Difference between revisions of "1979 USAMO Problems/Problem 2"
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<math>N</math> is the north pole. <math>A</math> and <math>B</math> are points on a great circle through <math>N</math> equidistant from <math>N</math>. <math>C</math> is a point on the equator. Show that the great circle through <math>C</math> and <math>N</math> bisects the angle <math>ACB</math> in the spherical triangle <math>ABC</math> (a spherical triangle has great circle arcs as sides). | <math>N</math> is the north pole. <math>A</math> and <math>B</math> are points on a great circle through <math>N</math> equidistant from <math>N</math>. <math>C</math> is a point on the equator. Show that the great circle through <math>C</math> and <math>N</math> bisects the angle <math>ACB</math> in the spherical triangle <math>ABC</math> (a spherical triangle has great circle arcs as sides). | ||
− | == | + | ==Solution== |
− | + | ||
+ | [[File:USAMO_1979_P2a.png|500px]] | ||
− | |||
Since <math>N</math> is the north pole, we define the Earth with a sphere of radius one in space with <math>N=(0,0,1)</math> and sphere center <math>O=(0,0,0)</math> | Since <math>N</math> is the north pole, we define the Earth with a sphere of radius one in space with <math>N=(0,0,1)</math> and sphere center <math>O=(0,0,0)</math> | ||
− | We then pick point <math>N</math> on the sphere and define the <math>xz | + | We then pick point <math>N</math> on the sphere and define the <math>xz</math>-plane as the plane that contains great circle points <math>A</math> , <math>B</math>, and <math>N</math> with the <math>x</math>-axis perpendicular to the <math>z</math>-axis and in the direction of <math>A</math>. |
− | Using this coordinate system and <math>x</math>, <math>y</math>, and <math>z</math> axes <math>A=(cos(\phi),0,sin(\phi))</math> where <math>\phi</math> is the angle from the <math>xy | + | Using this coordinate system and <math>x</math>, <math>y</math>, and <math>z</math> axes <math>A=(cos(\phi),0,sin(\phi))</math> where <math>\phi</math> is the angle from the <math>xy</math>-plane to <math>A</math> or latitude on this sphere with <math>\frac{-\pi}{2} < \phi < \frac{\pi}{2}</math> |
Since <math>A</math> and <math>B</math> are points on a great circle through <math>N</math> equidistant from <math>N</math>, then <math>B=(-cos(\phi),0,sin(\phi))</math> | Since <math>A</math> and <math>B</math> are points on a great circle through <math>N</math> equidistant from <math>N</math>, then <math>B=(-cos(\phi),0,sin(\phi))</math> | ||
− | Since <math>C</math> is a point on the equator, then <math>C=(cos(\theta),sin(\theta),0)</math> where <math>\theta</math> is the angle on the <math>xy | + | Since <math>C</math> is a point on the equator, then <math>C=(cos(\theta),sin(\theta),0)</math> where <math>\theta</math> is the angle on the <math>xy</math>-plane from the origin to <math>C</math> or longitude on this sphere with <math>-\pi < \phi \le \pi</math> |
We note that vectors from the origin to points <math>N</math>, <math>A</math>, <math>B</math>, and <math>C</math> are all unit vectors because all those points are on the unit sphere. | We note that vectors from the origin to points <math>N</math>, <math>A</math>, <math>B</math>, and <math>C</math> are all unit vectors because all those points are on the unit sphere. | ||
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We also define the following vectors as follows: | We also define the following vectors as follows: | ||
− | Vector | + | Vector <math>\overrightarrow{V_{CN}}</math> is the unit vector in the direction of arc <math>CN</math> and tangent to the great circle of <math>CN</math> at <math>C</math> |
+ | |||
+ | Vector <math>\overrightarrow{V_{CA}}</math> is the unit vector in the direction of arc <math>CA</math> and tangent to the great circle of <math>CA</math> at <math>C</math> | ||
+ | |||
+ | Vector <math>\overrightarrow{V_{CB}}</math> is the unit vector in the direction of arc <math>CB</math> and tangent to the great circle of <math>CB</math> at <math>C</math> | ||
+ | |||
+ | To calculate each of these vectors we shall use the cross product as follows: | ||
+ | |||
+ | <math>\overrightarrow{V_{CN}}=(\overrightarrow{C}\times\overrightarrow{N})\times\overrightarrow{C}</math> | ||
+ | |||
+ | <math>\overrightarrow{V_{CN}}=(\left\langle cos(\theta),sin(\theta),0 \right\rangle\times\left\langle 0,0,1 \right\rangle)\times \left\langle cos(\theta),sin(\theta),0 \right\rangle</math> | ||
+ | |||
+ | <math>\overrightarrow{V_{CN}}=\left\langle sin(\theta),-cos(\theta),0 \right\rangle\times \left\langle cos(\theta),sin(\theta),0 \right\rangle</math> | ||
+ | |||
+ | <math>\overrightarrow{V_{CN}}=\left\langle 0,0,sin^{2}(\theta)+cos^{2}(\theta) \right\rangle</math> | ||
+ | |||
+ | <math>\overrightarrow{V_{CN}}=\left\langle 0,0,1 \right\rangle</math> | ||
+ | |||
+ | Vector <math>\overrightarrow{V_{CA}}</math>: | ||
+ | |||
+ | <math>\overrightarrow{V_{CA}}=(\overrightarrow{C}\times\overrightarrow{A})\times\overrightarrow{C}</math> | ||
+ | |||
+ | <math>\overrightarrow{V_{CA}}=(\left\langle cos(\theta),sin(\theta),0 \right\rangle\times\left\langle cos(\phi),0,sin(\phi) \right\rangle)\times \left\langle cos(\theta),sin(\theta),0 \right\rangle</math> | ||
+ | |||
+ | <math>\overrightarrow{V_{CA}}=\left\langle sin(\theta)sin(\phi),-cos(\theta)sin(\phi),-sin(\theta)cos(\phi) \right\rangle\times \left\langle cos(\theta),sin(\theta),0 \right\rangle</math> | ||
+ | |||
+ | Since we're only interested in the <math>z</math> component of the vector | ||
+ | |||
+ | <math>\overrightarrow{V_{CA}}=\left\langle V_{CA_{x}},V_{CA_{y}},sin^{2}(\theta)sin(\phi)+cos^{2}(\theta)sin(\phi) \right\rangle</math> | ||
+ | |||
+ | <math>\overrightarrow{V_{CA}}=\left\langle V_{CA_{x}},V_{CA_{y}},sin(\phi) \right\rangle</math> | ||
+ | |||
+ | Vector <math>\overrightarrow{V_{CB}}</math>: | ||
+ | |||
+ | <math>\overrightarrow{V_{CB}}=(\overrightarrow{C}\times\overrightarrow{b})\times\overrightarrow{C}</math> | ||
+ | |||
+ | <math>\overrightarrow{V_{CB}}=(\left\langle cos(\theta),sin(\theta),0 \right\rangle\times\left\langle -cos(\phi),0,sin(\phi) \right\rangle)\times \left\langle cos(\theta),sin(\theta),0 \right\rangle</math> | ||
+ | |||
+ | <math>\overrightarrow{V_{CB}}=\left\langle sin(\theta)sin(\phi),-cos(\theta)sin(\phi),sin(\theta)cos(\phi) \right\rangle\times \left\langle cos(\theta),sin(\theta),0 \right\rangle</math> | ||
+ | |||
+ | Since we're only interested in the <math>z</math> component of the vector | ||
+ | |||
+ | <math>\overrightarrow{V_{CB}}=\left\langle V_{CB_{x}},V_{CB_{y}},sin^{2}(\theta)sin(\phi)+cos^{2}(\theta)sin(\phi) \right\rangle</math> | ||
+ | |||
+ | <math>\overrightarrow{V_{CB}}=\left\langle V_{CB_{x}},V_{CB_{y}},sin(\phi) \right\rangle</math> | ||
+ | |||
+ | Since we're working with unit vectors, then we can use dot products on the vectors with their angles as follows: | ||
+ | |||
+ | <math>cos(\angle ACN) = \overrightarrow{V_{CA}}\cdot \overrightarrow{V_{CN}}</math> | ||
+ | |||
+ | <math>cos(\angle ACN) = \left\langle V_{CA_{x}},V_{CA_{y}},sin(\phi) \right\rangle \cdot \left\langle 0,0,1 \right\rangle = 0*V_{CA_{x}}+0*V_{CA_{y}}+1*sin(\phi)=sin(\phi) </math> | ||
+ | |||
+ | Likewise, | ||
+ | |||
+ | <math>cos(\angle BCN) = \overrightarrow{V_{CB}}\cdot \overrightarrow{V_{CN}}</math> | ||
+ | |||
+ | <math>cos(\angle BCN) = \left\langle V_{CB_{x}},V_{CB_{y}},sin(\phi) \right\rangle \cdot \left\langle 0,0,1 \right\rangle = 0*V_{CB_{x}}+0*V_{CB_{y}}+1*sin(\phi)=sin(\phi) </math> | ||
+ | |||
+ | Therefore, | ||
+ | |||
+ | <math>cos(\angle ACN) = cos(\angle BCN)</math> and thus <math>\angle ACN = \angle BCN</math> | ||
+ | |||
+ | Since those angles are equal, it proves that the great circle through <math>C</math> and <math>N</math> bisects the <math>\angle ACB</math> in the spherical triangle <math>ABC</math> | ||
− | ~Tomas Diaz | + | ~Tomas Diaz. orders@tomasdiaz.com |
{{alternate solutions}} | {{alternate solutions}} |
Latest revision as of 22:09, 21 November 2023
Problem
is the north pole.
and
are points on a great circle through
equidistant from
.
is a point on the equator. Show that the great circle through
and
bisects the angle
in the spherical triangle
(a spherical triangle has great circle arcs as sides).
Solution
Since is the north pole, we define the Earth with a sphere of radius one in space with
and sphere center
We then pick point
on the sphere and define the
-plane as the plane that contains great circle points
,
, and
with the
-axis perpendicular to the
-axis and in the direction of
.
Using this coordinate system and ,
, and
axes
where
is the angle from the
-plane to
or latitude on this sphere with
Since and
are points on a great circle through
equidistant from
, then
Since is a point on the equator, then
where
is the angle on the
-plane from the origin to
or longitude on this sphere with
We note that vectors from the origin to points ,
,
, and
are all unit vectors because all those points are on the unit sphere.
So, we're going to define points ,
,
, and
as unit vectors with their coordinates.
We also define the following vectors as follows:
Vector is the unit vector in the direction of arc
and tangent to the great circle of
at
Vector is the unit vector in the direction of arc
and tangent to the great circle of
at
Vector is the unit vector in the direction of arc
and tangent to the great circle of
at
To calculate each of these vectors we shall use the cross product as follows:
Vector :
Since we're only interested in the component of the vector
Vector :
Since we're only interested in the component of the vector
Since we're working with unit vectors, then we can use dot products on the vectors with their angles as follows:
Likewise,
Therefore,
and thus
Since those angles are equal, it proves that the great circle through and
bisects the
in the spherical triangle
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1979 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.