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
12.05.2016 16:24
M=circumcenter of triangle ABC
12.05.2016 16:35
den_thewhitelion wrote: M=circumcenter of triangle ABC Can you prove that?
12.05.2016 16:38
Yes: MB=MC(triangle MBC has 2 20 angles) ∠BMC=2∠BAC
12.05.2016 17:34
where could you got the shortlist??
12.05.2016 18:23
See here http://www.mathematica.gr/forum/viewtopic.php?f=58&t=49341
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
06.09.2021 01:49
Note that M is the circumcenter of triangle ABC. Since triangle AMC is isosceles, ∠MAB=60∘. Thus, since triangle AMB is equilateral, AB=BM=MC, as desired.
06.09.2021 02:01
Since ∠ACM=10∘, we have that ∠MBC=20∘. Hence, △MBC is isosceles with MB=MC. Since M lies in the interior of △ABC, it is the circumcenter of △ABC. Additionally, since △AMB has one angle of 60∘, we have that △AMB is equilateral. Hence, AB=MB=MC=MA.
09.09.2021 00:57
16.02.2025 15:08
sina/sin60=sin(70-a)/sin10=m/n we find a =60 and triangle is equilateral