$\boxed{\text{G1}}$ Let $M$ be interior point of the triangle $ABC$ with <BAC=70and <ABC=80 If <ACM=10 and <CBM=20.Prove that $AB=MC$
Problem
Source: JBMO Shortlist 2007
Tags: geometry
den_thewhitelion
12.05.2016 16:24
M=circumcenter of triangle ABC
tenplusten
12.05.2016 16:35
den_thewhitelion wrote: M=circumcenter of triangle ABC Can you prove that?
den_thewhitelion
12.05.2016 16:38
Yes: MB=MC(triangle MBC has 2 20 angles) $\angle BMC=2\angle BAC$
MATH1945
12.05.2016 17:34
where could you got the shortlist??
tenplusten
12.05.2016 18:23
See here http://www.mathematica.gr/forum/viewtopic.php?f=58&t=49341
itslumi
29.11.2019 12:07
tenplusten wrote: den_thewhitelion wrote: M=circumcenter of triangle ABC Can you prove that? Because angle BMC=140 and we have that BAC=70 and also BM=MC
Taco12
06.09.2021 01:49
Note that $M$ is the circumcenter of triangle $ABC$. Since triangle $AMC$ is isosceles, $\angle MAB = 60^{\circ}$. Thus, since triangle $AMB$ is equilateral, $AB = BM = MC$, as desired.
Lamboreghini
06.09.2021 02:01
Since $\angle ACM=10^\circ,$ we have that $\angle MBC=20^\circ.$ Hence, $\triangle MBC$ is isosceles with $MB=MC.$ Since $M$ lies in the interior of $\triangle ABC,$ it is the circumcenter of $\triangle ABC.$ Additionally, since $\triangle AMB$ has one angle of $60^\circ,$ we have that $\triangle AMB$ is equilateral. Hence, $AB=MB=MC=MA.$
OlympusHero
09.09.2021 00:57
Construct $AM$, let $\angle MAC = x$, and angle chase. From Trig Ceva, we have $\frac{\sin(70-x)}{\sin x} \cdot \frac{\sin 20}{\sin 60} \cdot \frac{\sin 10}{\sin 20} = 1$, or $\frac{\sin(70-x)}{\sin x}=\frac{\sin 60}{\sin 10} \implies x =10$. But then $\angle BMA = \angle BAM$, so $AB=BM$. But $BM=MC$, so $AB=MC$ as desired.
nurlan2024
16.02.2025 15:08
sina/sin60=sin(70-a)/sin10=m/n we find a =60 and triangle is equilateral