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2006 AIME II Problems/Problem 9 - Revision history
2024-03-29T12:34:45Z
Revision history for this page on the wiki
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https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_9&diff=71783&oldid=prev
-- --: /* Solution */
2015-08-26T14:45:42Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
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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 14:45, 26 August 2015</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>The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</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>The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</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>From this information, the slope of each tangent can be uncovered. The slope of <math>t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{<del class="diffchange diffchange-inline">5</del>}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}</math>. The slope of <math>t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{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>From this information, the slope of each tangent can be uncovered. The slope of <math>t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{<ins class="diffchange diffchange-inline">4</ins>}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}</math>. The slope of <math>t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{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>The equation of <math>t_1</math> can be found by substituting the point <math>r (4,0)</math> into <math>y = \frac{1}{\sqrt{15}}x + b</math>, so <math>y = \frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}}</math>. The equation of <math>t_2</math>, found by substituting point <math>s (16,0)</math>, is <math>y = \frac{-1}{\sqrt{3}}x + \frac{16}{\sqrt{3}}</math>. Putting these two equations together results in the desired <math>\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}} </math> <math>\Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}</math> <math>= \frac{76 - 12\sqrt{5}}{4}</math> <math>= 19 - 3\sqrt{5}</math>. Thus, <math>p + q + r = 19 + 3 + 5 \Longrightarrow \boxed{027}</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>The equation of <math>t_1</math> can be found by substituting the point <math>r (4,0)</math> into <math>y = \frac{1}{\sqrt{15}}x + b</math>, so <math>y = \frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}}</math>. The equation of <math>t_2</math>, found by substituting point <math>s (16,0)</math>, is <math>y = \frac{-1}{\sqrt{3}}x + \frac{16}{\sqrt{3}}</math>. Putting these two equations together results in the desired <math>\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}} </math> <math>\Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}</math> <math>= \frac{76 - 12\sqrt{5}}{4}</math> <math>= 19 - 3\sqrt{5}</math>. Thus, <math>p + q + r = 19 + 3 + 5 \Longrightarrow \boxed{027}</math>.</div></td></tr>
</table>
-- --
https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_9&diff=69439&oldid=prev
Antler: /* Solution */
2015-03-24T00:20:05Z
<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" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 00:20, 24 March 2015</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l5" >Line 5:</td>
<td colspan="2" class="diff-lineno">Line 5:</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>[[Image:2006_II_AIME-9.png]]</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>[[Image:2006_II_AIME-9.png]]</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>Call the centers <math>O_1, O_2, O_3</math>, the points of tangency <math>r_1, r_2, s_1, s_2</math> (with <math>r</math> on <math>t_1</math> and <math>s</math> on <math>t_2</math>, and <math>s_2</math> on <math>\mathcal{C}_2</math>), and the intersection of each common internal tangent to the [[X-axis]] <math>r, s</math>.<!--Not very creative point names ..--> <math>\triangle O_1r_1r \sim \triangle O_2r_2r</math> since both triangles have a [[right angle]] and have vertical angles, and the same goes for <math>\triangle O_2s_2s \sim \triangle O_3s_1s</math>. By [[proportion|proportionality]], we find that <math>O_1r = 4</math>; solving <math>\triangle O_1r_1r</math> by the [[Pythagorean theorem]] yields <math>r_1r = \sqrt{15}</math>. On <math>\mathcal{C}_3</math>, we can do the same thing to get <math>O_3s_1 = 4</math> and <math>s_1s = 4\sqrt{<del class="diffchange diffchange-inline">15</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>Call the centers <math>O_1, O_2, O_3</math>, the points of tangency <math>r_1, r_2, s_1, s_2</math> (with <math>r</math> on <math>t_1</math> and <math>s</math> on <math>t_2</math>, and <math>s_2</math> on <math>\mathcal{C}_2</math>), and the intersection of each common internal tangent to the [[X-axis]] <math>r, s</math>.<!--Not very creative point names ..--> <math>\triangle O_1r_1r \sim \triangle O_2r_2r</math> since both triangles have a [[right angle]] and have vertical angles, and the same goes for <math>\triangle O_2s_2s \sim \triangle O_3s_1s</math>. By [[proportion|proportionality]], we find that <math>O_1r = 4</math>; solving <math>\triangle O_1r_1r</math> by the [[Pythagorean theorem]] yields <math>r_1r = \sqrt{15}</math>. On <math>\mathcal{C}_3</math>, we can do the same thing to get <math>O_3s_1 = 4</math> and <math>s_1s = 4\sqrt{<ins class="diffchange diffchange-inline">3</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>The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</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>The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</math>.  </div></td></tr>
</table>
Antler
https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_9&diff=69438&oldid=prev
Antler: /* Solution */
2015-03-24T00:19:30Z
<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 00:19, 24 March 2015</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l5" >Line 5:</td>
<td colspan="2" class="diff-lineno">Line 5:</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>[[Image:2006_II_AIME-9.png]]</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>[[Image:2006_II_AIME-9.png]]</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>Call the centers <math>O_1, O_2, O_3</math>, the points of tangency <math>r_1, r_2, s_1, s_2</math> (with <math>r</math> on <math>t_1</math> and <math>s</math> on <math>t_2</math>, and <math>s_2</math> on <math>\mathcal{C}_2</math>), and the intersection of each common internal tangent to the [[X-axis]] <math>r, s</math>.<!--Not very creative point names ..--> <math>\triangle O_1r_1r \sim \triangle O_2r_2r</math> since both triangles have a [[right angle]] and have vertical angles, and the same goes for <math>\triangle O_2s_2s \sim \triangle O_3s_1s</math>. By [[proportion|proportionality]], we find that <math>O_1r = 4</math>; solving <math>\triangle O_1r_1r</math> by the [[Pythagorean theorem]] yields <math>r_1r = \sqrt{15}</math>. On <math>\mathcal{C}_3</math>, we can do the same thing to get <math>O_3s_1 = <del class="diffchange diffchange-inline">8</del></math> and <math>s_1s = 4\sqrt{<del class="diffchange diffchange-inline">3</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>Call the centers <math>O_1, O_2, O_3</math>, the points of tangency <math>r_1, r_2, s_1, s_2</math> (with <math>r</math> on <math>t_1</math> and <math>s</math> on <math>t_2</math>, and <math>s_2</math> on <math>\mathcal{C}_2</math>), and the intersection of each common internal tangent to the [[X-axis]] <math>r, s</math>.<!--Not very creative point names ..--> <math>\triangle O_1r_1r \sim \triangle O_2r_2r</math> since both triangles have a [[right angle]] and have vertical angles, and the same goes for <math>\triangle O_2s_2s \sim \triangle O_3s_1s</math>. By [[proportion|proportionality]], we find that <math>O_1r = 4</math>; solving <math>\triangle O_1r_1r</math> by the [[Pythagorean theorem]] yields <math>r_1r = \sqrt{15}</math>. On <math>\mathcal{C}_3</math>, we can do the same thing to get <math>O_3s_1 = <ins class="diffchange diffchange-inline">4</ins></math> and <math>s_1s = 4\sqrt{<ins class="diffchange diffchange-inline">15</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>The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</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>The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</math>.  </div></td></tr>
</table>
Antler
https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_9&diff=66420&oldid=prev
Ryanyz10 at 16:53, 30 November 2014
2014-11-30T16:53:39Z
<p></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 16:53, 30 November 2014</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l5" >Line 5:</td>
<td colspan="2" class="diff-lineno">Line 5:</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>[[Image:2006_II_AIME-9.png]]</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>[[Image:2006_II_AIME-9.png]]</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>Call the centers <math>O_1, O_2, O_3</math>, the points of tangency <math>r_1, r_2, s_1, s_2</math> (with <math>r</math> on <math>t_1</math> and <math>s</math> on <math>t_2</math>, and <math>s_2</math> on <math>\mathcal{C}_2</math>), and the intersection of each common internal tangent to the [[<del class="diffchange diffchange-inline">x </del>axis]] <math>r, s</math>.<!--Not very creative point names ..--> <math>\triangle O_1r_1r \sim \triangle O_2r_2r</math> since both triangles have a [[right angle]] and have vertical angles, and the same goes for <math>\triangle O_2s_2s \sim \triangle O_3s_1s</math>. By [[proportion|proportionality]], we find that <math>O_1r = 4</math>; solving <math>\triangle O_1r_1r</math> by the [[Pythagorean theorem]] yields <math>r_1r = \sqrt{15}</math>. On <math>\mathcal{C}_3</math>, we can do the same thing to get <math>O_3s_1 = 8</math> and <math>s_1s = 4\sqrt{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>Call the centers <math>O_1, O_2, O_3</math>, the points of tangency <math>r_1, r_2, s_1, s_2</math> (with <math>r</math> on <math>t_1</math> and <math>s</math> on <math>t_2</math>, and <math>s_2</math> on <math>\mathcal{C}_2</math>), and the intersection of each common internal tangent to the [[<ins class="diffchange diffchange-inline">X-</ins>axis]] <math>r, s</math>.<!--Not very creative point names ..--> <math>\triangle O_1r_1r \sim \triangle O_2r_2r</math> since both triangles have a [[right angle]] and have vertical angles, and the same goes for <math>\triangle O_2s_2s \sim \triangle O_3s_1s</math>. By [[proportion|proportionality]], we find that <math>O_1r = 4</math>; solving <math>\triangle O_1r_1r</math> by the [[Pythagorean theorem]] yields <math>r_1r = \sqrt{15}</math>. On <math>\mathcal{C}_3</math>, we can do the same thing to get <math>O_3s_1 = 8</math> and <math>s_1s = 4\sqrt{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>The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</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>The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</math>.  </div></td></tr>
</table>
Ryanyz10
https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_9&diff=66419&oldid=prev
Ryanyz10 at 16:52, 30 November 2014
2014-11-30T16:52:59Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 16:52, 30 November 2014</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l5" >Line 5:</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>[[Image:2006_II_AIME-9.png]]</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>[[Image:2006_II_AIME-9.png]]</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>Call the centers <math>O_1, O_2, O_3</math>, the points of tangency <math>r_1, r_2, s_1, s_2</math> (with <math>r</math> on <math>t_1</math> and <math>s</math> on <math>t_2</math>, and <math>s_2</math> on <math>\mathcal{C}_2</math>), and the intersection of each common internal tangent to the [[x axis]] <math>r, s</math>.<!--Not very creative point names ..--> <math>\triangle O_1r_1r \sim \triangle O_2r_2r</math> since both triangles have a [[right angle]] and have <del class="diffchange diffchange-inline">[[</del>vertical <del class="diffchange diffchange-inline">angle]]s</del>, and the same goes for <math>\triangle O_2s_2s \sim \triangle O_3s_1s</math>. By [[proportion|proportionality]], we find that <math>O_1r = 4</math>; solving <math>\triangle O_1r_1r</math> by the [[Pythagorean theorem]] yields <math>r_1r = \sqrt{15}</math>. On <math>\mathcal{C}_3</math>, we can do the same thing to get <math>O_3s_1 = 8</math> and <math>s_1s = 4\sqrt{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>Call the centers <math>O_1, O_2, O_3</math>, the points of tangency <math>r_1, r_2, s_1, s_2</math> (with <math>r</math> on <math>t_1</math> and <math>s</math> on <math>t_2</math>, and <math>s_2</math> on <math>\mathcal{C}_2</math>), and the intersection of each common internal tangent to the [[x axis]] <math>r, s</math>.<!--Not very creative point names ..--> <math>\triangle O_1r_1r \sim \triangle O_2r_2r</math> since both triangles have a [[right angle]] and have vertical <ins class="diffchange diffchange-inline">angles</ins>, and the same goes for <math>\triangle O_2s_2s \sim \triangle O_3s_1s</math>. By [[proportion|proportionality]], we find that <math>O_1r = 4</math>; solving <math>\triangle O_1r_1r</math> by the [[Pythagorean theorem]] yields <math>r_1r = \sqrt{15}</math>. On <math>\mathcal{C}_3</math>, we can do the same thing to get <math>O_3s_1 = 8</math> and <math>s_1s = 4\sqrt{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>The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</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>The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</math>.  </div></td></tr>
</table>
Ryanyz10
https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_9&diff=66418&oldid=prev
Ryanyz10 at 16:51, 30 November 2014
2014-11-30T16:51:02Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 16:51, 30 November 2014</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l11" >Line 11:</td>
<td colspan="2" class="diff-lineno">Line 11:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>From this information, the slope of each tangent can be uncovered. The slope of <math>t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{5}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}</math>. The slope of <math>t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{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>From this information, the slope of each tangent can be uncovered. The slope of <math>t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{5}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}</math>. The slope of <math>t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{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>The equation of <math>t_1</math> can be found by substituting the point <math>r (4,0)</math> into <math>y = \frac{1}{\sqrt{15}}x + b</math>, so <math>y = \frac{1}{\sqrt{15}} - \frac{4}{\sqrt{15}}</math>. The equation of <math>t_2</math>, found by substituting point <math>s (16,0)</math>, is <math>y = \frac{-1}{\sqrt{3}} + \frac{16}{\sqrt{3}}</math>. Putting these two equations together results in the desired <math>\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}} </math> <math>\Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}</math> <math>= \frac{76 - 12\sqrt{5}}{4}</math> <math>= 19 - 3\sqrt{5}</math>. Thus, <math>p + q + r = 19 + 3 + 5 \Longrightarrow \boxed{027}</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>The equation of <math>t_1</math> can be found by substituting the point <math>r (4,0)</math> into <math>y = \frac{1}{\sqrt{15}}x + b</math>, so <math>y = \frac{1}{\sqrt{15}}<ins class="diffchange diffchange-inline">x </ins>- \frac{4}{\sqrt{15}}</math>. The equation of <math>t_2</math>, found by substituting point <math>s (16,0)</math>, is <math>y = \frac{-1}{\sqrt{3}}<ins class="diffchange diffchange-inline">x </ins>+ \frac{16}{\sqrt{3}}</math>. Putting these two equations together results in the desired <math>\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}} </math> <math>\Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}</math> <math>= \frac{76 - 12\sqrt{5}}{4}</math> <math>= 19 - 3\sqrt{5}</math>. Thus, <math>p + q + r = 19 + 3 + 5 \Longrightarrow \boxed{027}</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>
Ryanyz10
https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_9&diff=61162&oldid=prev
Anthony Cheng: /* Solution */
2014-03-24T17:31:01Z
<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 17:31, 24 March 2014</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l11" >Line 11:</td>
<td colspan="2" class="diff-lineno">Line 11:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>From this information, the slope of each tangent can be uncovered. The slope of <math>t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{5}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}</math>. The slope of <math>t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{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>From this information, the slope of each tangent can be uncovered. The slope of <math>t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{5}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}</math>. The slope of <math>t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{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>The equation of <math>t_1</math> can be found by substituting the point <math>r (4,0)</math> into <math>y = \frac{1}{\sqrt{15}}x + b</math>, so <math>y = \frac{1}{\sqrt{15}} - \frac{4}{\sqrt{15}}</math>. The equation of <math>t_2</math>, found by substituting point <math>s (16,0)</math>, is <math>y = \frac{-1}{\sqrt{3}} + \frac{16}{\sqrt{3}}</math>. Putting these two equations together results in the desired <math>\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}} </math> <math>\Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}</math> <math>= \frac{76 - 12\sqrt{5}}{4}</math> <math>= 19 - 3\sqrt{5}</math>. Thus, <math>p + q + r = 19 + 3 + 5 \Longrightarrow 027</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>The equation of <math>t_1</math> can be found by substituting the point <math>r (4,0)</math> into <math>y = \frac{1}{\sqrt{15}}x + b</math>, so <math>y = \frac{1}{\sqrt{15}} - \frac{4}{\sqrt{15}}</math>. The equation of <math>t_2</math>, found by substituting point <math>s (16,0)</math>, is <math>y = \frac{-1}{\sqrt{3}} + \frac{16}{\sqrt{3}}</math>. Putting these two equations together results in the desired <math>\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}} </math> <math>\Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}</math> <math>= \frac{76 - 12\sqrt{5}}{4}</math> <math>= 19 - 3\sqrt{5}</math>. Thus, <math>p + q + r = 19 + 3 + 5 \Longrightarrow <ins class="diffchange diffchange-inline">\boxed{</ins>027<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 class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 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>
Anthony Cheng
https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_9&diff=55426&oldid=prev
Nathan wailes at 03:25, 5 July 2013
2013-07-05T03:25:22Z
<p></p>
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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 03:25, 5 July 2013</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l19" >Line 19:</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>[[Category:Intermediate Trigonometry 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 Trigonometry Problems]]</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Intermediate Algebra 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 Algebra 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>
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Nathan wailes
https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_9&diff=19030&oldid=prev
Azjps: wik, long eqn breakup
2007-10-28T22:31:05Z
<p>wik, long eqn breakup</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 22:31, 28 October 2007</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
<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><del class="diffchange diffchange-inline">Circles </del><math> \mathcal{C}_1, \mathcal{C}_2, </math> and <math> \mathcal{C}_3 </math> have their <del class="diffchange diffchange-inline">centers </del>at (0,0), (12,0), and (24,0), and have radii 1, 2, and 4, respectively. Line <math> t_1 </math> is a common internal tangent to <math>\mathcal{C}_1</math> and <math>\mathcal{C}_2</math> and has a positive slope, and line <math>t_2</math> is a common internal tangent to <math>\mathcal{C}_2</math> and <math>\mathcal{C}_3</math> and has a negative slope. Given that lines <math>t_1</math> and <math>t_2</math> intersect at <math>(x,y),</math> and that <math>x=p-q\sqrt{r},</math> where <math>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime, find <math>p+q+r.</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">[[Circle]]s </ins><math> \mathcal{C}_1, \mathcal{C}_2, </math> and <math> \mathcal{C}_3 </math> have their <ins class="diffchange diffchange-inline">[[center]]s </ins>at (0,0), (12,0), and (24,0), and have <ins class="diffchange diffchange-inline">[[radius|</ins>radii<ins class="diffchange diffchange-inline">]] </ins>1, 2, and 4, respectively. Line <math> t_1 </math> is a common internal <ins class="diffchange diffchange-inline">[[</ins>tangent<ins class="diffchange diffchange-inline">]] </ins>to <math>\mathcal{C}_1</math> and <math>\mathcal{C}_2</math> and has a positive <ins class="diffchange diffchange-inline">[[</ins>slope<ins class="diffchange diffchange-inline">]]</ins>, and line <math>t_2</math> is a common internal tangent to <math>\mathcal{C}_2</math> and <math>\mathcal{C}_3</math> and has a negative slope. Given that lines <math>t_1</math> and <math>t_2</math> intersect at <math>(x,y),</math> and that <math>x=p-q\sqrt{r},</math> where <math>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime, find <math>p+q+r.</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 colspan="2" class="diff-lineno" id="mw-diff-left-l11" >Line 11:</td>
<td colspan="2" class="diff-lineno">Line 11:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>From this information, the slope of each tangent can be uncovered. The slope of <math>t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{5}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}</math>. The slope of <math>t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{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>From this information, the slope of each tangent can be uncovered. The slope of <math>t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{5}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}</math>. The slope of <math>t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{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>The equation of <math>t_1</math> can be found by substituting the point <math>r (4,0)</math> into <math>y = \frac{1}{\sqrt{15}}x + b</math>, so <math>y = \frac{1}{\sqrt{15}} - \frac{4}{\sqrt{15}}</math>. The equation of <math>t_2</math>, found by substituting point <math>s (16,0)</math>, is <math>y = \frac{-1}{\sqrt{3}} + \frac{16}{\sqrt{3}}</math>. Putting these two equations together results in the desired <math>\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}} \Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1} = \frac{76 - 12\sqrt{5}}{4} = 19 - 3\sqrt{5}</math>. Thus, <math>p + q + r = 19 + 3 + 5 \Longrightarrow 027</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>The equation of <math>t_1</math> can be found by substituting the point <math>r (4,0)</math> into <math>y = \frac{1}{\sqrt{15}}x + b</math>, so <math>y = \frac{1}{\sqrt{15}} - \frac{4}{\sqrt{15}}</math>. The equation of <math>t_2</math>, found by substituting point <math>s (16,0)</math>, is <math>y = \frac{-1}{\sqrt{3}} + \frac{16}{\sqrt{3}}</math>. Putting these two equations together results in the desired <math>\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}} <ins class="diffchange diffchange-inline"></math> <math></ins>\Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}<ins class="diffchange diffchange-inline"></math> <math></ins>= \frac{76 - 12\sqrt{5}}{4}<ins class="diffchange diffchange-inline"></math> <math></ins>= 19 - 3\sqrt{5}</math>. Thus, <math>p + q + r = 19 + 3 + 5 \Longrightarrow 027</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>
<tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l19" >Line 19:</td>
<td colspan="2" class="diff-lineno">Line 19:</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;"><div>[[Category:Intermediate Algebra 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 Algebra Problems]]</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">[[Category:Intermediate Number Theory Problems]]</del></div></td><td colspan="2"> </td></tr>
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Azjps
https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_9&diff=16978&oldid=prev
Azjps at 23:08, 25 September 2007
2007-09-25T23:08:54Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:08, 25 September 2007</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
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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">#REDIRECT </del>[[2006 <del class="diffchange diffchange-inline">AIME A </del>Problems<del class="diffchange diffchange-inline">/Problem 9</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">== Problem ==</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">Circles <math> \mathcal{C}_1, \mathcal{C}_2, </math> and <math> \mathcal{C}_3 </math> have their centers at (0,0), (12,0), and (24,0), and have radii 1, 2, and 4, respectively. Line <math> t_1 </math> is a common internal tangent to <math>\mathcal{C}_1</math> and <math>\mathcal{C}_2</math> and has a positive slope, and line <math>t_2</math> is a common internal tangent to <math>\mathcal{C}_2</math> and <math>\mathcal{C}_3</math> and has a negative slope. Given that lines <math>t_1</math> and <math>t_2</math> intersect at <math>(x,y),</math> and that <math>x=p-q\sqrt{r},</math> where <math>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime, find <math>p+q+r.</math></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">== Solution ==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>[[<ins class="diffchange diffchange-inline">Image:2006_II_AIME-9.png]]</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Call the centers <math>O_1, O_2, O_3</math>, the points of tangency <math>r_1, r_2, s_1, s_2</math> (with <math>r</math> on <math>t_1</math> and <math>s</math> on <math>t_2</math>, and <math>s_2</math> on <math>\mathcal{C}_2</math>), and the intersection of each common internal tangent to the [[x axis]] <math>r, s</math>.<!--Not very creative point names ..--> <math>\triangle O_1r_1r \sim \triangle O_2r_2r</math> since both triangles have a [[right angle]] and have [[vertical angle]]s, and the same goes for <math>\triangle O_2s_2s \sim \triangle O_3s_1s</math>. By [[proportion|proportionality]], we find that <math>O_1r = 4</math>; solving <math>\triangle O_1r_1r</math> by the [[Pythagorean theorem]] yields <math>r_1r = \sqrt{15}</math>. On <math>\mathcal{C}_3</math>, we can do the same thing to get <math>O_3s_1 = 8</math> and <math>s_1s = 4\sqrt{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> </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">The vertical [[altitude]] of each of <math>\triangle O_1r_1r</math> and <math>\triangle O_3s_1s</math> can each by found by the formula <math>c \cdot h = a \cdot b</math> (as both products equal twice of the area of the triangle). Thus, the respective heights are <math>\frac{\sqrt{15}}{4}</math> and <math>2\sqrt{3}</math>. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: <math>\sqrt{15 - \frac{15}{16}} = \frac{15}{4}</math>, and by 30-60-90: <math>6</math>. </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">From this information, the slope of each tangent can be uncovered. The slope of <math>t_1 = \frac{\Delta y}{\Delta x} = \frac{\frac{\sqrt{15}}{5}}{\frac{15}{4}} = \frac{1}{\sqrt{15}}</math>. The slope of <math>t_2 = -\frac{2\sqrt{3}}{6} = -\frac{1}{\sqrt{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> </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">The equation of <math>t_1</math> can be found by substituting the point <math>r (4,0)</math> into <math>y = \frac{1}{\sqrt{15}}x + b</math>, so <math>y = \frac{1}{\sqrt{15}} - \frac{4}{\sqrt{15}}</math>. The equation of <math>t_2</math>, found by substituting point <math>s (16,0)</math>, is <math>y = \frac{-1}{\sqrt{3}} + \frac{16}{\sqrt{3}}</math>. Putting these two equations together results in the desired <math>\frac{1}{\sqrt{15}}x - \frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{3}}x + \frac{16}{\sqrt{3}} \Longrightarrow x = \frac{16\sqrt{5} + 4}{\sqrt{5} + 1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1} = \frac{76 - 12\sqrt{5}}{4} = 19 - 3\sqrt{5}</math>. Thus, <math>p + q + r = 19 + 3 + 5 \Longrightarrow 027</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">== 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 class="diffchange diffchange-inline">{{AIME box|year=</ins>2006<ins class="diffchange diffchange-inline">|n=II|num-b=8|num-a=10}}</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">[[Category:Intermediate Geometry Problems]]</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">[[Category:Intermediate Trigonometry Problems]]</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">[[Category:Intermediate Algebra Problems]]</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">[[Category:Intermediate Number Theory </ins>Problems]]</div></td></tr>
</table>
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