Difference between revisions of "2016 USAJMO Problems/Problem 5"
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Since <math>AP = \frac{b^2 c}{4R^2}</math> and <math>AQ = \frac{bc^2}{4R^2},</math> | Since <math>AP = \frac{b^2 c}{4R^2}</math> and <math>AQ = \frac{bc^2}{4R^2},</math> | ||
− | < | + | <cmath>AP\cdot AQ = \frac{b^3 c^3}{16R^4} = \frac{16\sqrt{2}R^6}{16R^4} = \sqrt{2}R^2.</cmath> |
− | We see that < | + | We see that <math>AP\cdot AQ = AH\cdot AO.</math> |
− | Rearranging < | + | Rearranging <math>AP\cdot AQ = AH\cdot AO,</math> we get: <math>\frac{AP}{AH} = \frac{AO}{AQ}.</math> We also have <math>\angle PAH = \angle OAQ = \frac{\pi}{2} - \beta,</math> so <math>\triangle PAH\sim\triangle OAQ</math> by SAS similarity. Thus, <math>\angle AOQ = \angle APH,</math> so <math>\angle AOQ</math> is a right angle. |
− | Rearranging < | + | Rearranging <math>AP\cdot AQ = AH\cdot AO,</math> we get: <math>\frac{AP}{AO} = \frac{AO}{AH}.</math> We also have <math>\angle PAO = \angle HAQ = \frac{\pi}{2} - \gamma,</math> so <math>\triangle PAO\sim\triangle HAQ</math> by SAS similarity. Thus, <math>\angle AOP = \angle AQH,</math> so <math>\angle AOP</math> is a right angle. |
− | Since < | + | Since <math>\angle AOP</math> and <math>\angle AOQ</math> are both right angles, we get <math>\angle POQ = \pi,</math> so we conclude that <math>P, O, Q</math> are collinear, so we are done. We also obtain the extra fact that <math>AO\perp PQ.</math> |
Revision as of 03:07, 27 April 2016
Contents
[hide]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 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, so we are done. We also obtain the extra fact that
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 |