Difference between revisions of "2016 USAJMO Problems/Problem 5"
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== Solution 3 == | == 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> | 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> | ||
+ | |||
+ | Since the central angle <math>\angle AOB</math> subtends the same arc as the inscribed angle <math>\angle ACB</math> on the circumcircle of <math>\triangle ABC,</math> we have <math>\angle AOB = 2\gamma.</math> Note that <math>OA = OB,</math> so <math>\angle OAB = \angle OBA.</math> Thus, <math>\angle OAB = \frac{\pi}{2} - \gamma.</math> Similarly, one can show that <math>\angle OAC = \frac{\pi}{2} - \beta.</math> (One could probably cite this as well-known, but I have proved it here just in case.) | ||
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> | 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> | ||
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The Extended Law of Sines states that: | The Extended Law of Sines states that: | ||
<cmath>\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R.</cmath> | <cmath>\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R.</cmath> | ||
− | + | Therefore, <math>AP = \frac{b^2 c}{4R^2}</math> and <math>AQ = \frac{bc^2}{4R^2}.</math> It follows that: | |
− | |||
<cmath>AP\cdot AQ = \frac{b^3 c^3}{16R^4} = \frac{16\sqrt{2}R^6}{16R^4} = \sqrt{2}R^2.</cmath> | <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 <math>AP\cdot AQ = AH\cdot AO.</math> | We see that <math>AP\cdot AQ = AH\cdot AO.</math> | ||
− | Rearranging <math>AP\cdot AQ = AH\cdot AO,</math> we get | + | 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 | + | 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.</math> | 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 04:12, 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
Since the central angle subtends the same arc as the inscribed angle
on the circumcircle of
we have
Note that
so
Thus,
Similarly, one can show that
(One could probably cite this as well-known, but I have proved it here just in case.)
Clearly, Since
we have
Thus,
Note that Then, since
and
we have:
The Extended Law of Sines states that:
Therefore,
and
It follows that:
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 |