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Difference between revisions of "1983 IMO Problems/Problem 2"

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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>.
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==Problem==
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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. One of the common tangents to the circles touches <math>C_1</math> at <math>P_1</math> and <math>C_2</math> at <math>P_2</math>, while the other touches <math>C_1</math> at <math>Q_1</math> and <math>C_2</math> at <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>.
  
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{{IMO box|year=1983|before=1|num-a=3}}
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==Solution 1==
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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>.
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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.
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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>
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This solution was posted and copyrighted by Number1. The original thread for this problem can be found here: [https://aops.com/community/p444724]
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==Solution 2==
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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>.
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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]
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==Solution 3==
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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>.
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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</math>.
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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>.
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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>
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This solution was posted and copyrighted by matematikolimpiyati. The original thread for this problem can be found here: [https://aops.com/community/p3260115]
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==Solution 4==
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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>
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This solution was posted and copyrighted by spikerboy. The original thread for this problem can be found here: [https://aops.com/community/p3478185]
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==Solution 5==
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It can be easily solved with this lemma:
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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.
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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>.
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<math>Q. E. D.</math>
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This solution was posted and copyrighted by LeoLeon. The original thread for this problem can be found here: [https://aops.com/community/p18853122]
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{{IMO box|year=1983|num-b=1|num-a=3}}

Latest revision as of 22:35, 29 January 2021

Problem

Let $A$ be one of the two distinct points of intersection of two unequal coplanar circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively. One of the common tangents to the circles touches $C_1$ at $P_1$ and $C_2$ at $P_2$, while the other touches $C_1$ at $Q_1$ and $C_2$ at $Q_2$. Let $M_1$ be the midpoint of $P_1Q_1$ and $M_2$ the midpoint of $P_2Q_2$. Prove that $\angle O_1AO_2=\angle M_1AM_2$.

Solution 1

Let $A$ be one of the two distinct points of intersection of two unequal coplanar circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively. Let $S$ be such point on line $O_1O_2$ so that tangents on $C_1$ touches it at $P_1$ and $Q_1$ and tangents on $C_2$ touches it at $P_2$ and $Q_2$. Let $M_1$ be the midpoint of $P_1Q_1$ and $M_2$ the midpoint of $P_2Q_2$. Prove that $\angle O_1AO_2 = \angle M_1AM_2$.

Proof: Since $S$ is image of $M_1$ under inversion wrt circle $C_1$ we have:\[\angle O_1AM_1 = \angle O_1M_1'A'= \angle O_1SA\]Since $S$ is image of $M_2$ under inversion wrt circle $C_2$ we have:\[\angle O_2SA= \angle O_2A'S'= \angle O_2AM_2\]Image of $A$ is in both cases $A$ itself, since it lies on both circles. Since $\angle O_1SA = \angle O_2SA$ we have:\[\angle M_1AO_1=\angle M_2AO_2\]Now:\[\angle O_1AO_2 = \angle M_1AM_2-\angle M_1AO_1+\angle M_2AO_2 = \angle M_1AM_2\]

This solution was posted and copyrighted by Number1. The original thread for this problem can be found here: [1]

Solution 2

Let $P_1P_2$ and $Q_1Q_2$ meet at $R$. Let $RA$ meet $C_2$ at $B$. Now, it is well-known that $O_1O_2$, $P_1P_2$, and $Q_1Q_2$ are concurrent at $R$, the center of homothety between $C_1$ and $C_2$. Now, it is well-known that $O_1O_2$ bisects $\angle P_1RQ_1$. Since $P_1R = Q_1R$, we have that $O_1O_2R$ meets $P_1Q_1$ at its midpoint, $M_1$, and $P_1Q_1$ is perpendicular to $O_1O_2R$. Similarly, $O_1O_2R$ passes through $M_2$ and is perpendicular to $P_2Q_2$. Since $O_2P_2\perp P_2R$, we have that $RA\cdot RB = RP_2^2 = RM_2\cdot RO_2$, which implies that $ABM_2O_2$ is cyclic. Yet, since $A$ and $B$ lie on $C_1$ and $C_2$ respectively and are collinear with $R$, we see that the homothety that maps $C_1$ to $C_2$ about $R$ maps $A$ to $B$. Also, $O_1$ is mapped to $O_2$ by this homothety, and since $M_1$ and $M_2$ are corresponding parts in these circles, $M_1$ is mapped to $M_2$ by this homothety, so $\angle O_1AM_1 = \angle O_2BM_2 = \angle O_2AM_2$, from which we conclude that $\angle O_2AO_1 = \angle M_2AOM_1$.

This solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [2]

Solution 3

Let $B$ be the other intersection point of these two circles. Let $P_1P_2$, $Q_1Q_2$, $O_1O_2$ meet at $Q$. Let $AB$ meet $P_1P_2$ at $P$. Clearly, $P_1Q_1$ and $O_1O_2$ are perpendicular at $M_1$; $P_2Q_2$ and $O_1O_2$ are perpendicular at $M_2$. Since $\triangle O_1P_1M_1 \sim \triangle O_2P_2M_2$,\[\dfrac{O_1P_1}{O_1M_1} = \dfrac{O_2P_2}{O_2M_2} \Rightarrow \dfrac{O_1A}{O_1M_1} = \dfrac{O_2A}{O_2M_2} \qquad{(*)}\]Since $P$ is on the radical axis, $PP_1 = PP_2$, so in the right trapezoid $P_1P_2M_2M_1$, $AB$ is the midsegment. So we have $M_1A = AM_2$ and $\angle AM_1O_1 = \angle AM_2O_1$. Let $M$ be a point on $O_1O_2$ such that $\triangle AM_1O_1 \cong \triangle AM_2M$ (which means $AM=AO_1$ ve $MM_2 = M_1O_1$). So, from $(*)$, we get\[\dfrac{AM}{MM_2} = \dfrac{O_2A}{O_2M_2}\]This means, $AM_2$ is the angle bisector of $\angle MAO_2$. So,\[\angle O_2AM_2 = \angle M_2AM= \angle M_1AO_1 \Longrightarrow \angle M_1AM_2 = \angle O_1AO_2.\]

This solution was posted and copyrighted by matematikolimpiyati. The original thread for this problem can be found here: [3]

Solution 4

Let, $P_1P_2,Q_1Q_2,O_1O_2$ concur at $Q$.Then a homothety with centre $Q$ that sends $C_1$ to $C_2$.Let $QA \cap C_1 =B$.Under the homothety $A$ is the image of $B$.So, $\angle M_1BO_1 =\angle M_2AO_2$ and $\triangle QP_1O_1 \sim \triangle QM_1P_1 \implies \frac{QO_{1}}{QP_{1}} = \frac{QP_{1}}{QM_{1}}$.so $QO_1.QM_1 = Q.P^2_1 = QA .QB$ Hence $A,M_1,B,O_1$ are concyclic.so $\angle O_1BM_1=\angle O_1AM_1 \implies \angle O_1AM_1= \angle O_2AM_2 \implies \angle O_1AO_2=\angle M_1AM_2$

This solution was posted and copyrighted by spikerboy. The original thread for this problem can be found here: [4]

Solution 5

It can be easily solved with this lemma: Let $A$-symmedian of $\triangle ABC$ intersect its circumscribed circle $u$ at $D, P$ on $AD$ satisfying $AP = PD$. Let us define center of $u$ as $O$. Then $(B, C, P, O)$ are concyclic. Now let $S$ be intersection of tangents to circles from the problem, $AS$ intersects $C_1$ at $B$. One can prove that $\square AP_1BQ_1$ is harmonic qudrilateral, so $P_1Q_1$ is symmedian of $\triangle AP_1B$. By lemma we get that $(B, A, O_1, M_1)$ are concyclic, thus $\angle O_1AM_1$ is equivalent to $\angle O_1BM_1$. By homothety we get $\angle O_1BM_1 = \angle O_2AM_2$. $Q. E. D.$

This solution was posted and copyrighted by LeoLeon. The original thread for this problem can be found here: [5]


1983 IMO (Problems) • Resources
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