❤❤❤ thanks 💯🙏 keep going my dear teacher ❤️

@PreMath

8 ай бұрын

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.

@krishnaagarwal5163

6 ай бұрын

You are correct. If the sides can be any positive real number, there are infinite answers

@gregorywildie37

6 ай бұрын

So the answer provided is not actually the answer to the question as actually posed. An answer but not the answer

@lintelle2382

6 ай бұрын

I was thinking the same thing!

@michalswiderski507

6 ай бұрын

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

@costakapsalis7667

6 ай бұрын

The confusion would have been avoided if it was stated from the start that all sides are positive integers.

With the given information there are endless solutions. When a nears 0 , c nears 89+ . When a nears endles, c nears endles

@user-lj1nd8rq9w

6 ай бұрын

Not really - sides have to be positive integers and there is only one solution.

@Arqade38

3 ай бұрын

It's not, Since it's already stated that the side lengths must be positive integers.

@lnmukund6152

3 ай бұрын

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

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.

@taxidude

3 ай бұрын

Sorry but without any 2nd side or an angle , there are an infinite number of triangles.

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.

@patrickcorliss8878

5 ай бұрын

0:56 “Keep in your mind that the side lengths must be a positive integer”, See diagram: Sides ∈ Z+

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

@user-lj1nd8rq9w

6 ай бұрын

It is easy enough to prove your statement - just give us as least one more solution.

@pablomonroy332

5 ай бұрын

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.

@patrickcorliss8878

5 ай бұрын

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.

@roger7341

8 ай бұрын

Z^+ was given.

Amazing 👍 Thanks for sharing 😊

I'm definitely coming back to this to give it a try.

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.

@ybodoN

8 ай бұрын

​@@pluisjenijn to be exact, the funny property is n² + m² = (m + 1)² like (21, 220, 221) (201, 20200, 20201) (2001, 2002000, 2002001)

@sail2byzantium

8 ай бұрын

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!

@Ctrl_Alt_Sup

8 ай бұрын

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!

@ybodoN

8 ай бұрын

@@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).

@waheisel

7 ай бұрын

@@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.

Surely there are many possible solutions

@BruceArnold318

8 ай бұрын

He said they are integers.

@ybodoN

8 ай бұрын

Since 89 is a prime number, there is only one solution 🧐

@gayatrithanvi8901

8 ай бұрын

As they are positive integers and the number(89 Square) is PRIME having only one solution THERE IS ONLY ONE SOLUTION YOU FOOL

@MrPaulc222

8 ай бұрын

@@BruceArnold318 Ah, I missed that bit too. I was scratching my head thinking that the number of solutions is infinite.

@abefroman7393

8 ай бұрын

There’s only one….and stop calling me Shirley😂

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

@_Udo_Hammermeister

7 ай бұрын

Your formula is great. If b=3 than c=5 and a=4 . Fits best !

@douglasmiller1233

6 ай бұрын

"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.

@Ctrl_Alt_Sup

5 ай бұрын

@@douglasmiller1233 We are looking for sides belonging to Z+. In fact we are looking for a Pythagorean triple, and therefore only integers.

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.

@success762

5 ай бұрын

6.8.10 not like that

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)

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.

@simpleman283

8 ай бұрын

Go to 1:00 it shows him making a circle around the Z+. It means the side lengths can only be whole numbers.

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.

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?

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.

@pablomonroy332

5 ай бұрын

actually no, the prob says integer numbers on the sides, that narrow it down to only 1 solution.

@patrickcorliss8878

5 ай бұрын

0:56 “Keep in your mind that the side lengths must be a positive integer”, See diagram: Sides ∈ Z+

thank you

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.

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.

Okay Very thanks

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.

Respected Sir 🙏, I like the way of your answering

Obviously, the larger value is more acceptable here. Very easy

Solutions may not be full filled Pythagoras values Please check this sir.

I loved this question!

@PreMath

8 ай бұрын

❤️ Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

If the angle at A changes without changing the length of AB, will the answer be the same?

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".

Similar using for side b à prime number, à good idea for fun.

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.

the title page is misleading because it fails to say that the sides are integers.

That’s very nice Thanks Sir Thanks PreMath ❤❤❤❤❤

@PreMath

8 ай бұрын

Always welcome You are awesome. Keep it up 👍

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

@patrickcorliss8878

5 ай бұрын

0:56 “Keep in your mind that the side lengths must be a positive integer”, See diagram: Sides ∈ Z+

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".

@pablomonroy332

5 ай бұрын

the sides must be integer numbers..theres only 1 solution.

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.)

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

8 ай бұрын

As long as the angle A is less than 90°, we have a triangle, no matter how long is the hypotenuse 🤓

Aren't there infinite perimeters?

@ra15899550

6 ай бұрын

Yes, there are infinite solutions to the perimeter because of lack of information.

@pablomonroy332

5 ай бұрын

@@ra15899550 theres only 1 solution, the prob says the sides must be integer numbers, i had the same concern but thats the correct answer.

Probably could solve this much easier using trig to find side BC using arctan(). And then Pythagorean theorem to find side AC.

@northdallashs1

5 ай бұрын

So...arctan(89/BC) =

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.

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".

@patrickcorliss8878

5 ай бұрын

In the diagram: Sides ∈ Z+ is clumsy math talk for "sides are elements of the set of positive integers"

Well, that is one answer of many.

There are infinite solutions for this question due the given information.

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 ?"

@patrickcorliss8878

5 ай бұрын

0:56 “Keep in your mind that the side lengths must be a positive integer”, See diagram: Sides ∈ Z+

There are an infinite number of solutions to this problem depending on the slope of the hypotenuse

If the hypotenuse of the right triangle is 89 what is the perimeter and it's area?

C'est une possibilité, ça pourrait aussi être une infinité d'autres solutions, non ?

@ybodoN

8 ай бұрын

Comme le plus petit des trois nombres est premier, il n'y a qu'un seul triplet pythagoricien possible ! 😉

@olivierjosephdeloris8153

8 ай бұрын

@@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

@ybodoN

8 ай бұрын

@@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.

@olivierjosephdeloris8153

8 ай бұрын

​​​@@ybodoNd'accord, en effet la contrainte des nombres entiers, ça change tout. Le Z+ m'avait échappé

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

@pablomonroy332

5 ай бұрын

12

89^2=(c-a)(c+a), 89 is prime, 89^2=1×89^2 89×89 89^2×1, thus c-a=1, c+a=89^2, 2a+1=89^2, a=3960, c=3961, therefore the perimeter is 89+3960+3961=8010😊

@PreMath

8 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@BalakrishnaPerala

8 ай бұрын

no need to solve for a,c values.. we already got c+a=89^2 ; we need perimeter (c+a)+b = 89^2+89 = 8010

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!!!

@ybodoN

8 ай бұрын

There is an important detail in the upper right corner of the video: _sides_ ∈ ℤ⁺ 🧐

Line BC=89 line AC= 89*2^(1/2) is also an answer!

Will this solution satisfy Pythagorean solution.

@pietergeerkens6324

8 ай бұрын

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?

@pablomonroy332

5 ай бұрын

because a must be a integer and also c, so....

Il y a une infinité de valeurs de a et c, ainsi pour le périmètre

@BruceArnold318

8 ай бұрын

I thought so too but he said they are integers.

@ybodoN

8 ай бұрын

Comme le plus petit des trois nombres est premier, il n'y a qu'un seul triplet pythagoricien possible ! 😉

@gc1924

8 ай бұрын

​@@BruceArnold318merci, je ne suis pas très bon en anglais, je n'avais pas saisi

@gc1924

8 ай бұрын

​@@ybodoNje ne comprend pas vraiment bien l'anglais et je n'avais saisi : appartient à Z. Merci pour votre réponse

@AnthonyPierreLucien

8 ай бұрын

Je reste d'accord avec vous: il y a une infinité de solutions.

Would have been easier to plug into c-a=1, so a=c-1

@PreMath

8 ай бұрын

Thanks for your feedback! Cheers! 😀

The Perimeter is any value that is equal to or greater than 89!

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?

Assume the lengths are positive integers.

@patrickcorliss8878

5 ай бұрын

0:56 “Keep in your mind that the side lengths must be a positive integer”, See diagram: Sides ∈ Z+

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.

@darbyl3872

6 ай бұрын

So math lessons should be limited to what an engineer might see? What if he has poor eyesight? Mr. Magoo's Math 😂

Wow! I used trigonometry, but that got me nowhere. Great solution!

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 ?

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.

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.

P=89, b=3960, h=3961, May be

In real the problem has infinity answers

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!

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

As long as you eliminate a=0, P=89+a+c=89+7921 didn't really need to solve for the 2 sides.

@dawon7750

8 ай бұрын

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!

@MegaSuperEnrique

8 ай бұрын

​@@dawon7750pls watch video before commenting. 7:00

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。

anybody can see that given only the length of one side the problem has infinite solutions

But c^2=a^2 + b^2 Does not add up

It is not possible to solve that problem. It is necessary at least the values of two sides or the value of 2 angles and one side

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.

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

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

@patrickcorliss8878

5 ай бұрын

0:56 “Keep in your mind that the side lengths must be a positive integer”, See diagram: Sides ∈ Z+

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.

@pablomonroy332

5 ай бұрын

yhea but the problem says integer numbers...so...

Baba, you should tell the angle then only one solution will emerge

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.

@esunisen3862

8 ай бұрын

Only because it's in Z+ In R there is an infinity of solutions.

@sail2byzantium

8 ай бұрын

@@esunisen3862 I have no idea whatsoever as to what you just said. Thanks?

@simpleman283

8 ай бұрын

@@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..

@davidcoorey423

8 ай бұрын

@@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! :)

89 is prime number given, so the solution became possible

who says the sides have to be whole numbers?

@pablomonroy332

5 ай бұрын

emmm..the problem?

Simple steps... but a magnificent sequence!

208 ÷2

There is not enough information given to solve this one.

116, 145

He threw in an extra requirement that a and c differed by only1. That's cheating. Bad problem.

There can be numerous triangles with this information??? AC side can be anything more than 89? No?

Tudo errado, isso tem infinitas soluções , mas a solução proposta não é uma delas. Essa solução e inconsistente com o torema de Pitágoras.

There are many solutions for a and b

@ybodoN

8 ай бұрын

... but only one where a, b and c are integers 😉

@e1woqf

8 ай бұрын

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.

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

WHAT?!

Sorry, but it seems a little convincing "solution"

So, if you were asked to find the perimeter of the right triangle with limited information, like we have, you could do as the video did. It finds one possible perimeter. But if there are other values to the base, different hypotenuses will also result, leading to different perimeters. If the sides are integers (positive) there are many solutions. If the integer requirement is lifted, we have an infinite number of bases, hypotenuses and perimeters!

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....

يوجد اجوبة لا نهائية لكل من a. C حيث نعطي قيمة ل a ثم نحسب قيمة c. حسب نظرية فيثاغورث ولا داعي لكل هذا العمل 😂

I am not satisfy your solution

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 .