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> | + | |
+ | == See Also == {{IMO box|year=1985|num-b=4|num-a=6}} |
Latest revision as of 23:56, 29 January 2021
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 .
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 |