https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&feed=atom&action=history2011 USAJMO Problems/Problem 3 - Revision history2024-03-28T15:18:51ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=143790&oldid=prevTigerzhang: /* Work in progress: Solution 4 (Clean algebra) */2021-01-29T17:37:30Z<p><span dir="auto"><span class="autocomment">Work in progress: Solution 4 (Clean algebra)</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 17:37, 29 January 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l86" >Line 86:</td>
<td colspan="2" class="diff-lineno">Line 86:</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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath> We add these equations to get <cmath>\sqrt{3}(3+4(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}))=0.</cmath> We solve for <math>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}</math> to get <cmath>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=-\frac{3}{4}.</cmath> So, the y-coordinate of <math>\triangle</math> is <math>-\frac{1}{4}</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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath> We add these equations to get <cmath>\sqrt{3}(3+4(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}))=0.</cmath> We solve for <math>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}</math> to get <cmath>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=-\frac{3}{4}.</cmath> So, the y-coordinate of <math>\triangle</math> is <math>-\frac{1}{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="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>We will prove that the x-coordinate of <math>\triangle</math> can be any real number. If <math>2a_{1}</math> tends to infinity, then <math>2a_{2}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{3}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily large. Similarly, if we let <math>2a_{3}</math> tend to negative infinity, then <math>2a_{1}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{2}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily small. Since <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> is continuous, it can take any real value. So, the locus is <del class="diffchange diffchange-inline">all points on </del>the line <math>y=-\frac{1}{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>We will prove that the x-coordinate of <math>\triangle</math> can be any real number. If <math>2a_{1}</math> tends to infinity, then <math>2a_{2}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{3}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily large. Similarly, if we let <math>2a_{3}</math> tend to negative infinity, then <math>2a_{1}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{2}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily small. Since <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> is continuous, it can take any real value. So, the locus is the line <math>y=-\frac{1}{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>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{USAJMO newbox|year=2011|num-b=2|num-a=4}}</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>{{USAJMO newbox|year=2011|num-b=2|num-a=4}}</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td></tr>
</table>Tigerzhanghttps://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=143788&oldid=prevTigerzhang: /* Work in progress: Solution 4 (Clean algebra) */2021-01-29T17:36:26Z<p><span dir="auto"><span class="autocomment">Work in progress: Solution 4 (Clean algebra)</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 17:36, 29 January 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l84" >Line 84:</td>
<td colspan="2" class="diff-lineno">Line 84:</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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. Remember that the range of <math>\arctan(x)</math> is <math>(-90^{\circ},90^{\circ})</math>. We have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. Remember that the range of <math>\arctan(x)</math> is <math>(-90^{\circ},90^{\circ})</math>. We have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath> We add these equations to get <cmath>\sqrt{3}(3+4(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}))=0.</cmath> We solve for <math>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}</math> to get <cmath>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=-\frac{3}{4}</cmath><del class="diffchange diffchange-inline">. </del>So, the y-coordinate of <math>\triangle</math> is <math>-\frac{1}{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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath> We add these equations to get <cmath>\sqrt{3}(3+4(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}))=0.</cmath> We solve for <math>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}</math> to get <cmath>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=-\frac{3}{4}<ins class="diffchange diffchange-inline">.</ins></cmath> So, the y-coordinate of <math>\triangle</math> is <math>-\frac{1}{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="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>We will prove that the x-coordinate of <math>\triangle</math> can be any real number. If <math>2a_{1}</math> tends to infinity, then <math>2a_{2}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{3}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily large. Similarly, if we let <math>2a_{3}</math> tend to infinity, then <math>2a_{1}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{2}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily small. Since <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> is continuous, it can take any real value. So, the locus is all points on the line <math>y=-\frac{1}{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>We will prove that the x-coordinate of <math>\triangle</math> can be any real number. If <math>2a_{1}</math> tends to infinity, then <math>2a_{2}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{3}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily large. Similarly, if we let <math>2a_{3}</math> tend to <ins class="diffchange diffchange-inline">negative </ins>infinity, then <math>2a_{1}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{2}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily small. Since <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> is continuous, it can take any real value. So, the locus is all points on the line <math>y=-\frac{1}{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>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{USAJMO newbox|year=2011|num-b=2|num-a=4}}</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>{{USAJMO newbox|year=2011|num-b=2|num-a=4}}</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td></tr>
</table>Tigerzhanghttps://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=143784&oldid=prevTigerzhang: /* Work in progress: Solution 4 (Clean algebra) */2021-01-29T17:34:43Z<p><span dir="auto"><span class="autocomment">Work in progress: Solution 4 (Clean algebra)</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 17:34, 29 January 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l86" >Line 86:</td>
<td colspan="2" class="diff-lineno">Line 86:</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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath> We add these equations to get <cmath>\sqrt{3}(3+4(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}))=0.</cmath> We solve for <math>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}</math> to get <cmath>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=-\frac{3}{4}</cmath>. So, the y-coordinate of <math>\triangle</math> is <math>-\frac{1}{4}</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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath> We add these equations to get <cmath>\sqrt{3}(3+4(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}))=0.</cmath> We solve for <math>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}</math> to get <cmath>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=-\frac{3}{4}</cmath>. So, the y-coordinate of <math>\triangle</math> is <math>-\frac{1}{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="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>We will prove that the x-coordinate of <math>\triangle</math> can be <del class="diffchange diffchange-inline">anything</del>. If <math>2a_{1}</math> tends to infinity, then <math>2a_{2}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{3}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily large. Similarly, if we let <math>2a_{3}</math> tend to infinity, then <math>2a_{1}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{2}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily small. Since <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> is continuous, it can take any real value. So, the locus is all points on the line <math>y=-\frac{1}{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>We will prove that the x-coordinate of <math>\triangle</math> can be <ins class="diffchange diffchange-inline">any real number</ins>. If <math>2a_{1}</math> tends to infinity, then <math>2a_{2}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{3}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily large. Similarly, if we let <math>2a_{3}</math> tend to infinity, then <math>2a_{1}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{2}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily small. Since <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> is continuous, it can take any real value. So, the locus is all points on the line <math>y=-\frac{1}{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>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{USAJMO newbox|year=2011|num-b=2|num-a=4}}</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>{{USAJMO newbox|year=2011|num-b=2|num-a=4}}</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td></tr>
</table>Tigerzhanghttps://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=143781&oldid=prevTigerzhang: /* Work in progress: Solution 4 (Clean algebra) */2021-01-29T17:24:42Z<p><span dir="auto"><span class="autocomment">Work in progress: Solution 4 (Clean algebra)</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 17:24, 29 January 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l84" >Line 84:</td>
<td colspan="2" class="diff-lineno">Line 84:</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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. Remember that the range of <math>\arctan(x)</math> is <math>(-90^{\circ},90^{\circ})</math>. We have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. Remember that the range of <math>\arctan(x)</math> is <math>(-90^{\circ},90^{\circ})</math>. We have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath> We add these equations to get <cmath>\sqrt{3}(3+4(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}))=0.</cmath> <del class="diffchange diffchange-inline">Rearranging, we </del>get <<del class="diffchange diffchange-inline">math</del>>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=-\frac{3}{4}</<del class="diffchange diffchange-inline">math</del>>. So, the y-coordinate of <math>\triangle</math> is <math>-\frac{1}{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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath> We add these equations to get <cmath>\sqrt{3}(3+4(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}))=0.</cmath> <ins class="diffchange diffchange-inline">We solve for <math>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}</math> to </ins>get <<ins class="diffchange diffchange-inline">cmath</ins>>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=-\frac{3}{4}</<ins class="diffchange diffchange-inline">cmath</ins>>. So, the y-coordinate of <math>\triangle</math> is <math>-\frac{1}{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>We will prove that the x-coordinate of <math>\triangle</math> can be anything. If <math>2a_{1}</math> tends to infinity, then <math>2a_{2}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{3}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily large. Similarly, if we let <math>2a_{3}</math> tend to infinity, then <math>2a_{1}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{2}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily small. Since <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> is continuous, it can take any real value. So, the locus is all points on the line <math>y=-\frac{1}{4}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>We will prove that the x-coordinate of <math>\triangle</math> can be anything. If <math>2a_{1}</math> tends to infinity, then <math>2a_{2}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{3}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily large. Similarly, if we let <math>2a_{3}</math> tend to infinity, then <math>2a_{1}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{2}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily small. Since <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> is continuous, it can take any real value. So, the locus is all points on the line <math>y=-\frac{1}{4}</math>.</div></td></tr>
</table>Tigerzhanghttps://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=143778&oldid=prevTigerzhang: /* Work in progress: Solution 4 (Clean algebra) */2021-01-29T17:22:47Z<p><span dir="auto"><span class="autocomment">Work in progress: Solution 4 (Clean algebra)</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 17:22, 29 January 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l84" >Line 84:</td>
<td colspan="2" class="diff-lineno">Line 84:</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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. Remember that the range of <math>\arctan(x)</math> is <math>(-90^{\circ},90^{\circ})</math>. We have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. Remember that the range of <math>\arctan(x)</math> is <math>(-90^{\circ},90^{\circ})</math>. We have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath></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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath> <ins class="diffchange diffchange-inline">We add these equations to get <cmath>\sqrt{3}(3+4(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}))=0.</cmath> Rearranging, we get <math>a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=-\frac{3}{4}</math>. So, the y-coordinate of <math>\triangle</math> is <math>-\frac{1}{4}</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">We will prove that the x-coordinate of <math>\triangle</math> can be anything. If <math>2a_{1}</math> tends to infinity, then <math>2a_{2}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{3}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily large. Similarly, if we let <math>2a_{3}</math> tend to infinity, then <math>2a_{1}</math> tends to <math>\frac{\sqrt{3}}{2}</math> and <math>2a_{2}</math> tends to <math>-\frac{\sqrt{3}}{2}</math>. So, <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> can be arbitrarily small. Since <math>\frac{a_{1}+a_{2}+a_{3}}{3}</math> is continuous, it can take any real value. So, the locus is all points on the line <math>y=-\frac{1}{4}</math>.</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
<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>{{USAJMO newbox|year=2011|num-b=2|num-a=4}}</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>{{USAJMO newbox|year=2011|num-b=2|num-a=4}}</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td></tr>
</table>Tigerzhanghttps://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=143468&oldid=prevTigerzhang: /* Work in progress: Solution 4 (Clean algebra) */2021-01-27T21:08:48Z<p><span dir="auto"><span class="autocomment">Work in progress: Solution 4 (Clean algebra)</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 21:08, 27 January 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l82" >Line 82:</td>
<td colspan="2" class="diff-lineno">Line 82:</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>It can be easily shown that the center of <math>\triangle</math> has coordinates <math>\left(\frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}}{3}\right)</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>It can be easily shown that the center of <math>\triangle</math> has coordinates <math>\left(\frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}}{3}\right)</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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. Remember that the range of <math>\arctan</math> is <math>(-90^{\circ},90^{\circ})</math>. We have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. Remember that the range of <math>\arctan<ins class="diffchange diffchange-inline">(x)</ins></math> is <math>(-90^{\circ},90^{\circ})</math>. We have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath></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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath></div></td></tr>
</table>Tigerzhanghttps://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=143466&oldid=prevTigerzhang: /* Work in progress: Solution 4 (Clean algebra) */2021-01-27T21:06:15Z<p><span dir="auto"><span class="autocomment">Work in progress: Solution 4 (Clean algebra)</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 21:06, 27 January 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l82" >Line 82:</td>
<td colspan="2" class="diff-lineno">Line 82:</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>It can be easily shown that the center of <math>\triangle</math> has coordinates <math>\left(\frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}}{3}\right)</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>It can be easily shown that the center of <math>\triangle</math> has coordinates <math>\left(\frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}}{3}\right)</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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. Remember that the range of <math>\arctan</math> is <math>(-90,90)</math>. We have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. Remember that the range of <math>\arctan</math> is <math>(-90<ins class="diffchange diffchange-inline">^{\circ}</ins>,90<ins class="diffchange diffchange-inline">^{\circ}</ins>)</math>. We have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath></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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath></div></td></tr>
</table>Tigerzhanghttps://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=143465&oldid=prevTigerzhang: /* Work in progress: Solution 4 (Clean algebra) */2021-01-27T21:05:44Z<p><span dir="auto"><span class="autocomment">Work in progress: Solution 4 (Clean algebra)</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 21:05, 27 January 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l82" >Line 82:</td>
<td colspan="2" class="diff-lineno">Line 82:</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>It can be easily shown that the center of <math>\triangle</math> has coordinates <math>\left(\frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}}{3}\right)</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>It can be easily shown that the center of <math>\triangle</math> has coordinates <math>\left(\frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}}{3}\right)</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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. <del class="diffchange diffchange-inline">(</del>Remember that the range of <math>\arctan</math> is <math>(-90,90)</math>.<del class="diffchange diffchange-inline">) So, we </del>have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. Remember that the range of <math>\arctan</math> is <math>(-90,90)</math>. <ins class="diffchange diffchange-inline">We </ins>have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath></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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath></div></td></tr>
</table>Tigerzhanghttps://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=143464&oldid=prevTigerzhang: /* Work in progress: Solution 4 (Clean algebra) */2021-01-27T21:05:12Z<p><span dir="auto"><span class="autocomment">Work in progress: Solution 4 (Clean algebra)</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 21:05, 27 January 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l82" >Line 82:</td>
<td colspan="2" class="diff-lineno">Line 82:</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>It can be easily shown that the center of <math>\triangle</math> has coordinates <math>\left(\frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}}{3}\right)</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>It can be easily shown that the center of <math>\triangle</math> has coordinates <math>\left(\frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}}{3}\right)</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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. So, we have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. <ins class="diffchange diffchange-inline">(Remember that the range of <math>\arctan</math> is <math>(-90,90)</math>.) </ins>So, we have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath></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>Taking the tangent of both sides of each equation and rearranging, we get <cmath>\begin{align*}2a_{1}-2a_{2}&=\sqrt{3}(1+4a_{1}a_{2})\\2a_{2}-2a_{3}&=\sqrt{3}(1+4a_{2}a_{3})\\2a_{3}-2a_{1}&=\sqrt{3}(1+4a_{3}a_{1}).\end{align*}</cmath></div></td></tr>
</table>Tigerzhanghttps://artofproblemsolving.com/wiki/index.php?title=2011_USAJMO_Problems/Problem_3&diff=143462&oldid=prevTigerzhang: /* Work in progress: Solution 4 (Clean algebra) */2021-01-27T21:00:43Z<p><span dir="auto"><span class="autocomment">Work in progress: Solution 4 (Clean algebra)</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 21:00, 27 January 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l80" >Line 80:</td>
<td colspan="2" class="diff-lineno">Line 80:</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></asy></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></asy></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>It can be shown that the center of <math>\triangle</math> has coordinates <math>\left(\frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}}{3}\right)</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>It can be <ins class="diffchange diffchange-inline">easily </ins>shown that the center of <math>\triangle</math> has coordinates <math>\left(\frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}}{3}\right)</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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. So, we have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></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>Without loss of generality, let <math>a_{1}>a_{2}>a_{3}</math>. Notice that <math>\ell(P_2)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_1)</math>, <math>\ell(P_3)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_2)</math>, and <math>\ell(P_1)</math> is a <math>60^{\circ}</math> clockwise rotation of <math>\ell(P_3)</math>. By definition, <math>\arctan(2a_{i})</math> is the (directed) angle from the x-axis to <math>\ell(P_{i})</math>. So, we have <cmath>\begin{align*}\arctan(2a_{1})-\arctan(2a_{2})&=60^{\circ}\\\arctan(2a_{2})-\arctan(2a_{3})&=60^{\circ}\\\arctan(2a_{3})-\arctan(2a_{1})&=-120^{\circ}.\end{align*}</cmath></div></td></tr>
</table>Tigerzhang