The important extra information which is not emphasised is the requirement that sides must be positive integers. If sides can be any positive real number, there are an infinity of answers.
We need not find the values of a and c seperately, as the question is 'What is the perimeter? ' Perimeter is a + b + c we have got the value of a + c = 7921, just add a (89) to this to get the perimeter. ( a + c ) + b = a + b + c = 7921 + 89 = 8010, which is the answer you got by finding the values of a and c.
Find out 89^2=7921, decide the no into 2 consecutive nos 89^2=3960+3961, as per vedics,89^2= 3960^2+3961^2 implies all the 3 are sides, area is dead easy Mukund
This problem is incorrectly posed. If you move the point C either left or right the sides 'a' and 'c' will change and with them the perimeter. The problem is still solvable by making an additional assumption, which you actually do when you assign the values.
This solution only works if you assume all values are integers, which was not given as a condition. Introduce fractions, and there are an infinite number of possible solutions.
3-4-5, 5-12-13 and 7-24-25 are the three smallest Pythagorean triples where the the smallest side is listed first. There appears to be a pattern. That is c = b+1. The hypotenuse is one larger than the longer leg. Using a = 89, b, c = b+1, the Pythagorean Theorem and some algebra, you get b = 3960 and c = 3961. P = sum of three sides = 8010.
For any odd number n greater than 1, there is a Pythagorean triple (n, m, m + 1) where m = ½ (n² − 1). When n is a prime number, there is no other Pythagorean triple than this one and the perimeter is n² + n.
This is very good to know. For our PreMath problem above, are we just limited to Pythagorean triples? Or could PreMath's solution apply to all right triangles if missing two side lengths? Thank you!
I arrived at the same result because for any prime number b, the second scenario always leads to a=0. Only one solution is therefore possible for the perimeter p with c=(b²+1)/2 and a=(b²-1)/2 p = a+b+c = (b²-1)/2+b+(b²+1)/2 = (b²-1+2b+b²+1)/2 = (2b²+2b)/2 = b²+b We can deduce that for each prime number b, there exists a Pythagorean triplet (a, b, c) of non-zero natural integers verifying the Pythagorean relation a²+b²=c² with c=(b²+1)/2 and a=(b²-1)/2!
@@sail2byzantium since the _sides_ ∈ ℤ⁺ (as shown in the upper right corner of the video) we are limited to Pythagorean triples. But there could be multiple solutions: when b = 33 the solutions are (33, 44, 55), (33, 56, 65),, (33, 180, 183) and (33, 544, 545).
@@sail2byzantium Hello, when PreMath states the solutions are limited to those triangles with sides that are integers he is indeed limiting the answers to Pythagorean triples. And as @ybodoN alertly points out, if the given side is an odd prime number greater than 1 there will be one and only one Pythagorean triple solution.
At 1:00 of the video he does state the sides are positive integers. Otherwise it would be impossible to solve the problem. In the diagram it would have been better to state this in words rather than stating "sides E Z+". Also the the perimeter question is meaningless. It would have been better to just ask for the lengths of the other two sides.
Surely there are many integer possibilities for a and c. You just need to push the point opposite the 89 length and a and c will change whilst 89 remains the same. I think this is a possible solution but not THE solution as it cannot be defined.
b=89 is a prime number In fact for any prime number b, the second scenario always leads to a=0. Also there is only one possible solution: c=(b²+1)/2 and a=(b²-1)/2 And a perimeter p = a+b+c = (b²-1)/2+b+(b²+1)/2 = (b²-1+2b+b²+1)/2 = (2b²+2b)/2 = b²+b We check it with b=89, p=89²+89=7921+1=8010 We can deduce the following property... For each prime number b, there exists a Pythagorean triplet (a, b, c) of non-zero natural integers satisfying the Pythagorean relation a²+b²=c² with c=(b²+1)/2 and a=(b² -1)/2
"Also there is only one possible solution" FALSE. There is only one possible solution IN INTEGERS, but there are infinitely many non-integer solutions: a = 15, b = 89, c = sqrt(8146) = 90.255193756..., and P = 194.255193756... is a solution; a = 200, b = 89, c = sqrt(47921) = 218.9086567..., and P = 507.9086567... is another solution; etc.
Mathematical 'magic' was used here, because, in fact, as long as you don't have one more side or one more angle (except of the right one, of course), you have an infinite number of solutions.
Nothing need be prime, and the given value can be an irrational square root (for example) and so long as the number whose square root is taken is factorable, you will have a solution for every possible combination of the factors (BASED ON the factors, not the factors directly). And the given one, of course, with the two unknown sides being a single unit apart (for purists who will be apoplectic realizing I mean "one times a number" to be considered a prime factorization). So if the known value is the square root of 255, 1*255, 3*85, 5*51, and 15*17 will all generate solutions. (By the way, that last fact is why one uses a single pair of primes generating an encryption solution: using several gives the codebreaker several possible solutions.)) For example, from my last: 3*85. (85+3)/2 and (85-3)/2 are the two sides.
Something does not make sense. Can you not move the point C to the right keeping given side length at fixed 89 and thus change the sum of other two sides? By moving the point c anywhere on the line you would still keep the side length 'b' constant at 89 but change the perimeter of the triangle.
It looked like a 30, 60, 90 triangle so I took the given shortest side and formed three sides in the ratio of 1, 2, and root three. This produced sides of 89, 178, and 89 root 3. This checks with the Pythagorean Triple 7921 + 23763 = 31684.
Since the Triangle Inequality includes degenerate triangles it could be argued that a=0 does give an acceptable second answer for perimeter of 89+89+0=178.
When using the (89)(89) choice, the simultaneous equations can be solved in the same manner as with the (7921)(1) choice, namely by addition to eliminate "a".
When all are integers, a^2 = c^2 - b^2 c= (a^2 + 1)/2, b = c-1. Works for a=3, b=4, c=5; works for a=5, b=12, c=13. But it doesn't work for a=4. Works for a=21, b=220, c=221. I'm guessing it works for any a except if a itself is a square. Nope, a=9, b=40, c=41 works. I guess it works only when a is odd. Works for a=25, b=312, c=313.
If moves along the line BC, it’is obvious that the perimeter varies from zero to infinite. It should be specified that the solution is a set of Pythagorion numbers
You don't have enough info to calculate a and c . You either have to know 2 of the 3 sides or know the angle of one of the non right angle sides- you have neither. The side described would be a sliver and not look at all like the triangle drawn. So you can randomly find an infinite number of right triangles with one side of 89 units.
He does say the side lengths must be a positive integer. Otherwise, there would be an infinite number of answers. Also, the diagram does show Sides E Z+ although that's a clumsy way to indicate the sides are positive numbers greater than 0. I would have written, "The sides are integers".
Elegant way of solving the problem, but can a hypotenuse of 3961 be correct? It doesn't seem reasonable. That would make angle A about 88.7 degrees. BTW, love your videos. I try to solve several each day. (with your wonderful help, of course).
@@ybodoN admettons pour l'exemple avec un triangle particulier, mais je ne vois pas ce qui empêche d'avoir la base et l'hypoténuse de longueur quelconque
@@olivierjosephdeloris8153 on a un angle droit et les trois côtés doivent correspondre à des entiers naturels, ce qui implique un triplet pythagoricien. Quand le plus petit des trois nombres est impair, une des solutions est (n, m, m + 1) où m + 1 = ½ (n² + 1). Quand n est premier, c'est la seule solution.
Perimeter is greater than 178 If "a" was zero then "c "would be 89. Any value for "a" would increase "c" So ....the perimeter is 89+ (>89) + (>0) or >178
Un triangolo rettangolo con i lati : a - b - c (ipotenusa !) Conoscendo soltanto il valore di un solo lato a a = 3-5-7-9-11-13-15-fino all’infinito ! Come calcolare i lati : b e L’ ipotenusa : c ? a = 5 ; b = 12 ; c = 13 Prova : 5^2+12^2 =13^2 25 +144 = 169 Come calcolare : b e c ? Con a = 3-5-7-11-13 numero primo (una soluzione) Con a = 9-15 ( multiplo di 3) almeno due soluzioni ! a=9 ; b=40 ; c=41 9^2 + 40^2 = 41^2 81 + 1600 = 1681 Altra soluzione : a=9 ; b=12 ; c=15 9^2 + 12^2 =15^2 81 + 144 = 225 Pazzesco ! Con a = 33 (11x3) Esistono… 4 soluzioni ! 33^2+44^2 = 55^2 33^2+56^2 = 65^2 33^2+180^2=183^2 33^2+544^2=545^2 Potete spiegare perché ?
U are all read the prob carefully, sides are real nos, always dont try to pick up mistakes only, u fit 4 only that, develop positive attitude first, give suggestions like me better Mukundsir
A triangle can NOT be described/defined by one angle and one side. The given answer is correct but is one of many. I can not see the point of even attempting to solve it!!!
You forgot to mention your condition that only whole numbers apply. If not, 7921 can also be divided by any other number less than 7921 to produce a fraction, e.g. 7921=(100)(79.21). In that case c=89.605 and a=10.395. The circumference is then 189. This problem therefore gives an infinite number of answers. (You also don't have to calculate a and c separately. If you know that (a+c) is a value, you can add the known value b.)
Figuring out c+a = 7921 is enough to answer the question. It is not necessary to add c+a and c-a. Just add 89 to 7921 and you find the answer. Why bother calculating c and a individually? Besides, this problem has multiple solutions unless the length of the known side is not a prime number, and infinite solutions if c and/or a are not integers.
Seems that this is 'A" solution but not 'THE' solution because there are infinite valid solutions based on the scant data provided. Am I missing something hare?
I'm not a 'smart' man, but as I don't see explicitly where the sides & perimeter have to be all INTEGERS, then I'm postulating this triangle to be isosceles with the perimeter being ~ 303.8650070512055 But what do I know? I probably missed something.
This is just another one of those mathematical exercises that serve only as a mathematical curiosity but without any practical use. like something that exists just to make teachers horny in the classroom but we will never see an engineer having to solve a similar problem in their work.
That was amazing. I wouldn't have thought you could do it with TWO missing triangle sides. One, yes. But not two. Well, I stand corrected. Very memorable.
@@sail2byzantium Esunisen was trying to tell you Z+ is the little twist PreMath put on this puzzle. Go to 1:00 it shows him making a circle around the Z+. It means the side lengths can only be whole numbers. The clickbait picture only shows a right triangle with on side being 89. The true answers are infinite. You CAN NOT solve a triangle with only 2 pieces of information. One known angle of 90 deg. & one side measurement of 89 is not enough information. You always need at least 3 pieces of info to solve. PreMath gives the 3rd piece as (each side must be a whole number). It is kinda trichery, but it is a good math lesson..
@@simpleman283 Thanks for clarifying this point! I thought there surely can be no unique solution with only two pieces of information, but was then befuddled when he managed to produce an answer. Trickery, indeed! :)
I am confused Assume side a =1 then side c = square root of 7922 this gives a different Perimeter Assume side a =1oo then side c = square root of 17922 this gives a different Perimeter
So you just guessed it really! You have not actually found an answer just two whole numbers that fit Pythagorean theorem. Equally a could be 89 and so c 125.8.
That's what I thought as well when I saw the thumbnail. But in the video itself he added a second condition: a,b,c are positve integers. Therefore only one solution exists.
Your solution is only one of an infinite number of solutions. Side a could be 89, and we would have a right triangle with 2 45 degree angles. If I choose a value of 2 X 89 = 178 for c, then my right triangle would be a 30 60 90 right triangle. If you were one of my grade school math students, I would assign the following homework question for you: "How many angles and/or side lengths are required to uniquely specify any polygon ?"
There are many solutions The only limit is the possible maximum length that side c can take to remain a orthogonal triangle So my friend it seems to me you are out fishing Something is wrong with your geometry
Once again you assume interger values for the sides. If you take as a guess one of the sides is length 1 you will NOT get you calculated value of the perimeter. You are doing a disservice to mathematics by posting these solutions as it appears to the unsuspecting that this is the only possible result!
If side is not 89 , but 88 , the solution is different ! Sides are 88 , 105 and 137 . I think this problem is for high IQ people and not for standart people who beleive in everithing , even in politicians 😊 because in this triangle if one side is 89 , the other sides are 105,5 and 138,02625 .
I suppose 'math majors' will love this, but for the rest of us, it's a lesson in futile thinking. Hmm....has anyone done the trig to figure out how 'small' the opposite angle is?? NOT an integer, I presume..... hahaha....glad I never got this problem on an exam....
You’re quite mistaken. What you’ve shown is that there are some whole numbers (such as 8) that belong to more than one Pythagorean triple. But that doesn’t mean it’s true of EVERY whole number. Some whole numbers (including all odd primes, such as 89) belong to only one such triple.
❤❤❤ thanks 💯🙏 keep going my dear teacher ❤️
Thank you, I will ❤️
You are awesome. Keep it up 👍
The important extra information which is not emphasised is the requirement that sides must be positive integers. If sides can be any positive real number, there are an infinity of answers.
You are correct. If the sides can be any positive real number, there are infinite answers
So the answer provided is not actually the answer to the question as actually posed. An answer but not the answer
I was thinking the same thing!
yes now I got it - as was concluding that there are infinite number of solution as it depends on angle c which can be any between >o
The confusion would have been avoided if it was stated from the start that all sides are positive integers.
We need not find the values of a and c seperately, as the question is 'What is the perimeter? ' Perimeter is a + b + c we have got the value of a + c = 7921, just add a (89) to this to get the perimeter. ( a + c ) + b = a + b + c = 7921 + 89 = 8010, which is the answer you got by finding the values of a and c.
Sorry but without any 2nd side or an angle , there are an infinite number of triangles.
With the given information there are endless solutions. When a nears 0 , c nears 89+ . When a nears endles, c nears endles
Not really - sides have to be positive integers and there is only one solution.
It's not,
Since it's already stated that the side lengths must be positive integers.
Find out 89^2=7921, decide the no into 2 consecutive nos
89^2=3960+3961, as per vedics,89^2= 3960^2+3961^2 implies all the 3 are sides, area is dead easy
Mukund
This problem is incorrectly posed. If you move the point C either left or right the sides 'a' and 'c' will change and with them the perimeter. The problem is still solvable by making an additional assumption, which you actually do when you assign the values.
0:56 “Keep in your mind that the side lengths must be a positive integer”, See diagram: Sides ∈ Z+
This solution only works if you assume all values are integers, which was not given as a condition.
Introduce fractions, and there are an infinite number of possible solutions.
Z^+ was given.
3-4-5, 5-12-13 and 7-24-25 are the three smallest Pythagorean triples where the the smallest side is listed first. There appears to be a pattern. That is c = b+1. The hypotenuse is one larger than the longer leg. Using a = 89, b, c = b+1, the Pythagorean Theorem and some algebra, you get b = 3960 and c = 3961. P = sum of three sides = 8010.
6.8.10 not like that
It is arbitrary to say that, if xy = zt, then x=z andy=t.
As a matter of fact, there are infinite triangles having a side = 89
It is easy enough to prove your statement - just give us as least one more solution.
yes there are, but the sides must be integer numbers, and the only solution to that is the one that is shown on the vid.
0:56 “Keep in your mind that the side lengths must be a positive integer”, See diagram: Sides ∈ Z+
For any odd number n greater than 1, there is a Pythagorean triple (n, m, m + 1) where m = ½ (n² − 1).
When n is a prime number, there is no other Pythagorean triple than this one and the perimeter is n² + n.
@@pluisjenijn to be exact, the funny property is n² + m² = (m + 1)²
like (21, 220, 221) (201, 20200, 20201) (2001, 2002000, 2002001)
This is very good to know. For our PreMath problem above, are we just limited to Pythagorean triples? Or could PreMath's solution apply to all right triangles if missing two side lengths? Thank you!
I arrived at the same result because for any prime number b, the second scenario always leads to a=0.
Only one solution is therefore possible for the perimeter p with c=(b²+1)/2 and a=(b²-1)/2
p = a+b+c = (b²-1)/2+b+(b²+1)/2 = (b²-1+2b+b²+1)/2 = (2b²+2b)/2 = b²+b
We can deduce that for each prime number b, there exists a Pythagorean triplet (a, b, c) of non-zero natural integers verifying the Pythagorean relation a²+b²=c² with c=(b²+1)/2 and a=(b²-1)/2!
@@sail2byzantium since the _sides_ ∈ ℤ⁺ (as shown in the upper right corner of the video) we are limited to Pythagorean triples.
But there could be multiple solutions: when b = 33 the solutions are (33, 44, 55), (33, 56, 65),, (33, 180, 183) and (33, 544, 545).
@@sail2byzantium Hello, when PreMath states the solutions are limited to those triangles with sides that are integers he is indeed limiting the answers to Pythagorean triples. And as @ybodoN alertly points out, if the given side is an odd prime number greater than 1 there will be one and only one Pythagorean triple solution.
You don`t need to know what c and a equal. All you need is what c+a is equal to. You add b and you have the perimiter.
畢氏數(Pythagorean triple)有通解(General solutions) ：
(b,(b²-1/2),(b²+1)/2),當b為奇數(odd)，或(2b,b²-1,b²+1)
At 1:00 of the video he does state the sides are positive integers. Otherwise it would be impossible to solve the problem. In the diagram it would have been better to state this in words rather than stating "sides E Z+". Also the the perimeter question is meaningless. It would have been better to just ask for the lengths of the other two sides.
Amazing 👍
Thanks for sharing 😊
Since the given side is 89 which is prime.The other sides are 1/2(89^2+1),1/2(89^2-1)
Perimeter is 89+1/2(89^2+1+89^2-1)
=89+89^2=89×90=8100-90=8010
Surely there are many integer possibilities for a and c. You just need to push the point opposite the 89 length and a and c will change whilst 89 remains the same. I think this is a possible solution but not THE solution as it cannot be defined.
b=89 is a prime number
In fact for any prime number b, the second scenario always leads to a=0.
Also there is only one possible solution: c=(b²+1)/2 and a=(b²-1)/2
And a perimeter p = a+b+c = (b²-1)/2+b+(b²+1)/2 = (b²-1+2b+b²+1)/2 = (2b²+2b)/2 = b²+b
We check it with b=89, p=89²+89=7921+1=8010
We can deduce the following property...
For each prime number b, there exists a Pythagorean triplet (a, b, c) of non-zero natural integers satisfying the Pythagorean relation a²+b²=c² with c=(b²+1)/2 and a=(b² -1)/2
Your formula is great. If b=3 than c=5 and a=4 . Fits best !
"Also there is only one possible solution" FALSE. There is only one possible solution IN INTEGERS, but there are infinitely many non-integer solutions: a = 15, b = 89, c = sqrt(8146) = 90.255193756..., and P = 194.255193756... is a solution; a = 200, b = 89, c = sqrt(47921) = 218.9086567..., and P = 507.9086567... is another solution; etc.
@@douglasmiller1233
We are looking for sides belonging to Z+. In fact we are looking for a Pythagorean triple, and therefore only integers.
Thank you!
Mathematical 'magic' was used here, because, in fact, as long as you don't have one more side or one more angle (except of the right one, of course), you have an infinite number of solutions.
actually no, the prob says integer numbers on the sides, that narrow it down to only 1 solution.
0:56 “Keep in your mind that the side lengths must be a positive integer”, See diagram: Sides ∈ Z+
Nothing need be prime, and the given value can be an irrational square root (for example) and so long as the number whose square root is taken is factorable, you will have a solution for every possible combination of the factors (BASED ON the factors, not the factors directly). And the given one, of course, with the two unknown sides being a single unit apart (for purists who will be apoplectic realizing I mean "one times a number" to be considered a prime factorization). So if the known value is the square root of 255, 1*255, 3*85, 5*51, and 15*17 will all generate solutions.
(By the way, that last fact is why one uses a single pair of primes generating an encryption solution: using several gives the codebreaker several possible solutions.))
For example, from my last: 3*85. (85+3)/2 and (85-3)/2 are the two sides.
Something does not make sense. Can you not move the point C to the right keeping given side length at fixed 89 and thus change the sum of other two sides? By moving the point c anywhere on the line you would still keep the side length 'b' constant at 89 but change the perimeter of the triangle.
Go to 1:00 it shows him making a circle
around the Z+. It means the side lengths can only be whole numbers.
It looked like a 30, 60, 90 triangle so I took the given shortest side and formed three sides in the ratio of 1, 2, and root three. This produced sides of 89, 178, and 89 root 3. This checks with the Pythagorean Triple 7921 + 23763 = 31684.
Since the Triangle Inequality includes degenerate triangles it could be argued that a=0 does give an acceptable second answer for perimeter of 89+89+0=178.
Respected Sir 🙏, I like the way of your answering
When using the (89)(89) choice, the simultaneous equations can be solved in the same manner as with the (7921)(1) choice, namely by addition to eliminate "a".
I'm definitely coming back to this to give it a try.
Doubt this, what will happen if on the drawing BC is reduced by8 units? You do not have th angles of the BAC and ACB?
When all are integers, a^2 = c^2 - b^2 c= (a^2 + 1)/2, b = c-1. Works for a=3, b=4, c=5; works for a=5, b=12, c=13. But it doesn't work for a=4. Works for a=21, b=220, c=221. I'm guessing it works for any a except if a itself is a square. Nope, a=9, b=40, c=41 works. I guess it works only when a is odd. Works for a=25, b=312, c=313.
If moves along the line BC, it’is obvious that the perimeter varies from zero to infinite. It should be specified that the solution is a set of Pythagorion numbers
0:56 “Keep in your mind that the side lengths must be a positive integer”, See diagram: Sides ∈ Z+
Solutions may not be full filled
Pythagoras values
Please check this sir.
the title page is misleading because it fails to say that the sides are integers.
If the angle at A changes without changing the length of AB, will the answer be the same?
You don't have enough info to calculate a and c . You either have to know 2 of the 3 sides or know the angle of one of the non right angle sides- you have neither. The side described would be a sliver and not look at all like the triangle drawn. So you can randomly find an infinite number of right triangles with one side of 89 units.
Surely there are many possible solutions
He said they are integers.
Since 89 is a prime number, there is only one solution 🧐
As they are positive integers
and the number(89 Square) is PRIME having only one solution
THERE IS ONLY ONE SOLUTION YOU FOOL
@@BruceArnold318 Ah, I missed that bit too. I was scratching my head thinking that the number of solutions is infinite.
There’s only one….and stop calling me Shirley😂
If the hypotenuse of the right triangle is 89 what is the perimeter and it's area?
Obviously, the larger value is more acceptable here.
Very easy
Probably could solve this much easier using trig to find side BC using arctan(). And then Pythagorean theorem to find side AC.
So...arctan(89/BC) =
I'm no math guru but I it seems to me that there are infinite answers depending on the position of point "C" relative to "B".
the sides must be integer numbers..theres only 1 solution.
That’s very nice
Thanks Sir
Thanks PreMath
❤❤❤❤❤
Always welcome
You are awesome. Keep it up 👍
He does say the side lengths must be a positive integer. Otherwise, there would be an infinite number of answers. Also, the diagram does show Sides E Z+ although that's a clumsy way to indicate the sides are positive numbers greater than 0. I would have written, "The sides are integers".
In the diagram: Sides ∈ Z+ is clumsy math talk for "sides are elements of the set of positive integers"
Line BC=89 line AC= 89*2^(1/2) is also an answer!
Elegant way of solving the problem, but can a hypotenuse of 3961 be correct? It doesn't seem reasonable. That would make angle A about 88.7 degrees. BTW, love your videos. I try to solve several each day. (with your wonderful help, of course).
As long as the angle A is less than 90°, we have a triangle, no matter how long is the hypotenuse 🤓
C'est une possibilité, ça pourrait aussi être une infinité d'autres solutions, non ?
Comme le plus petit des trois nombres est premier, il n'y a qu'un seul triplet pythagoricien possible ! 😉
@@ybodoN admettons pour l'exemple avec un triangle particulier, mais je ne vois pas ce qui empêche d'avoir la base et l'hypoténuse de longueur quelconque
@@olivierjosephdeloris8153 on a un angle droit et les trois côtés doivent correspondre à des entiers naturels, ce qui implique un triplet pythagoricien.
Quand le plus petit des trois nombres est impair, une des solutions est (n, m, m + 1) où m + 1 = ½ (n² + 1). Quand n est premier, c'est la seule solution.
@@ybodoNd'accord, en effet la contrainte des nombres entiers, ça change tout. Le Z+ m'avait échappé
Aren't there infinite perimeters?
Yes, there are infinite solutions to the perimeter because of lack of information.
@@ra15899550 theres only 1 solution, the prob says the sides must be integer numbers, i had the same concern but thats the correct answer.
There are an infinite number of solutions to this problem depending on the slope of the hypotenuse
I loved this question!
❤️
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
thank you
Okay
Very thanks
Perimeter is greater than 178
If "a" was zero then "c "would be 89.
Any value for "a" would increase "c"
So ....the perimeter is 89+ (>89) + (>0) or >178
Will this solution satisfy Pythagorean solution.
Yes.
3961^2 - 3960^2
factors as
(3961- 3960) * (3961+3960) = 1 * 89^2.
Google "Euclid's gnomon".
Why do you assume (a+c) and (a-c) are integers?
because a must be a integer and also c, so....
The Perimeter is any value that is equal to or greater than 89!
Un triangolo rettangolo con i lati : a - b - c (ipotenusa !)
Conoscendo soltanto il valore di un solo lato a
a = 3-5-7-9-11-13-15-fino all’infinito !
Come calcolare i lati : b e
L’ ipotenusa : c ?
a = 5 ; b = 12 ; c = 13
Prova : 5^2+12^2 =13^2
25 +144 = 169
Come calcolare : b e c ?
Con a = 3-5-7-11-13 numero primo (una soluzione)
Con a = 9-15 ( multiplo di 3) almeno due soluzioni !
a=9 ; b=40 ; c=41
9^2 + 40^2 = 41^2
81 + 1600 = 1681
Altra soluzione :
a=9 ; b=12 ; c=15
9^2 + 12^2 =15^2
81 + 144 = 225
Pazzesco !
Con a = 33 (11x3)
Esistono…
4 soluzioni !
33^2+44^2 = 55^2
33^2+56^2 = 65^2
33^2+180^2=183^2
33^2+544^2=545^2
Potete spiegare perché ?
Well, that is one answer of many.
Similar using for side b à prime number, à good idea for fun.
U are all read the prob carefully, sides are real nos, always dont try to pick up mistakes only, u fit 4 only that, develop positive attitude first, give suggestions like me better
Mukundsir
AC can be any value that is greater than or equal to AB! Why did you assume that BC could not be zero? BC can be any postive value from 0 to infinity!
Il y a une infinité de valeurs de a et c, ainsi pour le périmètre
I thought so too but he said they are integers.
Comme le plus petit des trois nombres est premier, il n'y a qu'un seul triplet pythagoricien possible ! 😉
@@BruceArnold318merci, je ne suis pas très bon en anglais, je n'avais pas saisi
@@ybodoNje ne comprend pas vraiment bien l'anglais et je n'avais saisi : appartient à Z. Merci pour votre réponse
Je reste d'accord avec vous: il y a une infinité de solutions.
A triangle can NOT be described/defined by one angle and one side. The given answer is correct but is one of many.
I can not see the point of even attempting to solve it!!!
There is an important detail in the upper right corner of the video: _sides_ ∈ ℤ⁺ 🧐
You forgot to mention your condition that only whole numbers apply. If not, 7921 can also be divided by any other number less than 7921 to produce a fraction, e.g. 7921=(100)(79.21).
In that case c=89.605 and a=10.395.
The circumference is then 189.
This problem therefore gives an infinite number of answers.
(You also don't have to calculate a and c separately. If you know that (a+c) is a value, you can add the known value b.)
Figuring out c+a = 7921 is enough to answer the question. It is not necessary to add c+a and c-a. Just add 89 to 7921 and you find the answer. Why bother calculating c and a individually? Besides, this problem has multiple solutions unless the length of the known side is not a prime number, and infinite solutions if c and/or a are not integers.
yhea but the problem says integer numbers...so...
Et si on augmente l'angle BAC ?
A sera toujours de 89 mais les deux autres côtés auront augmenté...
Seems that this is 'A" solution but not 'THE' solution because there are infinite valid solutions based on the scant data provided.
Am I missing something hare?
P=89, b=3960, h=3961, May be
Yes but what if the hypotenuse is a gorilla. This is overlooked more often than we realize.
What is the 5 term of sequence given -8,-3,2,7,_,_,22
12
In real the problem has infinity answers
Assume the lengths are positive integers.
0:56 “Keep in your mind that the side lengths must be a positive integer”, See diagram: Sides ∈ Z+
I'm not a 'smart' man, but as I don't see explicitly where the sides & perimeter have to be all INTEGERS, then I'm postulating this triangle to be isosceles with the perimeter being ~ 303.8650070512055
But what do I know? I probably missed something.
Simple steps... but a magnificent sequence!
Wow! I used trigonometry, but that got me nowhere.
Great solution!
No need to find each side separately
We know that one of the sides is 89 and the sum of the other two sides is 89^2
This is just another one of those mathematical exercises that serve only as a mathematical curiosity but without any practical use. like something that exists just to make teachers horny in the classroom but we will never see an engineer having to solve a similar problem in their work.
So math lessons should be limited to what an engineer might see? What if he has poor eyesight? Mr. Magoo's Math 😂
But c^2=a^2 + b^2
Does not add up
一個方程式（畢氏定理）兩個未知數，故有無窮盡的解。需再加一條件，例如邊長是整數方可解出另二邊長，此視頻就是這樣設定的。
As long as you eliminate a=0, P=89+a+c=89+7921 didn't really need to solve for the 2 sides.
But side “a” couldn’t be equal to zero! It must have a value other than zero. If side “a” is zero, then the figure could not anymore be a triangle!
@@dawon7750pls watch video before commenting. 7:00
Would have been easier to plug into c-a=1, so a=c-1
Thanks for your feedback! Cheers! 😀
That was amazing. I wouldn't have thought you could do it with TWO missing triangle sides. One, yes. But not two. Well, I stand corrected. Very memorable.
Only because it's in Z+
In R there is an infinity of solutions.
@@esunisen3862
I have no idea whatsoever as to what you just said. Thanks?
@@sail2byzantium Esunisen was trying to tell you Z+ is the little twist
PreMath put on this puzzle. Go to 1:00 it shows him making a circle
around the Z+. It means the side lengths can only be whole numbers.
The clickbait picture only shows a right triangle with on side being 89.
The true answers are infinite. You CAN NOT solve a triangle with only 2 pieces of information. One known angle of 90 deg. & one side measurement of 89 is not enough information. You always need at least 3 pieces of info to solve. PreMath gives the 3rd piece as (each side must be a whole number).
It is kinda trichery, but it is a good math lesson..
@@simpleman283 Thanks for clarifying this point! I thought there surely can be no unique solution with only two pieces of information, but was then befuddled when he managed to produce an answer. Trickery, indeed! :)
I am confused
Assume side a =1 then side c = square root of 7922 this gives a different Perimeter
Assume side a =1oo then side c = square root of 17922 this gives a different Perimeter
0:56 “Keep in your mind that the side lengths must be a positive integer”, See diagram: Sides ∈ Z+
The 3rd Binomial formula was the key, very good.
anybody can see that given only the length of one side the problem has infinite solutions
So you just guessed it really! You have not actually found an answer just two whole numbers that fit Pythagorean theorem.
Equally a could be 89 and so c 125.8.
There are many solutions for a and b
... but only one where a, b and c are integers 😉
That's what I thought as well when I saw the thumbnail. But in the video itself he added a second condition: a,b,c are positve integers. Therefore only one solution exists.
Your solution is only one of an infinite number of solutions. Side a could be 89, and we would have a right triangle with 2 45 degree angles. If I choose a value of 2 X 89 = 178 for c, then my right triangle would be a 30 60 90 right triangle. If you were one of my grade school math students, I would assign the following homework question for you: "How many angles and/or side lengths are required to uniquely specify any polygon ?"
0:56 “Keep in your mind that the side lengths must be a positive integer”, See diagram: Sides ∈ Z+
There are many solutions
The only limit is the possible maximum length that side c can take to remain a orthogonal triangle
So my friend it seems to me you are out fishing
Something is wrong with your geometry
There can be numerous triangles with this information??? AC side can be anything more than 89? No?
Once again you assume interger values for the sides. If you take as a guess one of the sides is length 1 you will NOT get you calculated value of the perimeter. You are doing a disservice to mathematics by posting these solutions as it appears to the unsuspecting that this is the only possible result!
89 is prime number given, so the solution became possible
who says the sides have to be whole numbers?
emmm..the problem?
Baba, you should tell the angle then only one solution will emerge
208 ÷2
Tudo errado, isso tem infinitas soluções , mas a solução proposta não é uma delas. Essa solução e inconsistente com o torema de Pitágoras.
116, 145
If side is not 89 , but 88 , the solution is different ! Sides are 88 , 105 and 137 . I think this problem is for high IQ people and not for standart people who beleive in everithing , even in politicians 😊 because in this triangle if one side is 89 , the other sides are 105,5 and 138,02625 .
Wrong. The solution is infinite
You need another data point to make a finite solution problem
There is not enough information given to solve this one.
Sorry, but it seems a little convincing "solution"
I suppose 'math majors' will love this, but for the rest of us, it's a lesson in futile thinking. Hmm....has anyone done the trig to figure out how 'small' the opposite angle is?? NOT an integer, I presume..... hahaha....glad I never got this problem on an exam....
He threw in an extra requirement that a and c differed by only1. That's cheating. Bad problem.
a = 8, b = 15, c = 17 perimeter = 40; OR a = 8, b = 6, c = 10 perimeter = 24. This proves the fallacy of this video.
You’re quite mistaken. What you’ve shown is that there are some whole numbers (such as 8) that belong to more than one Pythagorean triple. But that doesn’t mean it’s true of EVERY whole number. Some whole numbers (including all odd primes, such as 89) belong to only one such triple.
There are infinite solutions for this question due the given information.