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2009 AMC 12A Problems/Problem 13 - Revision history
2024-03-28T17:55:38Z
Revision history for this page on the wiki
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Equationcrunchor: /* Solution */ added somewhat faster law of cosines solution
2015-02-09T14:24:53Z
<p><span dir="auto"><span class="autocomment">Solution: </span> added somewhat faster law of cosines solution</span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 14:24, 9 February 2015</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l134" >Line 134:</td>
<td colspan="2" class="diff-lineno">Line 134:</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><cmath> AC_1^2 = AD_1^2 + D_1C_1^2 = (10 + 20/\sqrt 2)^2 + (20\sqrt 2)^2 = 100 + 400/\sqrt 2 + 200 + 200 = 500 + 200\sqrt 2 < 500 + 200\cdot 1.5 = 800 </cmath></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><cmath> AC_1^2 = AD_1^2 + D_1C_1^2 = (10 + 20/\sqrt 2)^2 + (20\sqrt 2)^2 = 100 + 400/\sqrt 2 + 200 + 200 = 500 + 200\sqrt 2 < 500 + 200\cdot 1.5 = 800 </cmath></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>Therefore for any valid <math>C</math> the value <math>AC^2</math> is surely in the interval <math>\boxed{ [700,800] }</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 for any valid <math>C</math> the value <math>AC^2</math> is surely in the interval <math>\boxed{ <ins class="diffchange diffchange-inline">\textbf{(D)}</ins>[700,800] <ins class="diffchange diffchange-inline">}</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> </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 class="diffchange diffchange-inline">===Alternate Solution===</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 class="diffchange diffchange-inline">From the law of cosines, <math>500-400\cos120^\circ<AC^2<500-400\cos135^\circ\implies700<AC^2<500+200\sqrt{2}</math>.  This is essentially the same solution as above.  The answer is <math>\boxed{\textbf{(D)}</ins>}</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>== 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>
Equationcrunchor
https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_13&diff=53781&oldid=prev
Nathan wailes: /* See Also */
2013-07-04T01:43:00Z
<p><span dir="auto"><span class="autocomment">See Also</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 01:43, 4 July 2013</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l139" >Line 139:</td>
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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;"></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>{{AMC12 box|year=2009|ab=A|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>{{AMC12 box|year=2009|ab=A|num-b=12|num-a=14}}</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;">{{MAA Notice}}</ins></div></td></tr>
</table>
Nathan wailes
https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_13&diff=36847&oldid=prev
Bwy: /* Answering the question */
2011-02-11T06:05:11Z
<p><span dir="auto"><span class="autocomment">Answering the question</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 06:05, 11 February 2011</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l28" >Line 28:</td>
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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>=== Answering the question ===</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>=== Answering the question ===</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>To answer the question we are asked, it is enough to compute <math>AC^2</math> for two different angles, preferably for both extremes (<math>45</math> and <math>60</math> degrees).</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>To answer the question we are asked, it is enough to compute <math>AC^2</math> for two different angles, preferably for both extremes (<math>45</math> and <math>60</math> degrees)<ins class="diffchange diffchange-inline">. You can use the law of cosines to do so</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>Alternately, it is enough to compute <math>AC^2</math> for one of the extreme angles. In case it falls inside one of the given intervals, we are done. In case it falls on the boundary between two options, we also have to argue whether our <math>AC^2</math> is the minimal or the maximal possible value of <math>AC^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>Alternately, it is enough to compute <math>AC^2</math> for one of the extreme angles. In case it falls inside one of the given intervals, we are done. In case it falls on the boundary between two options, we also have to argue whether our <math>AC^2</math> is the minimal or the maximal possible value of <math>AC^2</math>.</div></td></tr>
</table>
Bwy
https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_13&diff=30276&oldid=prev
Misof at 02:14, 15 February 2009
2009-02-15T02:14:49Z
<p></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 02:14, 15 February 2009</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l25" >Line 25:</td>
<td colspan="2" class="diff-lineno">Line 25:</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 ==</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 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;">=== Answering the question ===</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;">To answer the question we are asked, it is enough to compute <math>AC^2</math> for two different angles, preferably for both extremes (<math>45</math> and <math>60</math> degrees).</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;">Alternately, it is enough to compute <math>AC^2</math> for one of the extreme angles. In case it falls inside one of the given intervals, we are done. In case it falls on the boundary between two options, we also have to argue whether our <math>AC^2</math> is the minimal or the maximal possible value of <math>AC^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;"></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;">Below we show a complete solution in which we also show that all possible values of <math>AC^2</math> do indeed lie in the given interval.</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;">=== Complete solution ===</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>Let <math>C_1</math> be the point the ship would reach if it turned <math>45^\circ</math>, and <math>C_2</math> the point it would reach if it turned <math>60^\circ</math>. Obviously, <math>C_1</math> is the furthest possible point from <math>A</math>, and <math>C_2</math> is the closest possible point to <math>A</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>C_1</math> be the point the ship would reach if it turned <math>45^\circ</math>, and <math>C_2</math> the point it would reach if it turned <math>60^\circ</math>. Obviously, <math>C_1</math> is the furthest possible point from <math>A</math>, and <math>C_2</math> is the closest possible point to <math>A</math>.</div></td></tr>
</table>
Misof
https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_13&diff=30160&oldid=prev
Misof: New page: == Problem == A ship sails <math>10</math> miles in a straight line from <math>A</math> to <math>B</math>, turns through an angle between <math>45^{\circ}</math> and <math>60^{\circ}</math...
2009-02-12T20:05:51Z
<p>New page: == Problem == A ship sails <math>10</math> miles in a straight line from <math>A</math> to <math>B</math>, turns through an angle between <math>45^{\circ}</math> and <math>60^{\circ}</math...</p>
<p><b>New page</b></p><div>== Problem ==<br />
A ship sails <math>10</math> miles in a straight line from <math>A</math> to <math>B</math>, turns through an angle between <math>45^{\circ}</math> and <math>60^{\circ}</math>, and then sails another <math>20</math> miles to <math>C</math>. Let <math>AC</math> be measured in miles. Which of the following intervals contains <math>AC^2</math>?<br />
<asy><br />
unitsize(2mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair B=(0,0), A=(-10,0), C=20*dir(50);<br />
<br />
draw(A--B--C);<br />
draw(A--C,linetype("4 4"));<br />
<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
label("$10$",midpoint(A--B),S);<br />
label("$20$",midpoint(B--C),SE);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
</asy><br />
<br />
<math>\textbf{(A)}\ [400,500] \qquad \textbf{(B)}\ [500,600] \qquad \textbf{(C)}\ [600,700] \qquad \textbf{(D)}\ [700,800]</math><br />
<math>\textbf{(E)}\ [800,900]</math><br />
<br />
== Solution ==<br />
<br />
Let <math>C_1</math> be the point the ship would reach if it turned <math>45^\circ</math>, and <math>C_2</math> the point it would reach if it turned <math>60^\circ</math>. Obviously, <math>C_1</math> is the furthest possible point from <math>A</math>, and <math>C_2</math> is the closest possible point to <math>A</math>.<br />
<br />
Hence the interval of possible values for <math>AC^2</math> is <math>[AC_2^2,AC_1^2]</math>.<br />
<br />
<asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair B=(0,0), A=(-10,0), C1=20*dir(45), C2=20*dir(60);<br />
<br />
draw(A--B--C1);<br />
draw(A--C1,linetype("4 4"));<br />
draw(A--B--C2);<br />
draw(A--C2,linetype("4 4"));<br />
<br />
draw( arc(A, length(C1-A), 0, 55 ), dotted );<br />
draw( arc(A, length(C2-A), 0, 55 ), dotted );<br />
draw( arc(B, C1, C2) );<br />
<br />
dot(A);<br />
dot(B);<br />
dot(C1);<br />
dot(C2);<br />
label("$10$",midpoint(A--B),S);<br />
label("$20$",midpoint(B--C1),SE);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C_1$",C1,NE);<br />
label("$C_2$",C2,N);<br />
</asy><br />
<br />
We can find <math>AC_1^2</math> and <math>AC_2^2</math> as follows:<br />
<br />
Let <math>D_1</math> and <math>D_2</math> be the feet of the heights from <math>C_1</math> and <math>C_2</math> onto <math>AB</math>. The angles in the triangle <math>BD_1C_1</math> are <math>45^\circ</math>, <math>45^\circ</math>, and <math>90^\circ</math>, hence <math>BD_1 = D_1C_1 = BC_1 / \sqrt 2</math>. Similarly, the angles in the triangle <math>BD_2C_2</math> are <math>30^\circ</math>, <math>60^\circ</math>, and <math>90^\circ</math>, hence <math>BD_2 = BC_2 / 2</math> and <math>D_2C_2 = BC_2 \sqrt 3 / 2</math>.<br />
<br />
<asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair B=(0,0), A=(-10,0), C1=20*dir(45), C2=20*dir(60);<br />
<br />
draw(A--B--C1);<br />
draw(A--C1,linetype("4 4"));<br />
<br />
dot(A);<br />
dot(B);<br />
dot(C1);<br />
<br />
pair D1 = (C1.x,0);<br />
dot(D1);<br />
draw(B--D1--C1);<br />
<br />
label("$10$",midpoint(A--B),S);<br />
label("$20$",midpoint(B--C1),SE);<br />
label("$20/\sqrt 2$",midpoint(B--D1),S);<br />
label("$20/\sqrt 2$",midpoint(D1--C1),E);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C_1$",C1,NE);<br />
label("$D_1$",D1,S);<br />
</asy><br />
<br />
<asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair B=(0,0), A=(-10,0), C1=20*dir(45), C2=20*dir(60);<br />
<br />
draw(A--B--C2);<br />
draw(A--C2,linetype("4 4"));<br />
<br />
dot(A);<br />
dot(B);<br />
dot(C2);<br />
<br />
pair D2 = (C2.x,0);<br />
dot(D2);<br />
draw(B--D2--C2);<br />
<br />
label("$10$",midpoint(A--B),S);<br />
label("$20$",midpoint(B--C2),SE);<br />
label("$10$",midpoint(B--D2),S);<br />
label("$10\sqrt 3$",midpoint(D2--C2),E);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C_2$",C2,NE);<br />
label("$D_2$",D2,S);<br />
</asy><br />
<br />
Hence we get:<br />
<br />
<cmath> AC_2^2 = AD_2^2 + D_2C_2^2 = 20^2 + (10\sqrt 3)^2 = 400 + 300 = 700 </cmath><br />
<br />
<cmath> AC_1^2 = AD_1^2 + D_1C_1^2 = (10 + 20/\sqrt 2)^2 + (20\sqrt 2)^2 = 100 + 400/\sqrt 2 + 200 + 200 = 500 + 200\sqrt 2 < 500 + 200\cdot 1.5 = 800 </cmath><br />
<br />
Therefore for any valid <math>C</math> the value <math>AC^2</math> is surely in the interval <math>\boxed{ [700,800] }</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2009|ab=A|num-b=12|num-a=14}}</div>
Misof