Tuesday, July 24, 2012

Problem 790: Right Triangle, Altitude, Angle Bisector, Incircle, Tangency Point, Inradius, Circles, Tangent Line, Congruence

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

Click the figure below to see the complete problem 790.

Online Geometry Problem 790: Right Triangle, Altitude, Angle Bisector, Incircle, Tangency Point, Inradius, Circles, Tangent Line, Congruence.

Posted by Antonio Gutierrez at 11:56 AM
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4 comments:

  1. Peter TranJuly 24, 2012 at 1:24 PM

    http://img269.imageshack.us/img269/4894/problem790.png

    Draw lines per attached sketch
    We have DE perpen. to AC
    angle(HBD)= angle(DBE)= angle(BDE) => EB=ED
    Triangle ABE congruence to tri. AED (Right triangle Case HL)
    So AB=AD => A, I1, I and E are collinear
    And IB=ID= r.sqrt(2)
    Right triangles AHB, HBC and ABC are similar
    And r1/r=AB/AC= cos(A)
    r2/r=BC/AC= sin(A)
    In right triangle I1HI2 we have (I1I2)^2=(I1H)^2+I2H)^2
    =2(r1)^2 +2(r2)^2 = 2.r^2 (cosA^2+sinA^2)=2.r^2
    So I1I2=r.sqrt(2)

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  2. Jacob HaJuly 24, 2012 at 5:31 PM

    BI=r*sqrt(2) is obvious since angle IBA=45 degrees.
    DI=r*sqrt(2) is followed from Problem 789 that "DE=r".

    Let AB=a, BC=b, AC=c.

    Then
    BH=ab/c
    AH=a^2/c
    CH=b^2/c
    r=ab/(a+b+c)

    Denoted the radius of circle I1 and I2 by r1, r2 respectively.

    Then
    r1=[a^3 b/c^2] / [a + ab/c + a^2/c] = (a/c)*r
    r2=[a b^3/c^2] / [b + ab/c + b^2/c] = (b/c)*r

    I1H = r1*sqrt(2) = (a/c)*r*sqrt(2)
    I2H = r2*sqrt(2) = (b/c)*r*sqrt(2)

    Since angle I1HI2 = 90 degrees, using Pythagoras theorem,
    I1I2 = r*sqrt(2)

    ReplyDelete
  3. W FungJuly 24, 2012 at 5:55 PM

    BI = r*sqrt2 is straight forward by right-angled triangle and the properties of incircle.

    DI = r*sqrt2 is also obvious by the previous problem 789.

    For I1I2, Let AC = b, angle BCA = x, then BH = b*sinx*cosx, AH = b*(sinx)^2, CH = b*(cosx)^2

    then using the relationship between inradius and triangle area of ABC, AHB and BHC, we have

    r = b*sinx*cosx / (1+cosx+sinx)

    r1 = (b)*(sinx)^2*cosx*sinx / (1+sinx+cosx) = r*(sinx+cosx)
    r2 = (b)*(cosx)^2*cosx*sinx / (1+sinx+cosx) = r*(cosx-sinx)

    Now the horizontal distance between two incenters would be
    x = r1 + r2 = r*(sinx+cosx)

    the vertical distance between two incenters would be
    y = r2 - r1 = r*(cosx-sinx)

    By pyth. theorem,
    I1I2 = sqrt(x^2 + y^2)
    = r*sqrt(2)

    q.e.d.

    ReplyDelete
  4. APOSTOLIS MANOLOUDISAugust 5, 2016 at 11:39 AM

    Problem 790
    Let <ABI1=I1BH=BCI2=I2CA=x, <CAI1=I1AB=<HBD=DBC=y.Then <I1AI2=x+y=45, <I1BC+<BCI2=x+2y+x=2(x+y)=2.45=90 so BI1 is perpendicular in CI2. Similar
    AI1 is perpendicular in BI2. But <AI1I=x+y=45.Si BI1 intersects the CI at M and
    BI2 intersects the AI1 at N,then I is orthocenter of triangleBI1I2 .Βut MI=MI1
    and BM=MI2.So triagle BMI=triangleI2MI1.So DI=BI=I1I2=r√2.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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