https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&feed=atom&action=history1997 AIME Problems/Problem 13 - Revision history2024-03-28T18:46:53ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=159601&oldid=prevBlehlivesonearth: /* Solution 1 */2021-08-04T23:56:02Z<p><span dir="auto"><span class="autocomment">Solution 1</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>===Solution 1===</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>===Solution 1 <ins class="diffchange diffchange-inline">(non-rigorous)</ins>===</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">:''This solution is non-rigorous.''</del></div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>f(x) = \Big|\big||x|-2\big|-1\Big|</math>, <math>f(x) \ge 0</math>. Then <math>f(x) + f(y) = 1 \Longrightarrow f(x), f(y) \le 1 \Longrightarrow x, y \le 4</math>. We only have a <math>4\times 4</math> area, so guessing points and graphing won't be too bad of an idea. Since <math>f(x) = f(-x)</math>, there's a symmetry about all four [[quadrant]]s, so just consider the first quadrant. We now gather some points:</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>f(x) = \Big|\big||x|-2\big|-1\Big|</math>, <math>f(x) \ge 0</math>. Then <math>f(x) + f(y) = 1 \Longrightarrow f(x), f(y) \le 1 \Longrightarrow x, y \le 4</math>. We only have a <math>4\times 4</math> area, so guessing points and graphing won't be too bad of an idea. Since <math>f(x) = f(-x)</math>, there's a symmetry about all four [[quadrant]]s, so just consider the first quadrant. We now gather some points:</div></td></tr>
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</table>Blehlivesonearthhttps://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=159226&oldid=prevMineric: /* Solution 3 (FASTEST) */2021-07-28T15:10:38Z<p><span dir="auto"><span class="autocomment">Solution 3 (FASTEST)</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>We make use of several consecutive substitutions.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>We make use of several consecutive substitutions.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>||x| - 2|= x_1</math> and similarly with <math>y</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>||x| - 2|= x_1</math> and similarly with <math>y</math>.</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Therefore, our graph is <math>|x_1 - 1| + |y_1 - 1| = 1</math>. This is a <del class="diffchange diffchange-inline">diamond </del>with perimeter <math>4\sqrt{2}</math>. Now, we make use of the following fact for a function of two variables <math>x</math> and <math>y</math>: Suppose we have <math>f(x, y) = c</math>. Then <math>f(|x|, |y|)</math> is equal to the graph of <math>f(x, y)</math> reflected across the y axis and x axis, and the reflection across the y axis across the x axis, therefore the perimeter of  of <math>f(|x|, |y|)</math> is 4 times the perimeter of <math>f(x, y)</math>. Now, we continue making substitutions at each absolute value sign (<math>|x| - 2 = x_2</math> and finally <math>x = x_3</math>, similarly for y as well. ),  noting that the constants don't matter as they just translate the graph and each absolute value sign increases the perimeter 4 times as much. Therefore, the length is <math>4^2 \times 4\sqrt{2} = 64\sqrt{2} = a\sqrt{b}</math>, and <math>a+b = \boxed{066}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Therefore, our graph is <math>|x_1 - 1| + |y_1 - 1| = 1</math>. This is a <ins class="diffchange diffchange-inline">rhombus </ins>with perimeter <math>4\sqrt{2}</math>. Now, we make use of the following fact for a function of two variables <math>x</math> and <math>y</math>: Suppose we have <math>f(x, y) = c</math>. Then <math>f(|x|, |y|)</math> is equal to the graph of <math>f(x, y)</math> reflected across the y axis and x axis, and the reflection across the y axis across the x axis, therefore the perimeter of  of <math>f(|x|, |y|)</math> is 4 times the perimeter of <math>f(x, y)</math>. Now, we continue making substitutions at each absolute value sign (<math>|x| - 2 = x_2</math> and finally <math>x = x_3</math>, similarly for y as well. ),  noting that the constants don't matter as they just translate the graph and each absolute value sign increases the perimeter 4 times as much. Therefore, the length is <math>4^2 \times 4\sqrt{2} = 64\sqrt{2} = a\sqrt{b}</math>, and <math>a+b = \boxed{066}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>- whatRthose</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>- whatRthose</div></td></tr>
</table>Minerichttps://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=133578&oldid=prevTempcm: /* Solution 1 */2020-09-13T17:43:26Z<p><span dir="auto"><span class="autocomment">Solution 1</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>| <math>f(0.5) = 0.5</math> || <math>f(2.1) = 0.9</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>| <math>f(0.5) = 0.5</math> || <math>f(2.1) = 0.9</math></div></td></tr>
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<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>| <math>f(1.5) = <del class="diffchange diffchange-inline">1</del>.5</math> || <math>f(2.9) = 0.1</math></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>| <math>f(1.5) = <ins class="diffchange diffchange-inline">0</ins>.5</math> || <math>f(2.9) = 0.1</math></div></td></tr>
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<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>| <math>f(2.5) = <del class="diffchange diffchange-inline">2</del>.5</math> || <math>f(3.1) = 0.1</math></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>| <math>f(2.5) = <ins class="diffchange diffchange-inline">0</ins>.5</math> || <math>f(3.1) = 0.1</math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|-  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|-  </div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>| <math>f(3.5) = <del class="diffchange diffchange-inline">3</del>.5</math> || <math>f(3.9) = 0.9</math></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>| <math>f(3.5) = <ins class="diffchange diffchange-inline">0</ins>.5</math> || <math>f(3.9) = 0.9</math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|}</div></td></tr>
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</table>Tempcmhttps://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=121044&oldid=prevAnellipticcurveoverq: /* Solution 4 (Exploiting intuitions about absolute value with linear functions) */2020-04-16T15:18:34Z<p><span dir="auto"><span class="autocomment">Solution 4 (Exploiting intuitions about absolute value with linear functions)</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Solution 4 (Exploiting intuitions about absolute value with linear functions)===</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Solution 4 (Exploiting intuitions about absolute value with linear functions)===</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>In a similar manner to the other posted solutions, we want to find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It's possible to go right for the smallest possible piece that'll help us by letting <math>a = \big||x|-2\big|-1</math> and <math>b = \big||y|-2\big|-1</math>. We then get that <math>b = 1 - a</math> (obviously the other way would work too). One specific "case" here is the one where all the numbers in the absolute value bars are positive, and thus where <math>b = 1 - x - 2 = -x - 1</math>. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes; the length of this piece also won't be affected by the constants that are hidden in <math>b</math>. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be <math>\sqrt{2}</math>.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>In a similar manner to the other posted solutions, we want to find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It's possible to go right for the smallest possible piece that'll help us by letting <math>a = \big||x|-2\big|-1</math> and <math>b = \big||y|-2\big|-1</math>. We then get that <math>b = 1 - a</math> (obviously the other way would work too). One specific "case" here is the one where all the numbers in the absolute value bars are positive, and thus where <math>b = 1 - x - 2 = -x - 1</math>. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes; the length of this piece also won't be affected by the constants that are hidden in <math>b</math>. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be <math>\sqrt{2}</math>.  </div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are <math>2^6</math> of these pieces of length <math>\sqrt{2}</math>, making our answer <math>2 + 64 = \boxed{066}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function<ins class="diffchange diffchange-inline">'s graph </ins>at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are <math>2^6</math> of these pieces of length <math>\sqrt{2}</math>, making our answer <math>2 + 64 = \boxed{066}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>~ anellipticcurveoverq</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>~ anellipticcurveoverq</div></td></tr>
</table>Anellipticcurveoverqhttps://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=121043&oldid=prevAnellipticcurveoverq: /* Solution 4 (Exploiting intuitions about absolute value with linear functions) */2020-04-16T15:17:53Z<p><span dir="auto"><span class="autocomment">Solution 4 (Exploiting intuitions about absolute value with linear functions)</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 15:17, 16 April 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l86" >Line 86:</td>
<td colspan="2" class="diff-lineno">Line 86:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Solution 4 (Exploiting intuitions about absolute value with linear functions)===</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Solution 4 (Exploiting intuitions about absolute value with linear functions)===</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>In a similar manner to the other posted solutions, we want to find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It's possible to go right for the smallest possible piece that'll help us by letting <math>a = \big||x|-2\big|-1</math> and <math>b = \big||y|-2\big|-1</math>. We then get that <math>b = 1 - a</math> (obviously the other way would work too). One specific "case" here is the one where all the numbers in the absolute value bars are positive, and thus where <math>b = 1 - x - 2 = -x - 1</math>. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be <math>\sqrt{2}</math>.  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>In a similar manner to the other posted solutions, we want to find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It's possible to go right for the smallest possible piece that'll help us by letting <math>a = \big||x|-2\big|-1</math> and <math>b = \big||y|-2\big|-1</math>. We then get that <math>b = 1 - a</math> (obviously the other way would work too). One specific "case" here is the one where all the numbers in the absolute value bars are positive, and thus where <math>b = 1 - x - 2 = -x - 1</math>. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes<ins class="diffchange diffchange-inline">; the length of this piece also won't be affected by the constants that are hidden in <math>b</math></ins>. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be <math>\sqrt{2}</math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are <math>2^6</math> of these pieces of length <math>\sqrt{2}</math>, making our answer <math>2 + 64 = \boxed{066}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are <math>2^6</math> of these pieces of length <math>\sqrt{2}</math>, making our answer <math>2 + 64 = \boxed{066}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
</table>Anellipticcurveoverqhttps://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=121042&oldid=prevAnellipticcurveoverq: /* Solution 4 (Exploiting intuitions about absolute value with linear functions) */2020-04-16T15:14:56Z<p><span dir="auto"><span class="autocomment">Solution 4 (Exploiting intuitions about absolute value with linear functions)</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 15:14, 16 April 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l86" >Line 86:</td>
<td colspan="2" class="diff-lineno">Line 86:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Solution 4 (Exploiting intuitions about absolute value with linear functions)===</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Solution 4 (Exploiting intuitions about absolute value with linear functions)===</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>In a similar manner to the other posted solutions, we want to find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It <del class="diffchange diffchange-inline">is </del>possible to go right for the smallest possible piece that'll help us by letting <math>a = \big||x|-2\big|-1</math> and <math>b = \big||y|-2\big|-1</math>. We then get that <math>b = 1 - a</math> (obviously the other way would work too). One specific "case" here is the one where all the numbers in the absolute value bars are positive, and thus where <math>b = 1 - x - 2 = -x - 1</math>. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be <math>\sqrt{2}</math>.  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>In a similar manner to the other posted solutions, we want to find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It<ins class="diffchange diffchange-inline">'s </ins>possible to go right for the smallest possible piece that'll help us by letting <math>a = \big||x|-2\big|-1</math> and <math>b = \big||y|-2\big|-1</math>. We then get that <math>b = 1 - a</math> (obviously the other way would work too). One specific "case" here is the one where all the numbers in the absolute value bars are positive, and thus where <math>b = 1 - x - 2 = -x - 1</math>. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be <math>\sqrt{2}</math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are <math>2^6</math> of these pieces of length <math>\sqrt{2}</math>, making our answer <math>2 + 64 = \boxed{066}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are <math>2^6</math> of these pieces of length <math>\sqrt{2}</math>, making our answer <math>2 + 64 = \boxed{066}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
</table>Anellipticcurveoverqhttps://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=121041&oldid=prevAnellipticcurveoverq: /* Solution 4 (Exploiting intuitions about absolute value with linear functions) */2020-04-16T15:14:29Z<p><span dir="auto"><span class="autocomment">Solution 4 (Exploiting intuitions about absolute value with linear functions)</span></span></p>
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<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 15:14, 16 April 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l86" >Line 86:</td>
<td colspan="2" class="diff-lineno">Line 86:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Solution 4 (Exploiting intuitions about absolute value with linear functions)===</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Solution 4 (Exploiting intuitions about absolute value with linear functions)===</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>In a similar manner to the other posted solutions, we want to <del class="diffchange diffchange-inline">just </del>find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It is possible to go right for the smallest possible piece that'll help us by letting <math>a = \big||x|-2\big|-1</math> and <math>b = \big||y|-2\big|-1</math>. We then get that <math>b = 1 - a</math> (obviously the other way would work too). One specific "case" here is the one where all the numbers in the absolute value bars are positive, and thus where <math>b = 1 - x - 2 = -x - 1</math>. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be <math>\sqrt{2}</math>.  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>In a similar manner to the other posted solutions, we want to find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It is possible to go right for the smallest possible piece that'll help us by letting <math>a = \big||x|-2\big|-1</math> and <math>b = \big||y|-2\big|-1</math>. We then get that <math>b = 1 - a</math> (obviously the other way would work too). One specific "case" here is the one where all the numbers in the absolute value bars are positive, and thus where <math>b = 1 - x - 2 = -x - 1</math>. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be <math>\sqrt{2}</math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are <math>2^6</math> of these pieces of length <math>\sqrt{2}</math>, making our answer <math>2 + 64 = \boxed{066}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are <math>2^6</math> of these pieces of length <math>\sqrt{2}</math>, making our answer <math>2 + 64 = \boxed{066}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
</table>Anellipticcurveoverqhttps://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=120871&oldid=prevAnellipticcurveoverq at 21:16, 12 April 20202020-04-12T21:16:22Z<p></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 21:16, 12 April 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l84" >Line 84:</td>
<td colspan="2" class="diff-lineno">Line 84:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>- whatRthose</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>- whatRthose</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">===Solution 4 (Exploiting intuitions about absolute value with linear functions)===</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">In a similar manner to the other posted solutions, we want to just find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It is possible to go right for the smallest possible piece that'll help us by letting <math>a = \big||x|-2\big|-1</math> and <math>b = \big||y|-2\big|-1</math>. We then get that <math>b = 1 - a</math> (obviously the other way would work too). One specific "case" here is the one where all the numbers in the absolute value bars are positive, and thus where <math>b = 1 - x - 2 = -x - 1</math>. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be <math>\sqrt{2}</math>. </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are <math>2^6</math> of these pieces of length <math>\sqrt{2}</math>, making our answer <math>2 + 64 = \boxed{066}</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">~ anellipticcurveoverq</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
</table>Anellipticcurveoverqhttps://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=118986&oldid=prevAnmol04: minor spacing fixes2020-03-10T05:21:02Z<p>minor spacing fixes</p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 05:21, 10 March 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>== Problem ==</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>==Problem==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>S</math> be the [[set]] of [[point]]s in the [[Cartesian plane]] that satisfy <center><math>\Big|\big||x|-2\big|-1\Big|+\Big|\big||y|-2\big|-1\Big|=1.</math></center> If a model of <math>S</math> were built from wire of negligible thickness, then the total length of wire required would be <math>a\sqrt{b}</math>, where <math>a</math> and <math>b</math> are positive integers and <math>b</math> is not divisible by the square of any prime number. Find <math>a+b</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>S</math> be the [[set]] of [[point]]s in the [[Cartesian plane]] that satisfy <center><math>\Big|\big||x|-2\big|-1\Big|+\Big|\big||y|-2\big|-1\Big|=1.</math></center> If a model of <math>S</math> were built from wire of negligible thickness, then the total length of wire required would be <math>a\sqrt{b}</math>, where <math>a</math> and <math>b</math> are positive integers and <math>b</math> is not divisible by the square of any prime number. Find <math>a+b</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">__TOC__</del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>==Solution==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>===Solution 1===</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>=== Solution 1 ===</div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:''This solution is non-rigorous.''</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:''This solution is non-rigorous.''</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>f(x) = \Big|\big||x|-2\big|-1\Big|</math>, <math>f(x) \ge 0</math>. Then <math>f(x) + f(y) = 1 \Longrightarrow f(x), f(y) \le 1 \Longrightarrow x, y \le 4</math>. We only have a <math>4\times 4</math> area, so guessing points and graphing won't be too bad of an idea. Since <math>f(x) = f(-x)</math>, there's a symmetry about all four [[quadrant]]s, so just consider the first quadrant. We now gather some points:</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>f(x) = \Big|\big||x|-2\big|-1\Big|</math>, <math>f(x) \ge 0</math>. Then <math>f(x) + f(y) = 1 \Longrightarrow f(x), f(y) \le 1 \Longrightarrow x, y \le 4</math>. We only have a <math>4\times 4</math> area, so guessing points and graphing won't be too bad of an idea. Since <math>f(x) = f(-x)</math>, there's a symmetry about all four [[quadrant]]s, so just consider the first quadrant. We now gather some points:</div></td></tr>
<tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l37" >Line 37:</td>
<td colspan="2" class="diff-lineno">Line 36:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Calculating the lengths is now easy; each rectangle has sides of <math>\sqrt{2}, 3\sqrt{2}</math>, so the answer is <math>4(\sqrt{2} + 3\sqrt{2}) = 16\sqrt{2}</math>. For all four quadrants, this is <math>64\sqrt{2}</math>, and <math>a+b=\boxed{066}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Calculating the lengths is now easy; each rectangle has sides of <math>\sqrt{2}, 3\sqrt{2}</math>, so the answer is <math>4(\sqrt{2} + 3\sqrt{2}) = 16\sqrt{2}</math>. For all four quadrants, this is <math>64\sqrt{2}</math>, and <math>a+b=\boxed{066}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>=== Solution 2 ===</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>===Solution 2===</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Since <math>0 \le \Big|\big||x| - 2\big| - 1\Big| \le 1</math> and <math>0 \le \Big|\big||y| - 2\big| - 1\Big| \le 1</math><br /></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Since <math>0 \le \Big|\big||x| - 2\big| - 1\Big| \le 1</math> and <math>0 \le \Big|\big||y| - 2\big| - 1\Big| \le 1</math><br /></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>- 1 \le \big||x| - 2\big| - 1 \le 1</math><br /></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>- 1 \le \big||x| - 2\big| - 1 \le 1</math><br /></div></td></tr>
<tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l79" >Line 79:</td>
<td colspan="2" class="diff-lineno">Line 78:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>So a total of <math>6</math> doublings = <math>2^6</math> = <math>64</math>, the total length = <math>64 \cdot \sqrt {2} = a\sqrt {b}</math>, and <math>a + b = 64 + 2 = \boxed{066}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>So a total of <math>6</math> doublings = <math>2^6</math> = <math>64</math>, the total length = <math>64 \cdot \sqrt {2} = a\sqrt {b}</math>, and <math>a + b = 64 + 2 = \boxed{066}</math>.</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>=== Solution 3 (FASTEST) ===</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>===Solution 3 (FASTEST)===</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>We make use of several consecutive substitutions.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>We make use of several consecutive substitutions.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>||x| - 2|= x_1</math> and similarly with <math>y</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>||x| - 2|= x_1</math> and similarly with <math>y</math>.</div></td></tr>
<tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l88" >Line 88:</td>
<td colspan="2" class="diff-lineno">Line 87:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{AIME box|year=1997|num-b=12|num-a=14}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{AIME box|year=1997|num-b=12|num-a=14}}</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Intermediate Algebra Problems]]</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Intermediate Algebra Problems]]</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td></tr>
</table>Anmol04https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=106192&oldid=prevWhatrthose: /* Solution 3 (FASTEST) */2019-06-09T02:15:34Z<p><span dir="auto"><span class="autocomment">Solution 3 (FASTEST)</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 02:15, 9 June 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l82" >Line 82:</td>
<td colspan="2" class="diff-lineno">Line 82:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>We make use of several consecutive substitutions.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>We make use of several consecutive substitutions.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>||x| - 2|= x_1</math> and similarly with <math>y</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>||x| - 2|= x_1</math> and similarly with <math>y</math>.</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Therefore, our graph is <math>|x_1 - 1| + |y_1 - 1| = 1</math>. This is a diamond with perimeter <math>4\sqrt{2}</math>. Now, we make use of the following fact for a function of two variables <math>x</math> and <math>y</math>: Suppose we have <math>f(x, y) = c</math>. Then <math>f(|x|, |y|)</math> is equal to the graph of <math>f(x, y)</math> reflected across the y axis and x axis, and the reflection across the y axis across the x axis, therefore the perimeter of  of <math>f(|x|, |y|)</math> is 4 times the perimeter of <math>f(x, y)</math>. Now, we continue making substitutions at each absolute value sign (<math>|x| - 2 = x_2</math> and <del class="diffchange diffchange-inline">so forth</del>),  noting that the constants don't matter and each absolute value sign increases the perimeter 4 times as much. Therefore, the length is <math>4^2 \times 4\sqrt{2} = 64\sqrt{2} = a\sqrt{b}</math>, and <math>a+b = \boxed{066}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Therefore, our graph is <math>|x_1 - 1| + |y_1 - 1| = 1</math>. This is a diamond with perimeter <math>4\sqrt{2}</math>. Now, we make use of the following fact for a function of two variables <math>x</math> and <math>y</math>: Suppose we have <math>f(x, y) = c</math>. Then <math>f(|x|, |y|)</math> is equal to the graph of <math>f(x, y)</math> reflected across the y axis and x axis, and the reflection across the y axis across the x axis, therefore the perimeter of  of <math>f(|x|, |y|)</math> is 4 times the perimeter of <math>f(x, y)</math>. Now, we continue making substitutions at each absolute value sign (<math>|x| - 2 = x_2</math> and <ins class="diffchange diffchange-inline">finally <math>x = x_3</math>, similarly for y as well. </ins>),  noting that the constants don't matter <ins class="diffchange diffchange-inline">as they just translate the graph </ins>and each absolute value sign increases the perimeter 4 times as much. Therefore, the length is <math>4^2 \times 4\sqrt{2} = 64\sqrt{2} = a\sqrt{b}</math>, and <math>a+b = \boxed{066}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>- whatRthose</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>- whatRthose</div></td></tr>
</table>Whatrthose