https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_23&feed=atom&action=history1994 AHSME Problems/Problem 23 - Revision history2024-03-28T21:46:04ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_23&diff=154371&oldid=prevLogsobolev: /* Solution 2 */2021-05-28T08:22:23Z<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 08:22, 28 May 2021</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>== Solution 2==</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 2==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></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>Consider the small rectangle between <math>x=3</math> and <math>x=5</math> with area <math>2</math>.  If we exclude that area then the shape becomes a much simpler, its just a rectangle.  In order to offset the exclusion of that area from the shape below the line, we must also exclude a shape with the same area above the line.  We want the remainder after excluding both areas to be  simple, so exclude a rectangle between <math>x=0</math> and <math>x=3</math> and <math>y=3</math> and <math>y=c</math> for some unknown <math>c</math>.  For the area to be the same we need <math>(3-c) \cdot 3 = 2</math>, or <math>c=7/3</math>.  After excluding our two offsetting areas, we're left with a rectangle from <math>x=0</math> to <math>x=3</math> and <math>y=0</math> to <math>y=c</math>.  The area of this region is clearly bisected by its diagonal line.  The slope <del class="diffchange diffchange-inline">of this line </del>is <math>c/3 = 7/9</math> and the answer is <math>\fbox{E}</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>Consider the small rectangle between <math>x=3</math> and <math>x=5</math> with area <math>2</math>.  If we exclude that area then the shape becomes a much simpler, its just a rectangle.  In order to offset the exclusion of that area from the shape below the line, we must also exclude a shape with the same area above the line.  We want the remainder after excluding both areas to be  simple, so exclude a rectangle between <math>x=0</math> and <math>x=3</math> and <math>y=3</math> and <math>y=c</math> for some unknown <math>c</math>.  For the area to be the same we need <math>(3-c) \cdot 3 = 2</math>, or <math>c=7/3</math>.  After excluding our two offsetting areas, we're left with a rectangle from <math>x=0</math> to <math>x=3</math> and <math>y=0</math> to <math>y=c</math>.  The area of this region is clearly bisected by its diagonal line.  The <ins class="diffchange diffchange-inline">line passes through <math>(0,0)</math> and <math>(3,c)</math> so its </ins>slope is <math>c/3 = 7/9</math> and the answer is <math>\fbox{E}</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> </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>==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>Logsobolevhttps://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_23&diff=154363&oldid=prevLogsobolev: /* Solution */2021-05-28T07:35:20Z<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 07:35, 28 May 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l10" >Line 10:</td>
<td colspan="2" class="diff-lineno">Line 10:</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 class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 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 the vertices be <math>A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)</math>. It is easy to see that the line must pass through <math>CD</math>. Let the line intersect <math>CD</math> at the point <math>G=(3,3-x)</math> (i.e. the point <math>x</math> units below <math>C</math>). Since the quadrilateral <math>ABCG</math> and pentagon <math>GDEFA</math> must have the same area, we have the equation <math>3\times\frac{1}{2}\times(x+3)=\frac{1}{2}\times3\times(3-x)+2</math>. This simplifies into <math>3x=2</math>, or <math>x=\frac{2}{3}</math>, so <math>G=(3,\frac{7}{3})</math>. Therefore the slope of the line is <math>\boxed{\textbf{(E)}\ \frac{7}{9}}</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let the vertices be <math>A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)</math>. It is easy to see that the line must pass through <math>CD</math>. Let the line intersect <math>CD</math> at the point <math>G=(3,3-x)</math> (i.e. the point <math>x</math> units below <math>C</math>). Since the quadrilateral <math>ABCG</math> and pentagon <math>GDEFA</math> must have the same area, we have the equation <math>3\times\frac{1}{2}\times(x+3)=\frac{1}{2}\times3\times(3-x)+2</math>. This simplifies into <math>3x=2</math>, or <math>x=\frac{2}{3}</math>, so <math>G=(3,\frac{7}{3})</math>. Therefore the slope of the line is <math>\boxed{\textbf{(E)}\ \frac{7}{9}}</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><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">== Solution 2==</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;">Consider the small rectangle between <math>x=3</math> and <math>x=5</math> with area <math>2</math>.  If we exclude that area then the shape becomes a much simpler, its just a rectangle.  In order to offset the exclusion of that area from the shape below the line, we must also exclude a shape with the same area above the line.  We want the remainder after excluding both areas to be  simple, so exclude a rectangle between <math>x=0</math> and <math>x=3</math> and <math>y=3</math> and <math>y=c</math> for some unknown <math>c</math>.  For the area to be the same we need <math>(3-c) \cdot 3 = 2</math>, or <math>c=7/3</math>.  After excluding our two offsetting areas, we're left with a rectangle from <math>x=0</math> to <math>x=3</math> and <math>y=0</math> to <math>y=c</math>.  The area of this region is clearly bisected by its diagonal line.  The slope of this line is <math>c/3 = 7/9</math> and the answer is <math>\fbox{E}</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;"></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;">==See Also==</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;">{{AHSME box|year=1994|num-b=22|num-a=24}}</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;">{{MAA Notice}}</ins></div></td></tr>
</table>Logsobolevhttps://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_23&diff=75976&oldid=prevEchoz: /* Solution */2016-02-15T17:51:14Z<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-marker" />
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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 17:51, 15 February 2016</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l9" >Line 9:</td>
<td colspan="2" class="diff-lineno">Line 9:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math> \textbf{(A)}\ \frac{2}{7} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{3}{4} \qquad\textbf{(E)}\ \frac{7}{9} </math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math> \textbf{(A)}\ \frac{2}{7} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{3}{4} \qquad\textbf{(E)}\ \frac{7}{9} </math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Let the vertices be <math>A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)</math>. It is easy to see that the line must pass through <math>CD</math>. Let the line intersect <math>CD</math> at the point <math>G=(3,3-x)</math> (i.e. the point <math>x</math> units below <math>C</math>). Since the quadrilateral <math>ABCG</math> and pentagon <math>GDEFA</math> must have the same area, we have the equation <math><del class="diffchange diffchange-inline">\frac{</del>3<del class="diffchange diffchange-inline">}</del>\times\frac{1}{2}\times(x+3)=\frac{1}{2}\times3\times(3-x)+2</math>. This simplifies into <math>3x=2</math>, or <math>x=\frac{2}{3}</math>, so <math>G=(3,\frac{7}{3})</math>. Therefore the slope of the line is <math>\boxed{\textbf{(E)}\ \frac{7}{9}}</math></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Let the vertices be <math>A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)</math>. It is easy to see that the line must pass through <math>CD</math>. Let the line intersect <math>CD</math> at the point <math>G=(3,3-x)</math> (i.e. the point <math>x</math> units below <math>C</math>). Since the quadrilateral <math>ABCG</math> and pentagon <math>GDEFA</math> must have the same area, we have the equation <math>3\times\frac{1}{2}\times(x+3)=\frac{1}{2}\times3\times(3-x)+2</math>. This simplifies into <math>3x=2</math>, or <math>x=\frac{2}{3}</math>, so <math>G=(3,\frac{7}{3})</math>. Therefore the slope of the line is <math>\boxed{\textbf{(E)}\ \frac{7}{9}}</math></div></td></tr>
</table>Echozhttps://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_23&diff=75975&oldid=prevEchoz: /* Solution */2016-02-15T17:50:41Z<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 17:50, 15 February 2016</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l9" >Line 9:</td>
<td colspan="2" class="diff-lineno">Line 9:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math> \textbf{(A)}\ \frac{2}{7} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{3}{4} \qquad\textbf{(E)}\ \frac{7}{9} </math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math> \textbf{(A)}\ \frac{2}{7} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{3}{4} \qquad\textbf{(E)}\ \frac{7}{9} </math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==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;">Let the vertices be <math>A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)</math>. It is easy to see that the line must pass through <math>CD</math>. Let the line intersect <math>CD</math> at the point <math>G=(3,3-x)</math> (i.e. the point <math>x</math> units below <math>C</math>). Since the quadrilateral <math>ABCG</math> and pentagon <math>GDEFA</math> must have the same area, we have the equation <math>\frac{3}\times\frac{1}{2}\times(x+3)=\frac{1}{2}\times3\times(3-x)+2</math>. This simplifies into <math>3x=2</math>, or <math>x=\frac{2}{3}</math>, so <math>G=(3,\frac{7}{3})</math>. Therefore the slope of the line is <math>\boxed{\textbf{(E)}\ \frac{7}{9}}</math></ins></div></td></tr>
</table>Echozhttps://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_23&diff=62383&oldid=prevTheMaskedMagician: Created page with "==Problem== In the <math>xy</math>-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at <math>(0,0), (0,3), (3,3), (3,1), (5,1)</math>..."2014-06-28T02:08:29Z<p>Created page with "==Problem== In the <math>xy</math>-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at <math>(0,0), (0,3), (3,3), (3,1), (5,1)</math>..."</p>
<p><b>New page</b></p><div>==Problem==<br />
In the <math>xy</math>-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at <math>(0,0), (0,3), (3,3), (3,1), (5,1)</math> and <math>(5,0)</math>. The slope of the line through the origin that divides the area of this region exactly in half is<br />
<asy><br />
Label l;<br />
l.p=fontsize(6);<br />
xaxis("$x$",0,6,Ticks(l,1.0,0.5),EndArrow);<br />
yaxis("$y$",0,4,Ticks(l,1.0,0.5),EndArrow);<br />
draw((0,3)--(3,3)--(3,1)--(5,1)--(5,0)--(0,0)--cycle,black+linewidth(2));</asy><br />
<math> \textbf{(A)}\ \frac{2}{7} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{3}{4} \qquad\textbf{(E)}\ \frac{7}{9} </math><br />
==Solution==</div>TheMaskedMagician