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| − | ==Problem 25==
| + | #redirect [[2011 AMC 12A Problems/Problem 22]] |
| − | Let <math>R</math> be a square region and <math>n\ge4</math> an integer. A point <math>X</math> in the interior of <math>R</math> is called <math>n\text{-}ray</math> partitional if there are <math>n</math> rays emanating from <math>X</math> that divide <math>R</math> into <math>n</math> triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?
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| − | <math>\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500</math>
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| − | == Solution ==
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| − | The domain of <math>f_{1}(x)=\sqrt{1-x}</math> is defined when <math>x\leq-1</math>. <math>f_{2}(x)=f_{1}(\sqrt{4-x})=\sqrt{1-\sqrt{4-x}}</math>. Applying the domain of <math>f_{1}(x)</math> and the fact that square roots must be positive, we get <math>0\leq\sqrt{4-x}\leq1</math>. Simplify this to arrive at the domain for <math>f_{2}(x)</math>, which is defined when <math>3\leq x\leq4</math>. Repeat this process for <math>f_{3}(x)=\sqrt{1-\sqrt{4-\sqrt{9-x}}}</math> to get a domain of <math>-7\leq x\leq0</math>. For <math>f_{4}(x)</math>, since square roots are positive, we can exclude the negative values of the previous domain to arrive at <math>\sqrt{16-x}=0</math> as the domain of <math>f_{4}(x)</math>. We now arrive at a domain with a single number that defines <math>x</math>, however, since we are looking for the largest value for <math>n</math> for which the domain of <math>f_{n}</math> is nonempty, we must continue until we arrive at a domain that is empty. We continue with <math>f_{5}(x)</math> to get a domain of <math>\sqrt{25-x}=16</math>. Solve for <math>x</math> to get <math>x=-231</math>. Since square roots cannot be negative, this is the last nonempty domain. We add to get <math>5-231=\boxed{\textbf{(A)}\ -226}</math>.
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