Difference between revisions of "2016 USAJMO Problems/Problem 5"
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Given that <cmath>AH^2=2\cdot AO^2,</cmath>prove that the points <math>O,P,</math> and <math>Q</math> are collinear. | Given that <cmath>AH^2=2\cdot AO^2,</cmath>prove that the points <math>O,P,</math> and <math>Q</math> are collinear. | ||
− | == Solution == | + | == Solution 1== |
+ | It is well-known that <math>AH\cdot 2AO=AB\cdot AC</math> (just use similar triangles or standard area formulas). Then by Power of a Point, | ||
+ | <cmath>AP\cdot AB=AH^2=AQ\cdot AC</cmath> Consider the transformation <math>X\mapsto \Psi(X)</math> which dilates <math>X</math> from <math>A</math> by a factor of <math>\dfrac{AB}{AQ}=\dfrac{AC}{AP}</math> and reflects about the <math>A</math>-angle bisector. Then <math>\Psi(O)</math> clearly lies on <math>AH</math>, and its distance from <math>A</math> is <cmath>AO\cdot\frac{AB}{AQ}=AO\cdot\frac{AB}{\frac{AH^2}{AC}}=AO\cdot\frac{AB\cdot AC}{AH^2}=\frac{AO\cdot AH\cdot 2AO}{AH^2}=\frac{2AO^2}{AH}=AH</cmath> so <math>\Psi(O)=H</math>, hence we conclude that <math>O,P,Q</math> are collinear, as desired. | ||
+ | |||
+ | == Solution 2== | ||
We will use barycentric coordinates with respect to <math>\triangle ABC.</math> The given condition is equivalent to <math>(\sin B\sin C)^2=\frac{1}{2}.</math> Note that <cmath>O=(\sin(2A):\sin(2B):\sin(2C)), P=(\cos^2B,\sin^2B,0), Q=(\cos^2C,0,\sin^2C).</cmath> Therefore, we must show that <cmath>\begin{vmatrix} | We will use barycentric coordinates with respect to <math>\triangle ABC.</math> The given condition is equivalent to <math>(\sin B\sin C)^2=\frac{1}{2}.</math> Note that <cmath>O=(\sin(2A):\sin(2B):\sin(2C)), P=(\cos^2B,\sin^2B,0), Q=(\cos^2C,0,\sin^2C).</cmath> Therefore, we must show that <cmath>\begin{vmatrix} |
Revision as of 19:18, 21 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.
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 |