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Em, whatever. I've finished it already

Thanks anyway!

Thank you!

I just need help with one more problem. (Please don't just give me a link to somewhere else, work me through some of it here.)

Two circles are externally tangent at point

, as shown. Segment is parallel to common external tangent Prove that the distance between the midpoints of and isEDIT: I extended lines AC and BD to meet at Q, then proved that QAB is similar to QCD. Then what?

And y duz the

What are the sign differences, may I ask? Just wanna make sure I didn't miss something.

(b) x^2=y^2+xy

(c)r^2+r=1, r = {sqrt(5)-1}/2

(d) cos 36 = {sqrt(5)+1}/4, cos 72 = {sqrt(5)-1}/4

Can you actually walk me through part (d)? I did (a), (b), and (c).

Many thanks!

What did you get as your cone volumes and final answer? Just checking my answers.

Thank you so much!

bob bundy wrote:

So you end up with

4x = 3y and

4x + 24 = 10 - 3y

Replacing 4x in the second equation with 3y (since 4x = 3y) and adding 3y to both sides gives 6y + 24 = 10. Subtracting 24 from both sides gives 6y = -14, or y=-7/3. We can't have negative lengths. I think the equation screwed up somewhere.

EDIT: Otherwise, your solution is nice and clear, and I feel that I can take the rest.

I get a negative fraction as y. X is equal to y too, for some reason.... Did you mess up somewhere?

bob bundy wrote:

hi !nval!d_us3rnam3

http://i.imgur.com/UBcgXOV.gif

If you 'drop' a perpendicular from A to the base you can work out FB using Pythagoras.

Then use similar triangles to work out EF and GH.

Then it should be straight forward to finish the task.

Well, I'll need to find the volume of the top cone. Either that, or I prove a rather confusing frustum volume formula. I know the height is 6, and FB is 8, and slant height is 10, so now what?

**!nval!d_us3rnam3**- Replies: 17

A sphere with radius 3 is inscribed in a conical frustum of slant height 10. (The sphere is tangent to both bases and the side of the frustum.) Find the volume of the frustum.

I haven't tried much, I'm just really stuck, can you give me a solution skeleton (i.e. where I prove a few things myself)? Thanks in advance!

~ !nval!d_us3rnam3

P.S. Please, don't give me another link to somewhere else on the website, because I've probably read it already. Just help me out here. Thanks again

P.S.S. If you have time, what's the list of working emojis on this website, other than , , , , , , and ? Thanks yet again!

P.S.S.S. How do people get the title of "Real Member"?

kk63353 wrote:

bob bundy wrote:SA = 10/3

From the second pair SA/SD = AQ/RD

=> RD = 1 x (5+10/3)/(10/3) = (15+10)/10 = 25/10 = 2.5

Can't we just say SA/SD=2/5=AQ/RD, so since AQ=1, RD=5/2?

I was thinking the exact same way. Why not use the slope of the line, and the height it gains as it crawls along the edge of the cube, rather than a big long solution that I partially didn't understand? This is my reasoning:

PA = 2, and AQ = 1, so line PQ has a slope of 1/2. Therefore, when the line QR, which is just a continuation of PQ, goes 5 units laterally, it climbs 2.5 units. Therefore, DR = 2.5.

Q.E.D.

Power of a point, in this diagram, looks like PA^2 = PD*PB = PC^2.

I've already proved similar triangles. Just need a little help going from there. Can you help me through the ratios? Most of the time I get the same result as using Power of a Point.

I'll be on this forum fairly actively if I need any homework help.

**!nval!d_us3rnam3**- Replies: 5

Let $\overline{PA}$ and $\overline{PC}$ be tangents from $P$ to a circle. Let $B$ and $D$ be points on the circle such that $B$, $D$, and $P$ are collinear. Prove that $AB \cdot CD = BC \cdot DA$.

I've already tried Power of a Point and similar triangles, but I can't seem to get anything to work. Can someone help me directly? By the way, don't post another link from this website, as I've probably checked that one already. Many thanks!

P.S. I need the answer before Saturday night.

Thank you so much for helping!

**!nval!d_us3rnam3**- Replies: 3

Let the incircle of triangle $ABC$ be tangent to sides $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$ at $D$, $E$, and $F$, respectively. Prove that triangle $DEF$ is acute.

I have tried proving that triangle DEF's angles were less than the opposite angles in triangle ABC, but that wasn't really complete. Can anyone help? Thanks!

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