AoPS - Problem

Problem

Source: 2020 RMM Shortlist G2

Tags: geometry, Euler Line, RMM, RMM 2020, RMM Shortlist

oVlad

08.10.2022 18:26

Let $ABC$ be an acute scalene triangle, and let $A_1, B_1, C_1$ be the feet of the altitudes from $A, B, C$. Let $A_2$ be the intersection of the tangents to the circle $ABC$ at $B, C$ and define $B_2, C_2$ similarly. Let $A_2A_1$ intersect the circle $A_2B_2C_2$ again at $A_3$ and define $B_3, C_3$ similarly. Show that the circles $AA_1A_3, BB_1B_3$, and $CC_1C_3$ all have two common points, $X_1$ and $X_2$ which both lie on the Euler line of the triangle $ABC$. United Kingdom, Joe Benton

AndreiVila

08.10.2022 23:38

Call $H$ and $O$ the orthocenter and circumcenter of $\triangle ABC$. Firstly notice that $HA\cdot HA_1=HB\cdot HB_1=HC\cdot HC_1$. Now we kill the problem with the following claim: $AA_3$ passes through the midpoint $M_A$ of arc $\widehat{B_2A_2C_2}$. Obviously, by the Angle Bisector theorem, it's enough to prove that $\frac{A_3B_2}{A_3C_2}=\frac{AB_2}{AC_2}$. However, $$\frac{A_3B_2}{A_3C_2}=\frac{\sin(\angle B_2A_2A_1)}{\sin(\angle C_2A_2A_1)}=\frac{[CA_2A_1]}{[BA_2A_1]}=\frac{CA_1}{BA_1}=\frac{\cot(\angle C)}{\cot (\angle B)}=\frac{\cot(\angle BAC_2)}{\cot(\angle CAB_2)}=\frac{\tan(\angle OC_2B_2)}{\tan(\angle OB_2C_2)}=\frac{AB_2}{AC_2},$$so the claim is proven. $\square$ Notice that $\angle A_2AO =|\angle B-\angle C|$. By the above claim, $$\angle A_2A_3A=\frac{\widehat{A_2M_A}}{2}=\frac{|\angle B_2-\angle C_2|}{2}=\frac{|(180^{\circ}-2\angle B)-(180^{\circ}-2\angle C)}{2}=|\angle B -\angle C|.$$This implies that $OA$ is tangent to $(AA_2A_3)$, and similarly to $(BB_2B_3)$ and $(CC_2C_3)$, obtaining that $O$ has the same power w.r.t. all three circles (namely $R^2$). We have obtained that both $H$ and $O$ have the same power w.r.t. the three circles, thus (since $H\neq O$) it must be the case that the three circles are coaxial, with the radical axis being $OH$, the Euler line of $\triangle ABC$.

VicKmath7

09.10.2022 11:12

Does this work (because it is much simpler imo)? To begin with, obviously $H$ lies on the radical axis of these circles as $HA.HA_1=HB.HB_1=HC.HC_1$. We will find another point on $OH$ that has equal powers wrt the circles $(AA_1A_3), (BB_1B_3), (CC_1C_3)$, which will be sufficient. By the configuration in Brazil 2013/6 for $\triangle A_2B_2C_2$, it is well-known that $A_2A_1, B_2B_1, C_2C_1$ meet on the Euler line of $\triangle ABC$, say at point $T$. We will prove that $TA_3.TA_1=TB_3.TB_1=TC_3.TC_1$ (it is sufficient to prove only the first one). But note that by PoP it is equivalent to $A_3B_1A_1B_3$ being cyclic, which is true by Reim's since $A_1B_1 \parallel A_2B_2$ (well-known, provable by angle chase), done.

Miku3D

09.10.2022 13:15

Cute config. Knowing it beforehand makes this problem much easier.

As mentioned by others, it suffices to show that $H$ and $O$ both have an equal power wrt all 3 circles, as that would imply that the three circles are coaxial with the Euler line of $\triangle ABC$ as a common radical axis. We have $AH \cdot HA_1 = BH \cdot HB_1 = CH \cdot HC_1$ for free, so we are left to handle $O$. To see things more clearly, view the problem wrt $\triangle A_2B_2C_2$ instead of $\triangle ABC$. We see that $O$ is the incenter and $ABC$ is the contact triangle. Let us also focus only on $A_1,A_3$ first. Let $A_4$ be the 2nd intersection of $(A_2BC)$ with $(A_2B_2C_2)$. Let $A,B,C$ denote $\angle C_2A_2B_2, \angle A_2B_2C_2, \angle B_2C_2A_2$ respectively Let $\measuredangle$ denote directed angles modulo $180^{\circ}$ and $\angle$ denote vanilla angles Claim 1: $\triangle A_1BC_2 \sim \triangle A_2CB_2$ ProofNote that $OAB_2C$ is cyclic by $\measuredangle OAB_2 = 90^{\circ} = \measuredangle OAC_2$, so $$\angle BAA_1 = 90 - \angle CBA = \angle ACO = \angle AB_2O$$Similarly, $\angle CAA_1 = \angle AC_2O$ $\therefore$ we have $$\frac{A_1B}{A_1C} = \frac{\frac{A_1B}{A_1A}}{\frac{A_1C}{A_1A}} = \frac{\tan \angle BAA_1}{\tan \angle CAA_1} = \frac{\tan \angle AB_2O}{\tan \angle AC_2O} = \frac{ \frac{OA}{AB_2} }{ \frac{OA}{AC_2} } = \frac{AC_2}{AB_2} = \frac{BC_2}{CB_2} $$Also, $$\angle A_1BC_2 = 180^{\circ} - \angle CBA_2 = 180^{\circ} - \angle A_2CB = \angle A_1CB_2$$$\therefore \triangle A_1BC_2 \sim \triangle A_2CB_2$$_{\blacksquare}$ Claim 2: $A_4$ is the center of the spiral similarity sending points $C,A_1,B$ to points $B_2,A,C_2$ ProofNote that $$\measuredangle A_4CB = \measuredangle A_4A_2B = \measuredangle A_4A_2C_2 = \measuredangle A_4B_2C_2$$Similarly, $$\measuredangle A_4BC = \measuredangle A_4C_2B_2$$$\therefore$, $A_4$ is the center of the spiral similarity sending segment $CB$ to $B_2C_2$ Meanwhile, Claim 1 implies that $$\frac{CA_1}{BA_1} = \frac{CB_2}{BC_2} = \frac{B_2A}{C_2A}$$so $A_1$ and $A$ are corresponding points in $\triangle A_4CB$ and $\triangle A_4B_2C_2$. $\therefore$, the spiral similarity sending segment $CB$ to segment $B_2C_2$ also sends $A_1$ to $A$$_\blacksquare$ Claim 3: Points $A_4, A_1, O$ are collinear ProofFrom the spiral similarity in Claim 2, we have $$\triangle A_4CB_2 \sim \triangle A_4BC_2 \implies \frac{A_4C}{A_4B} = \frac{CB_2}{BC_2}$$Combining this with Claim 1 gives $$\frac{A_4C}{A_4B} = \frac{CB_2}{BC_2} = \frac{CA_1}{BA_1}$$$\therefore$ $A_4A_1$ bisects $\angle CA_4B$. Note that $\measuredangle A_2BO = 90^{\circ} = \measuredangle A_2CO$, so $O \in (A_2BC)$ as well. Since $OB=OC$, then $O$ is the midpoint of arc $BOC$. $\therefore$ $A_4O$ also bisects $\angle CA_4B$. $\therefore$ lines $A_4A_1$ and $A_4O$ must coincide, so points $A_4, A_1, O$ are collinear$_\blacksquare$ Claim 4: $AA_1A_4A_3$ is a cyclic quadrilateral ProofBy the spiral similarity in Claim 2, we have $\triangle A_4CB_2 \sim \triangle A_4A_1A$ Therefore, $$\measuredangle A_4AA_1 = \measuredangle A_4B_2C = \measuredangle A_4B_2A_2 = \measuredangle A_4A_3A_2 = \measuredangle A_4A_3A_1$$which implies that $AA_1A_4A_3$ is cyclic$_\blacksquare$ Claim 5: The center of $AA_1A_4A_3$ lies on the line $B_2C_2$ ProofFrom Claim 1, we have $\frac{A_1B_2}{A_1C_2} = \frac{CB_2}{BC_2} = \frac{AB_2}{AC_2}$. From Claim 2, we have $$\triangle A_4CB_2 \sim \triangle A_4BC_2 \implies \frac{A_4B_2}{A_4C_2} = \frac{CB_2}{BC_2} = \frac{AB_2}{AC_2}$$ Since we have $\frac{AB_2}{AC_2} = \frac{A_1B_2}{A_1C_2} = \frac{A_4B_2}{A_4C_2}$, then $(AA_1A_4)$ is an Apollonius Circle wrt points $B_2$ and $C_2$. $\therefore$ its center must lie on the line $B_2C_2$$_\blacksquare$ Since the center of $(AA_1A_3)$ lies on line $B_2C_2$ and $OA \perp B_2C_2$, then $OA$ is tangent to $(AA_1A_3)$. $\therefore$ the power of $O$ wrt $(AA_1A_3)$ is $OA^2 = r^2$ where $r$ is the radius of $(ABC)$. By a similar argument, it can be shown that the power of $O$ wrt $(BB_1B_3)$ and $(CC_1C_3)$ is also $r^2$. $\therefore$ $O$ has equal power wrt to all 3 circles. Since $O$ and $H$ both have equal powers wrt all the 3 circles, then $OH$, which is also the Euler line of $\triangle ABC$, must be the radical axis of these 3 coaxial circles, so both $X_1$ and $X_2$ must lie on $OH$ too$_\blacksquare$

IAmTheHazard

23.01.2024 21:26

It suffices to exhibit two points on the Euler line with equal power to all three circles. It is obvious that $H$, the orthocenter of $ABC$, is one of these, since $HA\cdot HA_1=HB\cdot HB_1=HC\cdot HC_1$. Observe that $\measuredangle C_2AB=\measuredangle ACB=\measuredangle B_1C_1A$, so $\overline{B_1C_1} \parallel \overline{B_2C_2}$. Likewise, $\overline{C_1A_1} \parallel \overline{C_2A_2}$ and $\overline{A_1B_1} \parallel \overline{A_2B_2}$, so triangles $A_1B_1C_1$ and $A_2B_2C_2$ are homothetic. It follows that $\overline{A_1A_2A_3},\overline{B_1B_2C_3},\overline{C_1C_2C_3}$ concur at some point $X$, the homothetic center. Note that $X$ should lie on the line joining the circumcenters of $(A_1B_1C_1)$ and $(A_2B_2C_2)$. The former is the 9-point center of $\triangle ABC$, while the latter is its inverse about $(ABC)$, since $A_2,B_2,C_2$ are the inverses of the midpoints of the sides of $\triangle ABC$. Thus this line also contains the circumcenter $O$ of $\triangle ABC$, hence $X$ lies on the Euler line of $ABC$. To finish, note that $XA_2\cdot XA_3=XB_2\cdot XB_3=XC_2\cdot XC_3 \implies XA_1\cdot XA_3=XB_1\cdot XB_3=XC_1\cdot XC_3$. $\blacksquare$ Remark: I suppose to avoid losing 0.4 points (rather than 0.6, since I am not bashing!) the fact that the three circles actually intersect should be justified. This follows from the fact that $H$ lies in the interior of segments $\overline{AA_1},\overline{BB_1},\overline{CC_1}$ since $\triangle ABC$ is acute, hence lies in the interior of $(AA_1A_3),(BB_1B_3),(CC_1C_3)$.

CT17

26.02.2024 02:25

Clearly the orthocenter $H$ of $ABC$ has equal power WRT the three circles. Now observe that $A_3$ is the harmonic conjugate of the $A_2-$ Sharkydevil point in $(A_2B_2C_2)$ (which shows up in 2020 USA TSTST P2). Hence $A_3A$ passes through the insimilicenter $T$ of $(ABC)$ and $(A_2B_2C_2)$, and since a negative homothety at $T$ sends $(ABC)$ to $(A_2B_2C_2)$ the quantity $TA\cdot TA_3$ is symmetric, so $T$ also lies on the radical axis. Finally, it is well known that the Euler line of $ABC$ coincides with the line through the incenter and circumcenter of $(A_2B_2C_2)$, so line $HT$ is indeed the Euler line.