Friday, September 30, 2011

Problem 674: Triangle, Altitude, Angles

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Alejandro Astudillo.

Click the figure below to see the complete problem 674.

Online Geometry Problem 674: Triangle, Altitude, Angles

Posted by Antonio Gutierrez at 4:52 PM
Labels: , ,

9 comments:

  1. AlejandroSeptember 30, 2011 at 10:05 PM

    Thank you Antonio to put it in your site, this a really hard problem!

    ReplyDelete
  2. PravinOctober 1, 2011 at 4:01 AM

    AH:CH = tan 5x : tan 3x and also
    AH:CH = tan 2x : tan x

    tan 5x tan x = tan 3x tan 2x

    sin 5x cos 3x cos 2x sin x
    = cos 5x sin 3x sin 2x cos x

    (sin 8x + sin 2x)(sin 3x – sin x)
    = (sin 8x - sin 2x)(sin 3x + sin x)

    sin 8x : sin 2x = sin 3x : sin x
    (by componendo & dividendo)

    sin 8x sin x = sin 3x sin 2x

    sin 8x = 2 sin 3x cos x = sin 4x + sin 2x

    sin 8x – sin 4x = sin 2x

    2 cos 6x sin 2x = sin 2x

    cos 6x = 1/2

    6x = 60 deg

    x = 10 deg

    ReplyDelete
  3. AlejandroOctober 1, 2011 at 7:33 PM

    Of course, the trigonometry version is pretty easy, how about a synthetic solution?

    ReplyDelete
  4. AjitOctober 1, 2011 at 11:02 PM

    If trigonometry be allowed then even Ceva's Theorem in trigonometric form directly leads to sin(10x)= 4cos(x)sin(3x)cos(6x) and thence to sin(8x)= sin(2x)+ sin(4x) or cos(6x)=1/2 whence x=10. But the real challenge, I guess, is to find an elegant plane geometry solution. In this regard, I only have a suggestion to make: Extend CD to meet AB in E. Now /_ADE=3x while /_EAD=90-6x. If we can now prove that DE=AE then 90-6x=3x or x=10. However, so far I've not been successful in proving that Tr.ADE is isosceles.
    Ajit

    ReplyDelete
  5. AnonymousOctober 2, 2011 at 12:04 PM

    The altitude of triangle ABC is BH, not BD.

    ReplyDelete
  6. AnonymousOctober 10, 2011 at 2:37 AM

    Extend HC rightwards and construct AB=BI, I is on extension of HC. Connect DI; extend AD rightwards to intersect with BC at J; make DK intersect at AC so Angle CDK=Angle DAC;and connect JI.

    ReplyDelete
  7. AnonymousOctober 10, 2011 at 2:41 AM

    P.S. Connect JK.

    ReplyDelete
  8. AnonymousOctober 10, 2011 at 3:00 AM

    Since AB=BI in Iso Tri ABI, AD=DI (Angle DAI=DIA=x), Angle DCA-Angle DIC=Angle CDI=x, DC=CI. Angle JDC=Angle DBC=x, Angle JCD=DCB, Tri JDC~Tri DBC, DC^2=JC*CB; CI=DC, CI^2=JC*CB, Tri CJI~Tri CIB, Angle CIJ=Angle CBI=2x.Angle JID=Angle JIA-Angle DIA=x;DI bisects Angle JIA. Angle CDK=Angle CAD=x, Angle DCK=Angle ACD, Tri CDK~Angle CAD, DC^2=KC*AC. (ID bisects Angle JDK.)JC*CB=KC*AC, cyclic quadrilateral ABJK, Angle JKI=8x. 8x+x=90, x=10.

    ReplyDelete
  9. Antonio GutierrezOctober 2, 2019 at 10:34 AM

    Solucion enviada por Pedro Miranda de Peru
    Ver solucion aqui

    ReplyDelete

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