M stay on a circle with radius AB and center B 'cause $ \angle ABC = 2\angle AMC$. So $ \triangle BMA$ is an isosceles triangle and $ \angle MAB = \angle BMA = 20$ But it seems too simple

Nope that' the answer

The solution of ¬[ƒ(Gabriel)³²¹º]¼ is not tru. Because if $ \angle MAB = 20^0$ $ \Longrightarrow$ the suggestion. $ \bullet$ $ \angle KBC =\angle CBM = 30^0$ $ \Longrightarrow$ $ BKCM$ is circums cribed quadriangle. $ \bullet$ if $ \angle MAB = 20^0$ $ \Longrightarrow$ $ \angle BKM = 130^0$ but $ \angle BKM = \angle MCB$ $ \Longrightarrow$ $ \angle MCB = 130^0$. But $ \angle MCB + \angle MBC = 130^0$ $ \Longrightarrow$ the suggestion. real $ \angle MAB > 20^0$

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Real $ \angle MAB > 20^0$ Who can fine it. I will try and post it soon.

thanhnam2902 wrote: $ \bullet$ if $ \angle MAB = 20^0$ $ \Longrightarrow$ $ \angle BKM = 130^0$ Recheck that statement.

thanhnam2902 on your drawing where exactly is point M? i assume it's at the second A, but it makes your solution hard to follow

let $ \angle MAB=\alpha$, then $ \angle MAC=60-\alpha, \angle BMA=100-\alpha, \angle BCM=30+\alpha$. from law of sines we can get $ \sin \alpha\cdot \sin 20\cdot \sin 30\cdot \sin 60=\sin(60-\alpha)\cdot\sin 60\cdot \sin (100-\alpha)\cdot \sin (30+\alpha)$ this can be rewritten as $ \sin \alpha\sin 20=\sin(2\alpha+60)\sin(100-\alpha)$ or $ \cos(\alpha-20)-\cos(\alpha+20)=\cos(3\alpha-40)-\cos(\alpha+160)$ since $ \cos(\alpha-20)+\cos(\alpha+160)=0$, it follows that $ \cos(\alpha+20)+\cos(3\alpha-40)=0$, from where it's easy to conclude that $ \alpha=50$

let $ D$ be on line $ BC$ such that $ B$ lies between $ C$ and $ D$, and $ AB = BD$. since $ \angle ADC = \angle AMC = 30$ it follows that $ ADMC$ is cyclic. let $ O$ be the circumcenter of $ ADMC$. note that $ \angle AOC = 2\angle AMC = 60 = \angle ABC$, so $ A,B,O,C$ are cyclic. note that $ \angle AOD = 2\angle ACD = 120 = \angle ABD$, so $ A,O,B,D$ are cyclic. since the circumcircles of $ A,B,O,C$ and $ A,O,B,D$ cannot coincide it follows that $ B = O$, from where it follows easily that $ \angle BAM = 50$ edit: there's a much more easier solution... $ \angle CAD=90$, and since $ B$ is the midpoint of $ CD$ it follows that $ B$ is the circumcenter of $ ACD$, from where we conclude that $ B=O$... the rest follows easily..

Sory! My picture is wrong, solution of campos is very nice. Thank you very much.

Oh! Problem of Valentin Vornicu is very nice. Thanh you very much!

Actually this year where proposed very weak problems for 9th grade.This is also an easy one.

campos wrote: from law of sines we can get $ \sin \alpha\cdot \sin 20\cdot \sin 30\cdot \sin 60 = \sin(60 - \alpha)\cdot\sin 60\cdot \sin (100 - \alpha)\cdot \sin (30 + \alpha)$ can you explain this ? thank you !

I think you solution {CAMPOS} is right but yor result is wrong <<angle BAM=20>> I solve this problem with Ceve theorem . (angle BAM=x) sin(60-x)/sin(x) * sin(160-x)/sin(100-x) * sin(30+x)/sin(90+x) =1 & we now sin(60-x)=cos(30+x) & sin(90+x)= cos(x) & cos(30+x) * sin(30+x)=1/2 * sin(60+x) & cos(x) * sin(x)=1/2 * sin(2x) & sin(160-x)=sin(20+x) & sin(100-x)=sin(80+x) then we have sin(60+x) * sin(20+x) =sin(2x) * sin(80+x) then we have cos(40+x)=cos(80-x) then x=20

Atention, I modified the enunciation giving a degree of freedom to the point $ M$ ! Valentin Vornicu wrote: Let $ ABC$ be a equilateral triangle and let $ M$ be a point for which the line $ BC$ separates the points $ A$ , $ M$ and $ \{\begin{array}{c} m(\widehat {AMB}) = 20^{\circ} \\ \ m(\widehat {AMC}) = 30^{\circ}\end{array}$ . Ascertain $ m(\widehat {MAB})$ . Remark. Denote $ E\in BC\cap AM$ . CAUTION ! Appear two cases : $ E\in (BC)$ or $ B\in (EC)$ . Proof I. Denote the point $ N\in BC$ for which $ B\in (NC)$ and $ BN = BC$ . Observe that $ M(\widehat {ANC}) = m(\widehat {AMC}) = 30^{\circ}$ . Thus, $ ANMC$ is inscribed in the circle with the center $ B$ . Hence $ BM = BA$ , i.e. $ m(\widehat {MAB}) = m(\widehat {AMB}) = 20^{\circ}$ . Proof II. $ \{\begin{array}{c} m(\widehat {BAC}) = m(\widehat {ACB}) = m(\widehat {CBA}) = 60 \\ \ m(\widehat {CMA}) = 30\ \ ,\ \ m(\widehat {AMB}) = 20\end{array}$ . Denote $ m(\widehat {BAM}) = x$ . Case $ 1\blacktriangleright\ E\in (BC)$ . Observe that $ \{\begin{array}{c} m(\widehat {MAC}) = 60 - x \\ \ m(\widehat {BCM}) = 30 + x \\ \ m(\widehat {MBC}) = 100 - x\end{array}$ . Apply the well-known property in the quadrilateral $ ABMC$ : $ \boxed {\boxed {\ \begin{array}{c} \\ \ \sin\widehat {BAM}\cdot\sin\widehat {MBC}\cdot\sin\widehat {CMA}\cdot\sin\widehat {ACB} = \sin\widehat {MAC}\cdot\sin\widehat {CBA}\cdot\sin\widehat {AMB}\cdot\sin\widehat {BCM} \\ \\ \sin x\sin (100 - x)\sin 30\sin 60 = \sin (60 - x)\sin 60\sin 20\sin (30 + x) \\ \\ \sin x\sin (80 + x) = 2\cos (30 + x)\sin 20\sin (30 + x) \\ \\ \sin x\cos (10 - x) = \sin 20\sin (60 + 2x) \\ \\ \sin x\cos (10 - x) = \sin 20\cos (30 - 2x) \\ \\ \sin 10 + \sin (2x - 10) = \sin (50 - 2x) + \sin (2x - 10) \\ \\ \sin 10 = \sin (50 - 2x) \\ \\ 10 = 50 - 2x\ \ \vee\ \ 10 + (50 - 2x) = 180 \\ \\ x = 20\ \ \vee\ \ x = - 60\not\in \\ \\ \boxed {\ x = 20\ } \\ \\ \end{array}\ }}$ Case $ 2\blacktriangleright\ B\in (EC)$ . Observe that $ m(\widehat {BCM}) = 30 - x$ and $ m(\widehat {BMC}) = 10$ . Apply the trigonometrical form of the Ceva's theorem to the interior point $ B$ for the triangle $ AMC$ : $ \boxed {\boxed {\ \begin {array}{c} \\ \ \sin\widehat {BAM}\cdot\sin\widehat {BMC}\cdot\sin\widehat {BCA} = \sin\widehat {BAC}\cdot\sin\widehat {BMA}\cdot\sin\widehat {BCM} \\ \\ \sin x\sin 10\sin 60 = \sin 60\sin 20\sin (30 - x) \\ \\ \sin x\sin 10 = \sin 20\sin (30 - x) \\ \\ \cos (x - 10) - \cos (x + 10) = \cos (x - 10) - \cos (50 - x) \\ \\ \cos (x + 10) = \cos (50 - x) \\ \\ x + 10 = 50 - x \\ \\ \boxed {\ x = 20\ } \\ \\ \end{array}\ }}$ Virgil Nicula wrote: An easy extension. Let $ ABC$ be an $ B$ - isosceles triangle, i.e. $ BA = BC$ and let $ M$ be a point for which the line $ BC$ separates the points $ A$ , $ M$ and $ m(\widehat {AMC}) = 90^{\circ} - m(\widehat {BAC})$ . Prove that $ \widehat {AMB}\equiv\widehat {BAM}$ . Proof . Let $ N\in BC$ for which $ B\in (NC)$ and $ BN = BC$ . Observe that $ m(\widehat {ANC}) = m(\widehat {AMC}) = 90^{\circ} - m(\widehat {BAC})$ . Thus, $ ANMC$ is inscribed in the circle with the center $ B$ . Hence $ BM = BA$ , i.e. $ \widehat {AMB}\equiv\widehat {BAM}$ . Remark. Now you can generate and another similar problems !

Thank you M. Nicula ! Multsam fain ! Your proof is of a "biblical" simplicity ! I've made a figure for Proof 1 to thank you again !

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