I remember this problem was proposed in the TST of Sankt-Peterburg(Russia) by A.Smirnov two or tree years ago,I solved this problem with using complex numbers,I will post my solution,if noone find batter solution .

Let me first rewrite the problem Problem Let $ABC$ be a triangle and $A_1B_1C_1$ be the circumcevian triangle of a point $P$ wrt $ABC$. Let $A', B',C'$ be the reflections of $A_1, B_1, C_1$ through $BC, CA, AB$ respectively. Then the circumcircle of $A'B'C'$ passes through $H$. Proof Let $A_2B_2C_2$ be the circumcevian triangle of the isogonal conjugate $P'$ of $P$ wrt $ABC$. Let $D, E, F$ be the midpoints of $BC, CA, AB$. Call $P'_a, P'_b, P'_c$ the reflections of $A, B, C$ through $P$. Let $K$ be the reflection of $P'_a$ through $D$. Then $BP'_aCK$ is a parallelogram. Since $ABP'_aP'_b$ and $ACP'_aP'_c$ are parallelograms, it follows that $AP'_bCK$ and $AP'_cBK$ are parallelograms. This implies that $K$ is also the reflection of $P'_b$ through $E$, of $P'_c$ through $F$. Under the reflection through the midpoint $D$ of $BC$, $H$ is mapped the antipode $A_3$ of $A$ wrt circumcircle of $ABC$, $A'$ is mapped to $A_2$, $P'_a$ is mapped to $K$. Thus, $\measuredangle{HA'K} = \measuredangle{A_3A_2P'_a} = \measuredangle{A_3A_2A} = 90^0$. This yields that $A'$ lies on the circle with diameter $HK$. By similar argument , we obtain $B', C'$ also lie on the circle with diameter $HK$. Therefore, the circumcircle of $A'B'C'$ passes through $H$. And we are done.

Lemma: Let $P,Q,R$ be points on the circumcircles of $BHC,CHA,AHB$. Then $H,P,Q,R$ are concyclic (or collinear) if and only if $\frac{BP}{CP} \frac{CQ}{AQ} \frac{BR}{AR}=-1$, where the ratio is taken positive if if e.g. $HBPC$ is convex and negative otherwise. Proof: Invert wrt $H$ if $A,B,C,P,Q,R$ go to $A',B',C',P',Q',R'$ we must have $P',Q',R'$ collinear which by Menelaos is possible if an only if $\frac{B'P'}{C'P'} \frac{C'Q'}{A'Q'} \frac{B'R'}{A'R'}=-1$. However $\frac{B'P'}{C'P'}=\frac{BP}{CP} \frac{HB'}{HC'}$. By multiplying these relations we get the desired result. Note that $H$ is not necessarily the orthocenter. The problem follows clearly from this Lemma, since from Hamilton Theorem the symmetric of circumcircle of $ABC$ wrt $BC$ is the circumcircle of $HBC$ and the needed product of ratios is $1$ by Trigonometric Ceva Theorem.

Yes, Tiks, i'm interrested in your solution with complex numbers. I remeber that I have created a similar problem for RMNO, but this problem is nicer.

look here: http://www.mathlinks.ro/Forum/viewtopic.php?t=114907, there was part of a TST peru, too.

Cezar Lupu wrote: Yes, Tiks, i'm interrested in your solution with complex numbers. I remeber that I have created a similar problem for RMNO, but this problem is nicer. Only for you dear Cezar! Let that the circumcircle of the triangle $ABC$ is a unit one. And let's take complex plane with center in the circumcentr. I will write the small letter $x$ as a complex coordinate of the a point $X$. We have that $A_{1}\in{(O)}$ and $P\in{AA_{1}}$=> \[a_{1}\overline{a_{1}}=1\] and \[\frac{a-p}{\overline{a}-\overline{p}}=\frac{a-a_{1}}{\overline{a}-\overline{a_{1}}}=-aa_{1}\] => \[a_{1}=\frac{a-p}{a\overline{p}-1}\] Now as we have that $A_{2}$ is simetric of $A_{1}$ with respect to $BC$ then $a_{1}+a_{2}=b+c$=> \[a_{2}=b+c+\frac{p-a}{a\overline{p}-1}\] Similary \[b_{2}=a+c+\frac{p-b}{b\overline{p}-1}\] \[c_{2}=a+b+\frac{p-c}{c\overline{p}-1}\] But we also know that $h=a+b+c$(H is the orthocenter) So finally we just need to check that $\lambda$ is a real number,that is \[\lambda=\overline{\lambda}\] where \[\lambda=\frac{(a_{2}-c_{2})(b_{2}-h)}{(b_{2}-c_{2})(a_{2}-h)}\]

Tiks wrote: So finally we just need to check that $ \lambda$ is a real number,that is \[ \lambda = \overline{\lambda} \] where \[ \lambda = \frac {(a_{2} - c_{2})(b_{2} - h)}{(b_{2} - c_{2})(a_{2} - h)} \] Could you or somebody else explain this step in details?

Lemma: Let $W$, $X$, $Y$, and $Z$, be any points. Then if the reflection of $Y$ across the midpoint of $WZ$ is reflected across the midpoint of $XZ$, the point obtained is the same as the reflection of $W$ across the midpoint of $XY$. Proof: Let $w, x, y, z$ be the vectors corresponding to $W$, $X$, $Y$, and $Z$. For any vectors $x$, $p$, and $q$, the reflection of $x$ across the midpoint of $p$ and $q$ is given by $p+q-x$, since $\frac{(p+q-x)+x}{2} = \frac{p+q}{2}$. Hence, the reflection of $Y$ across the midpoint of $WZ$ is $w+z-y$, and the reflection of this point across the midpoint of $XZ$ is $x+z-(w+z-y) = x+y-w$, which is indeed the reflection of $W$ across the midpoint of $X$ and $Y$. Let $M_a$ be the midpoint of $AA_1$, $M_b$ be the midpoint of $BB_1$, $M_c$ be the midpoint of $CC_1$, and let $O$ be the circumcenter of $\triangle ABC$. $M_a$, $M_b$, and $M_c$ lie on the circle with diameter $OP$. Let the homothety centered at $A$ with factor 2 map $M_a$ to $R_a$, $M_b$ to $R_c$, and $O$ to $A'$. $M_a$ is mapped to $A_1$, so we can conclude that $R_b R_c A_1 A'$ is cyclic. If we reflect these points across the midpoint of $BC$, we find that $A_1$ is mapped to $A_2$, $A'$ is mapped to $H$ (a result of trivial angle chasing), and by the lemma, $R_b$ is mapped to $B_2$ and $R_c$ is mapped to $B_3$, so $A_2 B_2 C_2 H$ is indeed cyclic.

treegoner wrote: Let me first rewrite the problem Problem Let $ABC$ be a triangle and $A_1B_1C_1$ be the circumcevian triangle of a point $P$ wrt $ABC$. Let $A', B',C'$ be the reflections of $A_1, B_1, C_1$ through $BC, CA, AB$ respectively. Then the circumcircle of $A'B'C'$ passes through $H$. Proof Let $A_2B_2C_2$ be the circumcevian triangle of the isogonal conjugate $P'$ of $P$ wrt $ABC$. Let $D, E, F$ be the midpoints of $BC, CA, AB$. Call $P'_a, P'_b, P'_c$ the reflections of $A, B, C$ through $P$. Let $K$ be the reflection of $P'_a$ through $D$. Then $BP'_aCK$ is a parallelogram. Since $ABP'_aP'_b$ and $ACP'_aP'_c$ are parallelograms, it follows that $AP'_bCK$ and $AP'_cBK$ are parallelograms. This implies that $K$ is also the reflection of $P'_b$ through $E$, of $P'_c$ through $F$. Under the reflection through the midpoint $D$ of $BC$, $H$ is mapped the antipode $A_3$ of $A$ wrt circumcircle of $ABC$, $A'$ is mapped to $A_2$, $P'_a$ is mapped to $K$. Thus, $\measuredangle{HA'K} = \measuredangle{A_3A_2P'_a} = \measuredangle{A_3A_2A} = 90^0$. This yields that $A'$ lies on the circle with diameter $HK$. By similar argument , we obtain $B', C'$ also lie on the circle with diameter $HK$. Therefore, the circumcircle of $A'B'C'$ passes through $H$. And we are done. CAN SOMEBODY EXPLAIN WHY $\measuredangle{A_3A_2P'_a} = \measuredangle{A_3A_2A}$?

I have nice approach. I restate the problem like this, using Ceva: Let $\omega$ be a circle through $H$, the orthocenter of $ABC$. Let $\Omega$ be the circumcircle of $ABC$, and let $\Omega_A, \Omega_B, \Omega_C$ be the reflections of $\Omega$ through sides $BC, CA, AB$, respectively, with centers $O_A, O_B, O_C$, respectively. Let $\omega$ intersect these three circles in $H_A,H_B,H_C$, respectively. Prove that: $\left( \frac{BH_A}{H_AC} \right) \left( \frac{CH_B}{H_BA} \right) \left( \frac{AH_C}{H_CB} \right) = 1$. PROOF: Let $O$ be the center of $\omega$. I prove that $\displaystyle\prod_{cyc} \left( \frac{BH_A}{BH} \right) \left( \frac{CH}{CH_A} \right) = 1$. I use Sine Law in triangles $BHH_A$, $CHH_A$. Let $\alpha(a,c)$ be the sine of the angle formed by side $a$ (namely $BC$) with line $OO_C$, and define $\alpha(b,a), \alpha(a,b)$ and so on similarly. Then the problem reduces to proving $\alpha(a,b) \alpha(b,c) \alpha(c,a) = \alpha(b,a) \alpha(c,b) \alpha(a,c) $ Let $X$ be a point such that $XA || OO_A$. Notice that $\frac{ \alpha(b,a)}{ \alpha(c,a)} = \frac{ sinCAX}{sin XAB}$. Therefore, by trigonometric Ceva, the problem reduces to proving that the parallel lines to $OO_A,OO_B,OO_C$ through $A,B,C$ (respectively) concur. But this is trivial since triangles $ABC$ and $O_AO_BO_C$ are homothetic (the center is the 9-point center).

Dear Mathlinkers, this is the P-circle of Hagge. For a synthetic proof and more, http://perso.orange.fr/jl.ayme vol. 5 le P-cercle de Hagge. Sincerely Jean-Louis

This is a very nice generalization of Fuhrmann circle Let $ H $ be the orthocenter of $ \triangle ABC $ . Let $ \triangle A_1B_1C_1 $ be the circumcevian triangle of $ P $ . Let $ A_2, B_2, C_2 $ be the reflection of $ A_1, B_1, C_1 $ in $ BC, CA, AB $ . Let $ A_3=AH \cap (A_2B_2C_2), B_3=BH \cap (A_2B_2C_2), C_3=CH \cap (A_2B_2C_2) $ . Then (1) $ A_2, B_2, C_2, H $ are concyclic ( $ P $-Hagge circle ) . (2) $ \triangle ABC \cap \triangle A_1B_1C_1 \cap P $ ~ $ \triangle A_3B_3C_3 \cap \triangle A_2B_2C_2 \cap P $ . ( See Some properties of Hagge circle )

After inversion about $H$ with radius $1$, it would suffice to show that \[ \prod \frac{A_2'B'}{A_2'C'}=1 \] Now $A_2'B'=\frac{A_2B}{HA_2\cdot HB}$, but all factors $HA,HB,HC,HA_2,HB_2,HC_2$ cancel out in the product. So it suffices to show \[ \prod \frac{A_2B}{A_2C}=1 \] But of course $\frac{A_2B}{A_2C}=\frac{A_1B}{A_1C}$, and $\prod \frac{A_1B}{A_1C}=1$ iff $AA_1,BB_1,CC_1$ concur. So we are done.

i am sorry for reviving this thread .

P.S.this actually proves the existence of Hagge Circle.(if the solution is not wrong )

Here is a proof using Ptolemy's sine lemma. Unfortunately, we can't apply it straight away without running into configuration issues, so we have to go through some formalities. Hopefully this won't take away from the beauty of the solution. Let $A',B',C'$ be the antipodes of $A,B,C$. Of course, we will take $P\not\in OA\cup OB\cup OC$ (where $O$ is the circumcenter). We will choose an arbitrary direction of reference $\hat{k}$ perpendicular to our plane, and we will let $\varepsilon_a,\varepsilon_b$ and $\varepsilon_c$ be the signs of $\overrightarrow{OP}\times\overrightarrow{OA}$ (or equivalently of $\Im\left(a/p\right)$) and so on. Using barycentric coordinates, we know that $$\overrightarrow{OP}=\sum\frac{[BPC]}{[ABC]}\cdot\overrightarrow{OA},$$and therefore $$\overrightarrow{0}=\sum [BPC]\cdot\left(\overrightarrow{OP}\times\overrightarrow{OA}\right)=\hat{k}\sum [BPC]\cdot OP\cdot R\cdot\sin(\angle AOP)\cdot\varepsilon_a.$$Using the law of sines in $\triangle AOP$ and $\triangle AA'A_1$ we get $$0=\sum \frac{PB\cdot PC\cdot\sin(\angle BPC)}{2}\cdot\varepsilon_a\cdot R\cdot PA\cdot\sin(\angle PAO)=\sum \frac{PB\cdot PC\cdot\sin(\angle BPC)}{4}\cdot\varepsilon_a\cdot PA\cdot A_1A'.$$Thus, $$\sum \varepsilon_b\varepsilon_c\sin(\angle BPC)\cdot HA_2=0.$$ In fact, it turns out that $\varepsilon_b\varepsilon_c\sin(\angle BPC)=\sin\left(\angle\left(\overrightarrow{B_2B'},\overrightarrow{C_2C'}\right)\right)$. We can prove this in two ways. One would be to dissect the disc $(ABC)$ in four regions given by the diameters $OB$ and $OC$, and argue with respect to the position of $P$. The simpler way would be through complex numbers: $$\angle\left(\overrightarrow{B_2B'},\overrightarrow{C_2C'}\right)=\arg\left(\frac{c'-c_1}{b'-b_1}\right)=\arg\left(\frac{c+c_1}{b+b_1}\right)=\arg\left(\frac{c\overline{p}-p\overline{c}}{b\overline{p}-p\overline{b}}\cdot\frac{\overline{b}-\overline{p}}{\overline{c}-\overline{p}}\right)=\arg\left(\frac{\Im(c/p)}{\Im(b/p)}\right)+\arg\left(\frac{c-p}{b-p}\right),$$from which the claim is obvious. Since $HA_2A'A_1$ and the others are parallelograms, we have that $\overrightarrow{HA_2}=\overrightarrow{A_1A'}$, and thus $$\sum HA_2\cdot\sin\left(\angle\left(\overrightarrow{HB_2},\overrightarrow{HC_2}\right)\right)=0,$$which is exactly Ptolemy's sine lemma. Remark: I have to admit, the correct and without-config-issues proof is quite cumbersome. It's greatly overshadowed by the much nicer complex numbers solution. This is probably one of the problems where it's best to just sweep the many different cases under the rug and just base ourselves on the first figure we come across.