Proving SA > (5)^(0.5) for ABCD Tetrahedron w/ Inscribed Sphere

[klawesyn28]

There is ABCD tetrahedron with inscribed sphere. S is a center of the
sphere, radius of the sphere equates 1 and SA>=SB>=SC. Prove that
SA>(5)^(0,5).


I can't solve it. Could anybody help me?

 

[Tide]

How can SA be larger than the radius of the sphere?

 

[klawesyn28]

The sphere is inscribed, it is inside the tetrahedron. I don't understand your question.

 

[verty]

It's obvious that SA must be at least as long as the circumradius of a regular tetrahedron. I looked around and I see it can be shown that a regular tetrahedron has a circumradius 3x as long as the inradius. That would make SA >= 3.

This stuff is new to me so I don't know how you were supposed to do it, but following the logic here is one way.

Wait, maybe it's not obvious. One doesn't know how long SD is. Oh well, it's beyond me.

 

[AKG]

Why must SA be at least as long as the circumradius? Suppose the tetrahedron is really big compared to the sphere, and imagine the tetrahedron sitting upright. Place the sphere sitting at the center of the bottom face. Let A, B, C be the vertices of the bottom face. Then SA = SB = SC, which is approximately equal to distance from the center of the bottom face to A, which is much less than the circumradius, which is the distance from the center of the entire tetrahedron to A. Of course, in this case, since the sphere is so small in comparison, the desired result would still hold, but certainly not because SA is greater than the circumradius (since that's not true).

 

[verty]

I see, a shallow tetrahedron could have an arbitrarily large circumradius.

 

[AKG]

If I read mathworld.com right, then there are many circumradii, but we can speak of "the" circumradius as the least one. It doesn't really make sense, then, to say that a tetrahedron has arbitrarily large circumradius. It has a fixed circumradius. Also, the shallowness of the tetrahedron is irrelevant. Keep in mind that a circumscribed sphere needs simply to be a subset of the tetrahedron, so it can be a tiny speck at the center of the tetrahedron, it doesn't have to be "maximal" and touch all four faces. So just imagine a regular tetrahedron the size of a Great Pyramid. Imagine you have a tiny round marble on the floor of your Great Tetrahedron, in the middle of the floor. Let 2H be the height of your Tetrahedron, and h be the radius of your tiny marble. Then the circumradius, R is (SA² - h² + H²)^1/2, which is approximately (SA² + H²)^1/2 which is certainly greater than SA. So in this case SA < R, not at least R as you claimed.

 

[Tide]

The sphere is inscribed, it is inside the tetrahedron. I don't understand your question.

Sorry - I misread the question and had the tetrahedron inside the sphere.

 

[ivybond]

I have not proven this lemma (it'll be your contribution):
If pont D is a fixed distance from the ABC plane, SA is min when triangle ABC is equalateral and point D projects onto its center.
=======================
Projection of the inscribed sphere S onto the ABC plane is a circle
with center O.
Let triangle $$A_1 B_1 C_1$$ be circumscribed about circle O,
its sides respectively parallel to the sides of triangle $$ABC$$ .
The farther point D is from the $$ABC$$ plane,
the "closer" triangle $$ABC$$ is to triangle $$A_1 B_1 C_1$$ and the smaller SA is.
Triangle $$ABC$$ can get infinitely close to triangle $$A_1 B_1 C_1$$.
$$lim(ABC) = A_1 B_1 C_1$$
$$lim(SA) = SA_1$$ and $$SA > SA_1$$
Inradius of trianglle $$A_1 B_1 C_1$$ equates 1,
its circumradius $$OA_1 = 2$$.
From triangle $$S O A_1 :$$
$$SA_1 = \sqrt (SO^2 + OA_1 ^2) = \sqrt (1^2 + 2^2) = \sqrt 5$$
$$SA > \sqrt 5$$.
Q.E.D.

 

[klawesyn28]

why its circumradius $$OA_1 = 2$$?

 

[ivybond]

Triangle $$ABC$$ is equalateral according to the lemma, and so is triangle $$A_1 B_1 C_1$$ by constructing.

P.S. This problem (and another one you posted on the Calculus forum) strikes me as an olympiad kind. Where did you get them?
Art of Problem Solving is probably a better place to get help. But you do need to show SOME work.