Challenging Geometry Puzzle: Problem 1586. Share your solution by posting it in the comment box provided.
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Trigonometry Solution
ReplyDeleteLet AB = BC = CA = 2a
Let CD = DE = ED = 2b
Let AF = FD = DA = 2c
Let < FAB = t so that < BAD = 60 - t & < CAD = t
Let < AFB = @
Triangles AFB & CAD are congruent so
FB = CD = 2b & < ADC = < AFB = @
From Triangle AFB we have
Cos@ = (b^2 + c^2 - a^2)/2bc and cos t = (a^2 + c^2 - b^2)/2ac ......(2)
Also sin@ / a = sin t / b.............(3)
Now S(ADF) = 1/2 (2c)^2 sin60 = (V3/2).c^2.........(1)
Also S(AGDH) - X (say) = ac sin (60 - t) + bc sin (60 + @)
X / c = a.sin 60. cos t - a.sin t . cos 60 + b.sin 60 cos @ + b.cos60. sin @
2 X/c = V3. (a^2 + c^2 - b^2) + V3. (b^2 + c^2 - a^2) + (b.sin @ - a sint) from (2)
2X/c = V3. (2.c^2) / c + 0 (from (3))
Hence X = V3.c^2 = 2. S(ADF)
Sumith Peiris
Moratuwa
Sri Lanka