Sunday, September 22, 2013

Geometry Problem 927: Semicircle, Diameter, Radius, Equal Arcs, Chord, Congruence

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 927.

Online Geometry Problem 927: Semicircle, Diameter, Radius, Equal Arcs, Chord, Congruence

4 comments:

  1. FG = 2R sin80°tan10°
    PQ = 2R sin60°tan10°
    MN = 2R sin40°tan10°
    KL = 2R sin20°tan10°

    FG + PQ + MN + KL
    = 2R tan10° [sin20°+sin40°+sin60°+sin80°]
    = 2R tan10° [4 sin50°cos20°cos10°]
    = 8R cos20°cos40°cos80°
    = 8R [sin20°cos20°cos40°cos80°]/sin20°
    = 4R [sin40°cos40°cos80°]/sin20°
    = 2R [sin80°cos80°]/sin20°
    = R sin160°/sin20°
    = R

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    Replies
    1. To Jacob: Thanks for your solution. Now, try an elementary geometry (Euclid's Elements.) solution.

      Delete
  2. http://img849.imageshack.us/img849/7236/vn8u.png

    Define points C’, D’,E’,F’, C1, D1,E1, S, T and U as per sketch
    Due to sysmetrical proprety, EFF’E’ is a isoceles trapezoid with its diagonals intersect at U on diameter AB.
    ∠ (EUF)=1/2(Arc(EF)+Arc(E’F’))= 20°
    ∠ (FUO)= ∠ (EUA)=80°
    So UE//OF and UF//OG

    Similarly we have SC//TD//UE//OF and SD//TE//UF//OG
    From this proprety we have FG=OU, PQ=EE1=UT, MN=DD1=TS, KL=CC1=AS
    So FG+PQ+MN+KL=OU+UT+TS+SA= radius R

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  3. The following facts could be easily found so i skip the proof:
    EF//DG//CH//AI & FG//EH//DI//CJ//AB
    So FEPG,EDMQ,DCKN,CAOL are parallelogram.
    In addition, observe that
    ∆FEP~∆PDM~∆MCK~∆KAO~∆FOG~∆POQ~∆MON~∆KOL

    Now the proof starts:

    1.
    OA = OF = R
    So, ∆KAO = ∆FOG -> OK=FG

    2.
    EF=FG=OK (from 1)
    So, ∆FEP = ∆KOL -> FP=KL

    3.
    CK=CL-KL=OA-KL=OF-FP=OP
    So, ∆MCK=∆POQ -> MK = PQ

    4.
    ∠MOD=∠FOD=40=∠DOA=∠MDO, so DM=MO
    So, ∆PDM=∆MON -> PM = MN

    Summing up the result of 1,2,3,4,
    FG + PQ + MN + KL
    = OK + FP + MK + PM
    = OK + KM + MP + PF
    =R

    q.e.d.

    ReplyDelete