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2004 JBMO Problems/Problem 2 - Revision history
2024-03-19T01:15:34Z
Revision history for this page on the wiki
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https://artofproblemsolving.com/wiki/index.php?title=2004_JBMO_Problems/Problem_2&diff=99518&oldid=prev
Olivera: /* Solution */
2018-12-18T16:02:58Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 16:02, 18 December 2018</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;"><div>==Solution==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Let length of side <math>CB<del class="diffchange diffchange-inline"></math> </del>= <del class="diffchange diffchange-inline"><math></del>x</math> and length of <math>QM = a</math>. We shall first prove that <math>QM = QB</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Let length of side <math>CB = x</math> and length of <math>QM = a</math>. We shall first prove that <math>QM = QB</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>O</math> be the circumcenter of <math>\triangle ACB</math> which must lie on line <math>Z</math> as <math>Z</math> is a perpendicular bisector of isosceles <math>\triangle ACB</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>O</math> be the circumcenter of <math>\triangle ACB</math> which must lie on line <math>Z</math> as <math>Z</math> is a perpendicular bisector of isosceles <math>\triangle ACB</math>.</div></td></tr>
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Olivera
https://artofproblemsolving.com/wiki/index.php?title=2004_JBMO_Problems/Problem_2&diff=99515&oldid=prev
KRIS17 at 04:22, 18 December 2018
2018-12-18T04:22:17Z
<p></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 04:22, 18 December 2018</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l26" >Line 26:</td>
<td colspan="2" class="diff-lineno">Line 26:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>m . (2 a \cos C/2) = (3/2 x) . a = (3/2).(2Ra) \cos C/2 = 3Ra \cos C/2</math> or,</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>m . (2 a \cos C/2) = (3/2 x) . a = (3/2).(2Ra) \cos C/2 = 3Ra \cos C/2</math> or,</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math> R = 2/<del class="diffchange diffchange-inline">3m</del></math></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math> R = <ins class="diffchange diffchange-inline">(</ins>2/<ins class="diffchange diffchange-inline">3)m</ins></math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></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><math>Kris17</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>Kris17</math></div></td></tr>
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KRIS17
https://artofproblemsolving.com/wiki/index.php?title=2004_JBMO_Problems/Problem_2&diff=99514&oldid=prev
KRIS17: Created page with "==Problem== Let <math>ABC</math> be an isosceles triangle with <math>AC=BC</math>, let <math>M</math> be the midpoint of its side <math>AC</math>, and let <math>Z</math> be t..."
2018-12-18T04:21:47Z
<p>Created page with "==Problem== Let <math>ABC</math> be an isosceles triangle with <math>AC=BC</math>, let <math>M</math> be the midpoint of its side <math>AC</math>, and let <math>Z</math> be t..."</p>
<p><b>New page</b></p><div>==Problem==<br />
<br />
Let <math>ABC</math> be an isosceles triangle with <math>AC=BC</math>, let <math>M</math> be the midpoint of its side <math>AC</math>, and let <math>Z</math> be the line through <math>C</math> perpendicular to <math>AB</math>. The circle through the points <math>B</math>, <math>C</math>, and <math>M</math> intersects the line <math>Z</math> at the points <math>C</math> and <math>Q</math>. Find the radius of the circumcircle of the triangle <math>ABC</math> in terms of <math>m = CQ</math>.<br />
<br />
<br />
==Solution==<br />
Let length of side <math>CB</math> = <math>x</math> and length of <math>QM = a</math>. We shall first prove that <math>QM = QB</math>.<br />
<br />
Let <math>O</math> be the circumcenter of <math>\triangle ACB</math> which must lie on line <math>Z</math> as <math>Z</math> is a perpendicular bisector of isosceles <math>\triangle ACB</math>.<br />
<br />
So, we have <math>\angle ACO = \angle BCO = \angle C/2</math>.<br />
<br />
Now <math>MQBC</math> is a cyclic quadrilateral by definition, so we have:<br />
<math>\angle QMB = \angle QCB = \angle C/2</math> and,<br />
<math>\angle QBM = \angle QCM = \angle C/2</math>, thus <math>\angle QMB = \angle QBM</math>, so <math>QM = QB = a</math>.<br />
<br />
Therefore in isosceles <math>\triangle QMB</math> we have that <math>MB = 2 QB \cos C/2 = 2 a \cos C/2</math>.<br />
<br />
Let <math>R</math> be the circumradius of <math>\triangle ACB</math>.<br />
So we have <math>CM = x/2 = R \cos C/2</math> or <math>x = 2R \cos C/2</math><br />
<br />
Now applying Ptolemy's theorem in cyclic quadrilateral <math>MQBC</math>, we get:<br />
<br />
<math>m . MB = x . QM + (x/2) . QB</math> or,<br />
<br />
<math>m . (2 a \cos C/2) = (3/2 x) . a = (3/2).(2Ra) \cos C/2 = 3Ra \cos C/2</math> or,<br />
<br />
<math> R = 2/3m</math><br />
<br />
<br />
<math>Kris17</math></div>
KRIS17