The Mirror Triangle
Starting from an isosceles triangle, you draw a new triangle by reflecting each of the vertices about their opposite sides. If the base of the original triangle is fixed and the height is uniformly distributed on [ 0 , N ] , what is the limit of the expected ratio of the areas as N → ∞ ?
The answer is 3.
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2 solutions
There's no such thing as a uniform distribution over ( 0 , ∞ ) .
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If we replace infinity with A, does the result hold for any finite A?
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No, it only holds as a limit, which isn't what the question is asking.
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@D G – It's not particularly clear what the question is asking. That's the problem. So I'll assume that instead of inifinity, we have a very large finite number.
Starting with a triangle of area y , with vertices at ( ± 1 , 0 ) and ( 0 , y ) , the new vertices are ( 0 , − y ) and ( ± y 2 + 1 3 y 2 − 1 , y 2 + 1 4 y ) and the ratio of areas is y ( y 2 + 1 3 y 2 − 1 ) ( y 2 + 1 4 y + y ) = ( y 2 + 1 ) 2 ( y 2 + 5 ) ( 3 y 2 − 1 ) . For a fixed height range ( 0 , h ) , the average ratio is h 1 ∫ 0 h ( y 2 + 1 ) 2 ( y 2 + 5 ) ( 3 y 2 − 1 ) d y = 3 − h 2 + 1 8 . The limit of this as h → ∞ is 3 .
(Actually, I don't need to compute the integral to compute the limit, since L'Hopital and the Fundamental Theorem of Calculus say that the limit should be equal to the limit of ( h 2 + 1 ) 2 ( h 2 + 5 ) ( 3 h 2 − 1 ) , which is 3 . )
Suppose the height is uniformly distributed over ( 0 , ∞ ) and the base is fixed at some finite value (say 2 , for example). Clearly then, for the vast majority of cases, the triangle will resemble a rectangle whose height is much larger than its base.
Then, when we mirror the vertices about the opposite sides, we get the following:
The base becomes wider by a factor of 3, and the height remains the same as the triangle / rectangle flips over. Thus, the overall area is multiplied by 3.