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 $\lfloor d \rfloor$ ?

The answer is 782.

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An upvote because of the draws :)

Felipe Hofmann
- 7 years, 5 months ago

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Surely yes.

Piyushkumar Palan
- 7 years, 5 months ago

Thanks

Maharnab Mitra
- 7 years, 5 months ago

oh man!!!!!!!!!!!!!!.i made a calculation mistake and got 785.

Sanjay Mathai
- 7 years, 3 months ago

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

Kunal Gupta
- 6 years, 8 months ago

Centers of all 4 spheres form a pyramid with equilateral triangular base of side $200$ 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 $\frac{200}{\sqrt{3}}.$

By the Pythagorean Theorem, we have

$\left(\frac{200}{\sqrt{3}}\right)^2 + h^2 = 400^2 \implies h = 100\cdot\sqrt{\frac{44}{3}} \approx 382.97$

$d = 100 + 300 + 100\cdot\sqrt{\frac{44}{3}} \approx 782.97$

Answer: $\boxed{782}$

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Aww I put 482, forgot the extra 300...

Daniel Chiu
- 7 years, 5 months ago

I did the same way :) And the title of the problem is really cool!

jatin yadav
- 7 years, 5 months ago

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thanks

Piyushkumar Palan
- 7 years, 5 months ago

More problems of similar title coming up ...

Piyushkumar Palan
- 7 years, 5 months ago

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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 –

$\displaystyle d = 100\left( n + 1 + \frac{\sqrt{( (n + 1) \sin \frac{180}{n} )^2 - 1}}{\sin \frac{180}{n}} \right)$

So, no tension in future.

Sadman Sakib
- 7 years, 5 months ago

2004 AMC 12A #22 is very similar

Franklyn Wang
- 7 years, 5 months ago

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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? :D
Finn Hulse
- 7 years, 2 months ago

I got a 783!! Damn it!

shaurya gupta
- 7 years, 5 months ago

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You rounded instead of the symbol thingy

Patrick Chen
- 7 years, 5 months ago

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Well, it used to say find the integer part of d.

Daniel Chiu
- 7 years, 5 months ago

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@Daniel Chiu – DANG IT. I think I put 782, but I probably typoed.

Albert Xu
- 7 years, 4 months ago

Dang forgot the 300.

Patrick Chen
- 7 years, 5 months ago

i to forgot adding 300

Daanish bansal
- 7 years, 5 months ago

I think it should have stated that it was the Greatest integer function and not the Least integer function, don't you think?

Maharnab Mitra
- 7 years, 5 months ago

Hey, try this link and send reviews.

Maharnab Mitra
- 7 years, 5 months ago

![alt text] (http://s16.postimg.org/x4rmpmgxx/geom2.png)

So, $\lfloor AB \rfloor=\lfloor AO+TB \rfloor=\lfloor 400+382.79 \rfloor=\fbox{782}$

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*
400)-(200
*
200/3))
and this must be added to base height and radius of the large sphere which is approx..782.97

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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
$\sqrt{16-4/3}$
~ 382.97.

height of highest point on sphere = 382.97+100+300= 782.97

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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 $200sin60^o \times \frac{2}{3}= \frac{200}{\sqrt 3}$ . 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= $H+100+300=200 \sqrt\frac{11}{3} +400$ . GIF of the result is $782$ .

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.