2 2 2^2

Level 2

In the diagram, A B F G ABFG and B C D E BCDE are squares, given D G = 46 , E G = 34 , A D = x DG=46, EG=34, AD=\sqrt{x} , find x x .


The answer is 2018.

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1 solution

Let A B = F G = a |AB| = |FG| = a and B C = C D = b |BC| = |CD| = b . Then E F = a b ||EF| = a - b and A C = a + b |AC| = a + b . By Pythagoras, we then have that

F G 2 + E F 2 = 3 4 2 a 2 + ( a b ) 2 = 1156 |FG|^{2} + |EF|^{2} = 34^{2} \Longrightarrow a^{2} + (a - b)^{2} = 1156 and

E F 2 + A C 2 = 4 6 2 ( a b ) 2 + ( a + b ) 2 = 2116 ( a 2 2 a b + b 2 ) + ( a 2 + 2 a b + b 2 ) = 2116 a 2 + b 2 = 1058 |EF|^{2} + |AC|^{2} = 46^{2} \Longrightarrow (a - b)^{2} + (a + b)^{2} = 2116 \Longrightarrow (a^{2} - 2ab + b^{2}) + (a^{2} + 2ab + b^{2}) = 2116 \Longrightarrow a^{2} + b^{2} = 1058 .

Now ( ( a b ) 2 + ( a + b ) 2 ) ( a 2 + ( a b ) 2 ) = 2116 1156 ( a + b ) 2 a 2 = 960 2 a b + b 2 = 960 ((a - b)^{2} + (a + b)^{2}) - (a^{2} + (a - b)^{2}) = 2116 - 1156 \Longrightarrow (a + b)^{2} - a^{2} = 960 \Longrightarrow 2ab + b^{2} = 960 .

But by Pythagoras x = A D 2 = A C 2 + C D 2 = ( a + b ) 2 + b 2 = a 2 + 2 a b + b 2 + b 2 = ( a 2 + b 2 ) + ( 2 a b + b 2 ) = 1058 + 960 = 2018 x = |AD|^{2} = |AC|^{2} + |CD|^{2} = (a + b)^{2} + b^{2} = a^{2} + 2ab + b^{2} + b^{2} = (a^{2} + b^{2}) + (2ab + b^{2}) = 1058 + 960 = \boxed{2018} .

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