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#251 Re: Guestbook » Pi is fun » 2023-11-27 19:28:52
As the song of B.....s, "All I need is... pi=3.14" 
Kerim
#252 Re: Jai Ganesh's Puzzles » 10 second questions » 2023-11-27 05:11:59
#8984.
#253 Re: Dark Discussions at Cafe Infinity » Cared Quotes - II » 2023-11-26 17:05:39
Someone said (not me, to avoid getting another big zero ):
"The most precious thing that a human can give to another is sincerity".
#254 Re: Exercises » A Motion Exercise » 2023-11-26 16:40:10
If an object (A) is moving and another object (B) must aim to travel straight towards A then B's path is continuously changing direction. This is illustrated by the Guardian's Four Dogs Puzzle:
This may be applied in real life. But, as you know, the situation here is just a math exercise.
For instance, when I saw your first diagram, I thought you got quickly the right logical answer of the situation related to the 2 directions of W and the missile.
Unfortunately, you changed your mind 
As we will see, proving that your first diagram is right, and it gives indeed the minimum time of the mission is somehow simple if one discovers how to start it.
I believe you can prove it, right?
#255 Re: Exercises » A Motion Exercise » 2023-11-26 02:35:26
I've realised that WX and XY should not be part of the same straight line.
I couldn't get well the relation between this sentence and your diagram.
On your diagram, they do have the same direction, unless 'same straight line' and 'same direction' have different meaning.
#256 Re: Exercises » A Motion Exercise » 2023-11-26 02:21:41
hi KerimF
I have drawn a diagram. The question says "(while moving on a straight line towards the side of the other ship)."
That seems contradictory. I'm marking the two vessel's positions with S for ship and W for warship. If W heads directly towards S while S is moving forwards then W must continuously change the direction of travel so that it is once more heading at S. If W doesn't do this it will end up where S was rather than where S is. But that means W's movement is not in a straight line.
Sorry for the confusion. I thought that the meaning of 'towards the side of the other ship' is totally different from "towards the other ship'.
Let us assume that the two parallel straights of S and W are drawn horizontally.
The ordinary ship, S, was on the northern line and the warship, W, was on the southern one. Both were moving from west to east.
At t=0, the captain would have to change the direction of his warship, W, towards north-east. The angle formed by the new direction (straight line) of W and the horizontal line could be called θ for example. In theory, the angle θ could be any value between 0 and 90 degrees but the captain had to decide its exact value to achieve the minimum time for his mission. The trajectory of the missile is also a straight line (between the position of the warship, W, when the missile was launched and the position of the other ship, S, when it was hit by the missile).
Please don't hesitate to notify me if I missed something to clarify about the exercise.
Kerim
#257 Re: This is Cool » Miscellany » 2023-11-25 21:32:43
I liked to thank you for your great and continuous efforts in providing this collection of interesting info.
I wonder if you had the idea to make an index list of all your entries in this thread. Obviously, it will need to be updated continuously since it is related to a growing book (not a fixed one).
Best Regards,
Kerim
#258 Exercises » A Motion Exercise » 2023-11-25 20:58:57
- KerimF
- Replies: 7
An ordinary ship and a warship were moving, side by side, on two parallel straight lines. The distance between them was (D). Their speed was constant and equal to (S1).
At time zero, t=0, the captain of the warship was instructed to shut down the other ship.
Let us assume that the new constant speed of the warship is (S2), S2>S1 (while moving on a straight line towards the side of the other ship).
The captain can launch a missile whose constant speed is (S3) which is much higher than (S2). The maximum range of the missile is (R), R<D.
Find the minimum time (T) to hit the other ship, that is:
t_min = T = f(D,R,S1,S2,S3)
Kerim
Note1:
I heard of this exercise when I was in the final year of high school (Baccalauréat). At that time, math teachers used to skip it. So, it was out of question to me not to solve it, and naturally I did it (because I was very interested in doing it). Later, I heard that it was removed in the printings of newer math books of the final year.
Note2:
I am afraid that its solution needs to be done in two steps:
(1) Logical
(2) Mathematical
If the logical step is skipped, the mathematical solution will be somehow undetermined, instead of being obvious.
#259 Re: Exercises » An Unusual Tricky Exercise, Have Fun » 2023-11-25 20:50:35
It is clear, to me in the least, that you both, amnkb and you (Bob), are very good teachers.
#260 Re: Exercises » An Unusual Tricky Exercise, Have Fun » 2023-11-25 20:34:25
so i did it wrong b/c im not a teacher?
huh
twi
But, thanks to your first good work and my reply to you, our friend Bob was able to notice. more easily. the second right answer.
So, I guess, if Bob presented his solution before you did, you will likely be the teacher 
by noticing the second option as he did.
Anyway, this situation may look like a humble example of the saying:
"But many that are first shall be last; and the last shall be first."
So, as possible, I try always to be the last
#261 Re: Dark Discussions at Cafe Infinity » Cared Quotes - I » 2023-11-25 20:13:49
Very nice quotes, thank you.
This reminds me what happened to me about quoting when I was at school.
My French teacher used to ask us to write always some quotes which could be related to the topic of homework. My problem was (still is) that I used to be interested solely in math and applied science more than anything else (I seldom read books offered by famous writers). Once, I wrote a sentence very clearly and carefully to reflect my idea about the topic of interest. I followed it with ‘Kerim said’
On the next day, my grade was ‘a Big Zero’ followed with ‘Your teacher, Joseph, said’
#262 Re: Exercises » An Unusual Tricky Exercise, Have Fun » 2023-11-24 23:35:15
Bob
ps. Aren't we all still students?
Hi Bob,
Yes, we are all still students in life, since learning/discovering has no limit.
But, here, only a teacher may notice that 'k' is given twice in (2) and (5), though presented in different forms.
Following the norm, a student expects here that the right answer is (or has to be) given once on the list. So, when he finds the right answer, he just checks (x) on the expression (2) or (5), the one he got at his last step. But, by doing this, he also says, though without his knowledge, that none of the remaining 4 expressions is equal to the right value of 'k' which is wrong.
Kerim
#263 Re: Exercises » An Unusual Tricky Exercise, Have Fun » 2023-11-24 19:45:14
Hi amnkb,
Your work is right, and you got k = 9 / [4*sqrt(3)], thank you.
But the question of this specific exercise was: "Find the right value of ‘k’ in the 5 expressions below:"
I am afraid, you didn't specify exactly where you find it in these 5 expressions. You just said k = 9 / [4*sqrt(3)].
Please remember, it is an 'unusual' tricky exercise to all students, but not to skilled teachers.
Have fun.
Kerim
#264 Exercises » An Unusual Tricky Exercise, Have Fun » 2023-11-24 07:45:04
- KerimF
- Replies: 8
The perimeter of a square (Ps) is equal to the perimeter of an equilateral triangle (Pe).
And the area of the square (As) is ‘k’ times equal to the area of the equilateral triangle (Ae), that is As = k*Ae.
Find the right value of ‘k’ in the 5 expressions below:
(1) 6 / [3*sqrt(3)]
(2) 9 / [4*sqrt(3)]
(3) 2*sqrt(3) / 3
(4) 6 / [4*sqrt(3)]
(5) 3*sqrt(3) / 4
As we will see, the way this exercise is stated lets it be tricky to all students but not necessarily to skilled teachers.
Have fun.
Kerim
#265 Re: Exercises » F(x) Having no Local Limits » 2023-11-24 00:11:07
I thought that such exercise is for students, not teachers
Anyway, I guess you meant by your last line that the condition to be satisfied by the parameters A, B, C and D for f(x) doesn’t have local high and low limits is:
B^2 - 3*A*C ≤ 0
And the parameter (D) may have any value.
Thank you.
#266 Re: Guestbook » Is the photo editing profession going to leave the job market? » 2023-11-23 20:18:17
The way I, a realistic rational person, see it is that every neural network is created by the human's brain.
Therefore, these neural networks cannot be made cleverer than their human creators though some creators try their best to give the world the impression they are, at least in some fields (while in reality they are not). They are just advanced tools which speed up the world's progress for the good and bad as well.
Anyway, if it is possible someday to make something more intelligent than the human's brain then the human up-to-date profession in every field, not only of photographer, will disappear altogether
For instance, the profession of photographer has now a chance to achieve new things by the use of new advanced tools; things which were impossible to make/create in the past.
#267 Exercises » F(x) Having no Local Limits » 2023-11-23 19:37:57
- KerimF
- Replies: 2
We have f(x) = A*x^3 + B*x^2 + C*x + D
What is the condition to be satisfied by the parameters A, B, C and D for f(x) doesn’t have local high and low limits?
#268 Re: Help Me ! » Radicand Comes Out » 2023-11-23 19:21:37
Actually, you have the right idea, it's just that the ± symbol needs to come slightly earlier. Instead of:
you want to say:
And what amnkb is saying is that .
Just to give an idea about my nature, I respect always other's views and knowledge no matter if they sound right or wrong to me.
And when I contact others, I simply say what I know while I understand fully that they may have their good reasons not to accept everything I say/know.
#269 Re: Help Me ! » Find A Linear Equation » 2023-11-23 04:22:53
Perhaps you were expecting a solution like this:
The linear formula is of the form:
F = aC + b
where,
F= Fahrenheit (°F) degrees
C= Celsius (°C) degrees
'a' and 'b' are the unknown parameters to be found.
If C=0, F=32, then
32 = a*0 + b
b=32 , the intercept with the F, up, axis.
if C=100, F=212
212 = a*100 + 32 [was b]
212 - 32 = a*100
180 = a*100
a = 180/100 = 1.8 , the slope of the line (usually called m)
Note:
The slope m (or 'a' here) could be calculated directly if we know two points (x1,y1) and (x2,y2).
Here, x1=C1=0 and y1=F1=32, also x2=C2=100 and y2=F2=212
m= (y2-y1)/(x2-x1) = (F2-F1)/(C2-C1) = (212-32)/(100-0) = 180/100 = 1.8
C=1.8*F + 32
Sorry, I wrote the lines above while I am sure that you know them already.
#270 Re: Help Me ! » How many matches will be played in this tournament? » 2023-11-22 09:43:22
hi mrpace,
I draw an 18x18 table on Excel. I filled its diagonal with 18 zeros.
I let its below part be a replica of its upper one, since (P1,P2) = (P2,P1) for example.
By filling the rows of upper cells with 8 one's for each player, I got the total of matches equal to 144 (8*18).
And to check that every player plays 8 matches, I summed the cells of each row/player (horizontally) and of each column/player (vertically). It was possible to let every sum be 8.
Kerim
#271 Re: Exercises » Area of a Triangle » 2023-11-22 07:53:48
hi KerimF
Apologies for not responding to this sooner. I had an idea when I read your first post but it has knocked around in my brain for some days before immerging as a method. It's got lots of steps and, if you wanted an actual formula, it could be done but it would be messy. Instead I'll outline an algorithm that will work.
If you just knew the three angles you could draw a triangle with those. But there are an infinite number of such triangles, all similar in shape, each scalable to scale up its size, but only one will have the right perimeter.
There's a formula, called Heron's formula after it's discoverer, that will calculate the area of a triangle if you know its perimeter. The semi perimeter = P/2 , sidea a, b, and c, and the formula is:
This Wiki page has an article about it https://en.wikipedia.org/wiki/Heron%27s_formula but there P already stands for the semi perimeter.
So how to make use of this.
Choose a random value for a; I'll call it p.
You can then use the sine rule to work out the other two sides, q and r.
Work out P' the perimeter of this triangle by P' = p + q + r
If P' = P then you're done because you had a really lucky choice for a. Unlikely though. But what you have got is the sides of a similar triangle that is either a scaled down or scaled up version of the correct triangle.
So you can work out the scale factor P/P' for converting your triangle to the correct one.
So now you know a, b, and c and so can proceed with Heron's formula.
Bob
Very nice work, thank you.
As expected, there are more than one way to solve it.
Here is how I did it:
(back from step 1, to show below the complete solution)
The area's formula of a triangle ABC is:
S = base * height / 2
As a start, I chose the base AB. In this case:
height = CM (from C to AB, M on AB)
And,
S = AB * CM / 2
Now, we have two right triangles ACM and BCM (right angle at M)
From them, let us find:
AB = f(CM,A,B)
In the triangle ACM
AM = CM/tan(A)
And, in the triangle BCM
MB = CM/tan(B)
So AB=f(CM,A,B) is:
AB = AM+MB = CM*[1/tan(A) + 1/tan(B)]
===
Again, in the triangle ACM
AC=CM/sin(A)
And, in the triangle BCM
BC=CM/sin(B)
So,
P = AB+AC+BC = CM*[1/tan(A)+1/tan(B)+1/sin(A)+1/sin(B)]
P = CM*[cos(A)/sin(A)+cos(B)/sin(B)+1/sin(A)+1/sin(B)]
P = CM*{[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B)}
===
Now, it is time to substitude CM with S from the area's formula:
S=AB*CM/2 ==> CM=2*S/AB
P = 2*S/AB*{[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B)}
And we get:
AB = 2*S/P*{[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B)}
===
Similarly, we can say:
AC = 2*S/P*{[1+cos(A)]/sin(A) + [1+cos(C)]/sin(C)}
CB = 2*S/P*{[1+cos(C)]/sin(C) + [1+cos(B)]/sin(B)}
===
By adding the above three formulas of AB, AC and CB
P = AB+AC+CB = 2*S/P*{2*[1+cos(A)]/sin(A) + 2*[1+cos(B)]/sin(B) + 2*[1+cos(C)]/sin(C)}
P = 4*S/P*{[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B) + [1+cos(C)]/sin(C)}
And the required formula/function is:
S = f(P,A,B,C) = P^2 / {[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B) + [1+cos(C)]/sin(C)} / 4
Kerim
#272 Re: Exercises » Area of a Triangle » 2023-11-21 00:56:20
There may be more than one way to solve it.
My first step was to write the area's formula of a triangle ABC.
S = base * height / 2
I chose:
base = AB
height = CM (from C to AB, M on AB)
So,
S = AB * CM / 2
Now, we have two rectangular triangles ACM and BCM (or CMA and CMB, if you like).
The second step was to find at its end:
AB = f(CM,A,B)
...
#273 Re: Exercises » Area of a Triangle » 2023-11-20 10:54:09
KerimF wrote:The perimeter (P) of a triangle ABC is known, also its three angles (A), (B) and (C).
Find its surface (S).
S = f(P, A, B, C)a triangle is 2D
surface is 3D
do you maybe mean 'area'?
Sorry, I was French educated at school. You are right, I meant by surface the area of the triangle.
#274 Exercises » Area of a Triangle » 2023-11-20 09:44:19
- KerimF
- Replies: 6
The perimeter (P) of a triangle ABC is known, also its three angles (A), (B) and (C).
Find its area (S).
S = f(P, A, B, C)
#275 Re: Help Me ! » Rearranging formulae » 2023-11-20 09:37:26
Step 1
Mult.both sides by 4+t;
p(4+t)=3-2t
Step 1
Mult.both sides by 4+t, assuming 4+t≠0;
p(4+t)=3-2t