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>. |
+ | |||
+ | ==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 == | ||
+ | *[[Miquel's point]] | ||
+ | {{IMO box|year=1985|num-b=4|num-a=6}} |
Latest revision as of 15:19, 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 |