https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_14&feed=atom&action=history2003 AIME II Problems/Problem 14 - Revision history2024-03-29T01:26:33ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_14&diff=197146&oldid=prevPapermath: /* Solution 4 (No Trig) */2023-08-18T19:09:23Z<p><span dir="auto"><span class="autocomment">Solution 4 (No Trig)</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 (No Trig)==</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 (No Trig)==</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;"><asy> size(200); draw((0,0)--(10/sqrt(3),2)--(18/sqrt(3),6)--(10/sqrt(3),10)--(0,8)--(-8/sqrt(3),4)--cycle); dot((0,0));dot((10/sqrt(3),2));dot((18/sqrt(3),6));dot((10/sqrt(3),10));dot((0,8));dot((-8/sqrt(3),4)); label("$A (0,0)$",(0,0),SE);label("$B (b,2)$",(10/sqrt(3),2),SE);label("$C$",(18/sqrt(3),6),E);label("$D$",(10/sqrt(3),10),N);label("$E$",(0,8),NW);label("$F$",(-8/sqrt(3),4),W); xaxis("$x$");yaxis("$y$"); pair b=foot((10/sqrt(3),2),(0,0),(10,0)); pair f=foot((-8/sqrt(3),4),(0,0),(-10,0)); draw(b--(10/sqrt(3),2),dotted); draw(f--(-8/sqrt(3),4),dotted); label("$\theta$",(0,0),7*dir((0,0)--(10/sqrt(3),2)+(4*sqrt(21)/3,0))); </asy></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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solutions. We can draw a rectangle around the hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire rectangle is <math>10(x+y+z)</math>, since the opposite sides are parallel and thus the length of the rectangle is <math>4+4+2=10</math>. Then,  </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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solutions. We can draw a rectangle around the hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire rectangle is <math>10(x+y+z)</math>, since the opposite sides are parallel and thus the length of the rectangle is <math>4+4+2=10</math>. Then,  </div></td></tr>
</table>Papermathhttps://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_14&diff=176417&oldid=prevAmbriggs: /* Solution 4 (No Trig) */2022-07-30T21:25:51Z<p><span dir="auto"><span class="autocomment">Solution 4 (No Trig)</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>~ant08 and sky2025</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>~ant08 and sky2025</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;">==Video Solution by Sal Khan==</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;">https://www.youtube.com/watch?v=Ec-BKdC8vOo&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=4</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;">- AMBRIGGS</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>Ambriggshttps://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_14&diff=144197&oldid=prevRayfish at 23:14, 31 January 20212021-01-31T23:14:41Z<p></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:14, 31 January 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l56" >Line 56:</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>Hence the answer is <math>\boxed{51}</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>Hence the answer is <math>\boxed{51}</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 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===</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;">By symmetry the area of <math>ABCDEF</math> is twice the area of <math>ABCF</math>. Therefore, you only need to calculate the coordinates of <math>B</math>, <math>C</math>, and <math>F</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;"><div>== Solution 3 ==</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 3 ==</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>This is similar to solution 2 but faster and easier.</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 is similar to solution 2 but faster and easier.</div></td></tr>
</table>Rayfishhttps://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_14&diff=135949&oldid=prevHoyam at 00:12, 28 October 20202020-10-28T00:12:40Z<p></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 (No Trig)==</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 (No Trig)==</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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous  </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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solutions. We can draw a rectangle around <ins class="diffchange diffchange-inline">the </ins>hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire rectangle is <math>10(x+y+z)</math>, since the opposite sides are parallel and thus the length of the rectangle is <math>4+4+2=10</math>. Then,  </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>solutions. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call  </div></td><td colspan="2"> </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>the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the  </div></td><td colspan="2"> </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>bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire  </div></td><td colspan="2"> </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>rectangle is <math>10(x+y+z)</math>, since the opposite sides are parallel and thus the length of the rectangle is <math>4+4+2=10</math>. Then,  </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>if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).</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>if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).</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>From triangle ABX we have that <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^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>From triangle ABX we have that <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^2</math>.</div></td></tr>
</table>Hoyamhttps://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_14&diff=118144&oldid=prevAnt08: /* Solution 4 (No Trig) */2020-02-20T02:14:02Z<p><span dir="auto"><span class="autocomment">Solution 4 (No Trig)</span></span></p>
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</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l67" >Line 67:</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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous  </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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous  </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>solutions. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call  </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>solutions. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call  </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>the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the  </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>the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the  </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>bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire  </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>bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire  </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>rectangle is <math>10(x+y+z)</math>, since the opposite sides are parallel and thus the length of the rectangle is <math>4+4+2=10</math>. Then,  </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>rectangle is <math>10(x+y+z)</math>, since the opposite sides are parallel and thus the length of the rectangle is <math>4+4+2=10</math>. Then,  </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>if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).</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>if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).</div></td></tr>
</table>Ant08https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_14&diff=118143&oldid=prevAnt08: /* Solution 4 (No Trig) */2020-02-20T02:13:43Z<p><span dir="auto"><span class="autocomment">Solution 4 (No Trig)</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 02:13, 20 February 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l66" >Line 66:</td>
<td colspan="2" class="diff-lineno">Line 66:</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 (No Trig)==</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 (No Trig)==</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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solutions. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire rectangle is <math>10(x+y+z)</math>, since the opposite sides are parallel and thus the length of the rectangle is <math>4+4+2=10</math>. Then, if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).</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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous  </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>solutions. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call  </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>the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the  </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>bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire  </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>rectangle is <math>10(x+y+z)</math>, since the opposite sides are parallel and thus the length of the rectangle is <math>4+4+2=10</math>. Then,  </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>if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).</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>From triangle ABX we have that <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^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>From triangle ABX we have that <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^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>We can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us <math>BF=\sqrt{4+(x+z)^2}</math>. Setting our two values of BF equal and substituting <math>x^2</math> as <math>12+z^2</math> and simplifying, we get the equation <math>3z^4-16z^2-1024=0</math>. Now we can use the quadratic formula to get that <math>z^2=\frac{64}{3}</math> or <math>-18</math>, so <math>z^2=\frac{64}{3}</math>. Plugging this value back into the equation <math>x^2=12+z^2</math>, we get that <math>x^2=\frac{100}{3}</math>. Now we get that <math>x+z</math> is <math>6\sqrt{3}</math>, so the area of the hexagon is <math>8 \cdot 6\sqrt{3}=48\sqrt{3}</math>, so the answer is <math>48+3=\boxed{051}</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 can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us <math>BF=\sqrt{4+(x+z)^2}</math>. Setting our two values of BF equal and substituting <math>x^2</math> as <math>12+z^2</math> and simplifying, we get the equation <math>3z^4-16z^2-1024=0</math>. Now we can use the quadratic formula to get that <math>z^2=\frac{64}{3}</math> or <math>-18</math>, so <math>z^2=\frac{64}{3}</math>. Plugging this value back into the equation <math>x^2=12+z^2</math>, we get that <math>x^2=\frac{100}{3}</math>. Now we get that <math>x+z</math> is <math>6\sqrt{3}</math>, so the area of the hexagon is <math>8 \cdot 6\sqrt{3}=48\sqrt{3}</math>, so the answer is <math>48+3=\boxed{051}</math></div></td></tr>
</table>Ant08https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_14&diff=118131&oldid=prevAnt08: /* Solution 4 (No Trig) */2020-02-19T23:12:11Z<p><span dir="auto"><span class="autocomment">Solution 4 (No Trig)</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 23:12, 19 February 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l66" >Line 66:</td>
<td colspan="2" class="diff-lineno">Line 66:</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 (No Trig)==</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 (No Trig)==</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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solutions. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire rectangle is <math>10(x+y+z)</math>. Then, if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).</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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solutions. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire rectangle is <math>10(x+y+z)<ins class="diffchange diffchange-inline"></math>, since the opposite sides are parallel and thus the length of the rectangle is <math>4+4+2=10</ins></math>. Then, if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).</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>From triangle ABX we have that <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^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>From triangle ABX we have that <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^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>We can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us <math>BF=\sqrt{4+(x+z)^2}</math>. Setting our two values of BF equal and substituting <math>x^2</math> as <math>12+z^2</math> and simplifying, we get the equation <math>3z^4-16z^2-1024=0</math>. Now we can use the quadratic formula to get that <math>z^2=\frac{64}{3}</math> or <math>-18</math>, so <math>z^2=\frac{64}{3}</math>. Plugging this value back into the equation <math>x^2=12+z^2</math>, we get that <math>x^2=\frac{100}{3}</math>. Now we get that <math>x+z</math> is <math>6\sqrt{3}</math>, so the area of the hexagon is <math>8 \cdot 6\sqrt{3}=48\sqrt{3}</math>, so the answer is <math>48+3=\boxed{051}</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 can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us <math>BF=\sqrt{4+(x+z)^2}</math>. Setting our two values of BF equal and substituting <math>x^2</math> as <math>12+z^2</math> and simplifying, we get the equation <math>3z^4-16z^2-1024=0</math>. Now we can use the quadratic formula to get that <math>z^2=\frac{64}{3}</math> or <math>-18</math>, so <math>z^2=\frac{64}{3}</math>. Plugging this value back into the equation <math>x^2=12+z^2</math>, we get that <math>x^2=\frac{100}{3}</math>. Now we get that <math>x+z</math> is <math>6\sqrt{3}</math>, so the area of the hexagon is <math>8 \cdot 6\sqrt{3}=48\sqrt{3}</math>, so the answer is <math>48+3=\boxed{051}</math></div></td></tr>
</table>Ant08https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_14&diff=118130&oldid=prevAnt08: /* Solution 4 (No Trig) */2020-02-19T23:11:12Z<p><span dir="auto"><span class="autocomment">Solution 4 (No Trig)</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 23:11, 19 February 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l65" >Line 65:</td>
<td colspan="2" class="diff-lineno">Line 65:</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 (No Trig)==</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 (No Trig)==</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;">This is also similar to solution 2 but does not use any trig.</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="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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous <del class="diffchange diffchange-inline">solution</del>. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire rectangle is <math>10(x+y+z)</math>. Then, if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).</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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous <ins class="diffchange diffchange-inline">solutions</ins>. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire rectangle is <math>10(x+y+z)</math>. Then, if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).</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>From triangle ABX we have that <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^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>From triangle ABX we have that <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^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>We can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us <math>BF=\sqrt{4+(x+z)^2}</math>. Setting our two values of BF equal and substituting <math>x^2</math> as <math>12+z^2</math> and simplifying, we get the equation <math>3z^4-16z^2-1024=0</math>. Now we can use the quadratic formula to get that <math>z^2=\frac{64}{3}</math> or <math>-18</math>, so <math>z^2=\frac{64}{3}</math>. Plugging this value back into the equation <math>x^2=12+z^2</math>, we get that <math>x^2=\frac{100}{3}</math>. Now we get that <math>x+z</math> is <math>6\sqrt{3}</math>, so the area of the hexagon is <math>8 \cdot 6\sqrt{3}=48\sqrt{3}</math>, so the answer is <math>48+3=\boxed{051}</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 can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us <math>BF=\sqrt{4+(x+z)^2}</math>. Setting our two values of BF equal and substituting <math>x^2</math> as <math>12+z^2</math> and simplifying, we get the equation <math>3z^4-16z^2-1024=0</math>. Now we can use the quadratic formula to get that <math>z^2=\frac{64}{3}</math> or <math>-18</math>, so <math>z^2=\frac{64}{3}</math>. Plugging this value back into the equation <math>x^2=12+z^2</math>, we get that <math>x^2=\frac{100}{3}</math>. Now we get that <math>x+z</math> is <math>6\sqrt{3}</math>, so the area of the hexagon is <math>8 \cdot 6\sqrt{3}=48\sqrt{3}</math>, so the answer is <math>48+3=\boxed{051}</math></div></td></tr>
</table>Ant08https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_14&diff=118124&oldid=prevAnt08: /* Solution 4 (No Trig) */2020-02-19T20:14:52Z<p><span dir="auto"><span class="autocomment">Solution 4 (No Trig)</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 20:14, 19 February 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l69" >Line 69:</td>
<td colspan="2" class="diff-lineno">Line 69:</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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solution. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire rectangle is <math>10(x+y+z)</math>. Then, if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).</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>First, we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solution. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire rectangle is <math>10(x+y+z)</math>. Then, if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (let's call them ABX and AFY).</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>From triangle ABX we have that <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^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>From triangle ABX we have that <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^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>We can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us <math>BF=\sqrt{4+(x+z)^2}</math>. Setting our two values of BF equal and substituting <math>x^2</math> as <math>12+z^2</math> and simplifying, we get the equation <math>3z^4-16z^2-1024=0</math>. Now we can use the quadratic formula to get that <math>z^2=\frac{64}{3}</math> or -18, so <math>z^2=\frac{64}{3}</math>. Plugging this value back into the equation <math>x^2=12+z^2</math>, we get that <math>x^2=\frac{100}{3}</math>. Now we get that <math>x+z</math> is <math>6\sqrt{3}</math>, so the area of the hexagon is <math>8 \cdot 6\sqrt{3}=48\sqrt{3}</math>, so the answer is <math>48+3=\boxed{051}</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>We can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us <math>BF=\sqrt{4+(x+z)^2}</math>. Setting our two values of BF equal and substituting <math>x^2</math> as <math>12+z^2</math> and simplifying, we get the equation <math>3z^4-16z^2-1024=0</math>. Now we can use the quadratic formula to get that <math>z^2=\frac{64}{3}</math> or <ins class="diffchange diffchange-inline"><math></ins>-18<ins class="diffchange diffchange-inline"></math></ins>, so <math>z^2=\frac{64}{3}</math>. Plugging this value back into the equation <math>x^2=12+z^2</math>, we get that <math>x^2=\frac{100}{3}</math>. Now we get that <math>x+z</math> is <math>6\sqrt{3}</math>, so the area of the hexagon is <math>8 \cdot 6\sqrt{3}=48\sqrt{3}</math>, so the answer is <math>48+3=\boxed{051}</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>~ant08 and sky2025</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>~ant08 and sky2025</div></td></tr>
</table>Ant08https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_14&diff=118123&oldid=prevAnt08: /* Solution 4 (No Trig) */2020-02-19T20:14:13Z<p><span dir="auto"><span class="autocomment">Solution 4 (No Trig)</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 20:14, 19 February 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l66" >Line 66:</td>
<td colspan="2" class="diff-lineno">Line 66:</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 (No Trig)==</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 (No Trig)==</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>This is also similar to solution 2 but does not use any trig.</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 is also similar to solution 2 but does not use any trig.</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>First we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solution. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire rectangle is <math>10(x+y+z)</math>. Then, if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (<del class="diffchange diffchange-inline">lets </del>call them ABX and AFY).</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> </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>First<ins class="diffchange diffchange-inline">, </ins>we see that the y-coordinates of F, E, D, and C must be 4, 8, and 10, and 6, respectively, as in the previous solution. We can draw a rectangle around hexagon ABCDEF and use negative space to find the area of the hexagon. If we call the distance from the foot of the perpendiculars of B and F to A <math>x</math> and <math>z</math>, respectively, and the distance from the bottom left vertex of the rectangle to the foot of the perpendicular from B <math>y</math>. This tells us that the area of the entire rectangle is <math>10(x+y+z)</math>. Then, if we find the area of the extra triangles and subtract, we find that the area of hexagon ABCDEF as <math>6x+8z+2y</math>. However, noticing that <math>x=y</math>, the area of ABCDEF can also be expressed as <math>8(x+z)</math>. Now we just need to find <math>x+z</math>. Since <math>AB=AF</math> and <math>\angle BAF = 120</math> degrees, <math>BF=AB\sqrt{3}</math>. However, we can find AB by using the Pythagorean Theorem on either of the right triangles formed by dropping perpendiculars from B and F to the x-axis (<ins class="diffchange diffchange-inline">let's </ins>call them ABX and AFY).</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>From triangle ABX we have that <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^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>From triangle ABX we have that <math>AB=\sqrt{4+x^2}</math>, so <math>BF=\sqrt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^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>We can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us <math>BF=\sqrt{4+(x+z)^2}</math>. Setting our two values of BF equal and substituting <math>x^2</math> as <math>12+z^2</math> and simplifying, we get the equation <math>3z^4-16z^2-1024=0</math>. Now we can use the quadratic formula to get that <math>z^2=\frac{64}{3}</math> or -18, so <math>z^2=\frac{64}{3}</math>. Plugging this value back into the equation <math>x^2=12+z^2</math>, we get that <math>x^2=\frac{100}{3}</math>. Now we get that x+z is <math>6\sqrt{3}</math>, so the area of the hexagon is <math>8 \cdot 6\sqrt{3}=48\sqrt{3}</math>, so the answer is <math>48+3=\boxed{051}</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>We can also find BF by dropping a perpendicular from B to line FY and using the Pythagorean Theorem on the right triangle formed. This gives us <math>BF=\sqrt{4+(x+z)^2}</math>. Setting our two values of BF equal and substituting <math>x^2</math> as <math>12+z^2</math> and simplifying, we get the equation <math>3z^4-16z^2-1024=0</math>. Now we can use the quadratic formula to get that <math>z^2=\frac{64}{3}</math> or -18, so <math>z^2=\frac{64}{3}</math>. Plugging this value back into the equation <math>x^2=12+z^2</math>, we get that <math>x^2=\frac{100}{3}</math>. Now we get that <ins class="diffchange diffchange-inline"><math></ins>x+z<ins class="diffchange diffchange-inline"></math> </ins>is <math>6\sqrt{3}</math>, so the area of the hexagon is <math>8 \cdot 6\sqrt{3}=48\sqrt{3}</math>, so the answer is <math>48+3=\boxed{051}</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>~ant08 and sky2025</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>~ant08 and sky2025</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"> </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> </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>== 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=2003|n=II|num-b=13|num-a=15}}</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=2003|n=II|num-b=13|num-a=15}}</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>[[Category:Intermediate Geometry 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 Geometry 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>Ant08