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Source: JBMO Shortlist 2007

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#1 h May 12, 2016, 1:18 PM • 2 Y

Y by Adventure10, Mango247

$\boxed{\text{G1}}$ Let $M$ be interior point of the triangle $ABC$ with <BAC=70and <ABC=80 If <ACM=10 and <CBM=20.Prove that $AB=MC$

This post has been edited 2 times. Last edited by tenplusten, May 12, 2016, 1:20 PM

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den_thewhitelion

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#2 h May 12, 2016, 1:24 PM • 1 Y

Y by Adventure10

M=circumcenter of triangle ABC

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#3 h May 12, 2016, 1:35 PM • 2 Y

Y by Adventure10, Mango247

den_thewhitelion wrote:

M=circumcenter of triangle ABC

Can you prove that?

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den_thewhitelion

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#4 h May 12, 2016, 1:38 PM • 1 Y

Y by Adventure10

Yes:
MB=MC(triangle MBC has 2 20 angles)
$\angle BMC=2\angle BAC$

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#5 h May 12, 2016, 2:34 PM • 2 Y

Y by Adventure10, Mango247

where could you got the shortlist??

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#6 h May 12, 2016, 3:23 PM • 2 Y

Y by Adventure10, Mango247

See here http://www.mathematica.gr/forum/viewtopic.php?f=58&t=49341

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#7 h Nov 29, 2019, 9:07 AM • 1 Y

Y by Adventure10

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

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Taco12

#8 h Sep 5, 2021, 10:49 PM

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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.

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#9 h Sep 5, 2021, 11:01 PM • 3 Y

Y by Mango247, Mango247, Mango247

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, $AB=MB=MC=MA.$

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#10 h Sep 8, 2021, 9:57 PM

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Solution

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#11 h Feb 16, 2025, 12:38 PM

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sina/sin60=sin(70-a)/sin10=m/n we find a =60 and triangle is equilateral

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