Friday, September 12, 2008

Elearn Geometry Problem 178

Quadrilateral, Midpoint, Triangles, Area

See complete Problem 178 at:

Quadrilateral, Trisection of sides, Area. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. Here's a little theorem:
    Lines L1, L2 and L3 intercept respectively L4, L5 and L6 at A B C, D E F and G H I. If any 2 of the ratios AB:BC, DE:EF and GH:HI, and any 2 of the ratios AD:DG, BE:EH and CF:FI, are the same, then AB:BC=DE:EF=GH:HI and AD:DG=BE:EH=CF:FI.

    construct line L6'//L6 through A,
    DI intercepts L6' at K, by KD:DI=AD:DG=CF:FI, get KC//L5,
    DH intercepts L6' at L, by AL:LK=GH:HI=AB:BC, get LB//KC//L5,
    S_1=[IJKL] with I,J be nearest to A,B
    implying the above theorem, we get QJ=JK=KG and so are the others
    referring to Proposed Problem 150, S_1=[ENMF]/3=S/9

  2. iyi günler,
    çözümde geçen bazı harfler şekil üzerinde olmadığı için çözümü anlayamıyorum. mümkünse soruları yanıtlarken gerekli çizimleri ve harflendirmeleri yapabilir misiniz?

  3. We have that AN = NM and AP = PQ. Then,
    Let K = QG /\ MF
    Trace PJ // AK, J in QK ==> QJ = JK, PJ = 1/2 AK
    Trace NL // AK, L in MK ==> ML = LK, NL = 1/2 AK
    Then PJLN is a paralelogram and PL and JN intersect in its midpoint I.

    ==> PI = IL /\ NI = IJ
    We can aply the result to quadrilaterals NJGD and PBFL, and then to quadrilateral IECH.

    Then the four segments PH, QG, NE and MF are trisected

    By Geometry Problem 150 ( applied first to ABCD and then to PQGH we get S1 = 1/9 S.

    Also if little quadrilaterals are numbered in cyclic order around S1 as S2, .., S8, then S2 + S6 = S3 + S7 = S4 + S8 = S5 + S9 = 2/9 S

    The result can be extended to any quadrilateral whose sides are divided into n, even or odd, equal parts and the corresponding points are joined, with respect to little quadrilaterals symmetrically disposed. in each file and row of quadrilaterals, the areas are in arithmetic progression.

  4. Mejor:
    Los escaques están en una doble progresión aritmética, por lo que los simétricamente dispuestos suman 2/n^2 y el central, si lo hay, 1/n^2.
    Applet de #GeoGebra: