2016 USAMO Problems/Problem 5
Problem
An equilateral pentagon is inscribed in triangle such that and Let be the intersection of lines and Denote by the angle bisector of
Prove that is parallel to where is the circumcenter of triangle and is the incenter of triangle
Solution 1
Let be the intersection of line and the circumcircle of (other than ), then . Let be the point such that is a rhombus. It follows that .
Since , , or . It follows that .
Since , , , it follows that , so .
It is given that , and by basic properties of the incenter, . Therefore, , so . Since the rotation between the two triangles in 90 degrees, . However, is parallel to the bisector of , which is perpendicular to , so we are done.
Solution 2
Write for all chosen as distinct vertices of triangle . Define as sides opposite to angles , and , respectively. Place the triangle in the Euclidean plane with at the origin and on the positive x-axis. Assume without loss of generality that C is acute.
Consider the sides of the pentagon as vectors and note that
Define and as the angles made between the positive x-axis and and , respectively. Considering the x and y coordinates of the vectors in , it follows that
Suppose . Then , and the triangle is isosceles. In this case, it is clear by symmetry that is vertical. Further, since point exists, , so and must be vertical as well.
For the remainder of the proof, assume . Note that whenever and . Note further that the slope of the line defined by the vector formed by summing vectors and is this expression. Since is parallel to , the slope of can be formed by dividing expressions in and and inverting the sign:
Determine the coordinates of by drawing perpendiculars from to the sides and vertices of the triangle. By exploiting congruence between pairs of right triangles that share a vertex, one can partition into where are bases of these triangles that lie on the sides of triangle . From here it is clear that .
To find the coordinates of , note that and that in any acute triangle . It easily follows that . Note also that the perpendicular from to bisects . Hence, if triangle is acute.
If triangle is obtuse at , then it can be similarly shown that but that the remaining angles of this form are still and . It easily follows that holds if is obtuse. If is obtuse then and the coordinate of is . From this, follows in this case as well.
We can conclude the slope of is by the Law of Sines and rearrangement.
Setting is equivalent to
Since , this equation is equivalent to
This equation is equivalent to which is evident.
Solution 3
Let and be the arc midpoints of respectively. Let be crosspoint of and
Therefore is the circumcenter of triangle
Points and are collinear. is orthocenter of
is the Euler line of
Let be the centroid of lies on line is paraller to
Similarly rotation from to , from to and from to is in clockwise direction,
is perpendicular to
therefore in accordance with Claim is parallel to
This sum is parallel to , so we are done.
Claim
Let Then
Proof
vladimir.shelomovskii@gmail.com, vvsss
Video Solution by MOP 2024
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See also
2016 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.