Tuesday, December 16, 2008

Archimedes' Book of Lemmas, Proposition #3

Problem 642: Semicircle, Diameter, Perpendicular
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #3 (high school level) and lift up your geometry skills.
Click the figure to view details.

Archimedes' Book of Lemmas #3.

7 comments:

  1. We will refer to the length arc FB as x, and the length of EB will be called y.

    The line of the semicircle is partitioned into three parts of length arc AC, arc CF (=arc AC) and x. The diameter AB is partitioned into three parts of length AD, DE (=AD) and y.

    Now consider a congruent semicircle with a different partition: Now the line of the semicircle may consist of parts of length arc AC, x and again arc AC, such that the part of length x is the middle part. The partitioning points may be C and F'. The diameter may now consist of parts of lengths AD, y and again AD, such that y is the middle part. The partitioning points are called D and E'.

    So the triangle ACD is as in the original figure, but there is another triangle BF'E' at the other end of the semicircle, which is constructed in completely the same way, so they are congruent.

    Thus we have E'F'=CD and both of them are perpendicular to AB, so DE'F'C is a rectangle. Then DE'= F'C, and according to the construction DE'=y=EB and F'C, just like BF is the length of the chord belonging to the arc of the semicircle with length x.

    Therefore, EB=BF. (by Thomas)

    ReplyDelete
  2. http://mate-facil.co.cc/ by:TiNo

    ReplyDelete
  3. bjhopper bjhvash44@sbcglobal.netMarch 24, 2010 at 6:18 PM

    Draw CE,CE,FE,AF and the mirror image on the bottom of the full circle
    ang a = 1/2 of arc CF=CAF
    ang b= 1/2 of arc FB=FAB
    ang ACD = a=1/2 of arc AC lower
    trian ACD congru to ECD
    ang DCE = a
    triang ACE contains 4a+2b =180
    these same angles form a kite CEBF with CFE having 2b @C and 2a @ E and F.
    FBE has 2a @ B and a +b @ F and E
    BF=BF

    ReplyDelete
  4. AC = CE = CF
    Angles CFB = Pi - CAB = Pi - CEA = CEB
    Arcs AC, CF subtend equal angles at B
    Triangles CEB, CFB are congruent
    BE = BF

    ReplyDelete
  5. ABCF is a cyclic quad with the external angle at F (say @) = < A = < CEA

    But CE = CF since both are equal to AC hence < CEF = < CFA

    Hence it follows that < BEF = < BFE and hence BE = BF

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. C is the circumcenter of tr. AEF, thus <AFE = <ACD hence CD and EF concur onto the circle E at P, symmetrical of C about AB, thus APBF is cyclic; triangles APE and EBF are therefore similar and, since tr. APE is isosceles (CAPE being a rhombus), so is EBF, done.

    ReplyDelete
  7. Problem 642
    Set x=<ACD=<DCE=<ABC=<AFC=<CBF (AC=CE=CF) and y=<BCF=<BAF.Is <ECB=90-2x,but
    <BAF=y=90-2x(<AFB=90= <ACB).Then <ECB=y=<BCF. So triangle CEB=triangle FCB.
    Therefore EB=FB.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete