Sunday, September 8, 2013

Geometry Problem 920: Triangle, Angles, 10, 20, 30 Degrees, Congruence

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

This entry contributed by Erina-NJ.
Click the figure below to see the complete problem 920.

Online Geometry Problem 920: Triangle, Angles, 10, 20 Degrees, Congruence

Posted by Antonio Gutierrez at 12:39 AM
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10 comments:

  1. AjitSeptember 8, 2013 at 11:48 PM

    I think that Ceva's trigonometric theorem gives:sin(30)/sin(10)*sin(x)/sin(120-x)=1 which yields, x=20° and /_BCD=100° as may ascertained from the diagram above.
    In any case, what we are looking for is a synthetic solution, aren't we?

    ReplyDelete
  2. PravinSeptember 10, 2013 at 9:22 AM

    (Thank you once again Antony Gutierrez)
    Problem 920. Pl refer to my earlier post:
    By Ceva,
    sin DAB. sin DBC. sin DCA = sin DAC. sin DBA. sin DCA
    sin 10°.sin 30°.sin x = sin 10°.sin 10°.sin(120°-x).(This line is corrected)
    So sin 30°.sin x = sin 10°.sin(120° - x).
    Since sin 30° = 1/2 and sin(120°-x)= Sin(90°+30°-x)= cos(30°-x)
    sin x = 2 sin 10°.cos(30°-x)
    (OR)
    By sine rule:
    sin x / sin 10° = AD/DC = DB/DC = sin(120°-x)/sin 30° = 2 sin(120°-x)
    So sin x = 2 sin 10°.sin(120°-x)= 2 sin 10°.cos(30°-x)
    By inspection x = 20° satisfies the above equation.
    P.S.:Is there a simple way of solving the above trig equation?

    ReplyDelete
    Replies
    1. Antonio GutierrezSeptember 11, 2013 at 9:24 PM

      Problem 20 From Erina Nj To Pravin

      Sinx = 2 sin10 cos (30-x)
      sinx = sin (40-x)- sin (20-x)
      sin (40-x)- sinx= Sin (20-x)
      2sin (20-x) cos 20-sin (20-x) =0
      sin(20-x) [2cos20-1] =0
      2cos20-1 different from 0 => sin(20-x) = 0
      X=20

      Delete
    2. AjitOctober 4, 2013 at 3:52 AM

      @Anonymous, how about a synthetic solution to this exvellent problem?

      Delete
  3. Μιχάλης ΝάννοςOctober 14, 2013 at 8:28 PM

    http://www.mathematica.gr/forum/viewtopic.php?f=22&t=40350

    ReplyDelete
  4. UnknownJuly 1, 2014 at 4:58 PM

    can you solve using elemtery geomitry

    ReplyDelete
    Replies
    1. APOSTOLIS MANOLOUDISDecember 24, 2015 at 10:41 AM

      MANOLOUDIS APOSTOLIS
      4th HIGH SCHOOL OF KORYDALLOS -PIRAEUS-GREECE
      PROBLEM 920
      SOLUTION 1

      E is a point of AC. Then <EBC=10 degrees so <DBE=20 degrees .If O the pericenter of triangle ABE then OA=OE=OB=AE because <AOE=2×<ABE=2×30=60 degrees and <OAB=<OBA= 40 degrees. Also DA=DB so OD is perpendicular to AB and OD is vertical to the middle of AB hence <AOD=50 degrees= <OAD thus DO=DA,OE=AE so tri ADE=tri ODE.
      So <DEA = 60/2= 30 degrees = <DBC so DECB is inscribable hence <DCA=<DCE=<DBA= 20 degrees

      SECOND SOLUTION
      Let E, F the symmetrical of point A to DB and BC respectively. Then DA=DE=DB, AB=BE=BF, <CBF=<CBA= 40 degrees, <CFB= <CAB= 20 degrees. Then B is the pericenter of triangle AEF so <AFE =10 degrees so triangle EBF is equilateral. So DE=DB, EF=BF so DF is vertical to the middle of EB. We conclude that <DFA= 20 degrees, <DFC= 10 degrees= <DAC
      so ADCF is inscribable hence <DCA=<DFA= 20 degrees

      Delete
  5. Stan FulgerJanuary 1, 2016 at 1:42 AM

    It seems the following proof it's not among the given ones: Take E the reflection of C in AD; easily BE=CE. Construct the equilateral triangle CED' inside the Langley triangle ACE; D'E=BE and <ABD'=10 degs, hence D'=D. Now <ACD=<AED=20.

    Happy New Year!

    ReplyDelete
  6. Sumith PeirisApril 28, 2016 at 10:35 AM

    Excellent work Apostolis and Stan in finding these pure geometry proofs

    ReplyDelete
  7. APOSTOLIS MANOLOUDISOctober 5, 2016 at 10:06 AM

    Thank you very much for your nice words Sumith

    ReplyDelete

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