Tuesday, December 27, 2016

Geometry Problem 1299 Triangles, Interior Point, Angles, 30 degrees

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Apostolis Manoloudis, Korydallos, Piraeus, Greece.

Details: click on the figure below.

Geometry Problem 1299 Triangles, Interior Point, Angles, 30 degrees.

8 comments:

  1. Take D on AC so that <CMD=4, construct the equilateral triangle MDO, so that O is the circumcenter of triangle AMD. thus <AOM=34 and <MAO=73. Take now B' 2nd intersection of circle (CDM) with BC. Thus <MB'C=17 and, since <ACM=<BCM=13, we get B'M=MD. Since <B'MD=154, easy angle chase gives <AMB'=73, consequently tr. AMB' and AMO are congruent (s.a.s.), thus <MAB'=73 and <AB'C=51, consequently B'=B and therefore <BMC=<BMD-<CMD=154-4=150 degs.

    Best regards

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    Replies
    1. Awesome Solution Mr.Stan. Btw, Wondering how did you arrive at considering <CMD = 4 to start with ? Is there a technique to solve such problems ?

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    2. Good work Stan

      Antonio and Apostolis - do u have a different method?

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    3. The only technique I really know is the following: just to get the result!!

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  2. Here is my attempt
    Since Angle ACB is 26 deg, CM is bisector of angle ACB.
    Draw a line perpendicular from B to AM, lets assume it meets AM at E and AC at D.
    Locate a point M' on AM , such that AE=EM'.
    Angle ABD=Angle DBM'= Angle CBM'= 17 deg.
    Also angle ADB= Angle BDM' = Angle Angle CDM'=60 deg.
    We have BM' as bisector of angle DBC and DM' as bisector of angle BDC, hence incenter of triangle BDC is point M' and it lies on line AM. Since CM is bisector of Angle C, incentre should also lie on line CM. Hence M and M' must coincide and M must be incentre of triangle BDC, we have Angle MBC=17 deg. We get Angle CMB equal to 150 deg.

    It is true for any triangle with angles 2x,90+x and 90-3x. Here we have x=13 deg.

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    Replies
    1. Excellent work, really better than my own! Thank you!

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    2. Thanks Stan for your kind words!

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  3. Here is an approach which does not use 2 points assumed distinct at first, later shown to coincide.

    Construct the kite ABDC.

    Find N within Tr. BDC such that NC bisects < BCD & < NBC = 17. Let the bisector of < DBN meet DC at E.

    Now N is the incentre of Tr. BEC and so < DEB = <NEB = < NEC = 60.

    Hence Tr.s BED & BEN are congruent ASA and so BD = BN = AB.

    So Tr.s DNC & AMC are congruent ASA and so AM = DN.

    Therefore Tr.s ABM & BDN are congruent SAS and so < AMB = 73 and since < AMC = 137,
    x = 360-73-137 = 150.

    Sumith Peiris
    Moratuwa
    Sri Lanka

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