https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_11&feed=atom&action=history2004 AIME I Problems/Problem 11 - Revision history2024-03-28T18:44:52ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_11&diff=73751&oldid=prevJktkops: /* Solution 2 */2015-12-17T02:12:06Z<p><span dir="auto"><span class="autocomment">Solution 2</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 02:12, 17 December 2015</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l13" >Line 13:</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>Our original solid <math>V</math> has [[surface area]] <math>A_v = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone. Using the [[Pythagorean Theorem]] or Pythagorean Triple knowledge, we obtain <math>\ell = 5</math> and lateral area <math>A_\ell = 15\pi</math>. The area of the base is <math>A_B = 3^2\pi = 9\pi</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>Our original solid <math>V</math> has [[surface area]] <math>A_v = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone. Using the [[Pythagorean Theorem]] or Pythagorean Triple knowledge, we obtain <math>\ell = 5</math> and lateral area <math>A_\ell = 15\pi</math>. The area of the base is <math>A_B = 3^2\pi = 9\pi</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>V</math> and <math>C</math> are similar cones, because the plane that cut out <math>C</math> was parallel to the base of <math>V</math>. Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>\frac{A_c}{A_f}</math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8 - 5 x^2} = k</math> and the ratio of the volumes <math>\frac{V_c}{V_f}</math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8<del class="diffchange diffchange-inline">\pi </del>- 5 x^2} = \frac{x^3}{1 - x^3}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math>V</math> and <math>C</math> are similar cones, because the plane that cut out <math>C</math> was parallel to the base of <math>V</math>. Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>\frac{A_c}{A_f}</math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8 - 5 x^2} = k</math> and the ratio of the volumes <math>\frac{V_c}{V_f}</math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8 - 5 x^2} = \frac{x^3}{1 - x^3}</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>Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{387}{512}} = \frac{125}{387}</math>. And so we have <math>m + n = 125 + 387 = \boxed{512}</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>Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{387}{512}} = \frac{125}{387}</math>. And so we have <math>m + n = 125 + 387 = \boxed{512}</math>.</div></td></tr>
</table>Jktkopshttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_11&diff=73750&oldid=prevJktkops: /* Solution 2 */2015-12-17T02:11:31Z<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
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<col class="diff-marker" />
<col class="diff-content" />
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 02:11, 17 December 2015</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l13" >Line 13:</td>
<td colspan="2" class="diff-lineno">Line 13:</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>Our original solid <math>V</math> has [[surface area]] <math>A_v = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone. Using the [[Pythagorean Theorem]] or Pythagorean Triple knowledge, we obtain <math>\ell = 5</math> and lateral area <math>A_\ell = 15\pi</math>. The area of the base is <math>A_B = 3^2\pi = 9\pi</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>Our original solid <math>V</math> has [[surface area]] <math>A_v = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone. Using the [[Pythagorean Theorem]] or Pythagorean Triple knowledge, we obtain <math>\ell = 5</math> and lateral area <math>A_\ell = 15\pi</math>. The area of the base is <math>A_B = 3^2\pi = 9\pi</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>V</math> and <math>C</math> are similar cones, because the plane that cut out <math>C</math> was parallel to the base of <math>V</math>. Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>\frac{A_c}{A_f}</math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8<del class="diffchange diffchange-inline">\pi </del>- 5 x^2} = k</math> and the ratio of the volumes <math>\frac{V_c}{V_f}</math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8\pi - 5 x^2} = \frac{x^3}{1 - x^3}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math>V</math> and <math>C</math> are similar cones, because the plane that cut out <math>C</math> was parallel to the base of <math>V</math>. Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>\frac{A_c}{A_f}</math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8 - 5 x^2} = k</math> and the ratio of the volumes <math>\frac{V_c}{V_f}</math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8\pi - 5 x^2} = \frac{x^3}{1 - x^3}</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>Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{387}{512}} = \frac{125}{387}</math>. And so we have <math>m + n = 125 + 387 = \boxed{512}</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>Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{387}{512}} = \frac{125}{387}</math>. And so we have <math>m + n = 125 + 387 = \boxed{512}</math>.</div></td></tr>
</table>Jktkopshttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_11&diff=73749&oldid=prevJktkops: /* Solution 2 */2015-12-17T02:10:15Z<p><span dir="auto"><span class="autocomment">Solution 2</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 02:10, 17 December 2015</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l13" >Line 13:</td>
<td colspan="2" class="diff-lineno">Line 13:</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>Our original solid <math>V</math> has [[surface area]] <math>A_v = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone. Using the [[Pythagorean Theorem]] or Pythagorean Triple knowledge, we obtain <math>\ell = 5</math> and lateral area <math>A_\ell = 15\pi</math>. The area of the base is <math>A_B = 3^2\pi = 9\pi</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>Our original solid <math>V</math> has [[surface area]] <math>A_v = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone. Using the [[Pythagorean Theorem]] or Pythagorean Triple knowledge, we obtain <math>\ell = 5</math> and lateral area <math>A_\ell = 15\pi</math>. The area of the base is <math>A_B = 3^2\pi = 9\pi</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>V</math> and <math>C</math> are similar cones, because the plane that cut out <math>C</math> was parallel to the base of <math>V</math>. Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>A_c<del class="diffchange diffchange-inline"></math> to <math></del>A_f</math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8\pi - 5 x^2} = k</math> and the ratio of the volumes <math>V_c<del class="diffchange diffchange-inline"></math> to <math></del>V_f</math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8\pi - 5 x^2} = \frac{x^3}{1 - x^3}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math>V</math> and <math>C</math> are similar cones, because the plane that cut out <math>C</math> was parallel to the base of <math>V</math>. Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math><ins class="diffchange diffchange-inline">\frac{</ins>A_c<ins class="diffchange diffchange-inline">}{</ins>A_f<ins class="diffchange diffchange-inline">}</ins></math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8\pi - 5 x^2} = k</math> and the ratio of the volumes <math><ins class="diffchange diffchange-inline">\frac{</ins>V_c<ins class="diffchange diffchange-inline">}{</ins>V_f<ins class="diffchange diffchange-inline">}</ins></math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8\pi - 5 x^2} = \frac{x^3}{1 - x^3}</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>Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{387}{512}} = \frac{125}{387}</math>. And so we have <math>m + n = 125 + 387 = \boxed{512}</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>Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{387}{512}} = \frac{125}{387}</math>. And so we have <math>m + n = 125 + 387 = \boxed{512}</math>.</div></td></tr>
</table>Jktkopshttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_11&diff=73748&oldid=prevJktkops: /* Solution 2 */2015-12-17T02:08:32Z<p><span dir="auto"><span class="autocomment">Solution 2</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 02:08, 17 December 2015</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l13" >Line 13:</td>
<td colspan="2" class="diff-lineno">Line 13:</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>Our original solid <math>V</math> has [[surface area]] <math>A_v = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone. Using the [[Pythagorean Theorem]] or Pythagorean Triple knowledge, we obtain <math>\ell = 5</math> and lateral area <math>A_\ell = 15\pi</math>. The area of the base is <math>A_B = 3^2\pi = 9\pi</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>Our original solid <math>V</math> has [[surface area]] <math>A_v = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone. Using the [[Pythagorean Theorem]] or Pythagorean Triple knowledge, we obtain <math>\ell = 5</math> and lateral area <math>A_\ell = 15\pi</math>. The area of the base is <math>A_B = 3^2\pi = 9\pi</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>A_c</math> to <math>A_f</math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8\pi - 5 x^2} = k</math> and the ratio of the volumes <math>V_c</math> to <math>V_f</math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8\pi - 5 x^2} = \frac{x^3}{1 - x^3}</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><ins class="diffchange diffchange-inline"><math>V</math> and <math>C</math> are similar cones, because the plane that cut out <math>C</math> was parallel to the base of <math>V</math>. </ins>Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>A_c</math> to <math>A_f</math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8\pi - 5 x^2} = k</math> and the ratio of the volumes <math>V_c</math> to <math>V_f</math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8\pi - 5 x^2} = \frac{x^3}{1 - x^3}</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>Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{387}{512}} = \frac{125}{387}</math>. And so we have <math>m + n = 125 + 387 = \boxed{512}</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>Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{387}{512}} = \frac{125}{387}</math>. And so we have <math>m + n = 125 + 387 = \boxed{512}</math>.</div></td></tr>
</table>Jktkopshttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_11&diff=73747&oldid=prevJktkops: /* Solution 2 */2015-12-17T01:57:46Z<p><span dir="auto"><span class="autocomment">Solution 2</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 01:57, 17 December 2015</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l15" >Line 15:</td>
<td colspan="2" class="diff-lineno">Line 15:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>A_c</math> to <math>A_f</math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8\pi - 5 x^2} = k</math> and the ratio of the volumes <math>V_c</math> to <math>V_f</math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8\pi - 5 x^2} = \frac{x^3}{1 - x^3}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>A_c</math> to <math>A_f</math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8\pi - 5 x^2} = k</math> and the ratio of the volumes <math>V_c</math> to <math>V_f</math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8\pi - 5 x^2} = \frac{x^3}{1 - x^3}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{<del class="diffchange diffchange-inline">218</del>}{512}} = \frac{125}{387}</math>. And so we have <math>m + n = 125 + 387 = \boxed{512}</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>Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{<ins class="diffchange diffchange-inline">387</ins>}{512}} = \frac{125}{387}</math>. And so we have <math>m + n = 125 + 387 = \boxed{512}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
</table>Jktkopshttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_11&diff=73746&oldid=prevJktkops: /* Solution 2 */2015-12-17T01:57:07Z<p><span dir="auto"><span class="autocomment">Solution 2</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 01:57, 17 December 2015</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l15" >Line 15:</td>
<td colspan="2" class="diff-lineno">Line 15:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>A_c</math> to <math>A_f</math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8\pi - 5 x^2} = k</math> and the ratio of the volumes <math>V_c</math> to <math>V_f</math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8\pi - 5 x^2} = \frac{x^3}{1 - x^3}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>A_c</math> to <math>A_f</math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8\pi - 5 x^2} = k</math> and the ratio of the volumes <math>V_c</math> to <math>V_f</math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8\pi - 5 x^2} = \frac{x^3}{1 - x^3}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{<del class="diffchange diffchange-inline">343</del>}}{\frac{218}{<del class="diffchange diffchange-inline">343</del>}} = \frac{125}{<del class="diffchange diffchange-inline">218</del>}</math>. And so we have <math>m + n = 125 + <del class="diffchange diffchange-inline">218 </del>= \boxed{<del class="diffchange diffchange-inline">343</del>}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{<ins class="diffchange diffchange-inline">512</ins>}}{\frac{218}{<ins class="diffchange diffchange-inline">512</ins>}} = \frac{125}{<ins class="diffchange diffchange-inline">387</ins>}</math>. And so we have <math>m + n = 125 + <ins class="diffchange diffchange-inline">387 </ins>= \boxed{<ins class="diffchange diffchange-inline">512</ins>}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
</table>Jktkopshttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_11&diff=73745&oldid=prevJktkops: /* Solution */2015-12-17T01:55:24Z<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 01:55, 17 December 2015</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l2" >Line 2:</td>
<td colspan="2" class="diff-lineno">Line 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>A [[solid]] in the shape of a right circular [[cone]] is 4 inches tall and its base has a 3-inch radius. The entire [[surface]] of the cone, including its base, is painted. A [[plane]] [[parallel]] to the base of the cone divides the cone into two solids, a smaller cone-shaped solid <math> C </math> and a [[frustum]]-shaped solid <math> F, </math> in such a way that the [[ratio]] between the [[area]]s of the painted surfaces of <math> C </math> and <math> F </math> and the ratio between the [[volume]]s of <math> C </math> and <math> F </math> are both equal to <math> k</math>. Given that <math> k=\frac m n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </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>A [[solid]] in the shape of a right circular [[cone]] is 4 inches tall and its base has a 3-inch radius. The entire [[surface]] of the cone, including its base, is painted. A [[plane]] [[parallel]] to the base of the cone divides the cone into two solids, a smaller cone-shaped solid <math> C </math> and a [[frustum]]-shaped solid <math> F, </math> in such a way that the [[ratio]] between the [[area]]s of the painted surfaces of <math> C </math> and <math> F </math> and the ratio between the [[volume]]s of <math> C </math> and <math> F </math> are both equal to <math> k</math>. Given that <math> k=\frac m n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>== Solution <ins class="diffchange diffchange-inline"> ==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">===Solution 1=</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>Our original solid has volume equal to <math>V = \frac13 \pi r^2 h = \frac13 \pi 3^2\cdot 4 = 12 \pi</math> and has [[surface area]] <math>A = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone.  Using the [[Pythagorean Theorem]], we get <math>\ell = 5</math> and <math>A = 24\pi</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>Our original solid has volume equal to <math>V = \frac13 \pi r^2 h = \frac13 \pi 3^2\cdot 4 = 12 \pi</math> and has [[surface area]] <math>A = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone.  Using the [[Pythagorean Theorem]], we get <math>\ell = 5</math> and <math>A = 24\pi</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>Let <math>x</math> denote the [[radius]] of the small cone. Let <math>A_c</math> and <math>A_f</math> denote the area of the painted surface on cone <math>C</math> and frustum <math>F</math>, respectively, and let <math>V_c</math> and <math>V_f</math> denote the volume of cone <math>C</math> and frustum <math>F</math>, respectively.  Because the plane cut is parallel to the base of our solid, <math>C</math> is [[similar]] to the uncut solid and so the height and slant height of cone <math>C</math> are <math>\frac{4}{3}x</math> and <math>\frac{5}{3}x</math>, respectively. Using the formula for lateral surface area of a cone, we find that <math>A_c=\frac{1}{2}c\cdot \ell=\frac{1}{2}(2\pi x)\left(\frac{5}{3}x\right)=\frac{5}{3}\pi x^2</math>. By subtracting <math>A_c</math> from the surface area of the original solid, we find that <math>A_f=24\pi - \frac{5}{3}\pi x^2</math>.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>x</math> denote the [[radius]] of the small cone. Let <math>A_c</math> and <math>A_f</math> denote the area of the painted surface on cone <math>C</math> and frustum <math>F</math>, respectively, and let <math>V_c</math> and <math>V_f</math> denote the volume of cone <math>C</math> and frustum <math>F</math>, respectively.  Because the plane cut is parallel to the base of our solid, <math>C</math> is [[similar]] to the uncut solid and so the height and slant height of cone <math>C</math> are <math>\frac{4}{3}x</math> and <math>\frac{5}{3}x</math>, respectively. Using the formula for lateral surface area of a cone, we find that <math>A_c=\frac{1}{2}c\cdot \ell=\frac{1}{2}(2\pi x)\left(\frac{5}{3}x\right)=\frac{5}{3}\pi x^2</math>. By subtracting <math>A_c</math> from the surface area of the original solid, we find that <math>A_f=24\pi - \frac{5}{3}\pi x^2</math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Next, we can calculate <math>V_c=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi x^2 \left(\frac{4}{3}x\right)=\frac{4}{9}\pi x^3</math>.  Finally, we subtract <math>V_c</math> from the volume of the original cone to find that <math>V_f=12\pi - \frac{4}{9}\pi x^3</math>. We know that <math>\frac{A_c}{A_f}=\frac{V_c}{V_f}=k.</math> Plugging in our values for <math>A_c</math>, <math>A_f</math>, <math>V_c</math>, and <math>V_f</math>, we obtain the equation <math>\frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}=\frac{\frac{4}{9}\pi x^3}{12\pi - \frac{4}{9}\pi x^3}</math>.  We can take [[reciprocal]]s of both sides to simplify this [[equation]] to <math>\frac{72}{5x^2} - 1 = \frac{27}{x^3} - 1</math> and so <math>x = \frac{15}{8}</math>.  Then <math>k = \frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}= \frac{125}{387} = \frac mn</math> so the answer is <math>m+n=125+387=512</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>Next, we can calculate <math>V_c=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi x^2 \left(\frac{4}{3}x\right)=\frac{4}{9}\pi x^3</math>.  Finally, we subtract <math>V_c</math> from the volume of the original cone to find that <math>V_f=12\pi - \frac{4}{9}\pi x^3</math>. We know that <math>\frac{A_c}{A_f}=\frac{V_c}{V_f}=k.</math> Plugging in our values for <math>A_c</math>, <math>A_f</math>, <math>V_c</math>, and <math>V_f</math>, we obtain the equation <math>\frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}=\frac{\frac{4}{9}\pi x^3}{12\pi - \frac{4}{9}\pi x^3}</math>.  We can take [[reciprocal]]s of both sides to simplify this [[equation]] to <math>\frac{72}{5x^2} - 1 = \frac{27}{x^3} - 1</math> and so <math>x = \frac{15}{8}</math>.  Then <math>k = \frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}= \frac{125}{387} = \frac mn</math> so the answer is <math>m+n=125+387=<ins class="diffchange diffchange-inline">\boxed{</ins>512<ins class="diffchange diffchange-inline">}</ins></math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td 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;">Our original solid <math>V</math> has [[surface area]] <math>A_v = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone. Using the [[Pythagorean Theorem]] or Pythagorean Triple knowledge, we obtain <math>\ell = 5</math> and lateral area <math>A_\ell = 15\pi</math>. The area of the base is <math>A_B = 3^2\pi = 9\pi</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;">Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>A_c</math> to <math>A_f</math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8\pi - 5 x^2} = k</math> and the ratio of the volumes <math>V_c</math> to <math>V_f</math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8\pi - 5 x^2} = \frac{x^3}{1 - x^3}</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;">Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{343}}{\frac{218}{343}} = \frac{125}{218}</math>. And so we have <math>m + n = 125 + 218 = \boxed{343}</math>.</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== 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>Jktkopshttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_11&diff=55180&oldid=prevNathan wailes at 00:01, 5 July 20132013-07-05T00:01:52Z<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>[[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 colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">{{MAA Notice}}</ins></div></td></tr>
</table>Nathan waileshttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_11&diff=25651&oldid=prevI like pie: AIME box2008-04-27T20:20:25Z<p>AIME box</p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 20:20, 27 April 2008</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>== 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="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>2004 <del class="diffchange diffchange-inline">AIME </del>I <del class="diffchange diffchange-inline">Problems/Problem </del>10| <del class="diffchange diffchange-inline">Previous problem]]</del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">{{AIME box|year=</ins>2004<ins class="diffchange diffchange-inline">|n=</ins>I<ins class="diffchange diffchange-inline">|num-b=</ins>10|<ins class="diffchange diffchange-inline">num-a=</ins>12<ins class="diffchange diffchange-inline">}}</ins></div></td></tr>
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<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">* [[2004 AIME I Problems/Problem </del>12<del class="diffchange diffchange-inline">| Next problem]]</del></div></td><td colspan="2"> </td></tr>
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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>[[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>
</table>I like piehttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_11&diff=11900&oldid=prevJBL at 23:45, 30 November 20062006-11-30T23:45:10Z<p></p>
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<td colspan="2" class="diff-lineno">Line 1:</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>== Problem ==</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>== Problem ==</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>A solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid <math> C </math> and a frustum-shaped solid <math> F, </math> in such a way that the ratio between the <del class="diffchange diffchange-inline">areas </del>of the painted surfaces of <math> C </math> and <math> F </math> and the ratio between the <del class="diffchange diffchange-inline">volumes </del>of <math> C </math> and <math> F </math> are both equal to <math> k<del class="diffchange diffchange-inline">. </del></math> Given that <math> k=m<del class="diffchange diffchange-inline">/</del>n, </math> where <math> m </math> and <math> n </math> are relatively prime positive <del class="diffchange diffchange-inline">integers</del>, find <math> m+n. </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>A <ins class="diffchange diffchange-inline">[[</ins>solid<ins class="diffchange diffchange-inline">]] </ins>in the shape of a right circular <ins class="diffchange diffchange-inline">[[</ins>cone<ins class="diffchange diffchange-inline">]] </ins>is 4 inches tall and its base has a 3-inch radius. The entire <ins class="diffchange diffchange-inline">[[</ins>surface<ins class="diffchange diffchange-inline">]] </ins>of the cone, including its base, is painted. A <ins class="diffchange diffchange-inline">[[</ins>plane<ins class="diffchange diffchange-inline">]] [[</ins>parallel<ins class="diffchange diffchange-inline">]] </ins>to the base of the cone divides the cone into two solids, a smaller cone-shaped solid <math> C </math> and a <ins class="diffchange diffchange-inline">[[</ins>frustum<ins class="diffchange diffchange-inline">]]</ins>-shaped solid <math> F, </math> in such a way that the <ins class="diffchange diffchange-inline">[[</ins>ratio<ins class="diffchange diffchange-inline">]] </ins>between the <ins class="diffchange diffchange-inline">[[area]]s </ins>of the painted surfaces of <math> C </math> and <math> F </math> and the ratio between the <ins class="diffchange diffchange-inline">[[volume]]s </ins>of <math> C </math> and <math> F </math> are both equal to <math> k</math><ins class="diffchange diffchange-inline">. </ins>Given that <math> k=<ins class="diffchange diffchange-inline">\frac </ins>m n, </math> where <math> m </math> and <math> n </math> are <ins class="diffchange diffchange-inline">[[</ins>relatively prime<ins class="diffchange diffchange-inline">]] [[</ins>positive <ins class="diffchange diffchange-inline">integer]]s</ins>, find <math> m+n. </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>== 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 <math>x</math> denote the radius of the small cone. Let <math>A_c</math> and <math>A_f</math> denote the area of the painted surface on cone <math>C</math> and frustum <math>F</math>, respectively<del class="diffchange diffchange-inline">. Let </del><math>V_c</math> and <math>V_f</math> denote the volume of cone <math>C</math> and frustum <math>F</math>, respectively. <del class="diffchange diffchange-inline">Using </del>the <del class="diffchange diffchange-inline">Pythagorean Theorem</del>, <del class="diffchange diffchange-inline">we find that </del>the height and slant height of cone <math>C</math> are <math>\frac{4}{3}x</math> and <math>\frac{5}{3}x</math>, respectively. Using the formula for lateral surface area of a cone, we find that <math>A_c=\frac{1}{2}c\cdot \ell=\frac{1}{2}(2\pi x)\left(\frac{5}{3}x\right)=\frac{5}{3}\pi x^2</math>. By subtracting <math>A_c</math> from the <del class="diffchange diffchange-inline">lateral </del>surface area of the original <del class="diffchange diffchange-inline">cone and adding to this the are of the base</del>, we find that <math>A_f=<del class="diffchange diffchange-inline">12</del>\pi - \frac{<del class="diffchange diffchange-inline">4</del>}{<del class="diffchange diffchange-inline">9</del>}\pi x^<del class="diffchange diffchange-inline">3</del></math>. Next, we <del class="diffchange diffchange-inline">find that </del><math>V_c=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi x^2 \left(\frac{4}{3}x\right)=\frac{4}{9}\pi x^3<del class="diffchange diffchange-inline">.</del></math> Finally, we subtract <math>V_c</math> from the volume of the original cone to find that <math>V_f=12\pi - \frac{4}{9}\pi x^3</math>. We know that <math>\frac{A_c}{A_f}=\frac{V_c}{V_f}=k.</math> Plugging in our values for <math>A_c</math>, <math>A_f</math>, <math>V_c</math>, and <math>V_f</math>, we obtain the equation <math>\frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}=\frac{\frac{4}{9}\pi x^3}{12\pi - \frac{4}{9}\pi x^3}<del class="diffchange diffchange-inline">.</del></math> We <del class="diffchange diffchange-inline">solve for </del><math>x</math> <del class="diffchange diffchange-inline">to obtain </del><math>x=\frac{15}{8}</math> <del class="diffchange diffchange-inline">(Note that we have disregarded the trivial solution </del><math>x<del class="diffchange diffchange-inline">=0</math>). We plug this value of <math></del>x<del class="diffchange diffchange-inline"></math> into our expression for <math>k</math> and simplify to obtain <math>k</del>=\frac{125}{387}</math><del class="diffchange diffchange-inline">. Hence </del><math>m+n=125+387=512</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><ins class="diffchange diffchange-inline">Our original solid has volume equal to <math>V = \frac13 \pi r^2 h = \frac13 \pi 3^2\cdot 4 = 12 \pi</math> and has [[surface area]] <math>A = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone.  Using the [[Pythagorean Theorem]], we get <math>\ell = 5</math> and <math>A = 24\pi</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Let <math>x</math> denote the <ins class="diffchange diffchange-inline">[[</ins>radius<ins class="diffchange diffchange-inline">]] </ins>of the small cone. Let <math>A_c</math> and <math>A_f</math> denote the area of the painted surface on cone <math>C</math> and frustum <math>F</math>, respectively<ins class="diffchange diffchange-inline">, and let </ins><math>V_c</math> and <math>V_f</math> denote the volume of cone <math>C</math> and frustum <math>F</math>, respectively. <ins class="diffchange diffchange-inline"> Because the plane cut is parallel to </ins>the <ins class="diffchange diffchange-inline">base of our solid</ins>, <ins class="diffchange diffchange-inline"><math>C</math> is [[similar]] to the uncut solid and so </ins>the height and slant height of cone <math>C</math> are <math>\frac{4}{3}x</math> and <math>\frac{5}{3}x</math>, respectively. Using the formula for lateral surface area of a cone, we find that <math>A_c=\frac{1}{2}c\cdot \ell=\frac{1}{2}(2\pi x)\left(\frac{5}{3}x\right)=\frac{5}{3}\pi x^2</math>. By subtracting <math>A_c</math> from the surface area of the original <ins class="diffchange diffchange-inline">solid</ins>, we find that <math>A_f=<ins class="diffchange diffchange-inline">24</ins>\pi - \frac{<ins class="diffchange diffchange-inline">5</ins>}{<ins class="diffchange diffchange-inline">3</ins>}\pi x^<ins class="diffchange diffchange-inline">2</ins></math>.  </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Next, we <ins class="diffchange diffchange-inline">can calculate </ins><math>V_c=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi x^2 \left(\frac{4}{3}x\right)=\frac{4}{9}\pi x^3</math><ins class="diffchange diffchange-inline">.  </ins>Finally, we subtract <math>V_c</math> from the volume of the original cone to find that <math>V_f=12\pi - \frac{4}{9}\pi x^3</math>. We know that <math>\frac{A_c}{A_f}=\frac{V_c}{V_f}=k.</math> Plugging in our values for <math>A_c</math>, <math>A_f</math>, <math>V_c</math>, and <math>V_f</math>, we obtain the equation <math>\frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}=\frac{\frac{4}{9}\pi x^3}{12\pi - \frac{4}{9}\pi x^3}</math><ins class="diffchange diffchange-inline">.  </ins>We <ins class="diffchange diffchange-inline">can take [[reciprocal]]s of both sides to simplify this [[equation]] to </ins><math><ins class="diffchange diffchange-inline">\frac{72}{5x^2} - 1 = \frac{27}{</ins>x<ins class="diffchange diffchange-inline">^3} - 1</ins></math> <ins class="diffchange diffchange-inline">and so </ins><math>x = \frac{15}{8}</math><ins class="diffchange diffchange-inline">.  Then </ins><math><ins class="diffchange diffchange-inline">k = \frac{\frac{5}{3}\pi </ins>x<ins class="diffchange diffchange-inline">^2}{24\pi - \frac{5}{3}\pi </ins>x<ins class="diffchange diffchange-inline">^2}</ins>= \frac{125}{387} <ins class="diffchange diffchange-inline">= \frac mn</ins></math> <ins class="diffchange diffchange-inline">so the answer is </ins><math>m+n=125+387=512</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">Solution provided by 1337h4x.</del></div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== 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 colspan="2" class="diff-lineno" id="mw-diff-left-l13" >Line 13:</td>
<td colspan="2" class="diff-lineno">Line 16:</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>* [[2004 AIME I 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>* [[2004 AIME I Problems]]</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;">[[Category:Intermediate Geometry Problems]]</ins></div></td></tr>
</table>JBL