Monday, October 26, 2009

Problem 373: Square, Inscribed Circle, Diagonal, Perpendicular, Angle

Proposed Problem
Click the figure below to see the complete problem 373 about Square, Inscribed Circle, Diagonal, Perpendicular, Angle.

 Problem 373: Square, Inscribed Circle, Diagonal, Perpendicular, Angle.
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Complete Geometry Problem 373
Level: High School, SAT Prep, College geometry


  1. Let O be the centre of the circle 1 and join EF. Since EAFO is a square, angle EFO=45.
    Then, EAFH is concyclic, so that angle EFH=alpha. it is easy to find out that angle GOF=135 and hence angle OFH=22.5, so alpha=angle EFO-angle HFO=22.5

  2. let o be the centre of circle join O,F and E,F then angle EFA=45 so angleEFO=45 and GO=OF so angle OGF=GFObut angleDOF=45 so angleOGF=angleGFO=45/2 so angleEFH=45- 45/2=45/2. and angle EAF+angle EHF=180 so E,H,F,A are concyclic. hence angle EAH=angle EFH 45/2

  3. Tr.s EOF and GOF are both isoceles O being the centre. Hence < HFE = 22.5 which is = to alpha considering that AEHF is cyclic

    Sumith Peiris
    Sri Lanka

  4. Its more elegant using the cyclic quadrilateral like above but just for fun you can
    also use the ratio of the sides.

    1. Angle trace to show tr BEG is congruent to GHO and they are both 22.5/45/112.5
    2. Find the ratios of tr BEG in terms of the radius r:
    BG is [sqrt(2) - 1]r
    BE is 1r
    Draw a perpendicular line from BE to G and you can use the Pythagorean theorem twice to show EG = sqrt(2 = sqrt(2))r
    3. AEH is 112.5. through some angle tracing and
    AE is 1r
    EH is EG / sqrt(2) since EGH is a right isosceles.

    4. If triangles BEG and AEH are congruent AE / EH = EG / BE
    or 1 / EG/sqrt(2) = EG / (sqrt(2) - 1)
    That simplifies to EG^2 = 2 - sqrt(2) which is consistent with what we already found.

  5. Let J be mid-point of BC. JF Perpendicular to AD and m(EFJ)=45.
    G being mid-point of the arc EJ subtends 22.5 degrees at F with E
    Since AEHF is cyclic m(EFH)=m(EAH)=22.5 degrees