Sunday, February 1, 2009

Elearn Geometry Problem 239: Square, Midpoints, Congruence, Pythagoras

Try to use elementary geometry (Euclid's Elements).

Problem: Square, Midpoints, Lines, Congruence, Pythagoras

See complete Problem 239 at:

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. Let A be (0,0) then we've B;(0,a),C:(a,a) & D:(a,0). With routine analytical geometry one can determine DH:x+2y=a and CG:2x-y=a which gives us N:(3a/5,a/5). Likewise, we find M:(2a/5,4a/5).
    Hence, MN^2 = b^2 = (3a/5-2a/5)^2+(a/5-4a/5)^2
    = 2a^2/5 = 10a^2/5^2
    Hence, b=(a/5)*(10)^(1/2)

  2. Let MA and HN intersect at P.
    Let NC and MF intersect at Q.
    MPNQ is a square.
    We can show that MP = AM - BM
    MP*MP = AM*AM + BM*BM - 2*AM*BM
    = a*a - 2*AM*BM
    In triangle ABE,
    1/[AB*AB] + 1/[BE*BE] = 1/[BM*BM]
    5/ [a*a] = 1/[BM*BM]
    BM = a/ [square root of 5]
    AM * AE = AB*AB
    so, AM * [square root of 5]*a/2 = a*a
    AM = 2*a/[square root of 5]

    MP*MP = a*a - 4*a*a/5
    = a*a/5
    MN*MN = 2*a*a/5
    = 10*a*a/25
    MN = [square root of 10]*a /5

  3. The area of the newly formed inner square is 1/5 the area of the original square because:
    1-there are 4 copies of the same triange (AMB), 3 have been rotated by 90degrees, 180 degrees and 270degrees respectively
    2 - each of those has an area of (AM)(MB)/2
    3 - those triangles can be arranged so that the hypotenuse (a) becomes the side of the square
    4 - This leaves the original square with a square "hole" in the center with an area of (AM-MB)^2
    5 - The area of four triangles is 4*[(AM)(MB)/2] or 2(AM)(MB)
    6 - The area of the entire square is equal to the area of the small square plus the area of the four triangles, so a^2=2(AM)(MB)+(AM-MB)^2
    7- we also know that (AM)^2 + (MB)^2 = a^2 because the triangles are right triangles

    What I can't figure out is how to get to my final answer which should be (AM-MB)^2 = 1/5a^2??

    Any ideas or help?

  4. It is so simple Anonymous
    Since you said that the inner square is (1/5) the are of the outer square which has an area of a*a
    This means that the inner square has an area of

    Assuming that the left lower corner of the inner square is Q we have
    MQ = NQ = a/sqrt(5)

    So b = sqrt(MQ^2 + NQ^2) = sqrt(a^2/5 + a^2/5)
    = sqrt(2a^2/5).

    Multiplying both numerator and denominator within the sqrt by 5, yields a*sqrt(10)/5
    which is the answer.

  5. Joining HF, it is easy to see that
    Area of parallelogram BHDF
    = sum of areas of triangles AHD & BCF
    = (a^2)/2
    For the parallelogram
    Base = HD = a (sqrt 5)/2 ,
    Height = b/(sqrt 2) &
    Area = [a(sqrt 5)/ 2] x (b / sqrt 2)
    So ab(sqrt 5)/2(sqrt 2) =(a^2)/2,
    b = a(sqrt2)/(sqrt5) = a(sqrt 10)/5

  6. This comment has been removed by the author.

    1. See the drawing

      Define O intersection of AE and HD and P intersection of BF and GC
      ABE is congruent to CBF by rotation of pi/2
      => x = 2 y
      S= a^2 = 2 𝑥^2+ 3 𝑥^2 = 5 𝑥^2
      =>𝑥^2 = a^2/5
      b hypotenuse =>b^2= 2𝑥^2 = 2a^2/5
      Therefore b^2 = 10 a^2/25

  7. GN=(1/2)ND=>PN=(2/5)ND (H,P,N,D).=>b=((2v2)/5)HD

  8. 2 proofs by area method :

    P the intersection of BF and CG, Q that of AE and DH.

    1/ Rotating MBE clockwise around E by an angle of π, M->M’, B->C, E->E, yields a square MM’CP the same size as the central square MPNQ.
    Doing the same with PCF around F, NDG around G and QAH around H yields a total of 5 identical squares for ABCD.
    Therefore, a.a = 5.b.b/2 or b = a.sqrt(10)/5 QED

    2/ ABE = ABCD/4 = 5*EMB (since EMB = HQA = ABM/4)
    So square MPNQ = 4*EMB = ABCD/5 = a*a/5, has a side equal to a*sqrt(5)/5 and its diagonal b = a*sqrt(10)/5 QED

  9. Tr.s BCF & ABE are congruent SAS so AE is perpendicular to BM and similarly to HD

    Hence AM = a^2/ (sqrt5. a/2) = 2a/(sqrt5)

    Since HD also bisects AM, b = (AM/2). (sqrt2) = sqrt(2/5).a = a.(sqrt10)/5

    Sumith Peiris
    Sri Lanka

  10. Read V as Sqrt
    Area of //gm BHFD=BF.MN/V2=a^2/2

  11. Triangle BME ~ Triangle BCF
    ME=c, BM=2c
    By Pyth. Theorem (Triangle BME)
    By Pyth. Theorem (Triangle BAE)
    Let AM & HN meet at X and by mid-pt. theorem AX=XM=2c
    Similarly, NX=2c
    So by Pyth. Theorem on triangle MXN, b^2=8c^2