Inscribed Sphere

joml88

A sphere is inscribed in the tetrahedron whose vertices are $A=(6,0,0), B=(0,4,0), C=(0,0,2),$ and $D=(0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

**Posted:** Mon Dec 05, 2005 7:32 pm

Altheman

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zanttrang · Yang-Mills Theory Offline Joined: 02 Aug 2004 Posts: 645

**Posted:** Tue Dec 06, 2005 4:53 pm

Altheman

i checked too, it seemed to me that kalva has the aime II from that year...the problems were different

**Posted:** Tue Dec 06, 2005 6:27 pm

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zanttrang · Yang-Mills Theory Offline Joined: 02 Aug 2004 Posts: 645

**Posted:** Tue Dec 06, 2005 7:12 pm

matt276eagles

I used the same formula, altheman. Can you show how you found the area of the face that's not parallel to any axes? It's side lengths are really ugly.

(Sorry to revive an old topic)

**Posted:** Fri Mar 03, 2006 12:19 pm

JBL

A general result you can use: given a trirectangular tetrahedron (one in which three edges are mutually perpendicular, as is the case here), the sum of the squares of the areas of the three smaller faces equals the square of the area of the larger face. The only way I know to prove this is just plugging it in with Heron's formula, which (for the purposes of this problem) just takes all of the ugliness and changes it to ugliness with the letters a, b, c instead of 6, 4, 2.

**Posted:** Sat Mar 04, 2006 11:35 am

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Joel
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WarpedKlown1335 · Yang-Mills Theory Offline Joined: 17 Jun 2003 Posts: 638

Couldn't you prove it with vectors? I THINK it'd be a lot nicer than Heron's.

**Posted:** Sat Mar 04, 2006 1:02 pm

JBL

Quite possibly, yes, actually. Since the sides as vectors are easy to calculate and so on, that does seem quite nice.

**Posted:** Sat Mar 04, 2006 1:21 pm

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Joel
Hi Deeps! <3

matt276eagles

Could anyone post a link to a proof, or just show how to find the area of triangle $ABC$ ?

**Posted:** Sat Mar 04, 2006 3:41 pm

Farenhajt

Using Heron's isn't really that bad:

The sides are $2\sqrt{10},2\sqrt{13},2\sqrt{5}$ . Hence the semiperimeter is $\sqrt{10}+\sqrt{13}+\sqrt{5}$ , and now Heron's gives:

$A^2=(\sqrt{10}+\sqrt{13}+\sqrt{5})(\sqrt{10}+\sqrt{13}-\sqrt{5})(\sqrt{10}-\sqrt{13}+\sqrt{5})(-\sqrt{10}+\sqrt{13}+\sqrt{5})$

You can simplify this by using the difference of squares:

$A^2=((\sqrt{10}+\sqrt{13})^2-5)(5-(\sqrt{10}-\sqrt{13})^2)=\\=(10+2\sqrt{130}+13-5)(5-(10-2\sqrt{130}+13))$

or

$A^2=(18+2\sqrt{130})(-18+2\sqrt{130})$

Now again we get the difference of squares:

$A^2=4\cdot 130-18^2=520-324=196$ , hence $A=14$

GENERAL NOTE: Expanded Heron's for "ugly" side lengths is $A={1\over 4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$ (or cyclic, which yields the same result)

**Posted:** Sat Mar 04, 2006 4:13 pm

Elemennop

**Posted:** Sat Mar 04, 2006 7:40 pm

Farenhajt

Vector solution is also routine:

$\overrightarrow{AB}=\langle -6,4,0\rangle,\overrightarrow{AC}=\langle -6,0,2\rangle$

$\overrightarrow{AB}\times\overrightarrow{AC}=\left|\begin{matrix} \vec i &\vec j&\vec k\\ -6 & 4 & 0\\ -6 &am...$

$|\overrightarrow{AB}\times\overrightarrow{AC}|^2=64+144+576=784\implies |\overrightarrow{AB}\times\overrightarrow{AC}|=28$

$A={1\over 2}|\overrightarrow{AB}\times\overrightarrow{AC}|=14$

**Posted:** Sun Mar 05, 2006 5:17 am

djshowdown2

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**Posted:** Mon Mar 06, 2006 7:00 pm

deej21

I did the same as djshowdown... important to realize doing that, though, is that when using the point-plane distance formula to solve for the point, there are two possible solutions (same with point-line distance), because at first I kept getting 3 for an answer, until I realized to flip the part in the absolute value. Is 2/3 really the answer, though? I wouldn't think an AIME would have such a... simple fraction. They could have at least made the original points different to make it a more AIME-like ugly fraction.
Or maybe I'm rambling about nothing because 2/3 isn't really the answer...

**Posted:** Mon Mar 06, 2006 7:20 pm

chess64

2/3 is the answer. (Well, 2+3=5, actually Mr. Green

)

**Posted:** Tue Mar 07, 2006 1:44 pm

elephantmaster

Hi all,

Does the formula Volume=1/3*surface area*inradius hold just for tetrahedra or for all polyhedra?

Thanks a lot!

**Posted:** Fri Feb 08, 2008 4:12 pm

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${\left[\lim_{n \to \pm \infty}{\left(1+\frac{1}{n}\right)}^{ n}\right]}^{\sqrt{-x}}+1=0$

where $x$ is the volume of a torus with distance from center 2 and radius of circle $1/2$ .

abacadaea

another method

**Posted:** Sun Mar 15, 2009 5:25 pm

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