Difference between revisions of "1979 USAMO Problems/Problem 2"
(→Solution) |
|||
(16 intermediate revisions by the same user not shown) | |||
Line 3: | Line 3: | ||
<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. | ||
Line 34: | Line 34: | ||
<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 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}}= | + | <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 sin(\theta) | + | <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> | <math>\overrightarrow{V_{CN}}=\left\langle 0,0,1 \right\rangle</math> | ||
Line 70: | Line 70: | ||
Since we're working with unit vectors, then we can use dot products on the vectors with their angles as follows: | 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) = \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> | ||
− | <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 21:09, 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.