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** [https://bmos.ukmt.org.uk/home/bmo.shtml Past Problems] | ** [https://bmos.ukmt.org.uk/home/bmo.shtml Past Problems] | ||
** [https://bmos.ukmt.org.uk/solutions/ Video Solutions] | ** [https://bmos.ukmt.org.uk/solutions/ Video Solutions] | ||
+ | ==8-th British Mathematical Olympiad 1972 Problem 5== | ||
+ | In a right circular cone the semi-vertical angle of which is <math>\theta</math>, a cube is placed so that four of its vertices are on the base and four on the curved surface. Prove that as <math>\theta</math> varies the maximum value of the ratio of the volume of the cube to the volume of the cone occurs when <math>3\sin \theta = 1.</math> | ||
+ | ===Solution=== | ||
+ | [[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> | ||
+ | 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>\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 | ||
+ | <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>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> | ||
+ | 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''' | ||
+ | |||
+ | ==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=== | ||
+ | [[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 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 the hoop <math>\Omega</math> be <math>B.</math> The hoop is shown by red. | ||
+ | |||
+ | 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> | ||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | [[File:British 1985.png|350px|right]] | ||
+ | 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> | ||
+ | |||
+ | 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>\angle A'C'B = 2 \arcsin \frac {3}{5} \implies</cmath> | ||
+ | <cmath>A'A'' = 2(\frac {96}{25} - 3) = \frac {42}{25} = 1.68.</cmath> | ||
+ | The locus of the point <math>A</math> is the belt with a width of <math>1.68</math> cm located on a sphere with a radius of <math>3</math> cm symmetrically to the circumference of the great circle located parallel hoop. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Comments== | ||
{{stub}} | {{stub}} | ||
[[Category:Mathematics competitions]] | [[Category:Mathematics competitions]] |
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
Comments
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