2016 USAJMO Problems/Problem 5
Contents
Problem
Let be an acute triangle, with as its circumcenter. Point is the foot of the perpendicular from to line , and points and are the feet of the perpendiculars from to the lines and , respectively.
Given that prove that the points and are collinear.
Solution 1
It is well-known that (just use similar triangles or standard area formulas). Then by Power of a Point, Consider the transformation which dilates from by a factor of and reflects about the -angle bisector. Then clearly lies on , and its distance from is so , hence we conclude that are collinear, as desired.
Solution 2
We will use barycentric coordinates with respect to The given condition is equivalent to Note that Therefore, we must show that Expanding, we must prove
Let such that The left side is equal to The right side is equal to which is equivalent to the left hand side. Therefore, the determinant is and are collinear.
Solution 3
For convenience, let denote the lengths of segments respectively, and let denote the measures of respectively. Let denote the circumradius of
Since the central angle subtends the same arc as the inscribed angle on the circumcircle of we have Note that so Thus, Similarly, one can show that (One could probably cite this as well-known, but I have proved it here just in case.)
Clearly, Since we have Thus,
Note that The Extended Law of Sines states that: Therefore, Thus,
Since and we have: It follows that: We see that
Rearranging we get We also have so by SAS similarity. Thus, so is a right angle.
Rearranging we get We also have so by SAS similarity. Thus, so is a right angle.
Since and are both right angles, we get so we conclude that are collinear, and we are done. (We also obtain the extra interesting fact that )
Solution 4
Draw the altitude from to , and let the foot of this altitude be .
Then, by the Right Triangle Altitude Theorem on triangle , we have: .
Since is the perpendicular bisector of , .
Substituting this into our previous equation gives , which equals by the problem condition.
Thus, .
Again, by the Right Triangle Altitude Theorem, angle is right.
By dropping an altitude from to and using the same method, we can find that angle is right. Since , , , are collinear and we are done.
~champion999
Solution 5
We use complex numbers. Let lower case letters represent their respective upper case points, with . Spamming the foot from point to segment formula, we obtain and We now simplify the given length condition: We would like to show that , , are collinear, or After some factoring (or expanding) that takes about 15 minutes, this eventually reduces to which is true.
-MP8148
Solution 6
Claim:
Proof: We compute the area of using two methods. Let be the circumradius of .
First, by extended law of sines, . We are also given that . , so
Second, we compute the area using .
Equating these two expressions for the area of and reducing, we get But , so and .
Since , and both triangles share the angle , . This tells us or .
Substituting into the bolded equation, we get .
The original length condition can be written as .
We also have
Therefore, by SAS similarity, .
We can prove analogously that . We now have and . This implies that which tells us and are collinear, as desired.
-vvluo
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2016 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |