Friday, November 27, 2020

Geometry Problem 1486: Right Triangle, Altitude, Incircle, Tangent, Measurement

Geometry Problem 1486. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1486:Right Triangle, Altitude, Incircle, Tangent, Measurement.

7 comments:

  1. Tr.s ABD, BDC, ABC are similar.
    If r is the inradius, r/b = 3/c = 4/a ( = 1/n say) and since a^2 + c^2 = b^2, r = 5.

    So a = 4n, b = 5n and c = 3n

    Now also r = ac/(a+b+c) from which 5.(4n + 5n + 3n) =12.n^2
    so that n = 5 and hence a = 20, b = 25 and c = 15

    So DE = CD - CE = 20^2/25 - (20 + 25 - 15)/2 = 16 - 15 = 1

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  2. From D are drown vertical two equal tangential segments
    So FP (P tg point of red circle)=DE=4-3=1

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    Replies
    1. https://photos.app.goo.gl/9nhsrsHvmvmBh5HAA

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    2. So centers of three circles lie on a circle with center E
      => from one of righttriangles with side 3, 4 can calculate the R=5

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  3. Credits to Sumith sir for deriving InRadius = 5 and AB=15,BC=20,AC=25
    BD.25=15.20
    BD=12
    Also 3/AD=5/15
    AD=9
    let P,O,Q be the centers of Green,Blue and Red incircles.M,N be the foot of perpendiculars from P and Q on AC
    let AM=u, CN=v and we know DM=3, DN=4
    APM similar to AOE, so AE=5u/3
    CQN similar to COE, so CE=5v/4
    AE+CE=(20u+15v)/12=25
    4u+3v=60
    But u+3+v+4=25=> u+v=18
    hence u=6 & AE=10
    AD=9; it follows DE=1

    Another solution (using 1487):
    Let X be on AD such that BX is the angle bisector of ABD
    DX/XA=12/15=4/5
    XA=5DX/4
    AD=9DX/4=9
    So DX=4
    EX=Inradius=5 (O center of Blue circle. BOXA is concyclic and triangle OXE is right and isosceles)
    =>DE=EX-DX=1

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  4. ABD and BCD are similar and their sides are in the ratio 3:4
    Let BC=x,DC=y=>AB=3/4x,AD=9/16y
    Apply Pythagoras to ABC
    =>9/16x2+x2=25.25/16.16y2
    =>x=5/4y
    Substituting x with the above value and representing the sides of ABC in terms of y => a=5/4y, b=25/16y,c=15/16y
    Semi-perimeter of ABC P=15/8y
    AE=P-a=15/8y-5/4y=5/8y
    Hence DE=AE-AD=5/8y-9/16y=1/16y --------(1)
    To deduce the value of "y", lets first find the in-circle radius of ABC
    ADB is similar to ABC => AB/3=AC/r => 15y/16.3=25y/16.r => r=5
    r.P=S(ABC)
    =>5.15y/8=1/2.5y/4.15y/16
    =>y=16
    Substituting "y" in (1) => DE=1

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