Difference between revisions of "1983 IMO Problems/Problem 2"
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==Problem== | ==Problem== | ||
− | Let <math>A</math> be one of the two distinct points of intersection of two unequal coplanar circles <math> | + | 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>. |
+ | |||
+ | ==Solution 1== | ||
+ | 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>. | ||
+ | |||
+ | 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. | ||
+ | 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> | ||
+ | |||
+ | This solution was posted and copyrighted by Number1. The original thread for this problem can be found here: [https://aops.com/community/p444724] | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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>. | ||
+ | |||
+ | 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] | ||
+ | |||
+ | ==Solution 3== | ||
+ | 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>. | ||
+ | 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>. | ||
+ | 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>. | ||
+ | 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> | ||
+ | |||
+ | This solution was posted and copyrighted by matematikolimpiyati. The original thread for this problem can be found here: [https://aops.com/community/p3260115] | ||
+ | |||
+ | ==Solution 4== | ||
+ | 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> | ||
+ | |||
+ | This solution was posted and copyrighted by spikerboy. The original thread for this problem can be found here: [https://aops.com/community/p3478185] | ||
+ | |||
+ | ==Solution 5== | ||
+ | It can be easily solved with this lemma: | ||
+ | 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. | ||
+ | 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>. | ||
+ | <math>Q. E. D.</math> | ||
+ | |||
+ | This solution was posted and copyrighted by LeoLeon. The original thread for this problem can be found here: [https://aops.com/community/p18853122] | ||
+ | |||
+ | |||
{{IMO box|year=1983|num-b=1|num-a=3}} | {{IMO box|year=1983|num-b=1|num-a=3}} |
Latest revision as of 22:35, 29 January 2021
Problem
Let be one of the two distinct points of intersection of two unequal coplanar circles
and
with centers
and
respectively. One of the common tangents to the circles touches
at
and
at
, while the other touches
at
and
at
. Let
be the midpoint of
and
the midpoint of
. Prove that
.
Solution 1
Let be one of the two distinct points of intersection of two unequal coplanar circles
and
with centers
and
respectively. Let
be such point on line
so that tangents on
touches it at
and
and tangents on
touches it at
and
. Let
be the midpoint of
and
the midpoint of
. Prove that
.
Proof: Since is image of
under inversion wrt circle
we have:
Since
is image of
under inversion wrt circle
we have:
Image of
is in both cases
itself, since it lies on both circles.
Since
we have:
Now:
This solution was posted and copyrighted by Number1. The original thread for this problem can be found here: [1]
Solution 2
Let and
meet at
. Let
meet
at
. Now, it is well-known that
,
, and
are concurrent at
, the center of homothety between
and
. Now, it is well-known that
bisects
. Since
, we have that
meets
at its midpoint,
, and
is perpendicular to
. Similarly,
passes through
and is perpendicular to
. Since
, we have that
, which implies that
is cyclic. Yet, since
and
lie on
and
respectively and are collinear with
, we see that the homothety that maps
to
about
maps
to
. Also,
is mapped to
by this homothety, and since
and
are corresponding parts in these circles,
is mapped to
by this homothety, so
, from which we conclude that
.
This solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [2]
Solution 3
Let be the other intersection point of these two circles. Let
,
,
meet at
. Let
meet
at
.
Clearly,
and
are perpendicular at
;
and
are perpendicular at
.
Since
,
Since
is on the radical axis,
, so in the right trapezoid
,
is the midsegment. So we have
and
.
Let
be a point on
such that
(which means
ve
). So, from
, we get
This means,
is the angle bisector of
. So,
This solution was posted and copyrighted by matematikolimpiyati. The original thread for this problem can be found here: [3]
Solution 4
Let, concur at
.Then a homothety with centre
that sends
to
.Let
.Under the homothety
is the image of
.So,
and
.so
Hence
are concyclic.so
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 -symmedian of
intersect its circumscribed circle
at
on
satisfying
. Let us define center of
as
. Then
are concyclic.
Now let
be intersection of tangents to circles from the problem,
intersects
at
. One can prove that
is harmonic qudrilateral, so
is symmedian of
. By lemma we get that
are concyclic, thus
is equivalent to
. By homothety we get
.
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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