Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1985 IMO Problems/Problem 5"

(→‎Solution)
Line 5: Line 5:
 
== 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>.
 +
 +
== See Also == {{IMO box|year=1985|num-b=4|num-a=6}}

Revision as of 22:56, 29 January 2021

Problem

A circle with center $O$ passes through the vertices $A$ and $C$ of the triangle $ABC$ and intersects the segments $AB$ and $BC$ again at distinct points $K$ and $N$ respectively. Let $M$ be the point of intersection of the circumcircles of triangles $ABC$ and $KBN$ (apart from $B$). Prove that $\angle OMB = 90^{\circ}$.

Solution

$M$ is the Miquel Point of quadrilateral $ACNK$, so there is a spiral similarity centered at $M$ that takes $KN$ to $AC$. Let $M_1$ be the midpoint of $KA$ and $M_2$ be the midpoint of $NC$. Thus the spiral similarity must also send $M_1$ to $M_2$ and so $BMM_1 M_2$ is cyclic. $OM_1 B M_2$ is also cyclic with diameter $BO$ and thus $M$ must lie on the same circumcircle as $B$, $M_1$, and $M_2$ so $\angle OMB = 90^{\circ}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_IMO_Problems/Problem_5&oldid=144024"