Monday, July 23, 2012

Problem 789: Right Triangle, Altitude, Angle Bisector, Incircle, Tangency Point, Inradius, Circles, Tangent Line

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 789.

Online Geometry Problem 789: Right Triangle, Altitude, Angle Bisector, Incircle, Tangency Point, Inradius, Circles, Tangent Line.

Posted by Antonio Gutierrez at 9:55 AM
Labels: , , , , , ,

10 comments:

  1. Peter TranJuly 23, 2012 at 2:01 PM

    http://img138.imageshack.us/img138/1382/problem789.png

    ( complete solution)
    drawing lines and circle per attached sketch
    We have Arc B’F= Arc FC and Arc AB=Arc AB’
    ∠ABD=1/2(Arc AB’ +Arc B’F)
    ∠ADB=1/2(Arc AB +Arc FC) = ∠ABD
    So ABD is an isosceles triangle with AB=AD
    We also have ∠HBD=∠DBK=∠BDK.. ( alternate angles)
    So BKD is an isosceles triangle with KB=KD
    A,I and K are collinear ....(located on perpen. Bisector of BD)
    IK is an angle bisector of ∠ BKD
    So IM=IL=ED=r

    ReplyDelete
    Replies
    1. AnonymousJuly 23, 2012 at 7:48 PM

      Anonymous said...

      AB=AD so we put M and N the match points with AB and BC .

      IM=IN=BN=BM=r AM=AE---> AD-AE = AB-AM=r
      ED=r

      July 23, 2012 7:35 PM

      Delete
  2. AjitJuly 23, 2012 at 6:34 PM

    CD/HD = BC/BH or CD = CH *BC/(BC+BH) = a^2/(b+c). Now let the in-circle touch AB at F. AE =AF = c - r or CE = b-c+r which gives us:DE = CE-CD=b-c+r-a^2/(b+c)= (b^2-c^2-a^2)/(b+c) + r = r since b^2=a^2+c^2. Thus, DE = r

    ReplyDelete
  3. W FungJuly 23, 2012 at 6:40 PM

    Simply let AB = a, BC = b, AC = c

    By using the properties of inradius, r = ab/(a+b+c)
    Our target is DE = ab/(a+b+c)

    The altittude = ab/c
    CH = sqrt(b^2-(ab/c)^2) = (b^2)/c
    Using angle bisector theorem, CD = CH*(BC/(BC+CH)) = (b^2)/(a+c)
    By incircle properties, CE = (b+c-a)/2
    Then DE = CE - CD = (b+c-a)/2 - (b^2)/(a+c)
    By trigonometry, we can easily prove the remainings.

    ReplyDelete
  4. AnonymousJuly 23, 2012 at 7:30 PM

    Anonymous

    1. AB=AD We put M and N the points where they match with AB and BC

    IM=EN=NB=BM=r
    AM=AE AD-AE= AB-AM=r so in this case ED=r

    ReplyDelete
  5. Eder Contreras OrdenesJuly 24, 2012 at 12:44 PM

    http://img507.imageshack.us/img507/9789/p781resuelto.png
    (look at the picture)

    Let be P the intersection point of the ray AI and segment BC. Note that as triangle ABC is rectangle, we know that <BAC=<HBC, but AI and BD are bisectors, then <DAP=<DBP, i.e, the quadrilateral ADPB is cyclic. immediately <PDA=90° and BP=DP. As triangle DBP is isosceles, the incircle is tangent to DP too.
    Now let Q be the tangency point of incircle with DP. Note that IQED is a square of side r. Therefore DE=r

    Greetings :)

    ReplyDelete
    Replies
    1. Eder Contreras OrdenesJuly 24, 2012 at 5:04 PM

      Look at THIS picture! Sorry :P

      http://img27.imageshack.us/img27/8784/p789resuelto.png

      Delete
  6. Sumith PeirisFebruary 24, 2016 at 2:46 AM

    Let F be the tangency point on AB

    < IBC = 45 and < DBC = 45-C/2 so < IBD = C/2

    Hence BCDI is cyclic and BI = ID since IC bisects < C

    So Tr.s BFI and DEI are congruent (SS & right angle)

    Therefore DE = BF = r

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Sumith PeirisFebruary 24, 2016 at 2:51 AM

    Or because BCDI is cyclic and < IDE = IBC = 45 the result follows

    ReplyDelete
  8. APOSTOLIS MANOLOUDISAugust 5, 2016 at 10:11 AM

    Problem 789
    Let <ABH=<BCA=x and <HBD=<DBC=y then <ABD=x+y=<ADB so AB=AD.Si the incircle of radius r is tangent to AB at F then AF=AE,BF=ED.But <IBF=45 so ED=FB=IF=r.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete

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