Trapezoid Problem #1

Geometry Level pending

Given a trapezoid ABCD (AB is parallel to CD). AC meets BD at O. Draw a straight line through O which is parallel to AB and meets AD and BC at M and N, respectively.

Does the algebraic statement 1 A B + 1 C D = 2 M N \frac{1}{AB} + \frac{1}{CD} = \frac{2}{MN} always hold true ?

Note: The names of any lines mentioned in the statement above denote its length.

Paradoxical Answer It doesn't always hold true. Yes, always. No, never.

Tin Le by Tin Le , 15, Vietnam

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1 solution

Let the position coordinates of A , B , C A, B, C and D D be ( h , m h ) (h,mh) , ( k , m h ) (k, mh) , ( c , 0 ) (c, 0) and ( 0 , 0 ) (0,0) respectively. Then 1 A B ‾ + 1 C D ‾ = 1 k − h + 1 c = k − h + c c ( k − h ) \dfrac{1}{\overline {AB}}+\dfrac{1}{\overline {CD}}=\dfrac{1}{k-h}+\dfrac{1}{c}=\dfrac{k-h+c}{c(k-h)} . Also M N ‾ = c ( 2 k − h ) k − h + c − h c k − h + c = 2 ( k − h ) c k − h + c \overline {MN}=\dfrac{c(2k-h)}{k-h+c}-\dfrac{hc}{k-h+c}=\dfrac{2(k-h)c}{k-h+c} . So 2 M N ‾ = k − h + c ( k − h ) c \dfrac{2}{\overline {MN}}=\dfrac{k-h+c}{(k-h) c} . Therefore 1 A B ‾ + 1 C D ‾ = 2 M N ‾ \dfrac{1}{\overline {AB}}+\dfrac{1}{\overline {CD}}=\dfrac{2}{\overline {MN}}

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