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#1 Re: Ganesh's Puzzles » Level of Difficulty - III Questions. » 2009-04-20 16:56:34

#6 : cosC=1/2⇒b² -(b/2)²=c² -(a-b/2)² ⇒ab=a²+b²-c² ⇒3ab=(a+b)² -c²=(a+b+c)(a+b-c)=(a+b+c)(a+b+c-2c)=
(a+b+c)²-2c(a+b+c)⇒3ab+2c(a+b+c)=(a+b+c)²⇒3ab+3c(a+b+c)=(a+b+c)²+c(a+b+c)⇒
3(a+c)(b+c)=(a+b+c)(b+c+a+c)⇒[(b+c)+(a+c)]/(a+c)(b+c)=3/(a+b+c)⇒1/(a+c)+1/(b+c)=
3/(a+b+c).

#2 Re: Ganesh's Puzzles » Level of Difficulty - III Questions. » 2009-04-20 15:35:26

#1: tan(A/2) +tan(C/2) = (2/3)cot(B/2)

#4: x=pi/8 or 3pi/8

#3 Re: Puzzles and Games » Neat problem » 2008-12-30 08:53:42

Jane is right; the lazy and stupid ones are to be hated. (However, the rude and ill-bred ones are to be pitied). I apologize for not having realized the meaning of latex to female needs.

#4 Re: Ganesh's Puzzles » 10 second questions » 2008-12-30 07:03:30

Dear ganesh, I mean this: since  ( x+y+z ) ^ 2 = ( x^2+y^2+z^2 ) + 2 ( xy+yz+xz ) , then 20^2=33+2*18 or 400=69!!!

#9 Re: Ganesh's Puzzles » 10 second questions » 2008-12-07 04:35:55

#834

    #835. Are you sure about the text?   #836 
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#11 Re: Guestbook » maths product » 2008-12-06 08:49:56

Obviously, the product of the ages (or the number next door) must be a perfect square of a prime, let say n^2, so that the only possible combination of factors would be 1,n,n or 1,1,n^2. That's why the surveyor can't tell after the first answer but can after the second. So, n^2 can be 4,9,25 etc and the ages 1,1,4 , 1,1,9 , 1,1,25 etc.

#14 Re: Puzzles and Games » solve in integers » 2008-11-28 01:50:30

Yesterday's solution was incomplete (due to some excellent wine...) and, in fact, wrong. Here's the full stuff; First, (x,y) = (3,2) is valid and we get (a,b,c) = (a,a-4,a-5). BUT, there are four more combinations for ((xy),(x+y)) ; (2,-3) , (-2,3) , (3,-2) , (-3,2). Since we are looking for integers, we heve four more pairs for (x,y) ; (-1,3) , (3,-1) , (-2,-1) , (-1,-2) or (a,b,c) = (a,a+1,a) , (a,a,a-5) , (a,a,a) , (a,a-4,a-4). In fact, there are six groups of possible solutions, because six is the number of possible combinations for the three parenthesis (3!) to be 3,-1,-2 since (3)^3+(-1)^3+(-2)^3 = 18. Obviously, you can solve the problem starting from this last remark.

#15 Re: Puzzles and Games » solve in integers » 2008-11-27 01:58:56

Sorry about the smilies , I forgot to turn the thing off. So, here is the proper solytion; (a-b-1)^3+(b-c-2)^3+(c-a+3)^3=[(a)-(b+1)]^3+[(b+1)-(c+3)]^3+[(c+3)-(a)]^3. Let (b+1)=d & (c+3)=e. Then, (a-d)^3+(d-e)^3+(e-a)^3=18. Let a-d=x & d-e=y. Then, a=x+d & e-a=e-x-d=-x-d+e=-x-(d-e)=-x-y=-(x+y). So, x^3+y^3-(x+y)^3=18 or -3xy(x+y)=18 or xy(x+y)=-6. Since all numbers are integers, ((xy),(x+y)) = (1,-6), (-1,6),(6,-1),(-6,1) respectively and the only acceptable combination is (-6,1) resulting to (x,y) = (3,2) or (-2,3). From the first pair we get d-e=-1 & d-e=-2, so we select the second pair, thus x=-2 & y=3, so a-d=2, d-e=3, e-a=1 or b=a+1 & c=a-4. So, (a,b,c) = (a,a+1,a-4). There are infinite solutions, i.e. (a,b,c) = (2,3,-2),(3,4,-1),(50,51,46) etc.

#16 Re: Puzzles and Games » solve in integers » 2008-11-27 01:42:53

(a-b-1)^3+(b-c-2)^3+(c-a+3)^3=[(a)-(b+1)]^3+[(b+1)-(c+3)]^3+[(c+3)-(a)]^3. Let (b+1)=d & (c+3)=e. Then, (a-d)^3+(d-e)^3+(e-a)^3=18. Let a-d=x & d-e=y. Then, a=x+d & e-a=e-x-d=-x-d+e=-x-(d-e)=-x-y=-(x+y). So, x^3+y^3-(x+y)^3=18 or -3xy(x+y)=18 or xy(x+y)=-6. Since all numbers are integers, ((xy),(x+y))=(1,-6), (-1,6),(6,-1),(-6,1) respectively and the only acceptable combination is (-6,1) resulting to (x,y)=(3,2) or (-2,3). From the first pair we get d-e=-1 & d-e=-2, so we select the second pair, thus x=-2 & y=3, so a-d=2, d-e=3, e-a=1 or b=a+1 & c=a-4. So, (a,b,c)=(a,a+1,a-4). There are infinite solutions, i.e. (a,b,c)=(2,3,-2),(3,4,-1),(50,51,46) etc.

#17 Re: Guestbook » hey » 2008-11-26 21:54:32

As you will soon find out, the general idea is this; we suggest various problems, puzzles etc and Jane Fairfax solves them.

#18 Re: Jokes » A Mathematician, an Engineer and a Physicist Staying in a Hotel » 2008-11-20 07:46:07

If the physicist could calculate these, he would be an engineer. The point is that the engineer douses fires without calculating these.

#19 Re: Help Me ! » Inequality » 2008-11-10 06:57:17

Here's another suggestion; First we prove: if a,b,c,d positives, b<a<d<c, a-b>c-d, then a/b>c/d. Let a=b+k(k>0), c=d+m(m>0). Since a-b>c-d, b+k-b>d+m-d so k>m or k/m>1. Since b/d<1, k/m>b/d or k/b>m/d or a/b>c/d. Now; log(7)10=log10/log7 & log(11)13=log13/log11. Let log10=a, log7=b, log13=c & log11=d. Then, 10^a=10, 10^b=7, 10^c=13 & 10^d=11 or 10^(a-b)=10/7 & 10^(c-d)=13/11. You don't need a calculator to see that 10/7>13/11. So, 10^(a-b)> 10^(c-d) so, a-b>c-d (1). Since 0<log7<log10<log11<log13, 0<b<a<d<c (2). From (1) & (2) ; a/b>c/d or log10/log7>log13/log11 or log(7)10>log(11)13.

#20 Re: Dark Discussions at Cafe Infinity » Debate #001: Is math a science? » 2008-10-29 10:21:41

Mathematics is not only the Queen of the Sciences, is the ONLY Science. The others are  "something like Science". The most "un-scientific" elements are statistics, experiments and -sometimes- theories. In fact, the difference between Mathematics and other Sciences is the difference between a theorem and a theory. Many refer to Maths as "the pure science" to distinguish it from the others. Of course, it's an art too.

#21 Re: Help Me ! » Probability and Statistics » 2008-10-19 11:15:53

There can happen 7 deaths in a single day (7), in two (6-1 or 5-2 or 4-3),......, in six (2-1-1-1-1-1) or in seven different days (1-1-1-1-1-1-1). The probabilities are respectively p(1),p(2),...,p(6),p(7).The sum is obviously 1. The calculation is simple but messy and ends up to (appr.): p(1)= 4.2*10^-16,p(2)=9.7*10^-12,p(3)=1.1*10^-8,p(4)=3*10^-6,p(5)=7*10^-4,p(6)=0.055532694 and p(7)=0.943764295. Only p(5),p(6),p(7) are interesting since the others are too small. The answer to (b) is p(5)*360/365+p(6)*359/365+p(7)*358/365=0.981(appr.) or 98.1%. So the teacher was right. (c): a little under 1.9%. (d): the little!

#22 Re: Puzzles and Games » three sides of a triangle » 2008-10-17 10:47:00

......=(1989c^2-c^2)/2c^2=994    Note that cot(a) is negative!

#23 Re: Help Me ! » Weird Proof » 2008-10-17 07:37:19

Let n be:a(t+k)a(t+k-1)a(t+k-2)...a(t+1)a(t)a(t-1)...a(2)a(1). It has t+k digits. The last t digits form a t-digit number divided by 2^t. The first k digits form a k-digit number that can be written as : 10^[t+1]*[a(t+k)*10^[k]+a(t+k-1)*10^[k-1]+...+a(t+1)]. This number is divided by 2^t since 10^[t+1] is divited by 2^t: 10^[t+1]=[2*5]^[t+1]=2^t*[2*5^[t+1]]. So, the whole n is divited by 2^t. If the t-digit number is not divided by 2^t, then (n/2^t) is not an integer.

#24 Re: Puzzles and Games » Neat problem » 2008-10-16 05:02:33

1/2=sin(pi/6)=sin(3pi/18)=...=3sin(pi/18)cos^2(pi/18)-sin^3(pi/18)=3sin(pi/18)(1-sin^2(pi/18))-sin^3(pi/18)=...=3sin(pi/18)-4sin^3(pi/18)=sin(pi/18)(3-4sin^2(pi/18)). So, 2sin(pi/18)(3-4sin^2(pi/18))=1 & 4sin^2(pi/18)(3-4sin^2(pi/18))=1 giving the x=4sin^2(pi/18) since x(3-x)^2=1. The other two roots come out easily (2.347296177 & 3.532089061).

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