Three spheres of radius 100 are mutually tangent and rest on a horizontal plane H. A sphere of radius 300 rests over them.
Let d be the largest possible distance of a point on the larger sphere from the plane H. What is ⌊ d ⌋ ?
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An upvote because of the draws :)
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Surely yes.
Thanks
oh man!!!!!!!!!!!!!!.i made a calculation mistake and got 785.
gosh!! I didn't pay attention to "mutually" tangents and assumed that the circles are lined in horizontal row...then the answer would have been:916
Centers of all 4 spheres form a pyramid with equilateral triangular base of side 2 0 0 and say height h
The distance from the vertex of the equilateral triangle to its centroid can be found by 30-60-90 triangles to be 3 2 0 0 .
By the Pythagorean Theorem, we have
( 3 2 0 0 ) 2 + h 2 = 4 0 0 2 ⟹ h = 1 0 0 ⋅ 3 4 4 ≈ 3 8 2 . 9 7
d = 1 0 0 + 3 0 0 + 1 0 0 ⋅ 3 4 4 ≈ 7 8 2 . 9 7
Answer: 7 8 2
Aww I put 482, forgot the extra 300...
I did the same way :) And the title of the problem is really cool!
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thanks
More problems of similar title coming up ...
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I just solved your 400 on top of 4 100s and now this. So, shouldn’t we look for a generalization for n100 on top of n 100s ? Here’s one I came up with –
d = 1 0 0 ⎝ ⎛ n + 1 + sin n 1 8 0 ( ( n + 1 ) sin n 1 8 0 ) 2 − 1 ⎠ ⎞
So, no tension in future.
2004 AMC 12A #22 is very similar
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Hey! Congrats at States in the Countdown round, AND written round! 1st place! My name is Finn, I represented Berkeley, the 5th place team. I was disappointed at not making the top 16, but what are some tips for getting better? I saw your match against Daniel Chiu at Nationals.
You kind of whipped his butt, which I made sure to mention to him. My question, though, is how you got so incredible at math?! I have a note on the matter that you're welcome to post on, but seriously, you are amazing. I really want to go to Nationals next year, so do you have any tips? :DI got a 783!! Damn it!
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You rounded instead of the symbol thingy
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Well, it used to say find the integer part of d.
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@Daniel Chiu – DANG IT. I think I put 782, but I probably typoed.
Dang forgot the 300.
i to forgot adding 300
I think it should have stated that it was the Greatest integer function and not the Least integer function, don't you think?
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So, ⌊ A B ⌋ = ⌊ A O + T B ⌋ = ⌊ 4 0 0 + 3 8 2 . 7 9 ⌋ = 7 8 2
I would argue that you can balance a sphere on another sphere in a mathematical situation. Thus making it 1 0 0 + 1 0 0 + 3 0 0 + 3 0 0 = 8 0 0
the centres of three base spheres form an equilateral triangle of side =200 which inturn a tetrahedron with centre of large sphere with slant height 400 whose projection on the base is 200/sqrt(3) so the height of the tetrahedron is given by sqrt((400 400)-(200 200/3)) and this must be added to base height and radius of the large sphere which is approx..782.97
let O1, O2, O3, O4 be the centers of the circles(O4 is radius of larger circle).
O1, O2, O3, O4 form a pyramid with equilateral triangle(O1,O2,O3) of side 200 as base and and edges of length 400 connected to O4.
the median of the equilateral triangle(M) , one of O1,02,O3, and O4 forms a right triangle with MO4 as height of pyramid.
height of pyramid = 100
1
6
−
4
/
3
~ 382.97.
height of highest point on sphere = 382.97+100+300= 782.97
This is exactly the same as 2004 AMC 10 #25 except with different numbers:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&t=131337&
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First we have to observe the tetrahedron. IMAGE
The base is an equilateral triangle. The distance of the centre of the base from one of the vertices of the base is 2 0 0 s i n 6 0 o × 3 2 = 3 2 0 0 . Now apply Pythagorean theorem on the orange triangle to get H (heihht). IMAGE
Thus, the maximum distance of a point on the larger sphere from the plane= H + 1 0 0 + 3 0 0 = 2 0 0 3 1 1 + 4 0 0 . GIF of the result is 7 8 2 .
Note: I have used H to denote height. Please don't confuse it with the plane named H whose notation I have not used anywhere.