Friday, May 24, 2013

Problem 881: Triangle, three Squares, Centers, Areas

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

Click the figure below to see the complete problem 881.

Online Geometry Problem 881: Triangle, three Squares, Centers, Areas

Posted by Antonio Gutierrez at 1:00 PM
Labels: , , ,

8 comments:

  1. Peter TranMay 24, 2013 at 2:30 PM

    BM=4 , BN= SQRT(23), MN=SQRT(67)
    In triangle MBN we have MN^2=MB^2+NB^2- 2.MB.NB.cos(B+90)
    But cos(B+90)=- sin(B)
    So sin(B)=(MN^2-MB^2-NB^2)/2.MB.NB
    =(67-16-23)/(2 * 4 .*SQRT(23)= 7/(2 . SQRT(23))
    Area (ABC)=1/2 . AB* BC* Sin(B) = 14
    BM=4 , BN= SQRT(23), MN=SQRT(67)
    In triangle MBN we have MN^2=MB^2+NB^2- 2.MB.NB.cos(B+90)
    But cos(B+90)=- sin(B)
    So sin(B)=(MN^2-MB^2-NB^2)/2.MB.NB
    =(67-16-23)/(2 * 4 .*SQRT(23)= 7/(2 . SQRT(23))
    Area (ABC)=1/2 . AB* BC* Sin(B) = 14

    ReplyDelete
  2. Jacob HA (EMK2000)May 24, 2013 at 7:35 PM

    Join BM, BN.

    BM^2 = BD^2 / 2 = 16
    BN^2 = BF^2 / 2 = 23
    MN^2 = 67

    MN^2 = BM^2 + BN^2 − 2×BM×BN×cos(∠B+90°)
    67 = 16 + 23 + 2×BM×BN×sin∠B
    BM×BN×sin∠B = 14

    S(ABC) = 1/2×BA×BC×sin∠B
    = 1/2×(√2×BM)×(√2×BN)×sin∠B
    = BM×BN×sin∠B
    = 14

    ReplyDelete
  3. nakaMay 24, 2013 at 10:51 PM

    BD=AB=sqrt(32), BF=BC=sqrt(46), MN=sqrt(67)
    BM=sqrt(16)=4, BN=sqrt(23)
    By cosine law,
    MN^2 = BM^2 + BN^2 - 2*BM*BN*cosMBN
    MN^2 = BM^2 + BN^2 - 2*BM*BN*cos(90+ABC)
    MN^2 = BM^2 + BN^2 + 2*BM*BN*sinABC
    MN^2 = BM^2 + BN^2 + 2*1/sqrt2*1/sqrt2*(AB*BC*sinABC)
    67 - 16 - 23 = AB*BC*sinABC
    28 = AB*BC*sinABC
    14 = Area

    ReplyDelete
  4. Antonio GutierrezMay 27, 2013 at 3:07 PM

    Problem 881: solution by Michael Tsourakakis from Greece at
    www.gogeometry.com/school-college/p881-geometry-mixalis-greece.pdf

    Thanks,
    Mixalis

    ReplyDelete
  5. PravinMay 29, 2013 at 12:01 AM

    Prove that the center of the square MNPQ coincides with the midpoint of AC.

    ReplyDelete
    Replies
    1. Michael TsourakakisMay 29, 2013 at 9:51 AM

      this is very easy.
      Because MSP, is, diagonal of the square MNPQ ,and angle MSN=90 ,angle SMN=angle MNS = 45 ,angle MNP=90 ,then, angle SNP=45 ,so, angle NPS =45 .Therefore SP=SN=SM so,,S midpoint of the MP

      Delete
  6. PravinMay 29, 2013 at 7:19 PM

    What is S?. Kindly clarify.
    If you mean S is the point where diagonals of the square MNPQ meet, it is trivial to see that SP = SN = SM.
    If you mean S is the point where a diagonal, say MP, meets AC, you need to prove that S is the midpoint of the side AC of the triangle ABC.

    ReplyDelete
  7. Michael TsourakakisMay 30, 2013 at 7:12 AM

    Please read my solution.S, is midpoint of the AC

    ReplyDelete

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