A geometry problem by Sanyam Garg

Geometry Level 3

In the adjoining figure, A B C ABC and D E F DEF are equilateral triangles A B = 8 cm AB= 8\text{ cm} and D E = 3 cm DE = 3\text{ cm} . Find the possible value of A E + B D + C F AE+BD+CF to 2 decimal places.


The answer is 9.11.

Sanyam Garg by Sanyam Garg , 16, India

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2 solutions

Mahdi Raza
Jun 9, 2020
  • Using symmetry, AE = BD = CF = x x (say).
  • Then [ A B D ] = [ B C F ] = [ C A E ] [ABD] = [BCF] = [CAE]
  • Using Heron's formula, [ A B D ] = ( s ) ( s − a ) ( s − b ) ( s − c ) = ( x + 5.5 ) ( 5.5 ) ( 2.5 ) ( x − 2.5 ) [ABD] = \sqrt{(s)(s-a)(s-b)(s-c)} = \sqrt{(x+5.5)(5.5)(2.5)(x-2.5)}

[ A B C ] = [ D E F ] + 3 [ A B D ] 3 4 8 2 = 3 4 3 2 + 3 [ A B D ] 3 4 ( 8 2 − 3 2 ) = 3 [ A B D ] 3 4 ( 8 2 − 3 2 ) = 3 ( x + 5.5 ) ( 5.5 ) ( 2.5 ) ( x − 2.5 ) 55 3 4 = 3 220 x 2 + 660 x − 3025 16 ( 55 3 ) 2 = 9 ( 220 x 2 + 660 x − 3025 ) 3 x 2 + 9 x − 55 = 0 x ≈ 3.03 [ x > 0 ] \begin{aligned} [ABC] &= [DEF] + 3[ABD] \\ \dfrac{\sqrt{3}}{4}8^2 &= \dfrac{\sqrt{3}}{4}3^2 + 3[ABD] \\ \dfrac{\sqrt{3}}{4}(8^2-3^2) &= 3[ABD] \\ \dfrac{\sqrt{3}}{4}(8^2-3^2) &= 3\sqrt{(x+5.5)(5.5)(2.5)(x-2.5)} \\ \dfrac{55\sqrt{3}}{4}&= 3\sqrt{\dfrac{220x^2 + 660x - 3025}{16}} \\ (55\sqrt{3})^2 &= 9(220x^2 + 660x - 3025) \\ 3x^2 + 9x - 55 &= 0 \\ x &\approx 3.03 &[x >0] \end{aligned}

Thus, A E + B D + C F = 3 x ≈ 9.1 AE + BD + CF = 3x \approx \boxed{9.1}

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Sanyam Garg
Jun 8, 2020

Guys!.. that is D point inside the triangle not B.. it is misprinting... sorry for that

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