Difference between revisions of "2016 USAJMO Problems/Problem 5"
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2\cdot \frac{x-\frac{1}{x}}{2i}\cdot \frac{y-\frac{1}{y}}{2i}\cdot \frac{z-\frac{1}{z}}{2i}&=\frac{xyz-\frac{1}{xyz}-\frac{xy}{z}-\frac{yz}{x}-\frac{xz}{y}+\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}}{-4i}\\ | 2\cdot \frac{x-\frac{1}{x}}{2i}\cdot \frac{y-\frac{1}{y}}{2i}\cdot \frac{z-\frac{1}{z}}{2i}&=\frac{xyz-\frac{1}{xyz}-\frac{xy}{z}-\frac{yz}{x}-\frac{xz}{y}+\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}}{-4i}\\ | ||
&=\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-x^2-y^2-z^2}{-4i},\end{align*}</cmath> which is equivalent to the left hand side. Therefore, the determinant is <math>0,</math> and <math>O,P,Q</math> are collinear. <math>\blacksquare</math> | &=\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-x^2-y^2-z^2}{-4i},\end{align*}</cmath> which is equivalent to the left hand side. Therefore, the determinant is <math>0,</math> and <math>O,P,Q</math> are collinear. <math>\blacksquare</math> | ||
+ | |||
+ | |||
+ | == Solution 3 == | ||
+ | For convenience, let <math>a, b, c</math> denote the lengths of segments <math>BC, CA, AB,</math> respectively, and let <math>\alpha, \beta, \gamma</math> denote the measures of <math>\angle CAB, \angle ABC, \angle BCA,</math> respectively. Let <math>R</math> denote the circumradius of <math>\triangle ABC.</math> | ||
+ | |||
+ | Clearly, <math>AO = R.</math> Since <math>AH^2 = 2\cdot AO^2,</math> we have <math>AH = \sqrt{2}R.</math> Thus, <math>AH\cdot AO = \sqrt{2}R^2.</math> | ||
+ | |||
+ | Note that <math>AH = b\sin\gamma = c\sin\beta.</math> Then, since <math>\angle PHA = \beta</math> and <math>\angle QHA = \gamma,</math> we have: | ||
+ | <cmath>AP = AH\sin\beta = c\sin^2\beta</cmath> | ||
+ | <cmath>AQ = AH\sin\gamma = b\sin^2\gamma</cmath> | ||
+ | The Extended Law of Sines states that: | ||
+ | <cmath>\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R.</cmath> | ||
+ | |||
+ | Since <math>AP = \frac{b^2 c}{4R^2}</math> and <math>AQ = \frac{bc^2}{4R^2},</math> | ||
+ | <math></math>AP\cdot AQ = \frac{b^3 c^3}{16R^4} = \frac{16\sqrt{2}R^6}{16R^4} = \sqrt{2}R^2.<math> | ||
+ | We see that </math>AP\cdot AQ = AH\cdot AO.<math> | ||
+ | |||
+ | 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 </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 </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.$ | ||
+ | |||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 04:07, 27 April 2016
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, Q
AO\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 |