B
OK. Solve it yourself bro.
I CANNOT
B
Then go look up the answer wherever you found the question. I don't care about geometry puzzles tbh. I was only here for the problem you pretended needed solving and tagged me for. KEK
What puzzles you care about?
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B
Much more important ones concerning humanity and other big questions.
Again with the fucking wording of this shit. It's not clear if NP is supposed to rotate in 3D around to touch the point M (and line R) to form a cylinder, or if M is supposed to fold into the center of the triangle, or if this is supposed to be a 2D rotation all around R where R becomes the midpoint of a hexagon.
B
Did you find these from an old Putnam test?
No, it's from a local entrance exam for a university
B
Draw two lines from N and P straight up to touch R. Use Pythagorean to find an expression for these two right triangles. These lines are your circumference length for the cylinder. Put the extra triangles together and fold them into a cylinder. This is the volume you have to subtract for the cylindrical volume of the equilateral triangle.
Now do the work yourself.
What grade is this for?
High school? Uni?
Just curious.
High school? Uni?
Just curious.
High school
Thanks, found the answer
If T and R were at the edges of the cube the shape you'd have would be exactly half of a cube. Given how they are now each at half the sides of the cube my intuition says the volume should be 1/8 of the volume of the cube.
Damn. We didn't have shit like that in HS. Where are you from?
Brutal. What was the answer?found the answer
Thought Pi was for circles only.
Moggs Iranian education.
Death to all schools tho torturing you for useless shit like this.
Over
Thought Pi was for circles only.
Moggs Iranian education.
Death to all schools tho torturing you for useless shit like this.
GrAYcel country moment.
They take too long to answer the questions
B
Well done.Thanks, found the answer
I cheated on some subjects but I'm still in hs at 19Where do you fucking study that you have to answer shit like that? Thank fuck I cheated my way through HS
B
I got (3x^3)/16pi, where r = [(√3)x^2]/4pi.
I folded the NP side in a circle connected parallel to R.
You get a cilinder of height X, and R is x times the root of 3 divided by two, since it's an equilateral triangle.
Therefore pi . 3x³/4 minus the volume of the two cones.
(xr3/2)² times x/2 times 1/3 times 2, therefore pi. 3x³/12. then 3x³pi/4 minus pix³/4, therefore pix³/2
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B
Let A be the line that goes from vertex N or P and touches line R. Line A is perpendicular to R, since R is parallel to edge NP.
At point M, the height is x/2, since the triangle is equilateral. Using Pythagoras' Theorem, where x/2 is opposite, x is hypotenuse, and A is adjacent. We have A = (x√3)/2, for the new triangle formed with A.
Now you calculate the radius, r, of the circle formed by folding line A into a circle, where NP touches R in parallel. We have r = [(x√3)/2]/2pi = (x√3)/4pi. Note: Line A becomes the circumference.
We have four equal triangles here formed by line A at with parallel lines: N -> R, P -> R, and (x/2) -> M. Therefore, the volume formed by the equilateral triangle is exactly 1/2 (of the full cylinder), given by, V(triangle) = { pi [ (x√3)/4 pi ]^2} x/2 = (3x^3)/32pi, where h = x/2. I've spaced out the LHS a bit to make it more readable.
I made an error in the previous post. It should be 32pi in the denominator. Let me know if I've made any errors in the logic.
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I live from day to day, not worrying about the future
(x√3)/2 is already the radius of the cylinder, so there is no need to calculate a circle.
What about this one: ''In the coordinate system, let P(a, b) be the intersection point between the lines given by the real functions f and g defined by f(x) = 1/2 to the power of x and g(x) logx in the base 1/2.
It's correct to state that:''
a)a = log2( 1/log2(1/a) ) (log of 1/a in the base 2)
b)a = log2( log2 a ) (log of a in the base 2)
c)a = log1/2((log1/2(1/a) (log of 1/a in the base 1/2)
d)a = log2( log1/2 a ) (log of a in the 1/2 base)
Important, the first number after the log is its base
B
It's the circumference. Check your work again.
B
Mark your MNP, and line R on the cylinder on the right. I don't think you followed the instruction as written.
B
That's incorrect. This is where the line R passes through the vertex M and is PERPENDICULAR to the line NP. The question says parallel.
R is the radium, not the line where the rotation happens, as I'ts on the other draw
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B
WTF IS PAPPUS GULDINIS?
ROTATION OF THE TRIANGLE AROUND THE LINE R MEANS THE TRIANGLE FROM THE LINE NP BENDS IN A CIRCLE AROUND TO TOUCH THE VERTEX AND LINE R, THUS MAKING A CYLINDER.
FUCK YOUR POORLY WORDED QUESTIONS.
Two theorems that express the relation between the volume and area of the revolution solid. Since you tried to calculate the circumference and the distance between the triangle and the line r, I thought you were trying to use its formula.
The rotation will make a cylinder, but the question is not asking the volume of the cylinder
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B
No, I didn't use those theorems.
B
After further consideration, your drawing is correct. The wording of the question is still fucking confusing and implies a folding of the triangle. It needs to elaborate more and specify that vertex M becomes the centerpoint of the cylinder.
There's no future for inkwells, that is why we repeat: "it's over"
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