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.

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

2 comments:

  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

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  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)

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