Saturday, August 4, 2018

Geometry Problem 1373: Isosceles Triangle, Exterior Cevian, Inradius, Exradius, Altitude to the Base

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

Details: Click on the figure below.

Geometry Problem 1373: Isosceles Triangle, Exterior Cevian, Inradius, Exradius, Altitude to the Base.

5 comments:

  1. Denote center of small circle as P and large circle as Q
    Drop a perpendicular from P to BE,denote it as X and form a right-triangle BPX
    From Q, drop perpendicular to BC,denote it as Y and form a right-trianlge BQY
    We will prove that these two triangles are congruent and the result follows

    Let the foot of perpendicular from P to DA be G and similarly from Q to DC be H
    Similarly the foot of Perpendicular from P to DB be U and Q to DB be V
    Let AB=BC=a , AE=EC=b, GA=x, CH=y
    We know BV=BY=GE=PX=x+b ---------------(1)

    let m(BPA)=@ and m(BAC)=$
    =>m(XBP)=90-$+@ and m(BPX)=$-@ ------------(2)
    Similarly m(QBY)=$-@ and m(BQY)=90-$+@-------------(3)

    From (1),(2) and (3), BPX and QBY are congruent and BX=QY=r2
    Since XE=r1
    =>h=BX+XE=r2+r1 Q.E.D

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    Replies
    1. Correction :
      let m(ABP)=@ instead of m(BPA)=@

      Delete
    2. How do we know BY=GE?

      Delete
    3. Let BA=BC=a, UB=a1 and BV=a2 and r1 touch AB at Z
      then UB=BZ=a1 => AG=a-a1, ||ly BV=BY=a2 => CH=a-a2 -------(1)
      But UV=GH (since DV=DH)
      => a1+a2=GA+AE+EC+CH
      => a1+a2=a-a1+b+b+a-a2
      => a1+a2 = a+b
      => GH = a+b--------(2)

      Now GE= GH-EH
      => GE=(a+b)-(b+a-a2)
      => GE=a2=BY

      Delete
  2. Is the reverse true? I mean, if the height of the triangle is the sum of radii of the circles; is it always an isosceles triangle? I couldn't prove this.

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