Wednesday, May 14, 2014

Geometry Problem 1015. Right Triangle, Square, Altitude, Metric Relations

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

Click the figure below to enlarge the figure of problem 1015.

Online Geometry Problem 1015. Right Triangle, Square, Altitude, Metric Relations.


  1. Since ∠FEH=∠BAC=∠HBC, thus B,D,E,H,F are concyclic.
    Thus ∠BHF=∠BEF=45°, and so FH bisects ∠BHC.
    Hence, HE/HM=2/x.

    Since ΔEHM~ΔCHB~ΔBDM, so
    2/x = HE/HM = HC/HB = DB/DM
    2/x = 14/(x+16)
    14x = 2x + 32
    x = 8/3

  2. Note that points B, D,E,H,F are concyclic
    Power of point G to circle BDEF= GE.GD=GH.GF= 32….. (1)
    In right triangle GEF we have GF= 10.SQRT(2) … ( Pythagoras theorem)
    Replace in (1) we get GH= 8.SQRT(2)/5
    In similar triangles GHM and FHB we have
    Replace values of GF and HF=GF-GH we get x= 8/3

  3. Call GH=z, then HF=14z/x, and by Power point theorem 2(2+14)=z(z+14z/x), or z^2(x+14)/x=32
    tan(A)=2/x=sin(A)/sqrt(1-sin^2(A)) by similar triangles, so sin^2(A)=4/(x^2+4)
    From concyclic points already mentioned, HF^2=14^2*2*sin^2(A)=4*392/(x^2+4)
    But from similar triangles HF^2=196z^2/x^2, so setting the two equal we get z^2=8x^2/(x^2+4)
    From first equation, z^2=32x/(x+14), so setting the z^2 values equal we get equation 8x/(x^2+4)=32/(x+14)
    which simplifies to 3x^2-14x+16=(3x-8)(x-2). x=8/3 and not 2 because E and H are distinct.