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2000 JBMO Problems/Problem 3 - Revision history
2024-03-29T08:28:52Z
Revision history for this page on the wiki
MediaWiki 1.31.1
https://artofproblemsolving.com/wiki/index.php?title=2000_JBMO_Problems/Problem_3&diff=99269&oldid=prev
KRIS17 at 04:35, 5 December 2018
2018-12-05T04:35:09Z
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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 04:35, 5 December 2018</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l6" >Line 6:</td>
<td colspan="2" class="diff-lineno">Line 6:</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="background-color: #f8f9fa; color: #222; font-size: 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 begin by showing that <math>A</math> is the circumcenter of <math>\triangle KPQ</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 begin by showing that <math>A</math> is the circumcenter of <math>\triangle KPQ</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;">Consider the configuration where <math>\angle ACB</math> is obtuse and diameter <math>EF</math> lies on extended line <math>BC</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>Let <math>O</math> be the midpoint of diameter <math>EF</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 midpoint of diameter <math>EF</math>.</div></td></tr>
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KRIS17
https://artofproblemsolving.com/wiki/index.php?title=2000_JBMO_Problems/Problem_3&diff=99268&oldid=prev
KRIS17 at 04:21, 5 December 2018
2018-12-05T04:21:08Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<col class="diff-marker" />
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 04:21, 5 December 2018</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l5" >Line 5:</td>
<td colspan="2" class="diff-lineno">Line 5:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td></tr>
<tr><td colspan="2"> </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;">We begin by showing that <math>A</math> is the circumcenter of <math>\triangle KPQ</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>Let <math>O</math> be the midpoint of diameter <math>EF</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 midpoint of diameter <math>EF</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>Let <math>KA</math> meet <math>BC</math> at <math>G</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>KA</math> meet <math>BC</math> at <math>G</math>.</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">We begin by showing that <math>A</math> is the circumcenter of <math>\triangle KPQ</math>:</del></div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let us define <math>\angle KPA = \alpha</math> and <math>\angle KQA = \beta</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 us define <math>\angle KPA = \alpha</math> and <math>\angle KQA = \beta</math></div></td></tr>
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KRIS17
https://artofproblemsolving.com/wiki/index.php?title=2000_JBMO_Problems/Problem_3&diff=99264&oldid=prev
KRIS17: Created page with "==Problem 3== A half-circle of diameter <math>EF</math> is placed on the side <math>BC</math> of a triangle <math>ABC</math> and it is tangent to the sides <math>AB</math> an..."
2018-12-05T04:09:46Z
<p>Created page with "==Problem 3== A half-circle of diameter <math>EF</math> is placed on the side <math>BC</math> of a triangle <math>ABC</math> and it is tangent to the sides <math>AB</math> an..."</p>
<p><b>New page</b></p><div>==Problem 3==<br />
<br />
A half-circle of diameter <math>EF</math> is placed on the side <math>BC</math> of a triangle <math>ABC</math> and it is tangent to the sides <math>AB</math> and <math>AC</math> in the points <math>Q</math> and <math>P</math> respectively. Prove that the intersection point <math>K</math> between the lines <math>EP</math> and <math>FQ</math> lies on the altitude from <math>A</math> of the triangle <math>ABC</math>.<br />
<br />
<br />
== Solution ==<br />
<br />
Let <math>O</math> be the midpoint of diameter <math>EF</math>.<br />
Let <math>KA</math> meet <math>BC</math> at <math>G</math>.<br />
<br />
We begin by showing that <math>A</math> is the circumcenter of <math>\triangle KPQ</math>:<br />
<br />
Let us define <math>\angle KPA = \alpha</math> and <math>\angle KQA = \beta</math><br />
<br />
By applying Tangent Chord Angle theorem, we get:<br />
<math>\angle POE = 2\alpha</math> and <math>\angle QOF = 2\beta</math><br />
<br />
Now, <math>\angle POQ = 180 - 2(\alpha + \beta)</math>, and since <math>PAQO</math> is a cyclic quadrilateral,<br />
we have <math>\angle PAQ = 2(\alpha + \beta)</math><br />
<br />
Now <math>\angle POF = 180 - 2\alpha</math>, so <math>\angle PEF = 90 - \alpha</math><br />
<br />
Similarly, we have <math>\angle QOE = 180 - 2\beta</math>, so <math>\angle QFE = 90 - \beta</math><br />
<br />
From <math>\triangle EKF, \angle EKF = 180 - (\angle PEF + \angle QFE)</math><br />
<br />
<math>= 180 - (90 - \alpha + 90 - \beta) = (\alpha + \beta)</math><br />
<br />
Thus, we have <math>\angle PKQ = \angle EKF = \angle PAQ/2</math><br />
<br />
Also, <math>AP = AQ</math> (Since <math>AP</math> and <math>AQ</math> are tangents to the same circle)<br />
<br />
From the above 2 results, it readily follows that <math>A</math> is the circumcenter of <math>\triangle KPQ</math>.<br />
<br />
Thus, we have <math>AK = AP</math>, and so <math>\angle AKP = \angle KPA = \alpha</math><br />
<br />
So in <math>\triangle EKG, \angle KGE = 180 - (\angle AKP + \angle PEF)</math><br />
<br />
<math>= 180 - (\alpha + 90 - \alpha) = 90^{\circ}</math><br />
<br />
So <math>KA</math> is perpendicular to <math>BC</math>, hence <math>K</math> lies on the altitude from <math>A</math> of the triangle <math>ABC</math>.<br />
<br />
<br />
<math>Kris17</math></div>
KRIS17