Thursday, January 19, 2017

Geometry Problem 1307 Triangle, Incenter, Parallel line, Sides, Measurement

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 1307 Triangle, Incenter, Parallel line, Sides, Measurement.

Posted by Antonio Gutierrez at 4:22 PM
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5 comments:

  1. Sumith PeirisJanuary 19, 2017 at 7:38 PM

    Let h be the length of the altitude from B.
    From similar triangles,
    r/MI = h/c and area of Tr. ABC, S = r.s where r is the inradius and s the semi-perimeter and moreover,
    S = 1/2 hb
    These equations yield MI = 1/2bc/s = bc/(a+b+c)
    Similarly NI = ba/(a+b+c)

    So MN = MI+NI = b(a+c)/(a+b+c)

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. Sumith PeirisJanuary 19, 2017 at 10:02 PM

    Problem 1307 - Alternate Solution

    Let h be the height of the altitude from B and r the inradius and s the semi perimeter,
    Let S(ABC) = ∆.

    S(BMN)/∆= MN2/b2 = ½ MN(h-r)/(½bh)

    Since h=2∆/s so

    MN = b(h-r)/h = b(2∆/b - r)/(2∆/b)
    = b(2∆ - br)/2∆
    = b(2rs – br)/2rs
    = b(2s– b)/2s
    = b(a+c)/(a+b+c)

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. APOSTOLIS MANOLOUDISJanuary 19, 2017 at 10:58 PM

    Problem 1037
    Let (ABC) is area the triangle ABC then (ABC)=sr=bh/2.( 2s=a+b+c,r=radius of the inscribed circle and h=altitude of the triangle ABC).So h=2sr/b=(a+b+c)r/b. But MN//AC
    then from Thales' Theorem and triangle BMN with triangle ABC are similar we have
    MN/AC=h1/h=(h-r)/h=1-r/h=1-b/(a+b+c)=(a+c)/(a+b+c).Therefore MN=b(a+c)/(a+b+c).
    APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  4. Peter TranJanuary 19, 2017 at 11:41 PM

    Let h and r are height from B and inradius of triangle ABC
    We have MN/AC= (h-r)/h= 1- r/h
    2 * area of ABC= b.h=2p.r
    So r/h= b/2p= b/(a+b+c)
    So MN=b*(1-r/h)= b*(a+c)/(a+b+c)

    ReplyDelete
  5. ArifFebruary 11, 2018 at 6:26 AM

    Draw line from the point M perpendicular onto AC. The line meets AC at the point P
    Draw line from the point N perpendicular onto AC. The line meets AC at the point Q

    Let K be the point of tangency of the circle I with AB.
    Let H be the point of tangency of the circle I with BC.

    Then we see that the triangle AMP is congruent with MIK , and the the triangle CNQ is congruent with NIH.

    Therefore AM+CN =MN.

    Triangle BMN is similar to BAC therefore :

    c-AM/MN=c/b
    (1.)AMb=bc-cMN

    and

    a-CN/MN=a/b
    ab-CNb=MNa
    (2.)CNb=ab-MNa

    Adding equations 1 and 2 with CN+AM=MN

    b(CN+AM)=ab-MNa+bc-cMN
    bMN=ab+bc-MN(a+c)
    MN(a+b+c)=ab+bc
    MN=b(a+c)/(a+b+c)

    ReplyDelete

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