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Geometry Level 3

A trapezium AB exists such that A B / / C D AB//CD and C D = 4 A B CD = 4AB . If A D = 1 AD = 1 and B C = 2 BC = 2 and AC and BD meet perpendicularly, then A B = m n AB = \sqrt{\frac{m}{n}} where g c d ( m , n ) = 1 gcd(m, n) = 1 . Find m + n ? ? ? m+n???


The answer is 22.

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Noel Lo by Noel Lo , 24, Singapore

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1 solution

Noel Lo
Apr 23, 2015

Let AC and BD meet at an interior point E. A E 2 + D E 2 = A D 2 = 1 2 = 1 AE^2+DE^2 = AD^2 = 1^2=1 and B E 2 + C E 2 = B C 2 = 2 2 = 4 BE^2 + CE^2 = BC^2= 2^2 = 4 . Now as triangles ABE and CDE are similar, C D A B = D E B E = C E A E = 4 \frac{CD}{AB} = \frac{DE}{BE} = \frac{CE}{AE} = 4 .

Let A E = x , B E = y AE = x, BE = y . So C E = 4 x , D E = 4 y CE = 4x, DE = 4y and we have x 2 + ( 4 y ) 2 = 1 , y 2 + ( 4 x ) 2 = 4 x^2 + (4y)^2=1, y^2+(4x)^2=4 or x 2 + 16 y 2 = 1 , y 2 + 16 x 2 = 4 x^2+16y^2 = 1, y^2+16x^2 = 4 . Adding these 2 equations we have 17 ( x 2 + y 2 ) = 5 17(x^2+y^2) = 5 .

Now, A B = m n AB = \sqrt{\frac{m}{n}} so A B 2 = m n AB^2 = \frac{m}{n} . A B 2 = A E 2 + B E 2 = x 2 + y 2 = 5 17 AB^2 = AE^2+BE^2 = x^2+y^2 = \frac{5}{17} . So m + n = 5 + 17 = 22 m+n =5+17 =\boxed{22}

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