Difference between revisions of "2008 USAMO Problems/Problem 2"
(→Lemma 2) |
(→Solution 9 (analytical)) |
||
Line 175: | Line 175: | ||
We consider an [[inversion]] by an arbitrary [[radius]] about <math>A</math>. We want to show that <math>P', F',</math> and <math>N'</math> are [[collinear]]. Notice that <math>D', A,</math> and <math>P'</math> lie on a circle with center <math>B'</math>, and similarly for the other side. We also have that <math>B', D', F', A</math> form a cyclic quadrilateral, and similarly for the other side. By angle chasing, we can prove that <math>A B' F' C'</math> is a [[parallelogram]], indicating that <math>F'</math> is the midpoint of <math>P'N'</math>. {{incomplete|solution}} | We consider an [[inversion]] by an arbitrary [[radius]] about <math>A</math>. We want to show that <math>P', F',</math> and <math>N'</math> are [[collinear]]. Notice that <math>D', A,</math> and <math>P'</math> lie on a circle with center <math>B'</math>, and similarly for the other side. We also have that <math>B', D', F', A</math> form a cyclic quadrilateral, and similarly for the other side. By angle chasing, we can prove that <math>A B' F' C'</math> is a [[parallelogram]], indicating that <math>F'</math> is the midpoint of <math>P'N'</math>. {{incomplete|solution}} | ||
− | |||
− | |||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 18:29, 26 August 2012
Problem
(Zuming Feng) Let be an acute, scalene triangle, and let , , and be the midpoints of , , and , respectively. Let the perpendicular bisectors of and intersect ray in points and respectively, and let lines and intersect in point , inside of triangle . Prove that points , , , and all lie on one circle.
Contents
Solution
Solution 1 (synthetic)
Without loss of generality . The intersection of and is , the circumcenter of .
Let and . Note lies on the perpendicular bisector of , so . So . Similarly, , so . Notice that intercepts the minor arc in the circumcircle of , which is double . Hence , so is cyclic.
Lemma 1
is directly similar to since , , are collinear, is cyclic, and . Also because , and is the medial triangle of so . Hence .
Notice that since . . Then Hence .
Hence is similar to by AA similarity. It is easy to see that they are oriented such that they are directly similar.
Lemma 2
By the similarity in Lemma 1, . so by SAS similarity. Hence Using essentially the same angle chasing, we can show that is directly similar to . It follows that is directly similar to . So Hence , so is cyclic. In other words, lies on the circumcircle of . Note that , so is cyclic. In other words, lies on the circumcircle of . , , , , and all lie on the circumcircle of . Hence , , , and lie on a circle, as desired.
Solution 2 (synthetic)
Without Loss of Generality, assume . It is sufficient to prove that , as this would immediately prove that are concyclic. By applying the Menelaus' Theorem in the Triangle for the transversal , we have (in magnitude) Here, we used that , as is the midpoint of . Now, since and , we have Now, note that bisects the exterior and bisects exterior , making the -excentre of . This implies that bisects interior , making , as was required.
Solution 3 (synthetic)
Hint: consider intersection with ; show that the resulting intersection lies on the desired circle. Template:Incomplete
Solution 4 (synthetic)
This solution utilizes the phantom point method. Clearly, APON are cyclic because . Let the circumcircles of triangles and intersect at and .
Lemma. If are points on circle with center , and the tangents to at intersect at , then is the symmedian from to .
This is fairly easy to prove (as H, O are isogonal conjugates, plus using SAS similarity), but the author lacks time to write it up fully, and will do so soon.
It is easy to see (the intersection of ray and the circumcircle of ) is colinear with and , and because line is the diameter of that circle, , so is the point in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that satisfies the original construction for , showing ; we are done. Template:Incomplete
Solution 5 (trigonometric)
By the Law of Sines, . Since and similarly , we cancel to get . Obviously, so .
Then and . Subtracting these two equations, so . Therefore, (by AA similarity), so a spiral similarity centered at takes to and to . Therefore, it takes the midpoint of to the midpoint of , or to . So and is cyclic.
Solution 6 (isogonal conjugates)
Construct on such that . Then . Then , so , or . Then , so . Then we have
and . So and are isogonally conjugate. Thus . Then
.
If is the circumcenter of then so is cyclic. Then .
Then . Then is a right triangle.
Now by the homothety centered at with ratio , is taken to and is taken to . Thus is taken to the circumcenter of and is the midpoint of , which is also the circumcenter of , so all lie on a circle.
Solution 7 (symmedians)
Median of a triangle implies . Trig ceva for shows that is a symmedian. Then is a median, use the lemma again to show that , and similarly , so you're done. Template:Incomplete
Solution 8 (inversion)
We consider an inversion by an arbitrary radius about . We want to show that and are collinear. Notice that and lie on a circle with center , and similarly for the other side. We also have that form a cyclic quadrilateral, and similarly for the other side. By angle chasing, we can prove that is a parallelogram, indicating that is the midpoint of . Template:Incomplete
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
2008 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
- <url>viewtopic.php?t=202907 Discussion on AoPS/MathLinks</url>