Sphere inscribed in a tetrahedron
Thread Rating:
Poll
 | 1 vote (11.11%) | ||
| | 2 votes (22.22%) | ||
| | 2 votes (22.22%) | ||
| | 2 votes (22.22%) | ||
| | 2 votes (22.22%) | ||
| | 1 vote (11.11%) | ||
| | 1 vote (11.11%) | ||
| | 1 vote (11.11%) | ||
| | 2 votes (22.22%) | ||
| | 1 vote (11.11%) |
9 members have voted
April 18th, 2019 at 4:19:38 PM
permalink
A sphere is inscribed in a regular tetrahedron. What is the ratio of the volume of the sphere to the volume of the tetrahedron?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 19th, 2019 at 7:23:32 AM
permalink
How did the date go between oxygen and potassium?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 19th, 2019 at 9:40:07 AM
permalink
Quote: WizardHow did the date go between oxygen and potassium?
OK I guess
April 19th, 2019 at 10:23:54 AM
permalink
Quote: rsactuaryOK I guess
Correct!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 19th, 2019 at 10:54:21 AM
permalink
But not scientifically accurate. Turns out potassium likes to date with friends: K20. Lucky Oxygen dog.Quote: WizardCorrect!
(K2O2 also works, if you are into that type of thing.)
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
April 19th, 2019 at 11:09:52 AM
permalink
Quote: unJonBut not scientifically accurate. Turns out potassium likes to date with friends: K20. Lucky Oxygen dog.
(K2O2 also works, if you are into that type of thing.)
That is why I love this forum -- comments like that.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 19th, 2019 at 11:27:25 AM
permalink
ok i'll embarass myself again
4/3 pi sqrt(2)
the radius of the sphere is r
the point where the sphere touches one of the sides faces is in the center of the face
in a 30 x 60 x 90 triangle, the short leg is x, the long leg is x * sqrt (3), and the hypotenuse is x
the distance from the center of the face to one of the vertices is r * sqrt ( 3 )
now on the face we need to find the length of one of the sides
same situation, but this time we have the hypotenuse and need to solve for the long side times 2
I get 6r / 2, times 2 is 6r.
so the "a" length of the tetrehedron is 6r
volume of the sphere is 4/3 pi r^3
volume of the tetrahedron is a^3 / ( 6 * sqrt(2)) or (6r)^ 3 / ( 6 * sqrt(2) )
I seem to have lost some work here, but dividing one volume by the other
( 8 pi r^3 sqrt(2) ) / 6r^3
4/3 pi sqrt(2)
4/3 pi sqrt(2)
April 19th, 2019 at 12:18:18 PM
permalink
Quote: Dalex64ok i'll embarass myself again
the radius of the sphere is r
the point where the sphere touches one of the sides faces is in the center of the face
in a 30 x 60 x 90 triangle, the short leg is x, the long leg is x * sqrt (3), and the hypotenuse is x
the distance from the center of the face to one of the vertices is r * sqrt ( 3 )
now on the face we need to find the length of one of the sides
same situation, but this time we have the hypotenuse and need to solve for the long side times 2
I get 6r / 2, times 2 is 6r.
so the "a" length of the tetrehedron is 6r
volume of the sphere is 4/3 pi r^3
volume of the tetrahedron is a^3 / ( 6 * sqrt(2)) or (6r)^ 3 / ( 6 * sqrt(2) )
I seem to have lost some work here, but dividing one volume by the other
( 8 pi r^3 sqrt(2) ) / 6r^3
4/3 pi sqrt(2)
4/3 pi sqrt(2)
Maybe I'm the one in error, but I don't agree. In particular, I don't agree that the length of a side the triangle is six times the radius of the sphere, if that is what you were saying.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 19th, 2019 at 12:31:45 PM
permalink
Quote: unJonBut not scientifically accurate. Turns out potassium likes to date with friends: K20. Lucky Oxygen dog.
(K2O2 also works, if you are into that type of thing.)
Yep, one of those Kinky menage-a troises or foursomes.
This reminds me of the Wizard's joke about Silver Surfer and Iron Man being alloys (the problem being that silver does not alloy with iron or any other metal other than Mercury.)
So, I would speculate that, when dating, Wizard has a good sense of humor but poor chemistry.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
April 19th, 2019 at 4:06:17 PM
permalink
Quote: gordonm888So, I would speculate that, when dating, Wizard has a good sense of humor but poor chemistry.
Hah! You're at least correct on the poor chemistry part.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 20th, 2019 at 3:25:07 PM
permalink
Q: Did you hear about the snowman who got cooled to absolute zero?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 20th, 2019 at 3:58:55 PM
permalink
Quote: WizardQ: Did you hear about the snowman who got cooled to absolute zero?
Is he OK?
To make the calculations easy, I thought of a cube with these vertices: 000, 600, 060, 006, 660, 606, 066, and 666. (Of course, I should be typing the coordinates as (x,y,z) instead of xyz as in (0,0,0) instead of 000, for example.)
For the regular tetrahedron’s vertices, I selected these four of the cube’s vertices: 000, 660, 606, and 660.
The coordinates of the center of one face of the regular tetrahedron is the average of the corresponding coordinates of the face’s three vertices. The center of the tetrahedron’s face with vertices 660, 606, and 066 is thus 444.
The cube, the tetrahedron, and the inscribed sphere are all centered at 333. The inscribed sphere’s radius is the distance from 333 to 444 which is the square root of (4-3)2 + (4-3)2 + (4-3)2, or the square root of 3. So, the inscribed sphere has a volume of (4/3)(pi)(r^3) = 4(pi)sqrt(3).
The volume of our tetrahedron can be thought of as the volume of the cube minus four pyramids, each of height 6 and area of base (1/2)(6)(6), which are volumes of (1/3)(6)(18) = 36 each. So, the volume of the tetrahedron is 63 – 4(36) = 216 – 144 = 72.
The ratio of the volume of the inscribed sphere to tetrahedron’s volume is 4(pi)sqrt(3) / 72 = (pi) sqrt(3) / 18, or about 0.302300.
April 20th, 2019 at 5:34:11 PM
permalink
This may be the worst post of the month.
Actually, my answer to the snowman riddle is:
Is he 0K?
Chester used the letter O and I used the zero character.
How's that for trivial one-upmanship of the frivolous part of Chester's brilliant post?
Actually, my answer to the snowman riddle is:
Is he 0K?
Chester used the letter O and I used the zero character.
How's that for trivial one-upmanship of the frivolous part of Chester's brilliant post?
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
April 20th, 2019 at 5:38:20 PM
permalink
Last night I dreamt that I weighed much less than a thousandth of a gram. I was like 0mg.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
April 20th, 2019 at 7:48:31 PM
permalink
Quote: ChesterDogIs he OK?
As Gordon pointed out, the question should be is he 0K? Note the zero and not a letter O.
Quote:
To make the calculations easy, I thought of a cube with these vertices: 000, 600, 060, 006, 660, 606, 066, and 666. (Of course, I should be typing the coordinates as (x,y,z) instead of xyz as in (0,0,0) instead of 000, for example.)
For the regular tetrahedron’s vertices, I selected these four of the cube’s vertices: 000, 660, 606, and 660.
The coordinates of the center of one face of the regular tetrahedron is the average of the corresponding coordinates of the face’s three vertices. The center of the tetrahedron’s face with vertices 660, 606, and 066 is thus 444.
The cube, the tetrahedron, and the inscribed sphere are all centered at 333. The inscribed sphere’s radius is the distance from 333 to 444 which is the square root of (4-3)2 + (4-3)2 + (4-3)2, or the square root of 3. So, the inscribed sphere has a volume of (4/3)(pi)(r^3) = 4(pi)sqrt(3).
The volume of our tetrahedron can be thought of as the volume of the cube minus four pyramids, each of height 6 and area of base (1/2)(6)(6), which are volumes of (1/3)(6)(18) = 36 each. So, the volume of the tetrahedron is 63 – 4(36) = 216 – 144 = 72.
The ratio of the volume of the inscribed sphere to tetrahedron’s volume is 4(pi)sqrt(3) / 72 = (pi) sqrt(3) / 18, or about 0.302300.
I agree!!! Add one more to the tab of beers I owe you. Very well done, that was a tough problem.
With that one asked, please see the next math puzzle thread I'm about to start.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
