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#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).
#1: tan(A/2) +tan(C/2) = (2/3)cot(B/2)
#4: x=pi/8 or 3pi/8
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.
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!!!
Are you sure about #1047 ?
#1038. hint: 10=2+3+5
#834 #835. Are you sure about the text? #836 #837 #838 #839 #840 #841 #842 #843
#834
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.
#826.
#826.
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.
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.
(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.
As you will soon find out, the general idea is this; we suggest various problems, puzzles etc and Jane Fairfax solves them.
If the physicist could calculate these, he would be an engineer. The point is that the engineer douses fires without calculating these.
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.
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.
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!
......=(1989c^2-c^2)/2c^2=994 Note that cot(a) is negative!
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.
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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