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Difference between revisions of "2011 AMC 10A Problems/Problem 25"

(Created page with '==Problem 25== 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</ma…')
 
(Added in solution (Copy-pasted from 2011 AMC 12A #25))
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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>
 
<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>.

Revision as of 17:53, 23 February 2011

Problem 25

Let $R$ be a square region and $n\ge4$ an integer. A point $X$ in the interior of $R$ is called $n\text{-}ray$ partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?

$\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500$

Solution

The domain of $f_{1}(x)=\sqrt{1-x}$ is defined when $x\leq-1$. $f_{2}(x)=f_{1}(\sqrt{4-x})=\sqrt{1-\sqrt{4-x}}$. Applying the domain of $f_{1}(x)$ and the fact that square roots must be positive, we get $0\leq\sqrt{4-x}\leq1$. Simplify this to arrive at the domain for $f_{2}(x)$, which is defined when $3\leq x\leq4$. Repeat this process for $f_{3}(x)=\sqrt{1-\sqrt{4-\sqrt{9-x}}}$ to get a domain of $-7\leq x\leq0$. For $f_{4}(x)$, since square roots are positive, we can exclude the negative values of the previous domain to arrive at $\sqrt{16-x}=0$ as the domain of $f_{4}(x)$. We now arrive at a domain with a single number that defines $x$, however, since we are looking for the largest value for $n$ for which the domain of $f_{n}$ is nonempty, we must continue until we arrive at a domain that is empty. We continue with $f_{5}(x)$ to get a domain of $\sqrt{25-x}=16$. Solve for $x$ to get $x=-231$. Since square roots cannot be negative, this is the last nonempty domain. We add to get $5-231=\boxed{\textbf{(A)}\ -226}$.

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