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Geometry Difficulty 7.0 National Olympiad, round 2 Prove it Bulgaria

Problem:

Let HH be the orthocenter of △ABC\triangle ABC. The points A1≠AA_1 \neq A, B1≠BB_1 \neq B and C1≠CC_1 \neq C lie respectively on the circumcircles of △BCH\triangle BCH, △CAH\triangle CAH and △ABH\triangle ABH, and A1H=B1H=C1HA_1H = B_1H = C_1H. Denote by H1H_1, H2H_2 and H3H_3 the orthocenters of △A1BC\triangle A_1BC, △B1CA\triangle B_1CA and △C1AB\triangle C_1AB, respectively. Prove that △A1B1C1\triangle A_1B_1C_1 and △H1H2H3\triangle H_1H_2H_3 have the same orthocenter.

Solution

sin⁡∠C1AH=C1H2R1=A1H2R2=sin⁡∠A1BH \sin \angle C_1AH = \frac{C_1H}{2R_1} = \frac{A_1H}{2R_2} = \sin \angle A_1BH
and analogously
sin⁡∠C1AH=sin⁡∠A1BH=sin⁡∠A1CH=sin⁡∠C1BH=sin⁡∠B1AH. \sin \angle C_1AH = \sin \angle A_1BH = \sin \angle A_1CH = \sin \angle C_1BH = \sin \angle B_1AH.
Let ∠C1AH=∠A1BH=∠B1CH=φ\angle C_1AH = \angle A_1BH = \angle B_1CH = \varphi. Since ∠AHB=α+β=180∘−γ\angle AHB = \alpha + \beta = 180^\circ - \gamma, then ∠AC1B=γ\angle AC_1B = \gamma, and hence ∠AH3B=180∘−γ\angle AH_3B = 180^\circ - \gamma. Analogously, ∠BA1C=α\angle BA_1C = \alpha, ∠BH1C=180∘−α\angle BH_1C = 180^\circ - \alpha, ∠CB1A=β\angle CB_1A = \beta and ∠CH2A=180∘−β\angle CH_2A = 180^\circ - \beta. It follows that the points H1H_1, H2H_2 and H3H_3 belong to the circumcircle of △ABC\triangle ABC.

Now we shall use now vectors. Denote by [a⃗][\vec{a}] the image of the vector a⃗\vec{a} under rotation through 360∘−2φ360^\circ - 2\varphi. We get from the proved above that ∠C1AH3=90∘−γ=∠CAH\angle C_1AH_3 = 90^\circ - \gamma = \angle CAH. Hence ∠CAH3=φ\angle CAH_3 = \varphi and then ∠COH3=2φ\angle COH_3 = 2\varphi. Analogously, ∠AOH1=∠BOH2=2φ\angle AOH_1 = \angle BOH_2 = 2\varphi.

Let SS and TT be the orthocenters of △A1B1C1\triangle A_1B_1C_1 and △H1H2H3\triangle H_1H_2H_3, respectively. We shall prove that HS→=HT→\overrightarrow{HS} = \overrightarrow{HT}.

One has that
HS→=HA1→+HB1→+HC1→=HO→+OA1→+HO→+OB1→+HO→+OC1→=3HO→+OA1→+OB1→+OC1→=HO→+[OA1→+OB1→+OC1→+2HO→]=HO→+[OH1→+OH2→+OH3→]=HO→+[OH→] \begin{aligned} & \overrightarrow{HS} = \overrightarrow{HA_1} + \overrightarrow{HB_1} + \overrightarrow{HC_1} = \overrightarrow{HO} + \overrightarrow{OA_1} + \overrightarrow{HO} + \overrightarrow{OB_1} + \overrightarrow{HO} + \overrightarrow{OC_1} \\ & = 3\overrightarrow{HO} + \overrightarrow{OA_1} + \overrightarrow{OB_1} + \overrightarrow{OC_1} \\ & = \overrightarrow{HO} + [\overrightarrow{OA_1} + \overrightarrow{OB_1} + \overrightarrow{OC_1} + 2\overrightarrow{HO}] \\ & = \overrightarrow{HO} + [\overrightarrow{OH_1} + \overrightarrow{OH_2} + \overrightarrow{OH_3}] \\ & = \overrightarrow{HO} + [\overrightarrow{OH}] \end{aligned}

Similarly,
HT→=HO→+OT→=HO→+OH1→+OH2→+OH3→=HO→+[OH→] \overrightarrow{HT} = \overrightarrow{HO} + \overrightarrow{OT} = \overrightarrow{HO} + \overrightarrow{OH_1} + \overrightarrow{OH_2} + \overrightarrow{OH_3} = \overrightarrow{HO} + [\overrightarrow{OH}]
which means that S≡TS \equiv T.

The case ∠C1AH=∠A1CH=∠B1CH=φ\angle C_1AH = \angle A_1CH = \angle B_1CH = \varphi can be considered in a similar way and we omit the details.

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