Friday, May 16, 2014

Geometry Problem 1016. Triangle, 45 Degrees, Area, Quadrilateral, Altitude, Bisector

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

Click the figure below to enlarge.

Online Geometry Problem 1016. Triangle, 45 Degrees, Area, Triangle, Quadrilateral, Altitude, Bisector.


  1. Let Area(XYZ) denote area of triangle XYZ
    Draw altitudes AE’ and CF’ of triangle ABC
    Let N is orthocenter of triangle ABC
    Triangles ABE’ and CBF’ are right isosceles triangles
    Quadrilateral HNE’C is cyclic => ∠ (NHE)= ∠ (NCE)=45
    So E’ coincide to E
    Similarly F’ coincide to F
    Let BM and DL are altitudes of triangles BEF and DEF
    Triangle EBF similar to ABC … ( case SAS)
    So AB/BE=BC/BF = sqrt(2)= (BM+DL)/BM => DL/BM= sqrt(2)-1
    We have S= Area(ABC)= 2.Area(BEF)…. (1)
    And Area(DEF)/Area(BEF)=sqrt(2)-1.. ( both triangles have the same base)
    So S1=Area(BEF)+Area(DEF)= sqrt(2). Area(BEF)…(2)
    From (1) and (2) we have S/S1= sqrt(2) => S^2/S1^2= 2

  2. Denote AB+BC=s and AB*BC=p. 2*[FBED]=2*[ABC]-2*[AFD+CED]=p/sqrt(2)-(AF+EC)h, where h is equal height from D to AF and EC. Note that BFHC and BEHA are concyclic, so E and F are feet of altitudes of AE=AB/sqrt(2) and CF=BC/sqrt(2) respectively. Therefore (AF+EC)=s-s/sqrt(2)=s(sqrt(2)-1)/sqrt(2), and h=p/(sqrt(2)*s). Plugging these into first equation yields
    2*[FBED]=p/sqrt(2)-p(sqrt(2)-1)/2=sqrt(2)p/2-(sqrt(2)-1)p/2=p/2. 2*[ABC]=p/sqrt(2)
    so [ABC]/[FBED]=sqrt(2)

  3. Problem 1016
    Let S2 = S(ABD) and let S3 = S(CBD)

    < CHE = 45, so AHEB is concyclic.
    Hence < AEB = < AHB = 90
    and so BE = c/√2
    Similarly BF = a/√2

    Now AD = bc/(a+c) so S2/S = {bc/(a+c)}/b,
    Hence S2 = S{c/(a+c)}
    Similarly S3 = S{a/(a+c)}

    So S1 = S(BFD) + S(BDE) = {(a/ √2)/c}S2 + { (c/√2)/a}S3
    S1 = S{(a/√2/(a+c)} + S{ c/√2/(a+c)} = 1/√2
    Therefore S^2 /S1^2 = 2.

    Sumith Peiris
    Sri Lanka