Wednesday, August 1, 2018

Geometry Problem 1372: Equilateral Triangle, Exterior Cevian, Inradius, Exradius, Altitude, Sketch, iPad Apps

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 1372: Equilateral Triangle, Exterior Cevian, Inradius, Exradius, Altitude, Sketch, iPad Apps.

5 comments:

  1. Let:
    r_1 =sqrt(3)a
    r_2=sqrt(3)b
    h=sqrt(3)c

    Let R be the point on the line AB where the circle of radius r_1 is tangent.
    Let T be the point on the line AB where the circle of radius r_2 is tangent.
    Let N be the point where the circle of radius r_2 is tangent to the extension of DC
    Let O_1 be the center of the circle radius r_1.

    If we let M be the point on DN where the circle O_1 is tangent then O_1RAM is cyclical.
    Because ∠MAR =120 the triangle ARO_1 is a 30-60-90 triangle and AR=a

    If h=sqrt(3)c then the length of the sides of the equilateral triangle ABC =2c ,therefore AB =2c and BR =2c-a

    Be similar reasoning we find that AT=b and BT =2c-b

    Then let Q be the point where the circle of radius r_2 is tangent to the extension of line DB and let P be the point where the circle of radius r_1 is tangent to the line DB.

    Then PB = BR and BQ=BT therefore

    PB =2c -a
    BQ=2c-b

    and

    PQ=PB+BQ
    PQ= 4c-a-b

    We also have MA =a ,AC =2c and CN =b therefore

    MN=MA+AC+CN
    MN=2c+a+b

    MN=PQ (common tangent line) therefore

    4c-a-b=2c+a+b
    2c=2a+2b
    c=a+b

    With r_1=sqrt(3)a,r_2=sqrt(3)b and h=sqrt(3)c we have

    h/sqrt(3)=a/sqrt(3)+b/sqrt(3)
    h=r_1+r_2

    ReplyDelete
    Replies
    1. I noticed i made an error in the last two lines.
      Instead of:

      h/sqrt(3)=a/sqrt(3)+b/sqrt(3)


      it should have been :

      h/sqrt(3)=r_1/sqrt(3)+r_2/sqrt(3)

      Then we get

      h=r_1+r_2

      Delete
  2. Let in circle be P and excircle be Q.
    Let DAC touch circle P at X and circle Q at Y
    Let DB touch circle P at U and circle Q at V

    Let AB = BC = CA = a
    Let AX = b and CY = c

    UV = XY so a-b + a-c = b + a + c
    So b + c = a/2 ......(1)

    But b = r1/sqrt3 and c = r2/sqrt3
    Further a/2 = h/sqrt3

    So from (1),

    r1 + r2 = h

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. Such an elegant solution. After having tried for pretty long time I gave up and checked the solution. You make it look so simple and easy.
      Kudos to you Sir ......!!

      Delete
  3. Let the center of the circle with radius r1 be P and meet DA at E and similary the center of r2 be Q and meet at F
    Let the length of equilateral triangle be 'a'
    Let the length of DE be 'x'
    QCF is 30-60-90 triangle, hence CF=r2/sqrt(3) ------------(1)
    Also PAE is 30-60-90 triangle, hence EA=r1/sqrt(3) ------------(2) (Since m(c)=60,m(B)=60=>m(DAB)=120)
    As we know DF=Semi-perimeter of triangle DBC=DE+EA+AC+CF
    => S = x+r1/sqrt(3)+a+r2/sqrt(3)-------------------(3)

    However S=(DB+BC+CD)/2
    => S=[(x+a-r1/sqrt(3))+(a)+(a+r1/sqrt(3)+x)]/2
    => S=x+3a/2 ----------------(4)

    As (3)=(4)
    => r1/sqrt(3)+r2/sqrt(3)=a/2
    => r1+r2=h Q.E.D

    ReplyDelete