Difference between revisions of "1985 IMO Problems/Problem 5"
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== Solution == | == Solution == | ||
<math>M</math> is the Miquel Point of quadrilateral <math>ACNK</math>, so there is a spiral similarity centered at <math>M</math> that takes <math>KN</math> to <math>AC</math>. Let <math>M_1</math> be the midpoint of <math>KA</math> and <math>M_2</math> be the midpoint of <math>NC</math>. Thus the spiral similarity must also send <math>M_1</math> to <math>M_2</math> and so <math>BMM_1 M_2</math> is cyclic. <math>OM_1 B M_2</math> is also cyclic with diameter <math>BO</math> and thus <math>M</math> must lie on the same circumcircle as <math>B</math>, <math>M_1</math>, and <math>M_2</math> so <math>\angle OMB = 90^{\circ}</math>. | <math>M</math> is the Miquel Point of quadrilateral <math>ACNK</math>, so there is a spiral similarity centered at <math>M</math> that takes <math>KN</math> to <math>AC</math>. Let <math>M_1</math> be the midpoint of <math>KA</math> and <math>M_2</math> be the midpoint of <math>NC</math>. Thus the spiral similarity must also send <math>M_1</math> to <math>M_2</math> and so <math>BMM_1 M_2</math> is cyclic. <math>OM_1 B M_2</math> is also cyclic with diameter <math>BO</math> and thus <math>M</math> must lie on the same circumcircle as <math>B</math>, <math>M_1</math>, and <math>M_2</math> so <math>\angle OMB = 90^{\circ}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | [[File:1985 IMO.png|450px|right]] | ||
+ | Let <math>\Omega, \Omega', \omega</math> and <math>O,O',O''</math> be the circumcircles and circumcenters of <math>AKNC, ABC, BNKM,</math> respectively. | ||
+ | |||
+ | Let <math>\angle ACB = \gamma, AKNC</math> is cyclic <math>\implies \angle BKN = \gamma.</math> | ||
+ | |||
+ | The radius of <math>\omega</math> is <math>MO'' = BO'' = \frac {BN}{2 \sin \gamma}.</math> | ||
+ | |||
+ | Let <math>D</math> and <math>E</math> be midpoints of <math>BC</math> and <math>NC</math> respectively. | ||
+ | |||
+ | <math>OE \perp BC, OD \perp BC, OO' \perp AC, DE = \frac {BC}{2} - \frac {NC}{2} = \frac {BN}{2}</math> | ||
+ | <math>\implies OO' = \frac {DE}{\sin \gamma} = \frac {BN}{2 \sin \gamma} = MO''.</math> | ||
+ | |||
+ | <math>M</math> is the Miquel Point of quadrilateral <math>ACNK,</math> so <math>MO''O'O</math> is cyclic. | ||
+ | <math>MO''O'O</math> is trapezium <math>\implies O''O' || MO.</math> <math>O''O' \perp BM \implies MO\perp BM</math> as desired. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
== See Also == {{IMO box|year=1985|num-b=4|num-a=6}} | == See Also == {{IMO box|year=1985|num-b=4|num-a=6}} |
Revision as of 16:15, 4 December 2022
Contents
Problem
A circle with center passes through the vertices
and
of the triangle
and intersects the segments
and
again at distinct points
and
respectively. Let
be the point of intersection of the circumcircles of triangles
and
(apart from
). Prove that
.
Solution
is the Miquel Point of quadrilateral
, so there is a spiral similarity centered at
that takes
to
. Let
be the midpoint of
and
be the midpoint of
. Thus the spiral similarity must also send
to
and so
is cyclic.
is also cyclic with diameter
and thus
must lie on the same circumcircle as
,
, and
so
.
Solution 2
Let and
be the circumcircles and circumcenters of
respectively.
Let is cyclic
The radius of is
Let and
be midpoints of
and
respectively.
is the Miquel Point of quadrilateral
so
is cyclic.
is trapezium
as desired.
vladimir.shelomovskii@gmail.com, vvsss
See Also
1985 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |