Difference between revisions of "British Mathematical Olympiad"
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[[File:Ron Cone Cube D.png|400px|right]] | [[File:Ron Cone Cube D.png|400px|right]] | ||
Let the cube side be <math>a,</math> height of the cone be <math>h,</math> radius of the cone be <math>r = h \tan \theta.</math> | Let the cube side be <math>a,</math> height of the cone be <math>h,</math> radius of the cone be <math>r = h \tan \theta.</math> | ||
− | + | See diagram for the description of terms used. <math>A</math> is the vertex of the cone, <math>B</math> is the center of the cube upper face, <math>C</math> is the vertex of upper surface of the cube, <math>O</math> is the center of the base of the cone, <math>D</math> is the point where <math>AC</math> crosses the base of the cone. | |
<cmath>BO = a, BC = \frac {a}{\sqrt {2}}, AO = h, DO = r = h \tan \theta.</cmath> | <cmath>BO = a, BC = \frac {a}{\sqrt {2}}, AO = h, DO = r = h \tan \theta.</cmath> | ||
<cmath>\frac {AB}{AO} = \frac {BC}{OD} \implies \frac {h-a}{h} = \frac {a}{r \sqrt{2}} \implies \frac{r}{a} = \tan \theta + \frac {1}{\sqrt{2}}.</cmath> | <cmath>\frac {AB}{AO} = \frac {BC}{OD} \implies \frac {h-a}{h} = \frac {a}{r \sqrt{2}} \implies \frac{r}{a} = \tan \theta + \frac {1}{\sqrt{2}}.</cmath> | ||
The ratio of the volume of the cube to the volume of the cone is | The ratio of the volume of the cube to the volume of the cone is | ||
<cmath>\bar V = \frac {3a^3}{\pi r^2 h} = \frac {3r}{\pi h} \cdot \left( \frac {a}{r} \right)^3 = \frac {3}{\pi} \cdot \frac {\tan \theta}{(\tan \theta + \frac {1}{\sqrt{2}})^3} = \frac {3}{\pi f^3}.</cmath> | <cmath>\bar V = \frac {3a^3}{\pi r^2 h} = \frac {3r}{\pi h} \cdot \left( \frac {a}{r} \right)^3 = \frac {3}{\pi} \cdot \frac {\tan \theta}{(\tan \theta + \frac {1}{\sqrt{2}})^3} = \frac {3}{\pi f^3}.</cmath> | ||
− | + | Here we use <cmath>x^3 = \tan \theta, f = \frac {x^3 + \frac {1}{\sqrt{2}} }{x} = x^2 + \frac {1}{x \sqrt {2}} = x^2 + \frac {1}{2x \sqrt {2}} + \frac {1}{2x \sqrt {2}},</cmath> | |
<cmath>f \ge 3 \sqrt [3] {x^2 \cdot \frac {1}{2x \sqrt {2}} \cdot \frac {1}{2x \sqrt {2}}} = \frac {3}{2}</cmath> if | <cmath>f \ge 3 \sqrt [3] {x^2 \cdot \frac {1}{2x \sqrt {2}} \cdot \frac {1}{2x \sqrt {2}}} = \frac {3}{2}</cmath> if | ||
<cmath>x^2 = \frac {1}{2x \sqrt {2}} \implies \tan \theta = x^3 = \frac {1}{2 \sqrt {2}} \implies \sin \theta = \frac {1}{3}.</cmath> | <cmath>x^2 = \frac {1}{2x \sqrt {2}} \implies \tan \theta = x^3 = \frac {1}{2 \sqrt {2}} \implies \sin \theta = \frac {1}{3}.</cmath> | ||
<cmath> \bar V = \frac {3}{\pi f^3} \le \frac {8}{9 \pi} = 0.283.</cmath> | <cmath> \bar V = \frac {3}{\pi f^3} \le \frac {8}{9 \pi} = 0.283.</cmath> | ||
+ | The maximum ratio of the volume of the cube to the volume of the cone is <math>0.283.</math> | ||
+ | |||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
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[[File:British 1985 3D.png|350px|right]] | [[File:British 1985 3D.png|350px|right]] | ||
− | Let the centroid of the cylinder be the point <math>O.</math> The side surface of the cylinder is shown by | + | Let the centroid of the cylinder be the point <math>O.</math> The side surface of the cylinder is shown by blue. |
Let the center of one of the circular ends be the point <math>A.</math> This end is shown by green. Its edge is a purple circle <math>\omega, OA = 3.</math> | Let the center of one of the circular ends be the point <math>A.</math> This end is shown by green. Its edge is a purple circle <math>\omega, OA = 3.</math> | ||
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Let <math>\omega</math> cross <math>\Omega</math> at point <math>C.</math> Therefore <math>\omega</math> cross <math>\Omega</math> at second point symmetrical to <math>C</math> with respect to the plane <math>OAB.</math> | Let <math>\omega</math> cross <math>\Omega</math> at point <math>C.</math> Therefore <math>\omega</math> cross <math>\Omega</math> at second point symmetrical to <math>C</math> with respect to the plane <math>OAB.</math> | ||
<cmath>AC = BC = 4, BC \perp OA \implies AC = 5.</cmath> | <cmath>AC = BC = 4, BC \perp OA \implies AC = 5.</cmath> | ||
− | Let <math>\Theta</math> be the sphere of radius <math>5</math> centered at <math>O.</math> Part of this sphere is shown in the diagram by yellow. Let the cylinder | + | Let <math>\Theta</math> be the sphere of radius <math>5</math> centered at <math>O.</math> Part of this sphere is shown in the diagram by yellow. Let the cylinder be glued to the sphere and the point <math>O</math> be fixed. |
In this case <math>\omega</math> and <math>\Omega</math> both lie on <math>S,</math> and point <math>A</math> lies on the sphere centered at <math>O</math> with radius <math>OA = 3.</math> | In this case <math>\omega</math> and <math>\Omega</math> both lie on <math>S,</math> and point <math>A</math> lies on the sphere centered at <math>O</math> with radius <math>OA = 3.</math> | ||
− | The | + | The claim “The cylinder is free to move subject to the condition that each of its circular ends always touches the hoop at two points” has the equivalent form “The sphere is free to move with fixed center subject to the condition that <math>\omega</math> cross <math>\Omega.</math>” |
If the sphere rotates around an axis <math>OB,</math> then point <math>A</math> moves along circle with axis <math>OB.</math> | If the sphere rotates around an axis <math>OB,</math> then point <math>A</math> moves along circle with axis <math>OB.</math> | ||
[[File:British 1985.png|350px|right]] | [[File:British 1985.png|350px|right]] | ||
− | Let the sphere | + | Let the sphere rotate around an axis perpendicular <math>OA</math> and <math>OB.</math> |
Axis view is shown on the diagram. We rotate <math>\Theta </math> together with <math>\omega</math> in counterclockwise direction. Point <math>C</math> moves along <math>\Omega</math> till point <math>C'</math> in the plane <math>OAB</math> where <math>\omega</math> touch <math>\Omega.</math> The point <math>A</math> moves to extreme position <math>A'.</math> | Axis view is shown on the diagram. We rotate <math>\Theta </math> together with <math>\omega</math> in counterclockwise direction. Point <math>C</math> moves along <math>\Omega</math> till point <math>C'</math> in the plane <math>OAB</math> where <math>\omega</math> touch <math>\Omega.</math> The point <math>A</math> moves to extreme position <math>A'.</math> | ||
− | If one | + | If one rotates <math>\Theta</math> in clockwise direction, point <math>A</math> moves to position <math>A''</math> symmetric to <math>A'</math> with respect <math>OA.</math> |
<cmath>OA = OB = 3, BC' = A'C'= 4,</cmath> | <cmath>OA = OB = 3, BC' = A'C'= 4,</cmath> | ||
<cmath>\angle A'C'B = 2 \arcsin \frac {3}{5} \implies</cmath> | <cmath>\angle A'C'B = 2 \arcsin \frac {3}{5} \implies</cmath> |
Latest revision as of 09:12, 7 June 2023
The British Mathematical Olympiad is a national math competition held in the United Kingdom. Solvers who score over a certain threshold in the Senior Mathematical Challenge are automatically entered to the first round, but others can register for the first round.
Contents
[hide]Structure
The British Mathematical Olympiad is divided into two rounds. In the first round (BMO 1), solvers have 3.5 hours to solve 6 problems. High scorers can move on into the second round (BMO 2), where solvers have 3.5 hours to solve 4 problems.
For both rounds, each problem is worth 10 points. Like most Olympiads, complete solutions are required in order to get full credit.
Participants who submit a solution with the highest quality in BMO 2 can earn the Christopher Bradley elegance prize.
Resources
8-th British Mathematical Olympiad 1972 Problem 5
In a right circular cone the semi-vertical angle of which is , a cube is placed so that four of its vertices are on the base and four on the curved surface. Prove that as varies the maximum value of the ratio of the volume of the cube to the volume of the cone occurs when
Solution
Let the cube side be height of the cone be radius of the cone be See diagram for the description of terms used. is the vertex of the cone, is the center of the cube upper face, is the vertex of upper surface of the cube, is the center of the base of the cone, is the point where crosses the base of the cone. The ratio of the volume of the cube to the volume of the cone is Here we use if The maximum ratio of the volume of the cube to the volume of the cone is
vladimir.shelomovskii@gmail.com, vvsss
21-th Mathematical Olympiad 1985 Problem 5
A circular hoop of radius 4 cm is held fixed in a horizontal plane. A cylinder with radius 4 cm and length 6 cm rests on the hoop with its axis horizontal, and with each of its two circular ends touching the hoop at two points. The cylinder is free to move subject to the condition that each of its circular ends always touches the hoop at two points. Find, with proof, the locus of the centre of one of the cylinder’s circular ends.
Solution
Let the centroid of the cylinder be the point The side surface of the cylinder is shown by blue.
Let the center of one of the circular ends be the point This end is shown by green. Its edge is a purple circle
Let the center of the hoop be The hoop is shown by red.
Let cross at point Therefore cross at second point symmetrical to with respect to the plane Let be the sphere of radius centered at Part of this sphere is shown in the diagram by yellow. Let the cylinder be glued to the sphere and the point be fixed.
In this case and both lie on and point lies on the sphere centered at with radius
The claim “The cylinder is free to move subject to the condition that each of its circular ends always touches the hoop at two points” has the equivalent form “The sphere is free to move with fixed center subject to the condition that cross ”
If the sphere rotates around an axis then point moves along circle with axis
Let the sphere rotate around an axis perpendicular and Axis view is shown on the diagram. We rotate together with in counterclockwise direction. Point moves along till point in the plane where touch The point moves to extreme position
If one rotates in clockwise direction, point moves to position symmetric to with respect The locus of the point is the belt with a width of cm located on a sphere with a radius of cm symmetrically to the circumference of the great circle located parallel hoop.
vladimir.shelomovskii@gmail.com, vvsss
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