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 [quote="CrystyB"]I recently took an exam but failed at the last two geometry problems. If anyone would like to solve them for me i would apreciate it. 1) a) State the theorem of the sines in the triangle ABC. (this one i've done [img]http://www.greylabyrinth.com/Forums/wink.gif[/img]) b) Given a rhombus ABCD circumscribed around the circle C(O;R), prove that [[for any tangent to the circle ("on A's side") that intersects the edges AB and AD in points M and N]] the product |BM|*|DN| is a constant. (i've solved this after the exam using an analitical geometry less used). 2) VABCD is a pyramid with ABCD rectangle, O is the intersection of diagonals AC and BD and the angle between VO and ABC is pi/2. a) Compute the areas and the volume of the pyramid given that |VO|=h, |AB|=a and |BC|=b. (i've dont this too.) b) Given a random plane alpha that intersects the edges VA, VB, VC and VD in points A', B', C' and D' prove that
code:
1       1       1       1 ----- + ----- = ----- + ----- |VA'|   |VC'|   |VB'|   |VD'|

This one i've solved using SPACE analytical geometry, which i wasn't supposed to know before taking that exam. [b]SECOND PROBLEM[/b] I've a few ideas about general probability, but i never heard the explanation of the concept of "expected number of ...(i.e. tries) before ... (something happens)". The problem was this: Given an urn with four balls of diferent colours and the experiment "we take out of the urn a ball, place it on table as a model, take out a second ball and paint this second ball to the colour of the first", what is the expected number of times that the experiment is done before all the balls in the urn have the same colour. I solved just a related problem: "What is the probability that after the nth step we have all the balls of same colour?" (check out server3002.freeyellow.com/cristyb/results.txt and server3002.freeyellow.com/cristyb/distrib.gif ). After i read a related q&a (given 10 bottles in a line[[, each connected with the one on the right and with the first, leftmost, one; the last one is connected to the first and has a free exit; each choice has a 50-50% prob.;each going through a connection takes one minute]] what is the expected time until the ball gets out?) i came up with the next solution. My question is: [b]is it correct?[/b] Solution: the possible states are: "ABCD" (the beginning, state IV), "AABC" (the next step, state III), "AABB" (state II), "AAAB" (state I) and "AAAA" (the end, state "0"). Probability of outcomes of experiments: IV -> III 100% III -> III 50% (1/2) III -> II 16% (1/6) III -> I 33% (1/3) II -> II 33% (1/3) II -> I 66% (2/3) I -> II 25% (1/4) I -> I 50% (1/2) I -> 0 25% (1/4) We add the function N(from,to,through), which means the expected number of experiments to be run to go from state "from" to state "through" if the first outcome is "from -> through". So we have the formulae: N(from,to,to)=1 N(IV,0,all)=N(IV,III,all)+N(III,0,all) N(IV,III,all)=1 N(III,0,all)=N(III,I,all)+N(I,0,all) N(III,I,all)=33%*N(III,I,I)+16%*N(III,I,II)+50%*N(III,I,III) N(III,I,II)=1+N(II,I,all) N(II,I,all)=33%*N(II,I,II)+66%*N(II,I,I) N(II,I,II)=1+N(II,I,all) N(III,I,III)=1+N(III,I,all) N(I,0,all)=25%*N(I,0,0)+50%*N(I,0,I)+25%*N(I,II,all) N(I,0,I)=1+N(I,0,all) N(I,0,II)=1+N(II,0,all) N(II,0,all)=N(II,I,all)+N(I,0,all)
code:
from to III -II -I- --0- (through="all")  IV      1            9  III            5/2  II             3/2  I                  11/2

So the answer is 9. The funny thing is that after 9 experiments, the probability of all four being the same colour is 65.3%, and after only 7 experiments, it was already 50.5%. Anyway, that's why i ask you [b]is the above solution correct???[/b] THANKS TO ALL OF YOU TRYING TO HELP, CRISTIAN. [This message has been edited by CrystyB (edited 09-12-2000).][/quote]
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Author Message
CrystyB
Posted: Fri Mar 08, 2002 11:57 am    Post subject: 1

Next time you're joking, include : ) at the end. If you're not, please explain that post 12...
CzarJ
Posted: Sun Jan 13, 2002 2:26 am    Post subject: 0

Um... yes?

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"Blah blah blah. Blah blah blah. Blah blah blah."
--The Reverend Al Sharpton
CrystyB
Posted: Fri Jan 11, 2002 4:47 pm    Post subject: -1

CJ, what in the Hell's name are you talking about??? ??? ???
CzarJ
Posted: Thu Dec 20, 2001 4:16 pm    Post subject: -2

Um, 7? pi? plus i?

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x=(-bą(sqrt(b^2-4ac)))/2a

I seem to remember... a song...

extropalopakettle
Posted: Wed Dec 19, 2001 4:15 pm    Post subject: -3

Quote:
it's easy to come back a month later and find the topic again

Or 15 months later.
CrystyB
Posted: Wed Dec 19, 2001 3:43 pm    Post subject: -4

ok then to vsp it goes!
Vanyo
Posted: Sun Sep 24, 2000 3:09 am    Post subject: -5

I have to agree with daniel801. It seems the only reason it was posted here was because this forum is so seldom used that it's easy to come back a month later and find the topic again. Reminds me of the Yogi Berra quote someone posted in Off-Topic:
quote:

"Nobody goes there anymore; its too crowded."
-Yogi Berra

daniel801
Posted: Sun Sep 24, 2000 2:53 am    Post subject: -6

I know this is petty, but I suggested the post be moved because more people might be able to help in VSP. Now that you've tried to make an issue of it though, let me quote:
Educator Ideas: A large number of Grey Labyrinth visitors are teachers and educators. Reach out for classroom ideas.
Is this a classroom idea? Obiously not.

--versus--

Visitor Submitted Puzzles: Post your own puzzle and try to solve puzzles posted by others.
"I recently took an exam but failed at the last two geometry problems. If anyone would like to solve them for me i would apreciate it." Sounds more like it. 2 problems were posted in hopes that someone would try and solve them.

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Chuck
Posted: Thu Sep 21, 2000 6:33 am    Post subject: -7

After 37,000 trials I get 8.994+. I'd guess the answer is 9. I'll run the program over night and see what happens.
Alfie
Posted: Thu Sep 21, 2000 6:15 am    Post subject: -8

I can't help, but you are right about it being in the correct forum.
CrystyB
Posted: Wed Sep 20, 2000 6:05 pm    Post subject: -9

so how would you calculate the expected value in that computer program of yours. any answer would help me understand the idea better. Thnx.
Chuck
Posted: Wed Sep 20, 2000 1:39 am    Post subject: -10

I'd probably try a computer simulation to verify the result. There's nothing like a few million trials to give you a ballpark figure.
daniel801
Posted: Mon Sep 18, 2000 1:06 pm    Post subject: -11

Sorry, I was just trying to help; *cough*no responses yet*cough*.

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CrystyB
Posted: Sat Sep 16, 2000 1:59 pm    Post subject: -12

It's not a puzzle/off-topic-subject. This forum is in my oppinion the best place for this topic.

No offense intended, daniel, but i know what i'm talking about. I posted some uninteresting problems in VSP and i have to SEARCH my notebook for the names for which i should SEARCH. They're 4-5 screens away, if not more from the main page of VSP.

[This message has been edited by CrystyB (edited 09-20-2000).]
daniel801
Posted: Wed Sep 13, 2000 1:39 pm    Post subject: -13

i would probably move this thread to off-topic forums if i were you to get more attention to it

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CrystyB
Posted: Tue Sep 12, 2000 7:09 pm    Post subject: -14

I recently took an exam but failed at the last two geometry problems. If anyone would like to solve them for me i would apreciate it.
1) a) State the theorem of the sines in the triangle ABC. (this one i've done )
b) Given a rhombus ABCD circumscribed around the circle C(O;R), prove that [[for any tangent to the circle ("on A's side") that intersects the edges AB and AD in points M and N]] the product |BM|*|DN| is a constant. (i've solved this after the exam using an analitical geometry less used).
2) VABCD is a pyramid with ABCD rectangle, O is the intersection of diagonals AC and BD and the angle between VO and ABC is pi/2.
a) Compute the areas and the volume of the pyramid given that |VO|=h, |AB|=a and |BC|=b. (i've dont this too.)
b) Given a random plane alpha that intersects the edges VA, VB, VC and VD in points A', B', C' and D' prove that
code:

1 1 1 1
----- + ----- = ----- + -----
|VA'| |VC'| |VB'| |VD'|

This one i've solved using SPACE analytical geometry, which i wasn't supposed to know before taking that exam.

SECOND PROBLEM

I've a few ideas about general probability, but i never heard the explanation of the concept of "expected number of ...(i.e. tries) before ... (something happens)". The problem was this:
Given an urn with four balls of diferent colours and the experiment "we take out of the urn a ball, place it on table as a model, take out a second ball and paint this second ball to the colour of the first", what is the expected number of times that the experiment is done before all the balls in the urn have the same colour.

I solved just a related problem: "What is the probability that after the nth step we have all the balls of same colour?" (check out server3002.freeyellow.com/cristyb/results.txt and server3002.freeyellow.com/cristyb/distrib.gif ).
After i read a related q&a (given 10 bottles in a line[[, each connected with the one on the right and with the first, leftmost, one; the last one is connected to the first and has a free exit; each choice has a 50-50% prob.;each going through a connection takes one minute]] what is the expected time until the ball gets out?) i came up with the next solution. My question is: is it correct?
Solution:
the possible states are: "ABCD" (the beginning, state IV), "AABC" (the next step, state III), "AABB" (state II), "AAAB" (state I) and "AAAA" (the end, state "0").
Probability of outcomes of experiments:

IV -> III 100%

III -> III 50% (1/2)
III -> II 16% (1/6)
III -> I 33% (1/3)

II -> II 33% (1/3)
II -> I 66% (2/3)

I -> II 25% (1/4)
I -> I 50% (1/2)
I -> 0 25% (1/4)

We add the function N(from,to,through), which means the expected number of experiments to be run to go from state "from" to state "through" if the first outcome is "from -> through".

So we have the formulae:
N(from,to,to)=1
N(IV,0,all)=N(IV,III,all)+N(III,0,all)
N(IV,III,all)=1
N(III,0,all)=N(III,I,all)+N(I,0,all)
N(III,I,all)=33%*N(III,I,I)+16%*N(III,I,II)+50%*N(III,I,III)
N(III,I,II)=1+N(II,I,all)
N(II,I,all)=33%*N(II,I,II)+66%*N(II,I,I)
N(II,I,II)=1+N(II,I,all)
N(III,I,III)=1+N(III,I,all)
N(I,0,all)=25%*N(I,0,0)+50%*N(I,0,I)+25%*N(I,II,all)
N(I,0,I)=1+N(I,0,all)
N(I,0,II)=1+N(II,0,all)
N(II,0,all)=N(II,I,all)+N(I,0,all)
code:

from to III -II -I- --0- (through="all")
IV 1 9
III 5/2
II 3/2
I 11/2

So the answer is 9. The funny thing is that after 9 experiments, the probability of all four being the same colour is 65.3%, and after only 7 experiments, it was already 50.5%. Anyway, that's why i ask you is the above solution correct???

THANKS TO ALL OF YOU TRYING TO HELP,
CRISTIAN.

[This message has been edited by CrystyB (edited 09-12-2000).]