https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_11&feed=atom&action=history
2008 AIME II Problems/Problem 11 - Revision history
2024-03-29T15:58:39Z
Revision history for this page on the wiki
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https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_11&diff=176833&oldid=prev
Jeffersonj at 00:12, 10 August 2022
2022-08-10T00:12:39Z
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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 00:12, 10 August 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l27" >Line 27:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></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">2 </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>first let A = <PCB ; now connect the points as shown in the first solution's diagram ; realise that tanA = r/x = 16/y = r + 16/(x+y) where x = BY and y = CX ( the 2 tangents) ; then we have that QM = 64r = 56 - x - y => (x+y) = 56 - 64r</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>first let A = <PCB ; now connect the points as shown in the first solution's diagram ; realise that tanA = r/x = 16/y = r + 16/(x+y) where x = BY and y = CX ( the 2 tangents) ; then we have that QM = 64r = 56 - x - y => (x+y) = 56 - 64r</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>hence r/x = 16+r/(56-64r) ; now drop altitude AY to solve for tan2A ; now since we know tan2A we know tan A = r/x in terms of r hence solve the resulting equation in r</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>hence r/x = 16+r/(56-64r) ; now drop altitude AY to solve for tan2A ; now since we know tan2A we know tan A = r/x in terms of r hence solve the resulting equation in r</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 <del class="diffchange diffchange-inline">2</del>(pure synthetic) ==</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">3 </ins>(pure synthetic) ==</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>Refer to the above diagram. Let the larger circle have center <math>O_1</math>, the smaller have center <math>O_2</math>, and the incenter be <math>I</math>. We can easily calculate that the area of <math>\triangle ABC = 2688</math>, and <math>s = 128</math> and <math>R = 21</math>, where <math>R</math> is the inradius.  </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>Refer to the above diagram. Let the larger circle have center <math>O_1</math>, the smaller have center <math>O_2</math>, and the incenter be <math>I</math>. We can easily calculate that the area of <math>\triangle ABC = 2688</math>, and <math>s = 128</math> and <math>R = 21</math>, where <math>R</math> is the inradius.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l38" >Line 38:</td>
<td colspan="2" class="diff-lineno">Line 38:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Letting <math>r</math> be the desired radius and letting <math>Y</math> be the projection of <math>O_2</math> onto <math>\overline{BC}</math>, we find that <math>BY = \frac{4r}{3}</math>. Similarly, we find that the distance between the projection from <math>O_1</math> onto <math>\overline{BC}</math>, <math>W</math>, and <math>C</math>, is <math>\frac{64}{3}</math>. From there, we let the projection of <math>O_2</math> onto <math>\overline{O_1W}</math> be <math>Z</math>, and we have <math>O_2Z = 28 - \frac{4r}{3} + \frac{20}{3}</math>, <math>O_1Z = 16 - r</math>, and <math>O_1O_2 = 16 + r</math>. We finish with Pythagoras on <math>\triangle O_1O_2Z</math>, whence we get the desired answer of <math>\boxed{254}</math>. - Spacesam</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>Letting <math>r</math> be the desired radius and letting <math>Y</math> be the projection of <math>O_2</math> onto <math>\overline{BC}</math>, we find that <math>BY = \frac{4r}{3}</math>. Similarly, we find that the distance between the projection from <math>O_1</math> onto <math>\overline{BC}</math>, <math>W</math>, and <math>C</math>, is <math>\frac{64}{3}</math>. From there, we let the projection of <math>O_2</math> onto <math>\overline{O_1W}</math> be <math>Z</math>, and we have <math>O_2Z = 28 - \frac{4r}{3} + \frac{20}{3}</math>, <math>O_1Z = 16 - r</math>, and <math>O_1O_2 = 16 + r</math>. We finish with Pythagoras on <math>\triangle O_1O_2Z</math>, whence we get the desired answer of <math>\boxed{254}</math>. - Spacesam</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 <del class="diffchange diffchange-inline">3 </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>== Solution <ins class="diffchange diffchange-inline">4 </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>Let the incenter be O and the altitude from A to <math>\overline{BC}</math> be T. Note that by AA, <math>\triangle BQY \sim \triangle OBT</math> and <math>\triangle PXC \sim \triangle OTC.</math> Note that from <math>A = rs</math>, the inradius of the big triangle is <math>21</math> Using ravi substitution(or noticing that <math>\overline{AT}</math> is an altitude), we then have that <math>TB = TC = 28.</math> From similar triangles, we can now find <math>\overline{BY}.</math> We have <cmath>\frac{\overline{BY}}{QY} = \frac{7}{{21}} \rightarrow \overline{BY} = \frac{4}{3} r</cmath> Now, note that as in solution 1, drawing the perpendicular from Q to <math>\overline{PX}</math>(call it Z) yields <math>\overline{PZ} = 16 - r, \overline{ZX} = r.</math> Then, from this, <cmath>\overline{QZ} = \overline{YX} = \sqrt{(\overline{PQ})^2 - (\overline{PZ})^2} = \sqrt{(16+r)^2-(16-r)^2} = 8\sqrt{r}</cmath> Using similar similarity as was done to find <math>\overline{BY}</math> we have <math>\frac{\overline{PX}}{\overline{XC}} = \frac{\overline{OT}}{\overline{TC}} \rightarrow \frac{16}{\overline{XC}} = \frac{21}{28} \rightarrow \overline{XC} = \frac{64}{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 the incenter be O and the altitude from A to <math>\overline{BC}</math> be T. Note that by AA, <math>\triangle BQY \sim \triangle OBT</math> and <math>\triangle PXC \sim \triangle OTC.</math> Note that from <math>A = rs</math>, the inradius of the big triangle is <math>21</math> Using ravi substitution(or noticing that <math>\overline{AT}</math> is an altitude), we then have that <math>TB = TC = 28.</math> From similar triangles, we can now find <math>\overline{BY}.</math> We have <cmath>\frac{\overline{BY}}{QY} = \frac{7}{{21}} \rightarrow \overline{BY} = \frac{4}{3} r</cmath> Now, note that as in solution 1, drawing the perpendicular from Q to <math>\overline{PX}</math>(call it Z) yields <math>\overline{PZ} = 16 - r, \overline{ZX} = r.</math> Then, from this, <cmath>\overline{QZ} = \overline{YX} = \sqrt{(\overline{PQ})^2 - (\overline{PZ})^2} = \sqrt{(16+r)^2-(16-r)^2} = 8\sqrt{r}</cmath> Using similar similarity as was done to find <math>\overline{BY}</math> we have <math>\frac{\overline{PX}}{\overline{XC}} = \frac{\overline{OT}}{\overline{TC}} \rightarrow \frac{16}{\overline{XC}} = \frac{21}{28} \rightarrow \overline{XC} = \frac{64}{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;"><div>Now adding all these up and equating them to <math>\overline{BC}</math> yields  </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 adding all these up and equating them to <math>\overline{BC}</math> yields  </div></td></tr>
</table>
Jeffersonj
https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_11&diff=156489&oldid=prev
Awsomek: /* Solution 2(pure synthetic) */
2021-06-20T18:34:38Z
<p><span dir="auto"><span class="autocomment">Solution 2(pure synthetic)</span></span></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 18:34, 20 June 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l37" >Line 37:</td>
<td colspan="2" class="diff-lineno">Line 37:</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>Letting <math>r</math> be the desired radius and letting <math>Y</math> be the projection of <math>O_2</math> onto <math>\overline{BC}</math>, we find that <math>BY = \frac{4r}{3}</math>. Similarly, we find that the distance between the projection from <math>O_1</math> onto <math>\overline{BC}</math>, <math>W</math>, and <math>C</math>, is <math>\frac{64}{3}</math>. From there, we let the projection of <math>O_2</math> onto <math>\overline{O_1W}</math> be <math>Z</math>, and we have <math>O_2Z = 28 - \frac{4r}{3} + \frac{20}{3}</math>, <math>O_1Z = 16 - r</math>, and <math>O_1O_2 = 16 + r</math>. We finish with Pythagoras on <math>\triangle O_1O_2Z</math>, whence we get the desired answer of <math>\boxed{254}</math>. - Spacesam</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>Letting <math>r</math> be the desired radius and letting <math>Y</math> be the projection of <math>O_2</math> onto <math>\overline{BC}</math>, we find that <math>BY = \frac{4r}{3}</math>. Similarly, we find that the distance between the projection from <math>O_1</math> onto <math>\overline{BC}</math>, <math>W</math>, and <math>C</math>, is <math>\frac{64}{3}</math>. From there, we let the projection of <math>O_2</math> onto <math>\overline{O_1W}</math> be <math>Z</math>, and we have <math>O_2Z = 28 - \frac{4r}{3} + \frac{20}{3}</math>, <math>O_1Z = 16 - r</math>, and <math>O_1O_2 = 16 + r</math>. We finish with Pythagoras on <math>\triangle O_1O_2Z</math>, whence we get the desired answer of <math>\boxed{254}</math>. - Spacesam</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">== Solution 3 == </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 the incenter be O and the altitude from A to <math>\overline{BC}</math> be T. Note that by AA, <math>\triangle BQY \sim \triangle OBT</math> and <math>\triangle PXC \sim \triangle OTC.</math> Note that from <math>A = rs</math>, the inradius of the big triangle is <math>21</math> Using ravi substitution(or noticing that <math>\overline{AT}</math> is an altitude), we then have that <math>TB = TC = 28.</math> From similar triangles, we can now find <math>\overline{BY}.</math> We have <cmath>\frac{\overline{BY}}{QY} = \frac{7}{{21}} \rightarrow \overline{BY} = \frac{4}{3} r</cmath> Now, note that as in solution 1, drawing the perpendicular from Q to <math>\overline{PX}</math>(call it Z) yields <math>\overline{PZ} = 16 - r, \overline{ZX} = r.</math> Then, from this, <cmath>\overline{QZ} = \overline{YX} = \sqrt{(\overline{PQ})^2 - (\overline{PZ})^2} = \sqrt{(16+r)^2-(16-r)^2} = 8\sqrt{r}</cmath> Using similar similarity as was done to find <math>\overline{BY}</math> we have <math>\frac{\overline{PX}}{\overline{XC}} = \frac{\overline{OT}}{\overline{TC}} \rightarrow \frac{16}{\overline{XC}} = \frac{21}{28} \rightarrow \overline{XC} = \frac{64}{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;">Now adding all these up and equating them to <math>\overline{BC}</math> yields </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;"><cmath>\frac{4}{3}r + 8\sqrt{r}+ \frac{16}{3} = 56 \rightarrow r = 44 - 6\sqrt{35} \rightarrow 44 + 6\cdot 35 = \boxed{254}</cmath></ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== 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>
Awsomek
https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_11&diff=140080&oldid=prev
Quantomaticguy: /* Solution */
2020-12-21T11:54:58Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 11:54, 21 December 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l25" >Line 25:</td>
<td colspan="2" class="diff-lineno">Line 25:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>By the [[Pythagorean Theorem]], simplification, and the [[quadratic formula]], we can get <math>r = 44 - 6\sqrt {35}</math>, for a final answer of <math>\fbox{254}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>By the [[Pythagorean Theorem]], simplification, and the [[quadratic formula]], we can get <math>r = 44 - 6\sqrt {35}</math>, for a final answer of <math>\fbox{254}</math>.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">== Solution ==</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;">first let A = <PCB ; now connect the points as shown in the first solution's diagram ; realise that tanA = r/x = 16/y = r + 16/(x+y) where x = BY and y = CX ( the 2 tangents) ; then we have that QM = 64r = 56 - x - y => (x+y) = 56 - 64r</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;">hence r/x = 16+r/(56-64r) ; now drop altitude AY to solve for tan2A ; now since we know tan2A we know tan A = r/x in terms of r hence solve the resulting equation in r</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution 2(pure synthetic) ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution 2(pure synthetic) ==</div></td></tr>
</table>
Quantomaticguy
https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_11&diff=109555&oldid=prev
Spacesam at 00:32, 31 August 2019
2019-08-31T00:32:32Z
<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 00:32, 31 August 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l25" >Line 25:</td>
<td colspan="2" class="diff-lineno">Line 25:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>By the [[Pythagorean Theorem]], simplification, and the [[quadratic formula]], we can get <math>r = 44 - 6\sqrt {35}</math>, for a final answer of <math>\fbox{254}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>By the [[Pythagorean Theorem]], simplification, and the [[quadratic formula]], we can get <math>r = 44 - 6\sqrt {35}</math>, for a final answer of <math>\fbox{254}</math>.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">== Solution 2(pure synthetic) ==</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;">Refer to the above diagram. Let the larger circle have center <math>O_1</math>, the smaller have center <math>O_2</math>, and the incenter be <math>I</math>. We can easily calculate that the area of <math>\triangle ABC = 2688</math>, and <math>s = 128</math> and <math>R = 21</math>, where <math>R</math> is the inradius. </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, Line <math>\overline{AI}</math> is the perpendicular bisector of <math>\overline{BC}</math>, as <math>\triangle ABC</math> is isosceles. Letting the point of intersection be <math>X</math>, we get that <math>BX = 28</math> and <math>IX = 21</math>, and <math>B, O_2, I</math> are collinear as <math>O_2</math> is equidistant from <math>\overline{AB}</math> and <math>\overline{BC}</math>. By Pythagoras, <math>BI = 35</math>, and we notice that <math>\triangle BIX</math> is a 3-4-5 right triangle. </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;">Letting <math>r</math> be the desired radius and letting <math>Y</math> be the projection of <math>O_2</math> onto <math>\overline{BC}</math>, we find that <math>BY = \frac{4r}{3}</math>. Similarly, we find that the distance between the projection from <math>O_1</math> onto <math>\overline{BC}</math>, <math>W</math>, and <math>C</math>, is <math>\frac{64}{3}</math>. From there, we let the projection of <math>O_2</math> onto <math>\overline{O_1W}</math> be <math>Z</math>, and we have <math>O_2Z = 28 - \frac{4r}{3} + \frac{20}{3}</math>, <math>O_1Z = 16 - r</math>, and <math>O_1O_2 = 16 + r</math>. We finish with Pythagoras on <math>\triangle O_1O_2Z</math>, whence we get the desired answer of <math>\boxed{254}</math>. - Spacesam</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>
Spacesam
https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_11&diff=91023&oldid=prev
Da7687: /* Solution */
2018-02-11T15:42:21Z
<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 15:42, 11 February 2018</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l7" >Line 7:</td>
<td colspan="2" class="diff-lineno">Line 7:</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>pathpen=black;pointpen=black;pen f=fontsize(9);</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>pathpen=black;pointpen=black;pen f=fontsize(9);</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>real r=44-6*35^.5;</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>real r=44-6*35^.5;</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>pair A=(0,96),B=(-<del class="diffchange diffchange-inline">208</del>,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,P);</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>pair A=(0,96),B=(-<ins class="diffchange diffchange-inline">28</ins>,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,P);</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>path PC=CR(P,16),QC=CR(Q,r);</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>path PC=CR(P,16),QC=CR(Q,r);</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>D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed);</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>D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed);</div></td></tr>
</table>
Da7687
https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_11&diff=91022&oldid=prev
Da7687: /* Solution */
2018-02-11T15:42:07Z
<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 15:42, 11 February 2018</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l7" >Line 7:</td>
<td colspan="2" class="diff-lineno">Line 7:</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>pathpen=black;pointpen=black;pen f=fontsize(9);</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>pathpen=black;pointpen=black;pen f=fontsize(9);</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>real r=44-6*35^.5;</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>real r=44-6*35^.5;</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>pair A=(0,96),B=(-<del class="diffchange diffchange-inline">28</del>,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,<del class="diffchange diffchange-inline">A</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>pair A=(0,96),B=(-<ins class="diffchange diffchange-inline">208</ins>,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,<ins class="diffchange diffchange-inline">P</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>path PC=CR(P,16),QC=CR(Q,r);</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>path PC=CR(P,16),QC=CR(Q,r);</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>D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed);</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>D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed);</div></td></tr>
</table>
Da7687
https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_11&diff=91021&oldid=prev
Da7687: /* Solution */
2018-02-11T15:41:44Z
<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 15:41, 11 February 2018</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l7" >Line 7:</td>
<td colspan="2" class="diff-lineno">Line 7:</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>pathpen=black;pointpen=black;pen f=fontsize(9);</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>pathpen=black;pointpen=black;pen f=fontsize(9);</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>real r=44-6*35^.5;</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>real r=44-6*35^.5;</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>pair A=(0,96),B=(-28,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,<del class="diffchange diffchange-inline">P</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>pair A=(0,96),B=(-28,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,<ins class="diffchange diffchange-inline">A</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>path PC=CR(P,16),QC=CR(Q,r);</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>path PC=CR(P,16),QC=CR(Q,r);</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>D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed);</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>D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed);</div></td></tr>
</table>
Da7687
https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_11&diff=91020&oldid=prev
Da7687: /* Solution */
2018-02-11T15:40:16Z
<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 15:40, 11 February 2018</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l12" >Line 12:</td>
<td colspan="2" class="diff-lineno">Line 12:</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>D(PC); D(QC);  </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>D(PC); D(QC);  </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>MP("A",A,N,f);MP("B",B,f);MP("C",C,f);MP("X",X,f);MP("Y",Y,f);D(MP("P",P,NW,f));D(MP("Q",Q,NW,f));</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>MP("A",A,N,f);MP("B",B,f);MP("C",C,f);MP("X",X,f);MP("Y",Y,f);D(MP("P",P,NW,f));D(MP("Q",Q,NW,f));</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></asy></center</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></asy></center<ins class="diffchange diffchange-inline">></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>Let <math>X</math> and <math>Y</math> be the feet of the [[perpendicular]]s from <math>P</math> and <math>Q</math> to <math>BC</math>, respectively. Let the radius of <math>\odot Q</math> be <math>r</math>. We know that <math>PQ = r + 16</math>. From <math>Q</math> draw segment <math>\overline{QM} \parallel \overline{BC}</math> such that <math>M</math> is on <math>PX</math>. Clearly, <math>QM = XY</math> and <math>PM = 16-r</math>. Also, we know <math>QPM</math> is a right triangle.</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> and <math>Y</math> be the feet of the [[perpendicular]]s from <math>P</math> and <math>Q</math> to <math>BC</math>, respectively. Let the radius of <math>\odot Q</math> be <math>r</math>. We know that <math>PQ = r + 16</math>. From <math>Q</math> draw segment <math>\overline{QM} \parallel \overline{BC}</math> such that <math>M</math> is on <math>PX</math>. Clearly, <math>QM = XY</math> and <math>PM = 16-r</math>. Also, we know <math>QPM</math> is a right triangle.</div></td></tr>
</table>
Da7687
https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_11&diff=91019&oldid=prev
Da7687: /* Solution */
2018-02-11T15:39:38Z
<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 15:39, 11 February 2018</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l12" >Line 12:</td>
<td colspan="2" class="diff-lineno">Line 12:</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>D(PC); D(QC);  </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>D(PC); D(QC);  </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>MP("A",A,N,f);MP("B",B,f);MP("C",C,f);MP("X",X,f);MP("Y",Y,f);D(MP("P",P,NW,f));D(MP("Q",Q,NW,f));</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>MP("A",A,N,f);MP("B",B,f);MP("C",C,f);MP("X",X,f);MP("Y",Y,f);D(MP("P",P,NW,f));D(MP("Q",Q,NW,f));</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></asy></center<del class="diffchange diffchange-inline">></del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div></asy></center</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> and <math>Y</math> be the feet of the [[perpendicular]]s from <math>P</math> and <math>Q</math> to <math>BC</math>, respectively. Let the radius of <math>\odot Q</math> be <math>r</math>. We know that <math>PQ = r + 16</math>. From <math>Q</math> draw segment <math>\overline{QM} \parallel \overline{BC}</math> such that <math>M</math> is on <math>PX</math>. Clearly, <math>QM = XY</math> and <math>PM = 16-r</math>. Also, we know <math>QPM</math> is a right triangle.</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> and <math>Y</math> be the feet of the [[perpendicular]]s from <math>P</math> and <math>Q</math> to <math>BC</math>, respectively. Let the radius of <math>\odot Q</math> be <math>r</math>. We know that <math>PQ = r + 16</math>. From <math>Q</math> draw segment <math>\overline{QM} \parallel \overline{BC}</math> such that <math>M</math> is on <math>PX</math>. Clearly, <math>QM = XY</math> and <math>PM = 16-r</math>. Also, we know <math>QPM</math> is a right triangle.</div></td></tr>
</table>
Da7687
https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_11&diff=81081&oldid=prev
Peggy at 18:45, 13 November 2016
2016-11-13T18:45:44Z
<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 18:45, 13 November 2016</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l16" >Line 16:</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;"><div>Let <math>X</math> and <math>Y</math> be the feet of the [[perpendicular]]s from <math>P</math> and <math>Q</math> to <math>BC</math>, respectively. Let the radius of <math>\odot Q</math> be <math>r</math>. We know that <math>PQ = r + 16</math>. From <math>Q</math> draw segment <math>\overline{QM} \parallel \overline{BC}</math> such that <math>M</math> is on <math>PX</math>. Clearly, <math>QM = XY</math> and <math>PM = 16-r</math>. Also, we know <math>QPM</math> is a right triangle.</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> and <math>Y</math> be the feet of the [[perpendicular]]s from <math>P</math> and <math>Q</math> to <math>BC</math>, respectively. Let the radius of <math>\odot Q</math> be <math>r</math>. We know that <math>PQ = r + 16</math>. From <math>Q</math> draw segment <math>\overline{QM} \parallel \overline{BC}</math> such that <math>M</math> is on <math>PX</math>. Clearly, <math>QM = XY</math> and <math>PM = 16-r</math>. Also, we know <math>QPM</math> is a right triangle.</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>To find <math><del class="diffchange diffchange-inline">XY</del></math>, consider the right triangle <math>PCX</math>. Since <math>\odot P</math> is tangent to <math>\overline{AC},\overline{BC}</math>, then <math>PC</math> [[bisect]]s <math>\angle ACB</math>. Let <math>\angle ACB = 2\theta</math>; then <math>\angle PCX = \angle QBX = \theta</math>. Dropping the altitude from <math>A</math> to <math>BC</math>, we recognize the <math>7 - 24 - 25</math> [[right triangle]], except scaled by <math>4</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>To find <math><ins class="diffchange diffchange-inline">XC</ins></math>, consider the right triangle <math>PCX</math>. Since <math>\odot P</math> is tangent to <math>\overline{AC},\overline{BC}</math>, then <math>PC</math> [[bisect]]s <math>\angle ACB</math>. Let <math>\angle ACB = 2\theta</math>; then <math>\angle PCX = \angle QBX = \theta</math>. Dropping the altitude from <math>A</math> to <math>BC</math>, we recognize the <math>7 - 24 - 25</math> [[right triangle]], except scaled by <math>4</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 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>So we get that <math>\tan(2\theta) = 24/7</math>. From the [[trigonometric identity|half-angle identity]], we find that <math>\tan(\theta) = \frac {3}{4}</math>. Therefore, <math>XC = \frac {64}{3}</math>. By similar reasoning in triangle <math>QBY</math>, we see that <math>BY = \frac {4r}{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>So we get that <math>\tan(2\theta) = 24/7</math>. From the [[trigonometric identity|half-angle identity]], we find that <math>\tan(\theta) = \frac {3}{4}</math>. Therefore, <math>XC = \frac {64}{3}</math>. By similar reasoning in triangle <math>QBY</math>, we see that <math>BY = \frac {4r}{3}</math>.</div></td></tr>
</table>
Peggy