- #1

- 9

- 0

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?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter klawesyn28
- Start date

- #1

- 9

- 0

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?

- #2

Tide

Science Advisor

Homework Helper

- 3,089

- 0

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

- #3

- 9

- 0

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

- #4

verty

Homework Helper

- 2,182

- 198

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.

- #5

AKG

Science Advisor

Homework Helper

- 2,565

- 4

- #6

verty

Homework Helper

- 2,182

- 198

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

- #7

AKG

Science Advisor

Homework Helper

- 2,565

- 4

- #8

Tide

Science Advisor

Homework Helper

- 3,089

- 0

klawesyn28 said: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.

- #9

- 37

- 0

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 [tex]A_1 B_1 C_1[/tex] be circumscribed about circle O,

its sides respectively parallel to the sides of triangle [tex]ABC[/tex] .

The farther point D is from the [tex]ABC[/tex] plane,

the "closer" triangle [tex]ABC[/tex] is to triangle [tex]A_1 B_1 C_1[/tex] and the smaller SA is.

Triangle [tex]ABC[/tex] can get infinitely close to triangle [tex]A_1 B_1 C_1[/tex].

[tex]lim(ABC) = A_1 B_1 C_1 [/tex]

[tex]lim(SA) = SA_1[/tex] and [tex]SA > SA_1 [/tex]

Inradius of trianglle [tex]A_1 B_1 C_1[/tex] equates 1,

its circumradius [tex]OA_1 = 2[/tex].

From triangle [tex]S O A_1 :[/tex]

[tex]SA_1 = \sqrt (SO^2 + OA_1 ^2) =

\sqrt (1^2 + 2^2) = \sqrt 5 [/tex]

[tex]SA > \sqrt 5 [/tex].

Q.E.D.

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 [tex]A_1 B_1 C_1[/tex] be circumscribed about circle O,

its sides respectively parallel to the sides of triangle [tex]ABC[/tex] .

The farther point D is from the [tex]ABC[/tex] plane,

the "closer" triangle [tex]ABC[/tex] is to triangle [tex]A_1 B_1 C_1[/tex] and the smaller SA is.

Triangle [tex]ABC[/tex] can get infinitely close to triangle [tex]A_1 B_1 C_1[/tex].

[tex]lim(ABC) = A_1 B_1 C_1 [/tex]

[tex]lim(SA) = SA_1[/tex] and [tex]SA > SA_1 [/tex]

Inradius of trianglle [tex]A_1 B_1 C_1[/tex] equates 1,

its circumradius [tex]OA_1 = 2[/tex].

From triangle [tex]S O A_1 :[/tex]

[tex]SA_1 = \sqrt (SO^2 + OA_1 ^2) =

\sqrt (1^2 + 2^2) = \sqrt 5 [/tex]

[tex]SA > \sqrt 5 [/tex].

Q.E.D.

Last edited:

- #10

- 9

- 0

why its circumradius [tex]OA_1 = 2[/tex]?

- #11

- 37

- 0

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.

Share: