Friday, April 18, 2025

Geometry Problem 1598: Inradius Ratio with Median BM in a Right-Angled Triangle

Challenging Geometry Puzzle: Problem 1598. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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Illustration of Geometry Problem 1598

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5 comments:

  1. From the median theorem AB^2+BC^2=2(BM^2+MC^2)
    9+16=2(BM^2+25/4)
    From this, BM=5/2
    Also, from the area of ​​△ABM = area of ​​△CBM
    r1(AB+BM+MC)/2=r2(CB+BM+MC)
    r1(3+5/2+5/2)=r2(4+5/2+5/2)
    8r1=9r2
    Therefore, r1/r2=9/8

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  2. We have S(ABM)=S(BMC)= p1.r1= p2.r2 ( same height and equal bases)
    where p1 and p2 are half perimeter of triangles ABM and BMC
    we have AC=5 and AM=MC=MB= 2.5
    and 2p1= 8, and 2p2=9
    r1/r2= p2/p1= 9/8

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  3. S(ABM) = S(CBM) = (1/2) X (3 X 4 / 2 ) = 3

    r1 = S(ABM) / s1 = 3 / (8/2) = 3/4 and
    r2 = S(CBM) / s2 = 3 / (9 / 2) = 2/3

    Hence r1 / r2 = (3/4) / (2/3) = 9 / 8

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Let O1 and O2 be the centers of the incircles with radii r1 and r2, respectively. Let P and Q be the midpoints of sides AB and BC, respectively. Since ABC is a right triangle with legs 3 and 4, AC must have length 5, which means AM = BM = CM = 5/2. It follows that AMB and CMB are isosceles triangles so that the bisector of the angle between their congruent sides is also an altitude. Thus, MPBQ is a rectangle with sides 2 and 3/2. As a result, triangles AMB and CMB have equal area. Using the fact that the area of a triangle is the product of half its perimeter and the length of its inradius, (3 + 5/2 + 5/2)/2 * r1 = (4 + 5/2 + 5/2)/2 * r2, so that r1/r2 = 9/8.

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  5. ABC is 3,4,5 triangle
    ABM & BCM are is isosceles triangles
    AM = BM = CM = 2.5
    height of: ABM = 2, & BCM = 1.5
    r1 = AB/4 = 3/4, r2 = BC/6 = 4/6
    r1/r2 9/8

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