Fix a triangle $ABC$. The variable point $M$ in its interior is such that $\angle MAC = \angle MBC$ and $N$ is the reflection of $M$ with respect to the midpoint of $AB$. Prove that $|AM| \cdot |BM| + |CM| \cdot |CN|$ is independent of the choice of $M$.
Problem
Source: Bulgaria Autumn Tournament 2022
Tags: geometry, fixed
Assassino9931
21.11.2022 01:42
With $A = - 1$, $B = 1$, $T = 0$ (the midpoint of $AB$), $M = m$ and $N = -m$ we have $\angle MAC = \angle CBM$ (these have the same orientation) if and only if $\frac{c+1}{m-1} / \frac{m-1}{c-1} = \frac{c^2-1}{m^2-1} = \frac{1}{k}$ is a real number, i.e. $m^2 = kc^2 - k +1$. Hence $|AM| \cdot |BM| + |CM| \cdot |CN| = |m^2 - 1| + |m^2 - c^2| = (|k| + |k-1|)|c^2-1| = (|k| + |k|-1) \cdot |AC| \cdot |BC|$ and so it remains to argue that $k\in (0,1)$. For this, observe that $k>1$ implies $|c^2 - 1| < |m^2 - 1|$ implies $|AC| \cdot |BC| < |AM| \cdot |BM|$, contradicting $M$ to be in the interior, while $k < 0$ is impossible as $C$ and $M$ lie in the same halfpland with respect to $AB$.
Assassino9931
21.11.2022 01:43
Let $P$ be the reflection of $M$ with respect to the midpoint of $BC$. Angle chase to show that $CNBP$ is cyclic and so the desired expression is $|AC| \cdot |BC|$ by Ptolemy.
Assassino9931
21.11.2022 01:59
Denote $BC = a$, $AC = b$, $AM = m$, $BM = n$ Observe that $\angle CAM = \angle CBM$ and $\angle CAN = \angle CBN$ by the parallelogram $ANBM$. The Cosine Law in $\triangle CAN$ and $\triangle CBN$ implies $\frac{a^2 + AM^2 - CN^2}{2 \cdot a \cdot AM} = \frac{b^2 + BM^2 - CN^2}{2 \cdot b \cdot AN}$ and hence $CN^2 = \frac{(a^2 + AM^2) \cdot b \cdot BM - (b^2 + BM^2) \cdot a \cdot AM}{b \cdot BM - a \cdot AM} = \frac{(a^2 + m^2)bn - (b^2 + n^2)am}{bn - am} = \frac{(an-bm)(ab-mn)}{bn-am}$. Similarly, from triangles $CAM$ and $CBM$ we get $\frac{b^2 + AM^2 - CM^2}{2 \cdot b \cdot AM} = \frac{a^2 + BM^2 - CM^2}{2 \cdot a \cdot BM}$, i.e. $CM^2 = \frac{(b^2 + AM^2) \cdot a \cdot BM - (a^2 + BM^2) \cdot b \cdot AM}{a \cdot BM - b \cdot AM} = \frac{(b^2 + m^2)an - (a^2 + n^2)bm}{an - bm} = \frac{(bn - am)(ab - mn)}{an - bm}$. Hence $CM \cdot CN = |ab - mn| = ab - mn$ (as $M$ is in the interior of $ABC$) and hence adding $AM \cdot BM = mn$ gives overall $AC \cdot BC$, as desired.
Assassino9931
25.11.2022 21:07
Remark for the above: Actually, $AM \cdot BM < AC \cdot BC$ is not true for any $M$ in the interior - if $ABC$ is acute with a very small angle $\angle BAC$ and $M$ is the circumcenter, then $\frac{AC \cdot BC}{AM \cdot BM} = \frac{1}{4}\sin\angle BAC \sin\angle ABC$ can be arbitrarily small. To fix this, let $M$ be with the property in the problem condition and let $D$ be the reflection of $B$ with respect to $CM$ - then $DAMC$ is cyclic and $S_{DAM} < S_{DAC}$.
VicKmath7
25.11.2022 22:56
Here is my hybrid solution (though the trig bash part is quite clean imo). Let $MA \cap BC=D, MB \cap CA=E$; the angle condition implies that $ABDE$ is cyclic. Notice that $\triangle CDE \cup \{M\} \sim \triangle CAB \cup \{N\}$, which implies that $CM, CN$ are isogonal wrt $\angle CAB$. Let $\angle ACM= \angle BCN =\phi_1, \angle ACN= \angle BCM =\phi_2, \angle CME=\theta_1, \angle CMD = \theta_2$ (and obviously $\phi_1+\theta_1=\phi_2+\theta_2$ due to the cyclic quadrilateral). We will prove that $NA.NB+CM.CN=CA.CB \iff \frac {NA} {CA}.\frac {NB} {CB}+\frac {CN} {CB}.\frac {CM} {CA}=1$. Using the similarities and LoS, this rewrites as $\frac {\sin \phi_2} {\sin \theta_2}.\frac {\sin \phi_1} {\sin \theta_1}+\frac {\sin (\phi_1+\theta_1)}{\sin \theta_1}.\frac{\sin (\theta_2-\phi_1)}{\sin \theta_2}=1$. Multiplying by $\sin \theta_1. \sin \theta_2$ and applying $\sin \alpha. \sin \beta = \frac{\cos (\alpha-\beta)+\cos(\alpha+\beta)}{2}$ thrice makes the above expression equivalent to $\cos (\phi_1-\phi_2)-\cos(\phi_1+\phi_2)+\cos (2\phi_1 +\theta_1-\theta_2)-\cos(\theta_1+\theta_2)=\cos(\theta_1-\theta_2)-\cos(\theta_1+\theta_2)$ which is true, since $2\phi_1+\theta_1-\theta_2=\phi_1+\phi_2$ and $\cos (\phi_1-\phi_2)=\cos (\theta_1-\theta_2)$ (the last two equalities hold since $\theta_1+\phi_1=\theta_2+\phi_2$ and $\cos(x)=\cos(-x)$), so we are finally done. Remarks: -This solution was heavily inspired by ISL 2012 G2 (for the part with the similar triangles, which is sufficient to solve the ISL problem with a bit of additional angle chasing). In addition, the synthetic approach outlined in #3 can be also modified to solve the ISL problem (e.g. as in post #38 in the thread here: https://artofproblemsolving.com/community/c6h546176p3160578) but I wasn't aware of that approach before seeing the official solution of the current problem. -The isogonality of the lines $CM$ and $CN$ can be also showed by DDIT for the parallelogram $AMBN$.
Strudan_Borisov
01.12.2022 16:35
Here is my solution. We'll use standart notations for the angles of $\triangle ABC$, $\angle CAM = \angle MBC = \varphi$, $\angle ACM = \angle NCB = \theta$. We want to prove that $MA.MB+CM.CN=CA.CB$ As @VicKmath7 pointed out, $CM$ and $CN$ are isogonal. Let $L$ be a point on $BC$ such $\angle CAM = \angle CNL$. That gives us $\triangle CAM \sim \triangle CNL$, so $CM.CN = CA.CL$. To achieve our goal, we are left to prove that $BM.AM=CA.LB$. $AMBN$ is a parallelogram, so $\angle CAN = \angle NBC = \alpha+\beta-\varphi$. From $\triangle ANC$ we calculate $\angle ANC = \varphi + \theta$. But $\angle BLN = \angle LNC + \angle NCL = \varphi + \theta$. The last ones mean that $\triangle CAN \sim \triangle NBL \Rightarrow \frac {BN}{LB}=\frac {CA}{AN} \iff BN.AN=CA.LB \iff BM.AM = CA.LB$, as desired.
Helixglich
01.11.2024 16:45
Note that the locus of points $M$ satisfying the angle condition mod $180$ is a conic. Note further that for a fixed $ABC$ the $M$ satisfying $AM\cdot BM+CM\cdot CN=AC\cdot CB$ for $N=A+B-M$ is also a conic. Hence we need to show that the two have $5$ intersection points. Namely $M=A,B,C,B+A-C$ and the $C$- antipode are obvious.