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Difference between revisions of "1992 USAMO Problems/Problem 3"

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==Problem==
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== Problem ==
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Chords <math>AA'</math>, <math>BB'</math>, and <math>CC'</math> of a sphere meet at an interior point <math>P</math> but are not contained in the same plane.  The sphere through <math>A</math>, <math>B</math>, <math>C</math>, and <math>P</math> is tangent to the sphere through <math>A'</math>, <math>B'</math>, <math>C'</math>, and <math>P</math>.  Prove that <math>AA'=BB'=CC'</math>.
 
  
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==Solution==
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For a nonempty set <math>S</math> of integers, let <math>\sigma(S)</math> be the sum of the elements of <math>S</math>. Suppose that <math>A = \{a_1, a_2, \ldots, a_{11}\}</math> is a set of positive integers with <math>a_1 < a_2 < \cdots < a_{11}</math> and that, for each positive integer <math>n \le 1500</math>, there is a subset <math>S</math> of <math>A</math> for which <math>\sigma(S) = n</math>. What is the smallest possible value of <math>a_{10}</math>?
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Consider the plane through <math>A,A',B,B'</math>.  This plane, of course, also contains <math>P</math>. We can easily find the <math>\triangle APB</math> is isosceles because the base angles are equal. Thus, <math>AP=BP</math>. Similarly, <math>A'P=B'P</math>. Thus, <math>AA'=BB'</math>. By symmetry, <math>BB'=CC'</math> and <math>CC'=AA'</math>, and hence <math>AA'=BB'=CC'</math> as desired.  
 
  
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<math>\mathbb{QED.}</math>
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== Solution ==
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Let's a <math>n</math>-<math>p</math> set be a set <math>Z</math> such that <math>Z=\{a_1,a_2,\cdots,a_n\}</math>, where <math>\forall i<n</math>, <math>i\in \mathbb{Z}^+</math>, <math>a_i<a_{i+1}</math>, and for each <math>x\le p</math>, <math>x\in \mathbb{Z}^+</math>, <math>\exists Y\subseteq Z</math>, <math>\sigma(Y)=x</math>, <math>\nexists \sigma(Y)=p+1</math>.
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(For Example <math>\{1,2\}</math> is a <math>2</math>-<math>3</math> set and <math>\{1,2,4,10\}</math> is a <math>4</math>-<math>8</math> set)
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<br/>
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Furthermore, let call a <math>n</math>-<math>p</math> set a <math>n</math>-<math>p</math> good set if <math>a_n\le p</math>, and a <math>n</math>-<math>p</math> bad set if <math>a_n\ge p+1</math> (note that <math>\nexists \sigma(Y)=p+1</math> for any <math>n</math>-<math>p</math> set. Thus, we can ignore the case where <math>a_n=p+1</math>).
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<br/>
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Furthermore, if you add any amount of elements to the end of a <math>n</math>-<math>p</math> bad set to form another <math>n</math>-<math>p</math> set (with a different <math>n</math>), it will stay as a <math>n</math>-<math>p</math> bad set because <math>a_{n+x}>a_{n}>p+1</math> for any positive integer <math>x</math> and <math>\nexists \sigma(Y)=p+1</math>.
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<br/>
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<b>Lemma</b>) If <math>Z</math> is a <math>n</math>-<math>p</math> set, <math>p\leq 2^n-1</math>.
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<br/>
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For <math>n=1</math>, <math>p=0</math> or <math>1</math> because <math>a_1=1 \rightarrow p=1</math> and <math>a_1\ne1\rightarrow p=0</math>.
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Assume that the lemma is true for some <math>n</math>, then <math>2^n</math> is not expressible with the <math>n</math>-<math>p</math> set. Thus, when we add an element to the end to from a <math>n+1</math>-<math>r</math> set, <math>a_{n+1}</math> must be <math>\le p+1</math> if we want <math>r>p</math> because we need a way to express <math>p+1</math>. Since <math>p+1</math> is not expressible by the first <math>n</math> elements, <math>p+1+a_{n+1}</math> is not expressible by these <math>n+1</math> elements. Thus, the new set is a <math>n+1</math>-<math>r</math> set, where <math>r\leq p+1+a_{n+1} \leq 2^{n+1}-1</math>
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<b>Lemma Proven</b>
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<br/>
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The answer to this question is <math>\max{(a_{10})}=248</math>.
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The following set is a <math>11</math>-<math>1500</math> set:
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<math>\{1,2,4,8,16,32,64,128,247,248,750\}</math>
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Note that the first 8 numbers are power of <math>2</math> from <math>0</math> to <math>7</math>, and realize that any <math>8</math> or less digit binary number is basically sum of a combination of the first <math>8</math> elements in the set. Thus, <math>\exists Y\subseteq\{1,2,4,8,16,32,64,128\}</math>, <math>\sigma(Y)=x \forall 1\le x\leq 255</math>.
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<math>248\le\sigma(y)+a_9\le502</math> which implies that <math>\exists A\subseteq\{1,2,4,8,16,32,64,128,247\}</math>, <math>\sigma(A)=x \forall 1\le x\leq 502</math>.
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Similarly <math>\exists B\subseteq\{1,2,4,8,16,32,64,128,247,248\}</math>, <math>\sigma(A)=x \forall 1\le x\le750</math> and <math>\exists C\subseteq\{1,2,4,8,16,32,64,128,247,248,750\}</math>, <math>\sigma(A)=x \forall 1\leq x\leq 1500</math>.
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<br/>Thus, <math>\{1,2,4,8,16,32,64,128,247,248,750\}</math> is a <math>11</math>-<math>1500</math> set.
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<br/>
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<br/>
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Now, let's assume for contradiction that <math>\exists a_{10}\leq 247</math> such that <math>{a_1, a_2, \dots, a_{11}}</math> is a <math>11</math>-<math>q</math> set where <math>q\geq 1500</math>
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<math>{a_1, a_2, \dots a_8}</math> is a <math>8</math>-<math>a</math> set where <math>a\leq 255</math> (lemma).
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<math>max{(a_9)}=a_{10}-1\leq 246</math>
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Let <math>{a_1, a_2, \dots, a_{10}}</math> be a <math>10</math>-<math>b</math> set where the first <math>8</math> elements are the same as the previous set. Then, <math>256+a_9+a_{10}</math> is not expressible as <math>\sigma(Y)</math>. Thus, <math>b\leq 255+a_9+a_{10}\leq 748</math>.
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In order to create a <math>11</math>-<math>d</math> set with <math>d>748</math> and the first <math>10</math> elements being the ones on the previous set, <math>a_{11}\leq 749</math> because we need to make <math>749</math> expressible as <math>\sigma(Y)</math>. Note that <math>b+1+a_{11}</math> is not expressible, thus <math>d<b+1+a_{11}\leq 1498</math>.
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<br/>
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Done but not elegant...
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== See Also ==
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{{USAMO box|year=1992|num-b=2|num-a=4}}
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{{MAA Notice}}
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[[Category:Olympiad Combinatorics Problems]]

Latest revision as of 06:52, 19 July 2016

Problem

For a nonempty set $S$ of integers, let $\sigma(S)$ be the sum of the elements of $S$. Suppose that $A = \{a_1, a_2, \ldots, a_{11}\}$ is a set of positive integers with $a_1 < a_2 < \cdots < a_{11}$ and that, for each positive integer $n \le 1500$, there is a subset $S$ of $A$ for which $\sigma(S) = n$. What is the smallest possible value of $a_{10}$?

Solution

Let's a $n$-$p$ set be a set $Z$ such that $Z=\{a_1,a_2,\cdots,a_n\}$, where $\forall i<n$, $i\in \mathbb{Z}^+$, $a_i<a_{i+1}$, and for each $x\le p$, $x\in \mathbb{Z}^+$, $\exists Y\subseteq Z$, $\sigma(Y)=x$, $\nexists \sigma(Y)=p+1$.

(For Example $\{1,2\}$ is a $2$-$3$ set and $\{1,2,4,10\}$ is a $4$-$8$ set)


Furthermore, let call a $n$-$p$ set a $n$-$p$ good set if $a_n\le p$, and a $n$-$p$ bad set if $a_n\ge p+1$ (note that $\nexists \sigma(Y)=p+1$ for any $n$-$p$ set. Thus, we can ignore the case where $a_n=p+1$).


Furthermore, if you add any amount of elements to the end of a $n$-$p$ bad set to form another $n$-$p$ set (with a different $n$), it will stay as a $n$-$p$ bad set because $a_{n+x}>a_{n}>p+1$ for any positive integer $x$ and $\nexists \sigma(Y)=p+1$.


Lemma) If $Z$ is a $n$-$p$ set, $p\leq 2^n-1$.


For $n=1$, $p=0$ or $1$ because $a_1=1 \rightarrow p=1$ and $a_1\ne1\rightarrow p=0$.

Assume that the lemma is true for some $n$, then $2^n$ is not expressible with the $n$-$p$ set. Thus, when we add an element to the end to from a $n+1$-$r$ set, $a_{n+1}$ must be $\le p+1$ if we want $r>p$ because we need a way to express $p+1$. Since $p+1$ is not expressible by the first $n$ elements, $p+1+a_{n+1}$ is not expressible by these $n+1$ elements. Thus, the new set is a $n+1$-$r$ set, where $r\leq p+1+a_{n+1} \leq 2^{n+1}-1$

Lemma Proven


The answer to this question is $\max{(a_{10})}=248$.

The following set is a $11$-$1500$ set:

$\{1,2,4,8,16,32,64,128,247,248,750\}$

Note that the first 8 numbers are power of $2$ from $0$ to $7$, and realize that any $8$ or less digit binary number is basically sum of a combination of the first $8$ elements in the set. Thus, $\exists Y\subseteq\{1,2,4,8,16,32,64,128\}$, $\sigma(Y)=x \forall 1\le x\leq 255$.

$248\le\sigma(y)+a_9\le502$ which implies that $\exists A\subseteq\{1,2,4,8,16,32,64,128,247\}$, $\sigma(A)=x \forall 1\le x\leq 502$.

Similarly $\exists B\subseteq\{1,2,4,8,16,32,64,128,247,248\}$, $\sigma(A)=x \forall 1\le x\le750$ and $\exists C\subseteq\{1,2,4,8,16,32,64,128,247,248,750\}$, $\sigma(A)=x \forall 1\leq x\leq 1500$.


Thus, $\{1,2,4,8,16,32,64,128,247,248,750\}$ is a $11$-$1500$ set.



Now, let's assume for contradiction that $\exists a_{10}\leq 247$ such that ${a_1, a_2, \dots, a_{11}}$ is a $11$-$q$ set where $q\geq 1500$

${a_1, a_2, \dots a_8}$ is a $8$-$a$ set where $a\leq 255$ (lemma).

$max{(a_9)}=a_{10}-1\leq 246$

Let ${a_1, a_2, \dots, a_{10}}$ be a $10$-$b$ set where the first $8$ elements are the same as the previous set. Then, $256+a_9+a_{10}$ is not expressible as $\sigma(Y)$. Thus, $b\leq 255+a_9+a_{10}\leq 748$.

In order to create a $11$-$d$ set with $d>748$ and the first $10$ elements being the ones on the previous set, $a_{11}\leq 749$ because we need to make $749$ expressible as $\sigma(Y)$. Note that $b+1+a_{11}$ is not expressible, thus $d<b+1+a_{11}\leq 1498$.


Done but not elegant...

See Also

1992 USAMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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