\triangle A_1B_1C_1$ with $O_1$ circumcenter and rotate the figure of angle $\alpha$ around $O_1$ , getting the new triangle $\triangle A'B'C'$ ; next, let $A_1B_1\cap A'B'=C$ and, similarly $B$ and $C$ and we claim that $\triangle ABC$ is the subject triangle.
For this see that $A_1CO_1BA'$ is cyclic ( $\widehat{A_1CA'}=\widehat{A_1O_1A'}=\widehat{A_1BA'}=\alpha$ ), in a similar way $B'B_1AO_1C$ and $AC'C_1BO_1$ are cyclic.
Since $\triangle A_1O_1A'=\triangle B_1O_1B'=\triangle C_1O_1C'$ , the three circles are equal and we find ourselves within the so called 'the 5 lei coin problem', see post #8 :here.
If $O_a, O_b, O_c$ are the circumcenters of the three circles, see that $\triangle O_aO_bO_c=\triangle ABC$ and, as per the above problem, the orthocenter of each of them is the circumcenter of the other.
Since there is a spiral similarity sending $\triangle A_1B_1C_1$ to $\triangle O_aO_bO_c$ , we see, that the same sends $H_{A_1}$ to $O$ , inferring that $\triangle H_{A_1}O_1O \sim \triangle A_1O_1O_a$ , done.

N.B. there has been a wrong statement, now amended.

Best regards,
sunken rock

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#6 h Nov 28, 2012, 5:06 PM • 2 Y

Y by Adventure10, Mango247

This problem is almost the same as problem 81 IMO Longlist 1992.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005928&sid=9281bd38146e737a227c0109de6e1633#p2005928

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