Thursday, October 29, 2015

Geometry Problem 1160: Triangle, Three Squares, Area, Equivalent Triangles

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below for more details.

Online Math: Geometry Problem 1160: Triangle, Three Squares, Area, Equivalent Triangles.



    Denote S(XYZ)= area of XYZ
    Draw altitudes AR and CL and lines per sketch
    Observe that triangle B1HC congrurent to CLA ( case ASA) => B1H=CL
    We have S(B1BC)= ½ .(BC1 x B1K).... (1)
    But B1K= B1H+HK= CL+ BL
    Replace this in (1) we have S(B1BC)=1/2. BC1 x(CL+BL)= S(ABC)+1/2. S(BLMC1)
    Simillarly S(B2BA2)= S(ABC)+ ½. S(BRSA2)
    Note that quadrilateral ACRL is cyclic => BL. BA=BR.BC => S(BLMC1)= S(BRSA2)
    Replace it in above expressions we get S(B1BC)= S(B2BA2)

    1. I wonder where point H is...

    2. H is the projection of B1 over CL extended . See sketch for detail

  2. I doubt I would ever come up with such a brilliant proof like Peter's. I proved this with a somewhat dumber approach using algebra.
    Area(BB1C1) = BB1*B1C1*sin(∠BB1C1)/2
    Area(BB2A2) = BB2*A2B2*sin(∠BB2A2)/2
    My approach involves these two steps:
    (1) BB1*B1C1 == BB2*A2B2
    (2) ∠BB1C1 == ∠BB2A2
    They are not trivial but not too hard to reach either.