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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 15]] |
| − | Given that a sequence satisfies <math> x_0=0 </math> and <math> |x_k|=|x_{k-1}+3| </math> for all integers <math> k\ge 1, </math> find the minimum possible value of <math> |x_1+x_2+\cdots+x_{2006}|. </math>
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| − | == Solution ==
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| − | This solution requires you to disregard rigor. Additional solutions, or justification for the nonrigorous steps, would be appreciated.
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| − | Let <math>\triangle XYZ</math> be a [[triangle]] with sides of length <math>x, y</math> and <math>z</math>, and suppose this triangle is acute (so all [[altitude]]s are on the interior of the triangle).
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| − | Let the altitude to the side of length <math>x</math> be of length <math>h_x</math>, and similarly for <math>y</math> and <math>z</math>. Then we have by two applications of the [[Pythagorean Theorem]] that <math>x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}</math>. As a [[function]] of <math>h_x</math>, the [[RHS]] of this [[equation]] is strictly decreasing, so it takes each value in its [[range]] exactly once. Thus we must have that <math>h_x^2 = \frac1{16}</math> and so <math>h_x = \frac{1}4</math> and similarly <math>\displaystyle h_y = \frac15</math> and <math>h_z = \frac16</math>.
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| − | Since the [[area]] of the triangle must be the same no matter how we measure, <math>x\cdot h_x = y\cdot h_y = z \cdot h_z</math> and so <math>\frac x4 = \frac y5 = \frac z6 = 2A</math> and <math>x = 8A, y = 10A</math> and <math>z = 12A</math>. The [[semiperimeter]] of the triangle is <math>s = \frac{8A + 10A + 12A}{2} = 15A</math> so by [[Heron's formula]] we have <math>A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^2\sqrt{7}</math>. Thus <math>A = \frac{1}{15\sqrt{7}}</math> and <math>x + y + z = 30A = \frac2{\sqrt{7}}</math> and the answer is <math>2 + 7 = 009</math>.
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| − | Justification that there is a triangle with sides of length <math>x, y</math> and <math>z</math>:
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| − | Note that <math>x, y</math> and <math>z</math> are each the sum of two [[positive]] [[square root]]s of [[real number]]s, so <math>x, y, z \geq 0</math>. (Recall that, by [[AIME]] [[mathematical convention| convention]], all numbers (including square roots) are taken to be real unless otherwise indicated.) Also, <math>\sqrt{y^2-\frac{1}{16}} < \sqrt{y^2} = y</math>, so we have <math>x < y + z</math>, <math>y < z + x</math> and <math>z < x + y</math>. But these conditions are exactly those of the [[triangle inequality]], so there does exist such a triangle.
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| − | Justification that this triangle is an [[acute triangle]]:
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| − | Still needed.
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| − | == See also ==
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| − | *[[2006 AIME II Problems]]
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| − | [[Category:Intermediate Algebra Problems]]
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