Wednesday, January 21, 2015

Geometry Problem 1076: Square, 45 Degrees, Angle, Sum, Distance, Auxiliary Construction

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

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1076: Square, 45 Degrees, Angle, Sum, Distance, Auxiliary Construction.


  1. Draw circle centered A, radius=AD
    Locate point H on the circle such that ∠ (HAF)= ∠ (DAF)= alpha
    Triangles FAD congruence to HAF ( Case SAS) => FH=FD= b and ∠(AHF)=90
    We have ∠ (EAH)=45- alpha=∠ (BAE) => triangles EAH congruence to EAB
    So EH=EB=a and ∠ (EHA) 90
    So x= a+b

  2. Rotate anti-clockwise 90°, center at A.
    Then D→B. Let F→G.

    Then AF=AG and ∠EAF=∠EAG.
    So ΔEAF congruent to ΔEAG.

    Hence, x=EF=EG=a+b.

  3. Fold AB, AD along AE, AF respectively.
    B, D land at the same point P on EF.
    So x = EP + PF = EB + DF = a + b.

  4. Extend CD to G such that DG = a, then Tr.s ABE and ADG are congruent from which we can deduce that Tr.s AFE and AGF are also congruent.
    Hence x= FG = a+b

    Sumith Peiris
    Sri Lanka

  5. if you rotate the 45 degree wedge, while extending and shortening AF and AE so they F and E are always on the square, EF will form a quarter circle, kinda like what is happening here:

    so EF will be tangent to a circle centered at A with radius AB. let the point of tangency be n. because of tangent properties, BE = a = EN and DF = b = FN. FN + EN = x, so a + b = x

  6. See the drawing

    Define α = ∠BAF and β = ∠DAE
    α+β+45°=90° => α+β = 45°
    Define E’ rotation of E anticlockwise with an angle of 90°
    =>AE=AE’, β = ∠DAE = ∠BAE’
    ∠FAE’= α+β = 45°
    ΔFAE’ is congruent to ΔFAE (SAS) => FE’=FE
    Therefore a+b=x