2016 USAJMO Problems/Problem 5
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
Clearly, Since we have Thus,
Note that Then, since and we have: The Extended Law of Sines states that:
Since and $$ (Error compiling LaTeX. Unknown error_msg)AP\cdot AQ = \frac{b^3 c^3}{16R^4} = \frac{16\sqrt{2}R^6}{16R^4} = \sqrt{2}R^2.AP\cdot AQ = AH\cdot AO.AP\cdot AQ = AH\cdot AO,\frac{AP}{AH} = \frac{AO}{AQ}.\angle PAH = \angle OAQ = \frac{\pi}{2} - \beta,\triangle PAH\sim\triangle OAQ\angle AOQ = \angle APH,\angle AOQ$is a right angle.
Rearranging$ (Error compiling LaTeX. Unknown error_msg)AP\cdot AQ = AH\cdot AO,\frac{AP}{AO} = \frac{AO}{AH}.\angle PAO = \angle HAQ = \frac{\pi}{2} - \gamma,\triangle PAO\sim\triangle HAQ\angle AOP = \angle AQH,\angle AOP$is a right angle.
Since$ (Error compiling LaTeX. Unknown error_msg)\angle AOP\angle AOQ\angle POQ = \pi,P, O, QAO\perp PQ.$
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 |