https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_21&feed=atom&action=history
2018 AMC 12B Problems/Problem 21 - Revision history
2024-03-29T09:35:05Z
Revision history for this page on the wiki
MediaWiki 1.31.1
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_21&diff=177692&oldid=prev
MRENTHUSIASM at 08:12, 30 August 2022
2022-08-30T08:12:26Z
<p></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 08:12, 30 August 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l27" >Line 27:</td>
<td colspan="2" class="diff-lineno">Line 27:</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>dot("$M$",M,1.5*N,linewidth(4));</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>dot("$M$",M,1.5*N,linewidth(4));</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></asy></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></asy></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;">~MRENTHUSIASM</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>== 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>
</table>
MRENTHUSIASM
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_21&diff=177658&oldid=prev
Puffer13: /* Diagram */
2022-08-29T16:47:20Z
<p><span dir="auto"><span class="autocomment">Diagram</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 16:47, 29 August 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l27" >Line 27:</td>
<td colspan="2" class="diff-lineno">Line 27:</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>dot("$M$",M,1.5*N,linewidth(4));</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>dot("$M$",M,1.5*N,linewidth(4));</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></asy></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></asy></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;">~MRENTHUSIASM</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;"></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>
</table>
Puffer13
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_21&diff=163857&oldid=prev
MRENTHUSIASM: /* Solution */
2021-10-21T18:26:24Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 18:26, 21 October 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l41" >Line 41:</td>
<td colspan="2" class="diff-lineno">Line 41:</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>Solving this equation, we have <math>a=4,</math> from which <math>M=(4,4).</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>Solving this equation, we have <math>a=4,</math> from which <math>M=(4,4).</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>Finally, we apply the Shoelace Theorem to <del class="diffchange diffchange-inline">find </del><math><del class="diffchange diffchange-inline">[</del>MOI<del class="diffchange diffchange-inline">]</del>:</math> <cmath>[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{\textbf{(E)}\ \frac72}.</cmath></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>Finally, we apply the Shoelace Theorem to <math><ins class="diffchange diffchange-inline">\triangle </ins>MOI:</math> <cmath>[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{\textbf{(E)}\ \frac72}.</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;"><div><u><b>Remark</b></u></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><u><b>Remark</b></u></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>Alternatively, we can use <math>\overline{MI}</math> as the base and the distance from <math>O</math> to <math>\overleftrightarrow{MI}</math> as the height <del class="diffchange diffchange-inline">to find </del><math><del class="diffchange diffchange-inline">[</del>MOI<del class="diffchange diffchange-inline">]</del>:</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>Alternatively, we can use <math>\overline{MI}</math> as the base and the distance from <math>O</math> to <math>\overleftrightarrow{MI}</math> as the height <ins class="diffchange diffchange-inline">for </ins><math><ins class="diffchange diffchange-inline">\triangle </ins>MOI:</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>* By the Distance Formula, we have <math>MI=2\sqrt2.</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>* By the Distance Formula, we have <math>MI=2\sqrt2.</math></div></td></tr>
</table>
MRENTHUSIASM
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_21&diff=163856&oldid=prev
MRENTHUSIASM: /* Solution */
2021-10-21T18:15:52Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<col class="diff-content" />
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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 18:15, 21 October 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l44" >Line 44:</td>
<td colspan="2" class="diff-lineno">Line 44:</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><u><b>Remark</b></u></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><u><b>Remark</b></u></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>Alternatively, <del class="diffchange diffchange-inline">to find <math>[MOI],</math> </del>we can use <math>\overline{MI}</math> as the base and the distance from <math>O</math> to <math>\overleftrightarrow{MI}</math> as the height:</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>Alternatively, we can use <math>\overline{MI}</math> as the base and the distance from <math>O</math> to <math>\overleftrightarrow{MI}</math> as the height <ins class="diffchange diffchange-inline">to find <math>[MOI]</ins>:<ins class="diffchange diffchange-inline"></math></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>* By the Distance Formula, we have <math>MI=2\sqrt2.</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>* By the Distance Formula, we have <math>MI=2\sqrt2.</math></div></td></tr>
</table>
MRENTHUSIASM
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_21&diff=163855&oldid=prev
MRENTHUSIASM: /* Solution */
2021-10-21T18:09:35Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<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 18:09, 21 October 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l39" >Line 39:</td>
<td colspan="2" class="diff-lineno">Line 39:</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>Since <math>\odot M</math> is also tangent to both coordinate axes, its center is at <math>M=(a,a)</math> and its radius is <math>a</math> for some positive number <math>a.</math> Let <math>P</math> be the point of tangency of <math>\odot O</math> and <math>\odot M.</math> As <math>\overline{OP}</math> and <math>\overline{MP}</math> are both perpendicular to the common tangent line at <math>P,</math> we conclude that <math>O,M,</math> and <math>P</math> are collinear. It follows that <math>OM=OP-MP,</math> or <cmath>\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.</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>Since <math>\odot M</math> is also tangent to both coordinate axes, its center is at <math>M=(a,a)</math> and its radius is <math>a</math> for some positive number <math>a.</math> Let <math>P</math> be the point of tangency of <math>\odot O</math> and <math>\odot M.</math> As <math>\overline{OP}</math> and <math>\overline{MP}</math> are both perpendicular to the common tangent line at <math>P,</math> we conclude that <math>O,M,</math> and <math>P</math> are collinear. It follows that <math>OM=OP-MP,</math> or <cmath>\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.</cmath></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>Solving this equation, we have <math>a=4.</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>Solving this equation, we have <math>a=4<ins class="diffchange diffchange-inline">,</math> from which <math>M=(4,4)</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>Finally, we apply the Shoelace Theorem to find <math>[MOI]:</math> <cmath>[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{\textbf{(E)}\ \frac72}.</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>Finally, we apply the Shoelace Theorem to find <math>[MOI]:</math> <cmath>[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{\textbf{(E)}\ \frac72}.</cmath></div></td></tr>
</table>
MRENTHUSIASM
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_21&diff=163836&oldid=prev
MRENTHUSIASM: /* Solution */
2021-10-21T04:10:02Z
<p><span dir="auto"><span class="autocomment">Solution</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 04:10, 21 October 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l38" >Line 38:</td>
<td colspan="2" class="diff-lineno">Line 38:</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> denote the semiperimeter of <math>\triangle ABC.</math> The inradius of <math>\triangle ABC</math> is <math>\frac{[ABC]}{s}=\frac{30}{15}=2,</math> from which <math>I=(2,2).</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>Let <math>s</math> denote the semiperimeter of <math>\triangle ABC.</math> The inradius of <math>\triangle ABC</math> is <math>\frac{[ABC]}{s}=\frac{30}{15}=2,</math> from which <math>I=(2,2).</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;"></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>Since <math>\odot M</math> is tangent to both coordinate axes, its center is at <math>M=(a,a)</math> and its radius is <math>a</math> for some positive number <math>a.</math> Let <math>P</math> be the point of tangency of <math>\odot O</math> and <math>\odot M.</math> As <math>\overline{OP}</math> and <math>\overline{MP}</math> are both perpendicular to the common tangent line at <math>P,</math> we conclude that <math>O,M,</math> and <math>P</math> are collinear. It follows that <math>OM=OP-MP,</math> or <cmath>\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.</cmath></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>Since <math>\odot M</math> is <ins class="diffchange diffchange-inline">also </ins>tangent to both coordinate axes, its center is at <math>M=(a,a)</math> and its radius is <math>a</math> for some positive number <math>a.</math> Let <math>P</math> be the point of tangency of <math>\odot O</math> and <math>\odot M.</math> As <math>\overline{OP}</math> and <math>\overline{MP}</math> are both perpendicular to the common tangent line at <math>P,</math> we conclude that <math>O,M,</math> and <math>P</math> are collinear. It follows that <math>OM=OP-MP,</math> or <cmath>\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.</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;"><div>Solving this equation, we have <math>a=4.</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>Solving this equation, we have <math>a=4.</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>
MRENTHUSIASM
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_21&diff=163834&oldid=prev
MRENTHUSIASM: /* Solution */
2021-10-21T03:39:39Z
<p><span dir="auto"><span class="autocomment">Solution</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 03:39, 21 October 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l42" >Line 42:</td>
<td colspan="2" class="diff-lineno">Line 42:</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>Finally, we apply the Shoelace Theorem to find <math>[MOI]:</math> <cmath>[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{\textbf{(E)}\ \frac72}.</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>Finally, we apply the Shoelace Theorem to find <math>[MOI]:</math> <cmath>[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{\textbf{(E)}\ \frac72}.</cmath></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><u><b>Remark</b></u></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><u><b>Remark</b></u></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 colspan="2" class="diff-lineno" id="mw-diff-left-l52" >Line 52:</td>
<td colspan="2" class="diff-lineno">Line 51:</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>Therefore, we get <cmath>[MOI]=\frac12\cdot MI\cdot h_O=\frac72.</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>Therefore, we get <cmath>[MOI]=\frac12\cdot MI\cdot h_O=\frac72.</cmath></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>~pieater314159 ~MRENTHUSIASM</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>~pieater314159 ~MRENTHUSIASM</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>
MRENTHUSIASM
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_21&diff=163833&oldid=prev
MRENTHUSIASM at 03:24, 21 October 2021
2021-10-21T03:24:21Z
<p></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 03:24, 21 October 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l30" >Line 30:</td>
<td colspan="2" class="diff-lineno">Line 30:</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;">In this solution, let the brackets denote areas.</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 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 place the diagram in the coordinate plane: Let <math>A=(12,0),B=(0,5),</math> and <math>C=(0,0).</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>We place the diagram in the coordinate plane: Let <math>A=(12,0),B=(0,5),</math> and <math>C=(0,0).</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>Since <math>\triangle ABC</math> is a right triangle with <math>\angle ACB=90^\circ,</math> its circumcenter is the midpoint of <math>\overline{AB},</math> from which <math>O=\left(6,\frac52\right).</math> <del class="diffchange diffchange-inline">Let </del><math><del class="diffchange diffchange-inline">[</del>ABC<del class="diffchange diffchange-inline">]</del></math> <del class="diffchange diffchange-inline">and </del><math>s</math> denote <del class="diffchange diffchange-inline">the area and </del>the semiperimeter of <math>\triangle ABC<del class="diffchange diffchange-inline">,</del></math> <del class="diffchange diffchange-inline">respectively. </del>The inradius of <math>\triangle ABC</math> is <math>\frac{[ABC]}{s}=\frac{30}{15}=2,</math> from which <math>I=(2,2).</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>Since <math>\triangle ABC</math> is a right triangle with <math>\angle ACB=90^\circ,</math> its circumcenter is the midpoint of <math>\overline{AB},</math> from which <math>O=\left(6,\frac52\right).</math> <ins class="diffchange diffchange-inline">Note that the circumradius of </ins><math><ins class="diffchange diffchange-inline">\triangle </ins>ABC</math> <ins class="diffchange diffchange-inline">is <math>\frac{13}{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> </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">Let </ins><math>s</math> denote the semiperimeter of <math>\triangle ABC<ins class="diffchange diffchange-inline">.</ins></math> The inradius of <math>\triangle ABC</math> is <math>\frac{[ABC]}{s}=\frac{30}{15}=2,</math> from which <math>I=(2,2).</math></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">Since <math>\odot M</math> is tangent to both coordinate axes, its center is at <math>M=(a,a)</math> and its radius is <math>a</math> for some positive number <math>a.</math> Let <math>P</math> be the point of tangency of <math>\odot O</math> and <math>\odot M.</math> As <math>\overline{OP}</math> and <math>\overline{MP}</math> are both perpendicular to the common tangent line at <math>P,</math> we conclude that <math>O,M,</math> and <math>P</math> are collinear. It follows that <math>OM=OP-MP,</math> or <cmath>\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.</cmath></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">Solving this equation, we have <math>a=4.</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">Finally, we apply the Shoelace Theorem to find <math>[MOI]:</math> <cmath>[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{\textbf{(E)}\ \frac72}.</cmath></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"><u><b>Remark</b></u></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">Alternatively, to find <math>[MOI],</math> we can use <math>\overline{MI}</math> as the base and the distance from <math>O</math> to <math>\overleftrightarrow{MI}</math> as the height:</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="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">Since <math>\odot M</math> is tangent to both coordinate axes</del>, <del class="diffchange diffchange-inline">its center is at </del><math><del class="diffchange diffchange-inline">M</del>=<del class="diffchange diffchange-inline">(a,a)</math> and its radius is <math>a</math> for some positive number <math>a.</math> Let <math>P</math> be the point of tangency of <math>\odot O</math> and <math></del>\<del class="diffchange diffchange-inline">odot M</del>.</math> <del class="diffchange diffchange-inline">As <math>\overline{OP}</math> and <math>\overline{MP}</math> are both perpendicular to the common tangent line at <math>P,</math> it follows that <math>O,M,</math> and <math>P</math> are collinear.</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><ins class="diffchange diffchange-inline">* By the Distance Formula</ins>, <ins class="diffchange diffchange-inline">we have </ins><math><ins class="diffchange diffchange-inline">MI</ins>=<ins class="diffchange diffchange-inline">2</ins>\<ins class="diffchange diffchange-inline">sqrt2</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="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">Note that the circumradius </del>of <math>\<del class="diffchange diffchange-inline">triangle ABC</del></math> is <math><del class="diffchange diffchange-inline">\frac{13}{2}.</del></math> <del class="diffchange diffchange-inline">We have </del><math><del class="diffchange diffchange-inline">OM=OP-MP,</del></math> <del class="diffchange diffchange-inline">or </del><<del class="diffchange diffchange-inline">cmath</del>>\<del class="diffchange diffchange-inline">sqrt</del>{(<del class="diffchange diffchange-inline">a</del>-<del class="diffchange diffchange-inline">6</del>)^2+<del class="diffchange diffchange-inline">\left</del>(<del class="diffchange diffchange-inline">a</del>-<del class="diffchange diffchange-inline">\frac52\right</del>)^2}=\<del class="diffchange diffchange-inline">frac{13}{2}-a.</cmath></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><ins class="diffchange diffchange-inline">* The equation </ins>of <math>\<ins class="diffchange diffchange-inline">overleftrightarrow{MI}</ins></math> is <math><ins class="diffchange diffchange-inline">x-y+0=0,</math> so the distance from <math>O</ins></math> <ins class="diffchange diffchange-inline">to </ins><math><ins class="diffchange diffchange-inline">\overleftrightarrow{MI}</ins></math> <ins class="diffchange diffchange-inline">is </ins><<ins class="diffchange diffchange-inline">math</ins>><ins class="diffchange diffchange-inline">h_O=</ins>\<ins class="diffchange diffchange-inline">frac</ins>{<ins class="diffchange diffchange-inline">\left|1\cdot6+</ins>(-<ins class="diffchange diffchange-inline">1</ins>)<ins class="diffchange diffchange-inline">\cdot\frac52+0\right|}{\sqrt{1</ins>^2+(-<ins class="diffchange diffchange-inline">1</ins>)^2<ins class="diffchange diffchange-inline">}</ins>}=\<ins class="diffchange diffchange-inline">frac74\sqrt2</ins>.</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><del class="diffchange diffchange-inline">Solving this equation, we get <math>a=4</del>.</math></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;"></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">Finally</del>, we <del class="diffchange diffchange-inline">find the area of </del><<del class="diffchange diffchange-inline">math</del>><del class="diffchange diffchange-inline">\triangle{</del>MOI<del class="diffchange diffchange-inline">}</math> by the Shoelace Theorem: <cmath></del>\frac12\<del class="diffchange diffchange-inline">left|\left(4</del>\cdot<del class="diffchange diffchange-inline">\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|</del>=<del class="diffchange diffchange-inline">\boxed{\textbf{(E)}\ </del>\frac72<del class="diffchange diffchange-inline">}</del>.</cmath></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><ins class="diffchange diffchange-inline">Therefore</ins>, we <ins class="diffchange diffchange-inline">get </ins><<ins class="diffchange diffchange-inline">cmath</ins>><ins class="diffchange diffchange-inline">[</ins>MOI<ins class="diffchange diffchange-inline">]=</ins>\frac12\<ins class="diffchange diffchange-inline">cdot MI</ins>\cdot <ins class="diffchange diffchange-inline">h_O</ins>=\frac72.</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="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>~pieater314159 ~MRENTHUSIASM</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>~pieater314159 ~MRENTHUSIASM</div></td></tr>
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MRENTHUSIASM
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_21&diff=163819&oldid=prev
MRENTHUSIASM at 00:10, 21 October 2021
2021-10-21T00:10:10Z
<p></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 00:10, 21 October 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l39" >Line 39:</td>
<td colspan="2" class="diff-lineno">Line 39:</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>Solving this equation, we get <math>a=4.</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>Solving this equation, we get <math>a=4.</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>Finally, we find the area of <math>\triangle{MOI}</math> by the Shoelace Theorem: <cmath>=\boxed{}.</cmath></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>Finally, we find the area of <math>\triangle{MOI}</math> by the Shoelace Theorem: <cmath><ins class="diffchange diffchange-inline">\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|</ins>=\boxed{<ins class="diffchange diffchange-inline">\textbf{(E)}\ \frac72</ins>}.</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="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>~pieater314159 ~MRENTHUSIASM</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>~pieater314159 ~MRENTHUSIASM</div></td></tr>
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MRENTHUSIASM
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_21&diff=163813&oldid=prev
MRENTHUSIASM: Made the explanation more thorough (backed up with reasons). Also, showed other ways to calculate the area. Also, the answer choices should be vertical fractions. Source: https://www.scribd.com/document/527469246/2018-12B-AMC-test
2021-10-20T23:52:42Z
<p>Made the explanation more thorough (backed up with reasons). Also, showed other ways to calculate the area. Also, the answer choices should be vertical fractions. Source: https://www.scribd.com/document/527469246/2018-12B-AMC-test</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 23:52, 20 October 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l3" >Line 3:</td>
<td colspan="2" class="diff-lineno">Line 3:</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 <math>\triangle{ABC}</math> with side lengths <math>AB = 13</math>, <math>AC = 12</math>, and <math>BC = 5</math>, let <math>O</math> and <math>I</math> denote the circumcenter and incenter, respectively. A circle with center <math>M</math> is tangent to the legs <math>AC</math> and <math>BC</math> and to the circumcircle of <math>\triangle{ABC}</math>. What is the area of <math>\triangle{MOI}</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 <math>\triangle{ABC}</math> with side lengths <math>AB = 13</math>, <math>AC = 12</math>, and <math>BC = 5</math>, let <math>O</math> and <math>I</math> denote the circumcenter and incenter, respectively. A circle with center <math>M</math> is tangent to the legs <math>AC</math> and <math>BC</math> and to the circumcircle of <math>\triangle{ABC}</math>. What is the area of <math>\triangle{MOI}</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><math>\textbf{(A)}\ <del class="diffchange diffchange-inline">5/2</del>\qquad\textbf{(B)}\ 11<del class="diffchange diffchange-inline">/</del>4\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 13<del class="diffchange diffchange-inline">/</del>4\qquad\textbf{(E)}\ <del class="diffchange diffchange-inline">7/2</del></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>\textbf{(A)}\ <ins class="diffchange diffchange-inline">\frac52</ins>\qquad\textbf{(B)}\ <ins class="diffchange diffchange-inline">\frac{</ins>11<ins class="diffchange diffchange-inline">}{</ins>4<ins class="diffchange diffchange-inline">}</ins>\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ <ins class="diffchange diffchange-inline">\frac{</ins>13<ins class="diffchange diffchange-inline">}{</ins>4<ins class="diffchange diffchange-inline">}</ins>\qquad\textbf{(E)}\ <ins class="diffchange diffchange-inline">\frac72</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>== Diagram ==</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>== Diagram ==</div></td></tr>
<tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l30" >Line 30:</td>
<td colspan="2" class="diff-lineno">Line 30:</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;">We place the diagram in the coordinate plane: Let <math>A=(12,0),B=(0,5),</math> and <math>C=(0,0).</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;">Since <math>\triangle ABC</math> is a right triangle with <math>\angle ACB=90^\circ,</math> its circumcenter is the midpoint of <math>\overline{AB},</math> from which <math>O=\left(6,\frac52\right).</math> Let <math>[ABC]</math> and <math>s</math> denote the area and the semiperimeter of <math>\triangle ABC,</math> respectively. The inradius of <math>\triangle ABC</math> is <math>\frac{[ABC]}{s}=\frac{30}{15}=2,</math> from which <math>I=(2,2).</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;">Since <math>\odot M</math> is tangent to both coordinate axes, its center is at <math>M=(a,a)</math> and its radius is <math>a</math> for some positive number <math>a.</math> Let <math>P</math> be the point of tangency of <math>\odot O</math> and <math>\odot M.</math> As <math>\overline{OP}</math> and <math>\overline{MP}</math> are both perpendicular to the common tangent line at <math>P,</math> it follows that <math>O,M,</math> and <math>P</math> are collinear.</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;">Note that the circumradius of <math>\triangle ABC</math> is <math>\frac{13}{2}.</math> We have <math>OM=OP-MP,</math> or <cmath>\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.</cmath></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;">Solving this equation, we get <math>a=4.</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;">Finally, we find the area of <math>\triangle{MOI}</math> by the Shoelace Theorem: <cmath>=\boxed{}.</cmath></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>~pieater314159 ~MRENTHUSIASM</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>~pieater314159 ~MRENTHUSIASM</div></td></tr>
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MRENTHUSIASM