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 20:18, 21 April 2016
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.
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 |