whitley

08-25-2005, 10:44 PM

Help. if the hypotenuese = 3, and angle opposite the hypotnuese is 90 degrees how can I find the measurements of the legs????

I need help please

I need help please

View Full Version : Pythagoreum theory

whitley

08-25-2005, 10:44 PM

Help. if the hypotenuese = 3, and angle opposite the hypotnuese is 90 degrees how can I find the measurements of the legs????

I need help please

I need help please

08-26-2005, 12:18 AM

need more information to solve this one....

either 2 sides and one angle,

3 sides

or 1 side and 2 angles

....

either 2 sides and one angle,

3 sides

or 1 side and 2 angles

....

stapel

08-26-2005, 12:18 AM

With only the one measurement, the lengths of the legs of the right triangle cannot be determined. To use the Pythagorean Theorem, you need two of the sides; one isn't enough. Sorry.

Eliz.

Eliz.

TchrWill

08-26-2005, 09:36 AM

Given any positive integer, it is possible to derive Pythagorean Triples with that integer as the hypotenuse.

First, some general facts regarding hypotenuses.

The hypotenuse of a Pythagorean triangle must be of the form k(m^2 + n^2).

Only integers of the form k(m^2 + n^2) can be hypotenuses of Pythagorean triangles, e.g., 5, 10, 13, 15, etc.

To be primitive, k = 1, m and n must be co-prime integers, one odd, one even, m greater than n.

The hypotenuse of a primitive Pythagorean Triangle is always odd.

Every prime of the form 4n + 1 is a primitive hypotenuse.

Any number with at least one prime factor of the form 4n + 1 can be of the form k(m^2 + n^2).

A number N is a primitive hypotenuse if, and only if, all of its prime factors are of the form 4n + 1, e.g., 5, 13,

65, 85, etc. Every prime of the form 4n + 1 is a primitive hypotenuse.

Any prime number of the form 4n + 1 can be represented as the sum of two integral squares in one way.

Prime factors of the form 4n - 1 cannot be factors of a primitive hypotenuse

A prime number of the form 4n - 1 cannot be represented as the sum of two integral squares.

Any product whose factors are 2 and primes of the form 4n + 1 can be represented as the sum of two

integral squares.

If a number N has a total of n prime divisors of the form 4n + 1, N can be the hypotenuse of 2^(n-1) primitive Pythagorean triangles.

Viewing the factorization of N as 2^ao(p1^a1)p2^a2(p3^a3)....pn^an x q1^b1(q2^b2)q3^b3......qr^br, the p's being primes of the form 4n - 1 and the q's primes of the form 4n + 1. Then,

1) N is the hypotenuse of [(2b1 + 1)(2b2 + 1)(2b3 + 1).....(2br + 1)]/2 Pythagorean triangles

2) N is the sum of two unequal squares in [(b1 + 1)(b2 + 1)(b3 + 1).....(br + 1)]/2 ways for an even numerator or

[(b1 + 1)(b2 + 1)(b3 + 1).....(br + 1)]/2 - 1/2 for an odd numerator.

The method of verifying a hypotenuse makes use of two specific theorems:

1) A positive integer N can be represented by the sum of two squares if, and only if, its factorization into powers of distinct primes contains no odd powers of primes congruent to 3 modulo 4 (Two integers a and b are said to be congruent for the modulus m when their difference, a - b, is divisible by the integer m written as a = b (mod 4).)

2) By means of one of the most famous identities ever derived, the product of any two numbers, that are themselves the sums of two squares, can be represented as the sum of two other squares and often in two different ways.

.................................................. ..............= (ac + bd)^2 + (ad - bc)^2

.................................................. ............/

...............................(a^2 + b^2)(c^2 + d^2) or

.................................................. ............\

.................................................. ..............= (ac - bd)^2 + (ad + bc)^2

In the special case where c = d = 1, 2(a^2 + b^2) = (a + b)^2 + (b - a)^2 meaning that any product with factors of 2 and primes of the form 4n + 1 can be represented as the sum of two squares.

How do we make use of this information to solve problems of this type?

Out of the first 25 primes, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97, the following are of the form 4n + 1; 5, 13, 17, 29, 37, 41, 53, 61, 73, 79, 89, 97. The following are of the form 4n - 1; 3, 7, 19, 23, 31, 43, 59, 67, 71, 79, and 89.

Observe that 5 = (2^2 + 1^2), 13 = (3^2 + 2^2), 17 = (4^2 + 1^2), 29 = (5^2 + 2^2), 37 = (6^2 + 1^2), 41 = (5^2 + 4^2), and 53 = (7^2 + 2^2), for example.

Determine if any Pythagorean Triples exist with a hypotenuse of 65.

Consider N = x^2 + y^2 = 65.

............................................a..... ..b.....c.......d

.......................65 = 5x13 = (2^2 + 1^2)(3^2 + 2^2) with a, b, c, and d equaling 3, 2, 5, and 2. From our identities,

(2^2 + 1^2)(3^2 + 2^2) = (2x3 + 1x2)^2 + (2x2 - 1x3)^2 = (6 + 2)^2 + (4 - 3)^2 = (8^2 + 1^2) = 64 + 1 = 65 and

(2^2 + 1^2)(3^2 + 2^2) = (2x3 - 1x2)^2 + (2x2 + 1x3)^2 = (6 - 2)^2 + ( 4 + 3)^2 = (4^2 + 4^2) = 16 + 49 = 65.

Therefore, 65 = (8^2 + 1^2) = (4^2 + 7^2).

Thus, m = 8 with n = 1 or m = 7 with n = 4 will produce Pythagorean Triples with hypotenuses of 65.

Determine if any Pythagorean Triples exist with a hypotenuse of 377.

Consider N = x^2 + y^2 = 377

............................................a..... ..b.....c.......d

.....................377 = 13x29 = (3^2 + 2^2)(5^2 + 2^2) with a, b, c, and d equaling 3, 2, 5, and 2, we can make use of the identities to create

(3^2 + 2^2)(5^2 + 2^2) = (3x5 + 2x2)^2 + (3x2 - 2x5)^2 = (15 + 4)^2 + (6 - 10)^2 = (19^2 + 4^2) = 361 + 16 and ................................= (3x5 - 2x2)^2 + (3x2 + 2x5)^2 = (15 - 4)^2 + (6 + 10)^2 = (11^2 + 16^2) = 121 + 256 = 377.

Therefore, 377 = (19^2 + 4^2) = (11^2 + 16^2) and m = 19 with n = 4 and m = 16 with n = 11 will produce Pythagorean Triples with hypotenuse 377.

Determine if any Pythagorean Triples exist with a hypotenuse of 315.

Consider the number N = x^2 + y^2 = 315 = (3^2)x5x7. Since 315 is fo the form 4n - 1 and the factorization contains an odd power of the prime 7, 7 = 3 (mod 4), it cannot be represented as the sum of two squares.

Determine if any Pythagorean Triples exist with a hypotenuse of 3185.

Consider the number N = x^2 + y^2 = 3185. The prime factors of 3185 are 5x(7^2)x13. Since 3185 is of the form 4n + 1, and/or the factorization contains no odd power of a prime congruent to 3 modulo 4 (7^2 has an even power), the number is representable as a sum of two squares. Having 5 = (2^2 + 1^2), 7^2 = (7^2 + 0^2), and 13 = (3^2 + 2^2), we can write

................................a.......b.....c... ...d

3185 = 5x(7^2)x13 = (2^2 + 1^2)(7^2 + 0^2)(3^2 + 2^2)

............................= (14^2 + 7^2)(3^2 + 2^2)

............................= (14x3 + 7x2)^2 + (14x2 - 7x3)^2

............................=.(42 + 14)^2 + (28 - 21)^2

............................= (56^2 + 7^2) = 3136 + 49 = 3185.

Lets see how we can apply the method to a larger number.

Determine if any Pythagorean Triples exist with a hypotenuse of 1,395,182,880.

Consider N = x^2 + y^2 = 1,395,182,880.

The prime factors of this number are 2^5(3^4)5^1(7^2)13^3.

Being of the form 4n + 1, both 5 and 13 are the sums of two squares.

We already know that 5x13 can be written as (8^2 + 1^2) and (4^2 + 7^2).

We can rewrite our factorization of 2^5(3^4)5^1(7^2)13^3 as (2x5x13)x[2^4(3^4)7^2(13^2)].

From our third identity above, 2x5x13 = 9^2 + 7^2 and 11^2 + 3^2.

We can also rewrite [2^4(3^4)7^2(13^2)] as [2^2(3^2)7^1(13^1)]^2.

Then, 2^5(3^4)5^1(7^2)13^3 = [2^2(3^2)7^1(13^1)9^1]^2 + [2^2(3^2)7^1(13^1)7^1]^2

.......................................= 29,484^2 + 22,932^2

.......................................= 869,306,256 + 525,876,624

.......................................= 1,395,182,880 making m = 29,484 and n = 22,932

.................................................a nd

........2^5(3^4)5^1(7^2)13^3 = [2^2(3^2)7^1(13^1)11^1] + [2^2(3^2)7^1(13^1)3^1]

......................................= 36,036^2 + 9,828^2

......................................= 1,298,593,296 + 96,589,584

......................................= 1,395,182,880 making m = 36,036 and n = 9,828.

First, some general facts regarding hypotenuses.

The hypotenuse of a Pythagorean triangle must be of the form k(m^2 + n^2).

Only integers of the form k(m^2 + n^2) can be hypotenuses of Pythagorean triangles, e.g., 5, 10, 13, 15, etc.

To be primitive, k = 1, m and n must be co-prime integers, one odd, one even, m greater than n.

The hypotenuse of a primitive Pythagorean Triangle is always odd.

Every prime of the form 4n + 1 is a primitive hypotenuse.

Any number with at least one prime factor of the form 4n + 1 can be of the form k(m^2 + n^2).

A number N is a primitive hypotenuse if, and only if, all of its prime factors are of the form 4n + 1, e.g., 5, 13,

65, 85, etc. Every prime of the form 4n + 1 is a primitive hypotenuse.

Any prime number of the form 4n + 1 can be represented as the sum of two integral squares in one way.

Prime factors of the form 4n - 1 cannot be factors of a primitive hypotenuse

A prime number of the form 4n - 1 cannot be represented as the sum of two integral squares.

Any product whose factors are 2 and primes of the form 4n + 1 can be represented as the sum of two

integral squares.

If a number N has a total of n prime divisors of the form 4n + 1, N can be the hypotenuse of 2^(n-1) primitive Pythagorean triangles.

Viewing the factorization of N as 2^ao(p1^a1)p2^a2(p3^a3)....pn^an x q1^b1(q2^b2)q3^b3......qr^br, the p's being primes of the form 4n - 1 and the q's primes of the form 4n + 1. Then,

1) N is the hypotenuse of [(2b1 + 1)(2b2 + 1)(2b3 + 1).....(2br + 1)]/2 Pythagorean triangles

2) N is the sum of two unequal squares in [(b1 + 1)(b2 + 1)(b3 + 1).....(br + 1)]/2 ways for an even numerator or

[(b1 + 1)(b2 + 1)(b3 + 1).....(br + 1)]/2 - 1/2 for an odd numerator.

The method of verifying a hypotenuse makes use of two specific theorems:

1) A positive integer N can be represented by the sum of two squares if, and only if, its factorization into powers of distinct primes contains no odd powers of primes congruent to 3 modulo 4 (Two integers a and b are said to be congruent for the modulus m when their difference, a - b, is divisible by the integer m written as a = b (mod 4).)

2) By means of one of the most famous identities ever derived, the product of any two numbers, that are themselves the sums of two squares, can be represented as the sum of two other squares and often in two different ways.

.................................................. ..............= (ac + bd)^2 + (ad - bc)^2

.................................................. ............/

...............................(a^2 + b^2)(c^2 + d^2) or

.................................................. ............\

.................................................. ..............= (ac - bd)^2 + (ad + bc)^2

In the special case where c = d = 1, 2(a^2 + b^2) = (a + b)^2 + (b - a)^2 meaning that any product with factors of 2 and primes of the form 4n + 1 can be represented as the sum of two squares.

How do we make use of this information to solve problems of this type?

Out of the first 25 primes, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97, the following are of the form 4n + 1; 5, 13, 17, 29, 37, 41, 53, 61, 73, 79, 89, 97. The following are of the form 4n - 1; 3, 7, 19, 23, 31, 43, 59, 67, 71, 79, and 89.

Observe that 5 = (2^2 + 1^2), 13 = (3^2 + 2^2), 17 = (4^2 + 1^2), 29 = (5^2 + 2^2), 37 = (6^2 + 1^2), 41 = (5^2 + 4^2), and 53 = (7^2 + 2^2), for example.

Determine if any Pythagorean Triples exist with a hypotenuse of 65.

Consider N = x^2 + y^2 = 65.

............................................a..... ..b.....c.......d

.......................65 = 5x13 = (2^2 + 1^2)(3^2 + 2^2) with a, b, c, and d equaling 3, 2, 5, and 2. From our identities,

(2^2 + 1^2)(3^2 + 2^2) = (2x3 + 1x2)^2 + (2x2 - 1x3)^2 = (6 + 2)^2 + (4 - 3)^2 = (8^2 + 1^2) = 64 + 1 = 65 and

(2^2 + 1^2)(3^2 + 2^2) = (2x3 - 1x2)^2 + (2x2 + 1x3)^2 = (6 - 2)^2 + ( 4 + 3)^2 = (4^2 + 4^2) = 16 + 49 = 65.

Therefore, 65 = (8^2 + 1^2) = (4^2 + 7^2).

Thus, m = 8 with n = 1 or m = 7 with n = 4 will produce Pythagorean Triples with hypotenuses of 65.

Determine if any Pythagorean Triples exist with a hypotenuse of 377.

Consider N = x^2 + y^2 = 377

............................................a..... ..b.....c.......d

.....................377 = 13x29 = (3^2 + 2^2)(5^2 + 2^2) with a, b, c, and d equaling 3, 2, 5, and 2, we can make use of the identities to create

(3^2 + 2^2)(5^2 + 2^2) = (3x5 + 2x2)^2 + (3x2 - 2x5)^2 = (15 + 4)^2 + (6 - 10)^2 = (19^2 + 4^2) = 361 + 16 and ................................= (3x5 - 2x2)^2 + (3x2 + 2x5)^2 = (15 - 4)^2 + (6 + 10)^2 = (11^2 + 16^2) = 121 + 256 = 377.

Therefore, 377 = (19^2 + 4^2) = (11^2 + 16^2) and m = 19 with n = 4 and m = 16 with n = 11 will produce Pythagorean Triples with hypotenuse 377.

Determine if any Pythagorean Triples exist with a hypotenuse of 315.

Consider the number N = x^2 + y^2 = 315 = (3^2)x5x7. Since 315 is fo the form 4n - 1 and the factorization contains an odd power of the prime 7, 7 = 3 (mod 4), it cannot be represented as the sum of two squares.

Determine if any Pythagorean Triples exist with a hypotenuse of 3185.

Consider the number N = x^2 + y^2 = 3185. The prime factors of 3185 are 5x(7^2)x13. Since 3185 is of the form 4n + 1, and/or the factorization contains no odd power of a prime congruent to 3 modulo 4 (7^2 has an even power), the number is representable as a sum of two squares. Having 5 = (2^2 + 1^2), 7^2 = (7^2 + 0^2), and 13 = (3^2 + 2^2), we can write

................................a.......b.....c... ...d

3185 = 5x(7^2)x13 = (2^2 + 1^2)(7^2 + 0^2)(3^2 + 2^2)

............................= (14^2 + 7^2)(3^2 + 2^2)

............................= (14x3 + 7x2)^2 + (14x2 - 7x3)^2

............................=.(42 + 14)^2 + (28 - 21)^2

............................= (56^2 + 7^2) = 3136 + 49 = 3185.

Lets see how we can apply the method to a larger number.

Determine if any Pythagorean Triples exist with a hypotenuse of 1,395,182,880.

Consider N = x^2 + y^2 = 1,395,182,880.

The prime factors of this number are 2^5(3^4)5^1(7^2)13^3.

Being of the form 4n + 1, both 5 and 13 are the sums of two squares.

We already know that 5x13 can be written as (8^2 + 1^2) and (4^2 + 7^2).

We can rewrite our factorization of 2^5(3^4)5^1(7^2)13^3 as (2x5x13)x[2^4(3^4)7^2(13^2)].

From our third identity above, 2x5x13 = 9^2 + 7^2 and 11^2 + 3^2.

We can also rewrite [2^4(3^4)7^2(13^2)] as [2^2(3^2)7^1(13^1)]^2.

Then, 2^5(3^4)5^1(7^2)13^3 = [2^2(3^2)7^1(13^1)9^1]^2 + [2^2(3^2)7^1(13^1)7^1]^2

.......................................= 29,484^2 + 22,932^2

.......................................= 869,306,256 + 525,876,624

.......................................= 1,395,182,880 making m = 29,484 and n = 22,932

.................................................a nd

........2^5(3^4)5^1(7^2)13^3 = [2^2(3^2)7^1(13^1)11^1] + [2^2(3^2)7^1(13^1)3^1]

......................................= 36,036^2 + 9,828^2

......................................= 1,298,593,296 + 96,589,584

......................................= 1,395,182,880 making m = 36,036 and n = 9,828.

stapel

08-26-2005, 10:12 AM

Given any positive integer, it is possible to derive Pythagorean Triples with that integer as the hypotenuse.

However, the poster did not specify that the triangle's legs were integral.

Eliz.

However, the poster did not specify that the triangle's legs were integral.

Eliz.

pka

08-26-2005, 11:22 AM

We can give a general solution!

If L is the length of one leg then 0< L <3 and the other leg has length √(9−L<SUP>2</SUP>).

If L is the length of one leg then 0< L <3 and the other leg has length √(9−L<SUP>2</SUP>).

TchrWill

08-26-2005, 01:00 PM

History records that Pythagoras and Diophantus were probably the two most well known mathematicians that had anything to do with right triangles with integer sides. The Pythagoreans were known for their intense interest in relationships that could be expressed in whole numbers. The famous Pythagorean Theorem states that in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the two legs, or x^2 + y^2 = z^2. The Pythagoreans were thought to have been one of the first to provide a proof. Almost everyone has heard of the 3-4-5 right triangle where 3^2 + 4^2 = 5^2, the simplest, and most fundamental triangle based on the Pythagorean theorem. Not so surprisingly, fewer people know of the 5, 12, and 13 right triangle, the 7, 24, and 25 right triangle, the 20-21-29 right triangle, and infinitely more, which also satisfy the relationship and without any

proportional relationship to the 3, 4, and 5 triangle.

These infinitely many sets of three integers, all satisfying the Pythagorean theorem relationship, are traditionally referred to as Pythagorean Triples.

proportional relationship to the 3, 4, and 5 triangle.

These infinitely many sets of three integers, all satisfying the Pythagorean theorem relationship, are traditionally referred to as Pythagorean Triples.

Denis

08-26-2005, 02:58 PM

Help. if the hypotenuese = 3, and angle opposite the hypotnuese is 90 degrees how can I find the measurements of the legs????

I need help please

Geesh, why confuse this poor beginner...

whitley, one of the other 2 angles is required.

Label the triangle ABC (C at the 90 degree corner, thus AB = hypotenuse),

call the angles A, B and C (thus angle C = 90 degrees),

and let BC = a, AC = b and AB = c

(that's standard notation, so try and use it in the future)

We have c = 3; we need either angle A or angle B to solve a and b;

assume angle A is given; then (angle B = 90 - angle A) and:

a = c[SIN(A)]

b = sqrt(c^2 - a^2)

If angle B is given, same way:

b = c[SIN(B)]

a = sqrt(c^2 - b^2)

I need help please

Geesh, why confuse this poor beginner...

whitley, one of the other 2 angles is required.

Label the triangle ABC (C at the 90 degree corner, thus AB = hypotenuse),

call the angles A, B and C (thus angle C = 90 degrees),

and let BC = a, AC = b and AB = c

(that's standard notation, so try and use it in the future)

We have c = 3; we need either angle A or angle B to solve a and b;

assume angle A is given; then (angle B = 90 - angle A) and:

a = c[SIN(A)]

b = sqrt(c^2 - a^2)

If angle B is given, same way:

b = c[SIN(B)]

a = sqrt(c^2 - b^2)

TchrWill

08-26-2005, 05:27 PM

Hi Whitley,

I must admit I went off in a direction that you probably were not looking for. Upon reviewing your original question, "Help. if the hypotenuese = 3, and angle opposite the hypotnuese is 90 degrees how can I find the measurements of the legs?", I mistakenly went off in the direction of integral side measures, or Pythagorean Triples, as they are more typically referred to. It is clear that you were simply seeking the measures of the two sides having been given the hypotenuse of 3.

As has already been stated by other responders, you need the measure of one of the other two angles of the triangle or the measure of one of the sides itself. Given an angle µ, the two sides become 3sinµ and 3 cosµ.

Given one of the two sides, a, the third side is simply sqrt(3^2 - a^2).

My apologies for diverting your valuable time and possibly confusing you.

I must admit I went off in a direction that you probably were not looking for. Upon reviewing your original question, "Help. if the hypotenuese = 3, and angle opposite the hypotnuese is 90 degrees how can I find the measurements of the legs?", I mistakenly went off in the direction of integral side measures, or Pythagorean Triples, as they are more typically referred to. It is clear that you were simply seeking the measures of the two sides having been given the hypotenuse of 3.

As has already been stated by other responders, you need the measure of one of the other two angles of the triangle or the measure of one of the sides itself. Given an angle µ, the two sides become 3sinµ and 3 cosµ.

Given one of the two sides, a, the third side is simply sqrt(3^2 - a^2).

My apologies for diverting your valuable time and possibly confusing you.

pka

08-26-2005, 08:00 PM

TchrWill, your understanding “The Pythagoreans were known for their intense interest in relationships that could be expressed in whole numbers.” is correct.

In fact, they assigned whole number to all relations: man, woman, love, marriage, etc.

We owe them for the term irrational number because √2 is not the ratio of two whole numbers so it is not rational.

But I find their monastic rule, “EAT NO BEANS”, most memorable!

In fact, they assigned whole number to all relations: man, woman, love, marriage, etc.

We owe them for the term irrational number because √2 is not the ratio of two whole numbers so it is not rational.

But I find their monastic rule, “EAT NO BEANS”, most memorable!

Denis

08-27-2005, 12:52 AM

But I find their monastic rule, “EAT NO BEANS”, most memorable!

...and quiet :)

...and quiet :)

Denis

08-27-2005, 01:20 AM

Something for you to "ponder", pka and TchrWill:

what does right triangle 135-352-377 and 2 right triangles 66-360-366

stuck together to form an isosceles triangle have in common?

(my own discovery)

what does right triangle 135-352-377 and 2 right triangles 66-360-366

stuck together to form an isosceles triangle have in common?

(my own discovery)

mathmike

08-27-2005, 01:25 AM

Something for you to "ponder", pka and TchrWill:

what does right triangle 135-352-377 and 2 right triangles 66-360-366

stuck together to form an isosceles triangle have in common?

(my own discovery)

Area (47520)?

what does right triangle 135-352-377 and 2 right triangles 66-360-366

stuck together to form an isosceles triangle have in common?

(my own discovery)

Area (47520)?

TchrWill

08-27-2005, 03:00 PM

What does right triangle 135-352-377 and 2 right triangles 66-360-366

stuck together to form an isosceles triangle have in common?

They both have the same area of 23,760 sq.units.

stuck together to form an isosceles triangle have in common?

They both have the same area of 23,760 sq.units.

Denis

08-27-2005, 04:37 PM

What does right triangle 135-352-377 and 2 right triangles 66-360-366

stuck together to form an isosceles triangle have in common?

They both have the same area of 23,760 sq.units.

Yes, as a start; but that wouldn't be much of a "discovery"...

What else, and what else ?

stuck together to form an isosceles triangle have in common?

They both have the same area of 23,760 sq.units.

Yes, as a start; but that wouldn't be much of a "discovery"...

What else, and what else ?

TchrWill

08-27-2005, 05:29 PM

What does right triangle 135-352-377 and 2 right triangles 66-360-366

stuck together to form an isosceles triangle have in common?

They both have the same area of 23,760 sq.units.

Yes, as a start; but that wouldn't be much of a "discovery"...

What else, and what else ?

Could it be that they are both Heronian Triangles with the same perimeter?

2(66) + 2(366) = 864 (360 is the altitude)

135 + 352 + 377 = 864.

stuck together to form an isosceles triangle have in common?

They both have the same area of 23,760 sq.units.

Yes, as a start; but that wouldn't be much of a "discovery"...

What else, and what else ?

Could it be that they are both Heronian Triangles with the same perimeter?

2(66) + 2(366) = 864 (360 is the altitude)

135 + 352 + 377 = 864.

Denis

08-28-2005, 12:27 AM

Yes, TchrWill. I sort of lucked into this one, trying to devise a puzzle.

My real "discovery" was that this is (apparently) the only primitive case.

I thought you'd be interested, knowing the huge number of pytagorean

triplets.

There are 99745 of them with short leg below 10000.

My little "get 'em" program (Ubasic) finds them all in 14 seconds.

I tested up to 400000 being the smallest right triangle leg.

It was accepted as such by mathworld:

http://mathworld.wolfram.com/HeronianTriangle.html

However, there is no proof; perhaps you'll find the next one ? :)

My real "discovery" was that this is (apparently) the only primitive case.

I thought you'd be interested, knowing the huge number of pytagorean

triplets.

There are 99745 of them with short leg below 10000.

My little "get 'em" program (Ubasic) finds them all in 14 seconds.

I tested up to 400000 being the smallest right triangle leg.

It was accepted as such by mathworld:

http://mathworld.wolfram.com/HeronianTriangle.html

However, there is no proof; perhaps you'll find the next one ? :)

TchrWill

08-28-2005, 09:47 AM

Denis,

You might be interested in this.

How many triangles can you find whose area is equal to its perimeter?

This can be explored from a few viewpoints. We can explore integer right triangles, i.e., Pythagorean Triple triangles, right triangles with non-integer sides, scalene triangles with integer sides (Heronian Triangles) and scalene triangles with non-integer sides. Integer sided right triangles, i.e., Pythagorean Triples, can be pursued in two ways resulting in only two triangles that satisfy the criteria.

Deriving the specific integer answers for right triangles:

The three expressions for deriving the integer sides of a Pythagorean triangle, Pythagorean Triples, are x = m^2 - n^2, y = 2mn, and z = m^2 + n^2. The area can be expressed by A = 2mn(m^2 - n^2)/2. The perimeter can be expressed by P = m^2 - n^2 + 2mn + m^2 + n^2. Equating the two and simplifying we get m^2 - (2/n)m - (n^2 + 2) = 0. Solving for m using the quadratic formula, we derive m = [(2/n) +/- sqrt[(4/n^2) + 4n^2 + 8]]/2. The 4/n^2 allows only n = 1 or 2 in order to result in integer answers which results in m = 3 in both cases. Therefore, there are only 2 Pythagorean triangles whose areas and perimeters are equal. One derived from m = 3 and n = 1 and one from m = 3 and n = 2 which turn out to be the two you have already identified, the 6-8-10 and the 5-12-13 triangles.

Deriving all possible answers for right triangles:

Consider the right triangle with sides a, b, and c. The area A = ab/2 and the perimeter p = a + b + c. With a right triangle, c = sqrt(a^2 + b^2) making ab/2 = a + b + sqrt(a^2 + b^2). This leads to ab/2 - a - b = sqrt(a^2 + b^2). Squaring both sides, collecting and simplifying leads to ab/4 +2 -a -b = 0 which leads to b = (a - 2)/(a/4 - 1) or b = 4(a - 2)/(a - 4). Thus, there is a "b" for every value of "a" greater than 4. Values of a = 5 and 6 lead us to the same integer triangles already identified, 5-12-13 and 6-8-10. Values beyond a = 5 produce an infinite number of right triangle answers where at least one side is an integer and at most two.

Deriving integer answers for scalene triangles:

Consider the Heronian Triangle, a scalene triangle with integer sides, integer area, and at least one integer altitude. Let the three sides be a, b, and c. The area can be expressed, using Heron's famous area formula, by A = sqrt[s(s-a)(s-b)(s-c)] where s = (a + b + c)/2 and the perimeter p = (a + b + c) = 2s. Equating, we have sqrt[s(s-a)(s-b)(s-c)] = 2s. Squaring both sides yields [s(s-a)(s-b)(s-c)] = 4s^2 or (s-a)(s-b)(s-c)] = 4s.

I did not expand this to see if a closed solution exists, though I doubt it. Better still, I did some quick calculations with the basic low end Pythagorean Triples derived from m and n equaling 4, 3, 2, and 1.

As best as I can determine from these calculations and the literature, there are only 3 non-right-angled Heronian triangles that meet the criteria. To repeat, a Heronian triangle is one with integer sides, integer area, and at least one integer altitude. They are typically created from two Pythagorean Triple triangles joined at the common side. For instance, the 5-12-13 and 9-12-15 Pythagorean triangles can be joined, side by side, at the 12 leg to create the 13-14-15 Heronian Triangle with altitude 12. The area is 84. (By the way, the only triangle with consecutive integer altitude and sides.)

The 3 Heronian Triangles (HT) meeting the equal area and perimeter criteria are:

1--The 6-25-29 HT made up from the 15-20-25 Pythagorean Triangle being folded inside of the 20-21-29 Pythagorean Triangle along the common altitude leg of 20, the 15 leg being subtracted from the 21 leg to produce the 6 leg. The area and perimeter are both 60.

2--The 7-15-20 HT made up from the 9-12-15 Pythagorean Triangle being folded inside of the 12-16-20 Pythagorean Triangle along the common altitude leg of 12, the 9 leg being subtracted from the 16 leg to produce the 7 leg. The area and perimeter are both 42.

3--The 9-10-17 HT made up from the 6-8-10 Pythagorean Triangle being folded inside of the 8-15-17 Pythagorean Triangle along the common altitude leg of 8, the 6 leg being subtracted from the 15 leg to produce the 9 leg. The area and perimeter are both 36.

Deriving non-integer answers for non-right angled triangles:

The approach from above would have to be taken to its logical conclusion. The area can again be expressed, using Heron's area formula, by A = sqrt[s(s-a)(s-b)(s-c)] where s = (a + b + c)/2 and the perimeter p = (a + b + c) = 2s. Equating, we have sqrt[s(s-a)(s-b)(s-c)] = 2s. Squaring both sides yields [s(s-a)(s-b)(s-c)] = 4s^2 or (s-a)(s-b)(s-c)] = 4s. This expands to s^3 - (a+b+c)s^2 + (ab+ac+bc)s - abc = 4s which leads to (ab+ac+bc) - (abc)/s - s^2 = 4. This can be extended to an expression entirely in terms of a, b, and c, a rather long and involved expression, which I do not see a solution for at the moment but I have not given up.

You might be interested in this.

How many triangles can you find whose area is equal to its perimeter?

This can be explored from a few viewpoints. We can explore integer right triangles, i.e., Pythagorean Triple triangles, right triangles with non-integer sides, scalene triangles with integer sides (Heronian Triangles) and scalene triangles with non-integer sides. Integer sided right triangles, i.e., Pythagorean Triples, can be pursued in two ways resulting in only two triangles that satisfy the criteria.

Deriving the specific integer answers for right triangles:

The three expressions for deriving the integer sides of a Pythagorean triangle, Pythagorean Triples, are x = m^2 - n^2, y = 2mn, and z = m^2 + n^2. The area can be expressed by A = 2mn(m^2 - n^2)/2. The perimeter can be expressed by P = m^2 - n^2 + 2mn + m^2 + n^2. Equating the two and simplifying we get m^2 - (2/n)m - (n^2 + 2) = 0. Solving for m using the quadratic formula, we derive m = [(2/n) +/- sqrt[(4/n^2) + 4n^2 + 8]]/2. The 4/n^2 allows only n = 1 or 2 in order to result in integer answers which results in m = 3 in both cases. Therefore, there are only 2 Pythagorean triangles whose areas and perimeters are equal. One derived from m = 3 and n = 1 and one from m = 3 and n = 2 which turn out to be the two you have already identified, the 6-8-10 and the 5-12-13 triangles.

Deriving all possible answers for right triangles:

Consider the right triangle with sides a, b, and c. The area A = ab/2 and the perimeter p = a + b + c. With a right triangle, c = sqrt(a^2 + b^2) making ab/2 = a + b + sqrt(a^2 + b^2). This leads to ab/2 - a - b = sqrt(a^2 + b^2). Squaring both sides, collecting and simplifying leads to ab/4 +2 -a -b = 0 which leads to b = (a - 2)/(a/4 - 1) or b = 4(a - 2)/(a - 4). Thus, there is a "b" for every value of "a" greater than 4. Values of a = 5 and 6 lead us to the same integer triangles already identified, 5-12-13 and 6-8-10. Values beyond a = 5 produce an infinite number of right triangle answers where at least one side is an integer and at most two.

Deriving integer answers for scalene triangles:

Consider the Heronian Triangle, a scalene triangle with integer sides, integer area, and at least one integer altitude. Let the three sides be a, b, and c. The area can be expressed, using Heron's famous area formula, by A = sqrt[s(s-a)(s-b)(s-c)] where s = (a + b + c)/2 and the perimeter p = (a + b + c) = 2s. Equating, we have sqrt[s(s-a)(s-b)(s-c)] = 2s. Squaring both sides yields [s(s-a)(s-b)(s-c)] = 4s^2 or (s-a)(s-b)(s-c)] = 4s.

I did not expand this to see if a closed solution exists, though I doubt it. Better still, I did some quick calculations with the basic low end Pythagorean Triples derived from m and n equaling 4, 3, 2, and 1.

As best as I can determine from these calculations and the literature, there are only 3 non-right-angled Heronian triangles that meet the criteria. To repeat, a Heronian triangle is one with integer sides, integer area, and at least one integer altitude. They are typically created from two Pythagorean Triple triangles joined at the common side. For instance, the 5-12-13 and 9-12-15 Pythagorean triangles can be joined, side by side, at the 12 leg to create the 13-14-15 Heronian Triangle with altitude 12. The area is 84. (By the way, the only triangle with consecutive integer altitude and sides.)

The 3 Heronian Triangles (HT) meeting the equal area and perimeter criteria are:

1--The 6-25-29 HT made up from the 15-20-25 Pythagorean Triangle being folded inside of the 20-21-29 Pythagorean Triangle along the common altitude leg of 20, the 15 leg being subtracted from the 21 leg to produce the 6 leg. The area and perimeter are both 60.

2--The 7-15-20 HT made up from the 9-12-15 Pythagorean Triangle being folded inside of the 12-16-20 Pythagorean Triangle along the common altitude leg of 12, the 9 leg being subtracted from the 16 leg to produce the 7 leg. The area and perimeter are both 42.

3--The 9-10-17 HT made up from the 6-8-10 Pythagorean Triangle being folded inside of the 8-15-17 Pythagorean Triangle along the common altitude leg of 8, the 6 leg being subtracted from the 15 leg to produce the 9 leg. The area and perimeter are both 36.

Deriving non-integer answers for non-right angled triangles:

The approach from above would have to be taken to its logical conclusion. The area can again be expressed, using Heron's area formula, by A = sqrt[s(s-a)(s-b)(s-c)] where s = (a + b + c)/2 and the perimeter p = (a + b + c) = 2s. Equating, we have sqrt[s(s-a)(s-b)(s-c)] = 2s. Squaring both sides yields [s(s-a)(s-b)(s-c)] = 4s^2 or (s-a)(s-b)(s-c)] = 4s. This expands to s^3 - (a+b+c)s^2 + (ab+ac+bc)s - abc = 4s which leads to (ab+ac+bc) - (abc)/s - s^2 = 4. This can be extended to an expression entirely in terms of a, b, and c, a rather long and involved expression, which I do not see a solution for at the moment but I have not given up.

Denis

08-28-2005, 10:40 AM

Yes; the 5 triangles you show are known as the only "perfect triangles".

Definition: Perfect Triangles are triangles with integer sides,

integer area, and numerically equal area and perimeter.

It is easy to show that ALL triangles with numerically equal perimeter

and area have an incircle radius = 2.

Perhaps this can "ease" the search for non-integer triangles?

Definition: Perfect Triangles are triangles with integer sides,

integer area, and numerically equal area and perimeter.

It is easy to show that ALL triangles with numerically equal perimeter

and area have an incircle radius = 2.

Perhaps this can "ease" the search for non-integer triangles?

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