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Round 4: Problem 2 Solution

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Here’s a geometric construction proof:

  • Extend AF and BC so that they intersect at an exterior point \text{X}.
  • Extend FE and CD in a similar fashion so they intersect at another exterior point, \text{Y}, forming quadrilateral XFYC.
  • \angle XAB \cong \angle XBA \cong \angle YED \cong \angle YDE = 60^{\circ} since each external angle is supplementary with the 60 degree internal angle.
  • \angle X \cong \angle Y = 60^{\circ}, so \triangle XAB, \triangle YDE are equilateral triangles.
  • AB = XB, DE = YE in equilateral triangles.
  • XFYC is a parallelogram (both pairs of opposite angles are congruent), so FY = XC.
  • FE + EY = AB + BC, so FE + ED = AB + BC by equilateral triangles as shown above.

Written by Administrator

April 27, 2009 at 00:02

Posted in Round 4

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