Consider a triangle $ABC$ and two congruent triangles $A_1B_1C_1$ and $A_2B_2C_2$ which are respectively similar to $ABC$ and inscribed in it: $A_i,B_i,C_i$ are located on the sides of $ABC$ in such a way that the points $A_i$ are on the side opposite to $A$, the points $B_i$ are on the side opposite to $B$, and the points $C_i$ are on the side opposite to $C$ (and the angle at A are equal to angles at $A_i$ etc.). The circumcircles of $A_1B_1C_1$ and $A_2B_2C_2$ intersect at points $P$ and $Q$. Prove that the line $PQ$ passes through the orthocenter of $ABC$.
Problem
Source: ISR 2021 February TST p.3
Tags: geometry, congruent triangles
27.05.2022 13:02
dendimon18 wrote: The circumcircles of $A_1B_1C_1$ and $A_2B_2C_2$ intersect at points $P$ and $Q$. Prove that the line $PQ$ passes through the orthocenter of $ABC$. I think the problem is wrong. I proved that $P, Q, H, O$ are concyclic instead, where $H, O$ is the orthocenter and circumcenter of $\triangle ABC$, respectively. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.633877495300174, xmax = 17.779576775296935, ymin = -9.760364602021225, ymax = 6.3108914652527455; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((8.094977605842807,1.4909032287359212)--(5.46752128609946,-1.1018613679794371)--(3.4130063778572524,4.321241876280442)--cycle, linewidth(1) + zzttqq); draw((7.395677845458938,2.1013505721602757)--(6.9938375475226415,-1.182432290441179)--(2.531633267099551,1.4825352699118055)--cycle, linewidth(1) + zzttqq); /* draw figures */ draw((8.094977605842807,1.4909032287359212)--(5.46752128609946,-1.1018613679794371), linewidth(1) + zzttqq); draw((5.46752128609946,-1.1018613679794371)--(3.4130063778572524,4.321241876280442), linewidth(1) + zzttqq); draw((3.4130063778572524,4.321241876280442)--(8.094977605842807,1.4909032287359212), linewidth(1) + zzttqq); draw((7.395677845458938,2.1013505721602757)--(6.9938375475226415,-1.182432290441179), linewidth(1) + zzttqq); draw((6.9938375475226415,-1.182432290441179)--(2.531633267099551,1.4825352699118055), linewidth(1) + zzttqq); draw((2.531633267099551,1.4825352699118055)--(7.395677845458938,2.1013505721602757), linewidth(1) + zzttqq); draw(circle((5.127752634899763,1.8701418685184636), 2.9913618928662027), linewidth(1)); draw(circle((5.100574330040565,0.7157267758154495), 2.680942642333629), linewidth(1)); draw(circle((4.989634058226925,-3.9965290044377677), 5.259332338269737), linewidth(1) + linetype("4 4")); draw((1.7895012785770321,-0.9077066751882414)--(3.72,5.31), linewidth(1)); draw((3.72,5.31)--(11.425407619675013,-1.4163652555097912), linewidth(1)); draw((11.425407619675013,-1.4163652555097912)--(1.7895012785770321,-0.9077066751882414), linewidth(1)); /* dots and labels */ dot((6.72,0.97),dotstyle); label("$O$", (6.807069460330694,1.1713527229265517), NE * labelscalefactor); dot((3.72,5.31),dotstyle); label("$B$", (3.809005193973747,5.515486659892739), NE * labelscalefactor); dot((1.7895012785770321,-0.9077066751882414),dotstyle); label("$A$", (1.8714806680968086,-0.7049868179226938), NE * labelscalefactor); dot((11.425407619675013,-1.4163652555097912),dotstyle); label("$C$", (11.497918312453809,-1.214861693153467), NE * labelscalefactor); dot((3.494908898252047,1.0459280693019677),linewidth(4pt) + dotstyle); label("$H$", (3.584660248872207,1.2121427129450135), NE * labelscalefactor); dot((8.094977605842807,1.4909032287359212),dotstyle); label("$A_2$", (8.173534125949166,1.7016225931665556), NE * labelscalefactor); dot((5.46752128609946,-1.1018613679794371),linewidth(4pt) + dotstyle); label("$B_2$", (5.257049839629143,-1.520786618291931), NE * labelscalefactor); dot((3.4130063778572524,4.321241876280442),linewidth(4pt) + dotstyle); label("$C_2$", (2.8708354235491242,4.556921894458886), NE * labelscalefactor); dot((7.395677845458938,2.1013505721602757),dotstyle); label("$A_1$", (7.663659250718393,2.170707478378867), NE * labelscalefactor); dot((6.9938375475226415,-1.182432290441179),linewidth(4pt) + dotstyle); label("$B_1$", (7.051809400441465,-1.6227615933380855), NE * labelscalefactor); dot((2.531633267099551,1.4825352699118055),linewidth(4pt) + dotstyle); label("$C_1$", (2.1774055932352727,1.6812275981573248), NE * labelscalefactor); dot((7.770026076290646,0.4677735021423332),linewidth(4pt) + dotstyle); label("$Q$", (7.847214205801471,0.6410828526865474), NE * labelscalefactor); dot((2.42241151246695,0.593671976446567),linewidth(4pt) + dotstyle); label("$P$", (2.5037255133829674,0.763452822741933), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]
27.05.2022 23:00
Seicchi28 wrote: I think the problem is wrong. I proved that $P, Q, H, O$ are concyclic instead, where $H, O$ is the orthocenter and circumcenter of $\triangle ABC$, respectively. Could it be that you missed that $A_1B_1C_1$ and $A_2B_2C_2$ should be congruent? On your picture they look similar but not congruent.
28.05.2022 04:03
dendimon18 wrote: Could it be that you missed that $A_1B_1C_1$ and $A_2B_2C_2$ should be congruent? On your picture they look similar but not congruent. Ah, yes! Sorry for missing that Denote $O, H, R, a, b, c$ as usual. Let $t_1 = A_1B_1 / AB$ and $t_2 = A_2B_2/AB$. I will prove that $\text{Pow}(H, \odot(A_1B_1C_1)) = Ct_1^2$ for some constant $C$ (In fact, $C = 8R^2-a^2-b^2-c^2$). This immediately solves the original problem, because $$\triangle A_1B_1C_1 \cong \triangle A_2B_2C_2 \implies t_1 = t_2 \implies \text{Pow}(H, \odot(A_1B_1C_1)) = \text{Pow}(H, \odot(A_2B_2C_2)) \implies H \in \overline{PQ},$$since $\overline{PQ}$ is the radical axis of $\odot(A_1B_1C_1)$ and $\odot(A_2B_2C_2)$. Claim 01. $\odot(AB_1C_1), \odot(BA_1C_1), \odot(CA_1B_1)$ concur at the orthocenter of $\triangle A_1B_1C_1$, and this point is $O$. Proof. By Miquel, these circles concur at a point $X$. We know that $\measuredangle B_1XC_1 = \measuredangle B_1AC_1 = \measuredangle CAB = \measuredangle C_1A_1B_1$. Similarly, $\measuredangle C_1XA_1 = \measuredangle A_1C_1B_1$ and $\measuredangle A_1XB_1 = \measuredangle B_1C_1A_1$. If we let $\triangle A_0B_0C_0$ as the anti-complementary triangle of $\triangle A_1B_1C_1$, this means $X$ is the intersection of the three circles $\odot(A_0B_1C_1), \odot(B_0A_1C_1),$ and $\odot(C_0A_1B_1)$ i.e. $X$ is the orthocenter of $\triangle A_1B_1C_1$ as well as the circumcenter of $\triangle A_0B_0C_0$. Since $A, A_0, B_1, C_1, X$ lie on a circle with diameter $A_2X$, therefore \[ \frac{XA}{XA_0} = \sin \measuredangle(XC_1, AB) = \sin \measuredangle(XB_1, AC) = \sin \measuredangle(XA_1, BC), \]which means $XA/XA_0 = XB/XB_0 = XC/XC_0$. Since $XA_0 = XB_0 = XC_0$, therefore $XA=XB=XC$. The conclusion follows. $\blacksquare$ Claim 02. Let $O_1$ be the circumcenter of $\triangle A_1B_1C_1$. Then, $O_1O = O_1H$. Proof. It's easy to see that $O$ is the center of spiral that maps $\triangle A_0B_0C_0$ to $\triangle ABC$. Let $\alpha = \measuredangle(A_0B_0, AB)$. Therefore, $\measuredangle(OC_1, AB) = \measuredangle(OC_1, A_0B_0) + \measuredangle(A_0B_0, AB) = 90^{\circ} + \alpha$. We have \[ \cos \alpha = \sin(90^{\circ} + \alpha) = \frac{A_0O}{AO} = \frac{R}{2R_{\odot(A_1B_1C_1)}} = \frac{R}{2Rt_1} = \frac{1}{2t_1}. \]Let $T$ be the center of spiral similarity which maps $\triangle ABC \mapsto \triangle A_1B_1C_1$. This also maps $H \mapsto O$ and $O \mapsto O_1$, which means $\triangle TOH \cong \triangle TO_1O$. The rotation angle of this map is $\measuredangle(TA_1, TA) = \measuredangle (A_1B_1, AB) = \measuredangle(A_0B_0, AB) = \alpha$. Therefore \[ \measuredangle(HOO_1) = \measuredangle(TOO_1)+\measuredangle(HOT) = \measuredangle(THO)+\measuredangle(HOT) = \measuredangle(HTO) = \alpha.\]Let $Y$ be the projection of $O_1$ to $\overline{OH}$. We have \[ OY = OO_1 \cos \measuredangle(HTO) = OO_1 \cos \alpha = \frac{OO_1}{2t_1} = \frac{OH}{2}. \]Therefore, $Y$ is the midpoint of $OH$, i.e. $O_1O = O_1H$, as desired. $\blacksquare$ We can now finish the proof as follows: \begin{align*} \text{Pow}(H, \odot(A_1B_1C_1)) = O_1H^2 - R_{\odot(A_1B_1C_1)}^2 &\stackrel{\text{Claim 02}}{=} OO_1^2 - R_{\odot(A_1B_1C_1)}^2 \\ &= (OH^2)t_1^2 - (Rt_1)^2 \\ &= (OH^2 - R^2)t_1^2 = (8R^2-a^2-b^2-c^2)t_1^2, \end{align*}as desired (We use the fact that $OH^2=9R^2-a^2-b^2-c^2$). $\square$ Remark. From claim 02, $O_1$ is on the perpendicular bisector of $\overline{OH}$. But it's also on the perpendicular bisector of $\overline{PQ}$. Similarly, $O_2$, the center of $\odot(A_2B_2C_2)$, is also on the perpendicular bisectors of $\overline{OH}$ and $\overline{PQ}$. So, their perpendicular bisectors are the same (i.e. the line $O_1O_2$), so $PQOH$ is an isosceles trapezoid and therefore cyclic.