Emergency problems #3

Emergency problems #3 - MATH 348 Solved Emergency Problems...

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Leonid Chindelevitch August 24, 2008 1 Monday, July 21, 2008 1. Problem: Let ABC be an equilateral triangle inscribed in a circle ω , and let P be a point on ω , with A the vertex farthest away from P . Show that d ( A,P ) = d ( B,P ) + d ( C,P ). Solution: This follows immediately from a more general result known as Ptolemy’s theorem , http://en.wikipedia.org/wiki/Ptolemaios’ theorem , which says that the product of the diagonals of a cyclic quadrilateral equals the sum of the products of its opposite sides. In other words, if PQRS is a cyclic quadrilateral, then PR × QS = PQ × RS + PS × QR. Using Q = B,R = A,S = C , with AB = AC = BC , gives PA × BC = PB × AC + PC × BA = PA = PB + PC. 2. Problem: Show that the squares of side lengths 1 / 2 , 1 / 3 ,..., 1 /n can be ﬁt inside a unit square for every n N ,n > 1. Solution: Group the squares of sides 1 / 2 , 1 / 3 together, those of sides 1 / 4 , 1 / 5 , 1 / 6 , 1 / 7 together, and so on. In general, group the squares of sides 1 / 2 m , 1 / (2 m + 1) ,..., 1 / (2 m +1 - 1) together, for m N . Let us call this the n -th group. Allocate a slot of width 1 and height 1 / 2 m to the m -th group, for m N . Clearly, the slots will always ﬁt into the square because the total width is always 1 and the total height is at most X m =1 1 2 m = 1 . To prove that the squares of the m -th group ﬁt into the allocated slot, note that he maximum height of an element in the m -th group is 1 / 2 m , and 1 2 m + 1 2 m + 1 + ··· + 1 2 m +1 - 1 < 2 m 1 2 m = 1 , so placing the squares one next to the other inside a slot will never ﬁll the slot completely. 1

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Emergency problems #3 - MATH 348 Solved Emergency Problems...

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