pythagorean theorem

Geometry Level 3

In the triangle shown above, the medians D A DA and E B EB are perpendicular to each other. Find A B AB . If your answer is of the form a b a\sqrt{b} where a a and b b are coprimes, input your answer as a a .


The answer is 2.

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2 solutions

let D A = 3 x DA=3x , then A G = 2 x AG=2x and D G = x DG=x

let E B = 3 y EB=3y , then B G = 2 y BG=2y and G E = y GE=y

apply pythagorean theorem on △ D G B : \triangle DGB: x 2 + ( 2 y ) 2 = 9 x^2+(2y)^2=9 ⟹ \implies x 2 + 4 y 2 = 9 x^2+4y^2=9

apply pythagorean theorem on △ E G A : \triangle EGA: y 2 + ( 2 x ) 2 = 16 y^2+(2x)^2=16 ⟹ \implies y 2 + 4 x 2 = 16 y^2+4x^2=16

by addition, we get, 5 x 2 + 5 y 2 = 25 5x^2+5y^2=25 or x 2 + y 2 = 5 x^2+y^2=5

however in △ B G A \triangle BGA ,

( 2 y ) 2 + ( 2 x ) 2 = ( A B ) 2 (2y)^2+(2x)^2=(AB)^2 or 4 y 2 + 4 x 2 = ( A B ) 2 4y^2+4x^2=(AB)^2

since, x 2 + y 2 = 5 x^2+y^2=5 , 4 x 2 + 4 y 2 = 20 4x^2+4y^2=20

so,

( A B ) 2 = 20 (AB)^2=20 ⟹ \implies A B = 20 AB=\sqrt{20} ⟹ \implies A B = 2 5 AB=2\sqrt{5}

Finally, a = 2 \boxed{a=2}

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Ahmad Saad
May 15, 2017

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