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1983 IMO Problems/Problem 2 - Revision history
2024-03-28T18:26:14Z
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Hamstpan38825 at 03:35, 30 January 2021
2021-01-30T03:35:49Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 03:35, 30 January 2021</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Problem==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Problem==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Let <math>A</math> be one of the two distinct points of intersection of two unequal coplanar circles <math><del class="diffchange diffchange-inline">C1</del></math> and <math><del class="diffchange diffchange-inline">C2</del></math> with centers <math><del class="diffchange diffchange-inline">O1</del></math> and <math><del class="diffchange diffchange-inline">O2</del></math><del class="diffchange diffchange-inline">, </del>respectively. One of the common tangents to the circles touches <math><del class="diffchange diffchange-inline">C1</del></math> at <math><del class="diffchange diffchange-inline">P1</del></math> and <math><del class="diffchange diffchange-inline">C2</del></math> at <math><del class="diffchange diffchange-inline">P2</del></math>, while the other touches <math><del class="diffchange diffchange-inline">C1</del></math> at <math><del class="diffchange diffchange-inline">Q1</del></math> and <math><del class="diffchange diffchange-inline">C2</del></math> at <math><del class="diffchange diffchange-inline">Q2</del></math>. Let <math><del class="diffchange diffchange-inline">M1</del></math> be the midpoint of <math><del class="diffchange diffchange-inline">P1</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>Let <math>A</math> be one of the two distinct points of intersection of two unequal coplanar circles <math><ins class="diffchange diffchange-inline">C_1</ins></math> and <math><ins class="diffchange diffchange-inline">C_2</ins></math> with centers <math><ins class="diffchange diffchange-inline">O_1</ins></math> and <math><ins class="diffchange diffchange-inline">O_2</ins></math> respectively. One of the common tangents to the circles touches <math><ins class="diffchange diffchange-inline">C_1</ins></math> at <math><ins class="diffchange diffchange-inline">P_1</ins></math> and <math><ins class="diffchange diffchange-inline">C_2</ins></math> at <math><ins class="diffchange diffchange-inline">P_2</ins></math>, while the other touches <math><ins class="diffchange diffchange-inline">C_1</ins></math> at <math><ins class="diffchange diffchange-inline">Q_1</ins></math> and <math><ins class="diffchange diffchange-inline">C_2</ins></math> at <math><ins class="diffchange diffchange-inline">Q_2</ins></math>. Let <math><ins class="diffchange diffchange-inline">M_1</ins></math> be the midpoint of <math><ins class="diffchange diffchange-inline">P_1Q_1</math> and <math>M_2</math> the midpoint of <math>P_2Q_2</math>. Prove that <math>\angle O_1AO_2=\angle M_1AM_2</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">==Solution 1==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Let <math>A</math> be one of the two distinct points of intersection of two unequal coplanar circles <math>C_1</math> and <math>C_2</math> with centers <math>O_1</math> and <math>O_2</math> respectively. Let <math>S</math> be such point on line <math>O_1O_2</math> so that tangents on <math>C_1</math> touches it at <math>P_1</math> and <math>Q_1</math> and tangents on <math>C_2</math> touches it at <math>P_2</math> and <math>Q_2</math>. Let <math>M_1</math> be the midpoint of <math>P_1Q_1</math> and <math>M_2</math> the midpoint of <math>P_2Q_2</math>. Prove that <math>\angle O_1AO_2 = \angle M_1AM_2</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">Proof: Since <math>S</math> is image of <math>M_1</math> under inversion wrt circle <math>C_1</math> we have:<cmath> \angle O_1AM_1 = \angle O_1M_1'A'= \angle O_1SA  </cmath>Since <math>S</math> is image of <math>M_2</math> under inversion wrt circle <math>C_2</math> we have:<cmath> \angle O_2SA= \angle O_2A'S'= \angle O_2AM_2  </cmath>Image of <math>A</math> is in both cases <math>A</math> itself, since it lies on both circles.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Since <math>\angle O_1SA = \angle O_2SA</math> we have:<cmath> \angle M_1AO_1=\angle M_2AO_2  </cmath>Now:<cmath> \angle O_1AO_2 = \angle M_1AM_2-\angle M_1AO_1+\angle M_2AO_2 = \angle M_1AM_2  </cmath></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">This solution was posted and copyrighted by Number1. The original thread for this problem can be found here: [https://aops.com/community/p444724]</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">==Solution 2==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Let <math> P_1P_2</math> and <math> Q_1Q_2</math> meet at <math> R</math>. Let <math> RA</math> meet <math> C_2</math> at <math> B</math>. Now, it is well-known that <math> O_1O_2</math>, <math> P_1P_2</math>, and <math> Q_1Q_2</math> are concurrent at <math> R</math>, the center of homothety between <math> C_1</math> and <math> C_2</math>. Now, it is well-known that <math> O_1O_2</math> bisects <math> \angle P_1RQ_1</math>. Since <math> P_1R = Q_1R</math>, we have that <math> O_1O_2R</math> meets <math> P_1Q_1</math> at its midpoint, <math> M_1</math>, and <math> P_1Q_1</math> is perpendicular to <math> O_1O_2R</math>. Similarly, <math> O_1O_2R</math> passes through <math> M_2</math> and is perpendicular to <math> P_2Q_2</math>. Since <math> O_2P_2\perp P_2R</math>, we have that <math> RA\cdot RB = RP_2^2 = RM_2\cdot RO_2</math>, which implies that <math> ABM_2O_2</math> is cyclic. Yet, since <math> A</math> and <math> B</math> lie on <math> C_1</math> and <math> C_2</math> respectively and are collinear with <math> R</math>, we see that the homothety that maps <math> C_1</math> to <math> C_2</math> about <math> R</math> maps <math> A</math> to <math> B</math>. Also, <math> O_1</math> is mapped to <math> O_2</math> by this homothety, and since <math> M_1</math> and <math> M_2</math> are corresponding parts in these circles, <math> M_1</math> is mapped to <math> M_2</math> by this homothety, so <math> \angle O_1AM_1 = \angle O_2BM_2 = \angle O_2AM_2</math>, from which we conclude that <math> \angle O_2AO_1 = \angle M_2AOM_1</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">This solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [https://aops.com/community/p1247276]</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">==Solution 3==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Let <math>B</math> be the other intersection point of these two circles. Let <math>P_1P_2</math>, <math>Q_1Q_2</math>, <math>O_1O_2</math> meet at <math>Q</math>. Let <math>AB</math> meet <math>P_1P_2</math> at <math>P</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Clearly, <math>P_1Q_1</math> and <math>O_1O_2</math> are perpendicular at <math>M_1</math>; <math>P_2Q_2</math> and <math>O_1O_2</math> are perpendicular at <math>M_2</ins></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 class="diffchange diffchange-inline">Since <math>\triangle O_1P_1M_1 \sim \triangle O_2P_2M_2</math>,<cmath>\dfrac{O_1P_1}{O_1M_1} = \dfrac{O_2P_2}{O_2M_2} \Rightarrow \dfrac{O_1A}{O_1M_1} = \dfrac{O_2A}{O_2M_2} \qquad{(*)}</cmath>Since <math>P</math> is on the radical axis, <math>PP_1 = PP_2</math>, so in the right trapezoid <math>P_1P_2M_2M_1</math>, <math>AB</math> is the midsegment. So we have <math>M_1A = AM_2</math> and <math>\angle AM_1O_1 = \angle AM_2O_1</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Let <math>M</math> be a point on <math>O_1O_2</math> such that <math>\triangle AM_1O_1 \cong \triangle AM_2M</math> (which means <math>AM=AO_1</math> ve <math>MM_2 = M_1O_1</math>). So, from <math>(*)</math>, we get<cmath>\dfrac{AM}{MM_2} = \dfrac{O_2A}{O_2M_2}</cmath>This means, <math>AM_2</math> is the angle bisector of <math>\angle MAO_2</math>. So,<cmath>\angle O_2AM_2 = \angle M_2AM= \angle M_1AO_1 \Longrightarrow \angle M_1AM_2 = \angle O_1AO_2.</cmath></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">This solution was posted and copyrighted by matematikolimpiyati. The original thread for this problem can be found here: [https://aops.com/community/p3260115]</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">==Solution 4==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Let, <math>P_1P_2,Q_1Q_2,O_1O_2</math> concur at <math>Q</math>.Then a homothety with centre <math>Q</math> that sends <math>C_1</math> to <math>C_2</math>.Let <math>QA \cap C_1 =B</math>.Under the homothety <math>A</math> is the image of <math>B</math>.So, <math>\angle M_1BO_1 =\angle M_2AO_2</math> and <math>\triangle QP_1O_1 \sim \triangle QM_1P_1 \implies \frac{QO_{1}}{QP_{1}} = \frac{QP_{1}}{QM_{1}}</math>.so <math>QO_1.QM_1 = Q.P^2_1 = QA .QB</math> Hence <math>A,M_1,B,O_1</math> are concyclic.so <math>\angle O_1BM_1=\angle O_1AM_1 \implies \angle O_1AM_1= \angle O_2AM_2 \implies \angle O_1AO_2=\angle M_1AM_2 </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">This solution was posted and copyrighted by spikerboy. The original thread for this problem can be found here: [https://aops.com/community/p3478185]</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">==Solution 5==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">It can be easily solved with this lemma:</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Let <math>A</math>-symmedian of <math>\triangle ABC</math> intersect its circumscribed circle <math>u</math> at <math>D, P</math> on <math>AD</math> satisfying <math>AP = PD</math>. Let us define center of <math>u</math> as <math>O</math>. Then <math>(B, C, P, O)</math> are concyclic.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Now let <math>S</math> be intersection of tangents to circles from the problem, <math>AS</math> intersects <math> C_1</math> at <math>B</math>. One can prove that <math>\square AP_1BQ_1</math> is harmonic qudrilateral, so <math>P_1Q_1</math> is symmedian of <math>\triangle AP_1B</math>. By lemma we get that <math>(B, A, O_1, M_1)</math> are concyclic, thus <math>\angle O_1AM_1</math> is equivalent to <math>\angle O_1BM_1</math>. By homothety we get <math>\angle O_1BM_1 = \angle O_2AM_2</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline"><math>Q. E. D.</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">This solution was posted and copyrighted by LeoLeon. The original thread for this problem can be found here: [https://aops.com/community/p18853122]</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> </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>{{IMO box|year=1983|num-b=1|num-a=3}}</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>{{IMO box|year=1983|num-b=1|num-a=3}}</div></td></tr>
</table>
Hamstpan38825
https://artofproblemsolving.com/wiki/index.php?title=1983_IMO_Problems/Problem_2&diff=75005&oldid=prev
Ryanyz10 at 02:36, 1 February 2016
2016-02-01T02:36:33Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 02:36, 1 February 2016</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</td></tr>
<tr><td 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;">==Problem==</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>A</math> be one of the two distinct points of intersection of two unequal coplanar circles <math>C1</math> and <math>C2</math> with centers <math>O1</math> and <math>O2</math>, respectively. One of the common tangents to the circles touches <math>C1</math> at <math>P1</math> and <math>C2</math> at <math>P2</math>, while the other touches <math>C1</math> at <math>Q1</math> and <math>C2</math> at <math>Q2</math>. Let <math>M1</math> be the midpoint of <math>P1</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>A</math> be one of the two distinct points of intersection of two unequal coplanar circles <math>C1</math> and <math>C2</math> with centers <math>O1</math> and <math>O2</math>, respectively. One of the common tangents to the circles touches <math>C1</math> at <math>P1</math> and <math>C2</math> at <math>P2</math>, while the other touches <math>C1</math> at <math>Q1</math> and <math>C2</math> at <math>Q2</math>. Let <math>M1</math> be the midpoint of <math>P1</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>{{IMO box|year=1983|num-b=1|num-a=3}}</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>{{IMO box|year=1983|num-b=1|num-a=3}}</div></td></tr>
</table>
Ryanyz10
https://artofproblemsolving.com/wiki/index.php?title=1983_IMO_Problems/Problem_2&diff=75002&oldid=prev
Ryanyz10 at 02:35, 1 February 2016
2016-02-01T02:35:26Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 02:35, 1 February 2016</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>A</math> be one of the two distinct points of intersection of two unequal coplanar circles <math>C1</math> and <math>C2</math> with centers <math>O1</math> and <math>O2</math>, respectively. One of the common tangents to the circles touches <math>C1</math> at <math>P1</math> and <math>C2</math> at <math>P2</math>, while the other touches <math>C1</math> at <math>Q1</math> and <math>C2</math> at <math>Q2</math>. Let <math>M1</math> be the midpoint of <math>P1</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>A</math> be one of the two distinct points of intersection of two unequal coplanar circles <math>C1</math> and <math>C2</math> with centers <math>O1</math> and <math>O2</math>, respectively. One of the common tangents to the circles touches <math>C1</math> at <math>P1</math> and <math>C2</math> at <math>P2</math>, while the other touches <math>C1</math> at <math>Q1</math> and <math>C2</math> at <math>Q2</math>. Let <math>M1</math> be the midpoint of <math>P1</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>{{IMO box|year=1983|<del class="diffchange diffchange-inline">before</del>=1|num-a=3}}</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>{{IMO box|year=1983|<ins class="diffchange diffchange-inline">num-b</ins>=1|num-a=3}}</div></td></tr>
</table>
Ryanyz10
https://artofproblemsolving.com/wiki/index.php?title=1983_IMO_Problems/Problem_2&diff=75001&oldid=prev
Ryanyz10: Created page with "Let <math>A</math> be one of the two distinct points of intersection of two unequal coplanar circles <math>C1</math> and <math>C2</math> with centers <math>O1</math> and <math..."
2016-02-01T02:34:49Z
<p>Created page with "Let <math>A</math> be one of the two distinct points of intersection of two unequal coplanar circles <math>C1</math> and <math>C2</math> with centers <math>O1</math> and <math..."</p>
<p><b>New page</b></p><div>Let <math>A</math> be one of the two distinct points of intersection of two unequal coplanar circles <math>C1</math> and <math>C2</math> with centers <math>O1</math> and <math>O2</math>, respectively. One of the common tangents to the circles touches <math>C1</math> at <math>P1</math> and <math>C2</math> at <math>P2</math>, while the other touches <math>C1</math> at <math>Q1</math> and <math>C2</math> at <math>Q2</math>. Let <math>M1</math> be the midpoint of <math>P1</math>.<br />
<br />
{{IMO box|year=1983|before=1|num-a=3}}</div>
Ryanyz10