Difference between revisions of "2018 AMC 8 Problems/Problem 1"
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We conclude that the real numbers correspond bijectively to decimal expansions not terminating in an infinite string of nines. <math>\square</math> | We conclude that the real numbers correspond bijectively to decimal expansions not terminating in an infinite string of nines. <math>\square</math> | ||
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− | Using the constructive method given in [b] [color=#f00]Lemma 3[/color][/b], it is straightforward (and left as an exercise) to compute <cmath>\frac{275}{19} = 14.4736842\dots \approx 14 \rightarrow \boxed{\textbf{(A) }14}</cmath> and we are done. <math>\blacksquare</math> | + | Using the constructive method given in [b] [color=#f00]Lemma 3[/color][/b], it is straightforward (and left as an exercise) to compute <cmath>\frac{275}{19} = 14.4736842\dots \approx 14 \rightarrow \boxed{\textbf{(A) }14}</cmath> and we are done. <math>\blacksquare</math> |
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==See Also== | ==See Also== |
Revision as of 16:32, 18 April 2020
Problem 1
An amusement park has a collection of scale models, with a ratio , of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its duplicate to the nearest whole number?
Solution 1
You can set up a ratio: . Cross multiplying, you get . You divide by on each side to get . The closest integer is
Solution 2
You can just do and round your answer to get . It is basically Solution 1 without the ratio calculation, which might not be necessary.
Solution 3
Solved with [b] tworigami, JetC, pretzel, realquarterb, AIME12345, jskalarickal, niyu, ntboiii, kevinmathz, lilypads, ddonk, sadkid, aaa12, byungshinfighters, mrsyallapragada, franchester, ingenio, GameMaster402, anish9876, mira74, SD2014, stroller, AOPS12142015, huangyi_99, sriraamster, TheUltimate123, vedant10, kothasuhas, budu, One-And-A-Half-Genius, Radio2, and mathfun5[/b]
A beautiful problem.
Recall a [b]contraction[/b] from a metric space is a function such that for some constant and all , we have
[b][color=#f00]Lemma 1:[/color][/b] Suppose that is a contraction and that the metric space is complete. Then has a unique fixed point , and for any , the iterate converges to as .
[i]Proof:[/i] Choose any and define . We claim that for all we have which is immediate by the well ordering principle on . Now, we claim that the sequence is Cauchy. For any , choose so that If , we have \begin{align*} d(x_m, x_n) &\leq d(x_m, x_{m+1}) + d(x_{m+1}, x_{m+2}) + \dots + d(x_{n-1}, x_n)\\ &\leq k^md(x_0, x_1) + k^{m+1}d(x_0, x+1) + \dots + k^{n-1}d(x_0, x_1)\\ &= k^m(1+k + \dots + k^{n-m-1})d(x_0, x_1) \\ &\leq k^N \sum_{l =0}^{\infty} k^ld(x_0, x_1) = \frac{k^N}{1-k}d(x_0, x_1) < \varepsilon \end{align*} It is left as an exercise for the reader to show that this is equivalent to the usual definition of Cauchy, and we conclude is Cauchy. As is complete, the limit of the sequence is some which is contained in . For large, and lie in the neighborhood of , and since was arbitrary, this gives that as desired. Uniqueness is clear.
[b][color=#f00]Lemma 2:[/color][/b] We claim that the function defined by is a contraction. [i]Proof:[/i] Let and let . We have as desired.
We thus find that has a unique fixed point, in particular when . Because by Legendre symbol, we find that . We need one final lemma about the existence and uniqueness of decimal representations of the reals:
[b][color=#f00]Lemma 3:[/color][/b] The real numbers correspond bijectively to decimal expansions not terminating in an infinite string of nines.
[i]Proof:[/i] We outline the proof as follows: Given (with defined using Dedekind cuts), we define a decimal expansion , where is the largest integer , is the largest integer , is the largest integer , and so on.
We show
[i][color=#00f](i)[/color][/i] Each is a digit between 0 and 9, [i][color=#00f](ii)[/color][/i] For each there is so that , and then [i][color=#00f](iii)[/color][/i] Conversely, that for any such expansion not terminating in an infinite string of nines, the set is bounded and its least upper bound is a real number with decimal expansion
[i][color=#00f]Proof of (i):[/color][/i] Consider the decimal expansion . Because is the largest number , , so , and . Assume for . Now consider the number . By the above, this implies .
[i][color=#00f]Proof of (ii):[/color][/i] We want each real number to have a unique decimal representation. Let such that . Assume for the sake of contradiction that this set is nonempty. Then take as the minimal element of by WOP. Then, \begin{align*}
x &= N.x_1x_2\dots x_{k-1}999\dots \\ x' &= N.x_1x_2\dots (x_{k-1}+1)000 \dots
\end{align*} where . By continuity we can pick some such that . But there clearly cannot be a decimal between and , so . This contradicts the definition of (ie is , so is empty and such that .
[i][color=#00f]Proof of (iii):[/color][/i] We want to show such that . This is equivalent to . We can take for example, so .
We conclude that the real numbers correspond bijectively to decimal expansions not terminating in an infinite string of nines.
Using the constructive method given in [b] [color=#f00]Lemma 3[/color][/b], it is straightforward (and left as an exercise) to compute and we are done.
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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