What is the smallest number that can be used in all three positions (as the hypotenuse, as the longer leg, and

Question

and as the shorter leg) in three different right triangles?

Answer 1

Integer solutions only, presumably

Are you limiting us to primitive Pythagoran triples only?

If no, then 15:
as hypoteneuse: 9-12-15
as longer leg: 8-15-17
as shorter leg: 15-112-113 or 15-36-39

If yes, then 221:
as hypoteneuse: 21-220-221 or 140-171-221
as longer leg: 60-221-229
as shorter leg: 221-24420-24421

For all three triples to be primitive (i.e. the three sides have no common factor) the numbers have to be somewhat bigger and it will need systematic searching through lists of triples with the following methodology:

(a) Find numbers that appear as as a hypoteneuse, (relatively rare) (i.e. look in column 3)

(b) Find a triple in which such a number appears as a longer leg (i.e. look lower down in column 2)

(c) Generate a triple in which such a number appears as a shorter leg

if it is odd, a, (a^2-1)/2, (a^2+1)/2
if it is even, a, (a^2-4)/4, (a^2+4)/4

From a list of such low triples (see link) I identified 221 as the lowest number fitting the spec: the list that follows is included as indicative content.

LIST OF PRIMITIVE TRIPLES WITH HYPOTENEUSES UNDER 250

3 4 5
5 12 13
8 15 17
7 24 25
20 21 29
12 35 37
9 40 41
28 45 53
11 60 61
16 63 65
33 56 65
48 55 73
13 84 85
36 77 85
39 80 89
65 72 97

20-99-101
60-91-109
15-112-113
44-117-125
88-105-137
17-144-145
24-143-145
51-140-149
85-132-157
119-120-169
52-165-173
19-180-181
57-176-185
104-153-185
95-168-193
28-195-197

84-187-205
133-156-205
21-220-221
140-171-221
60-221-229
105-208-233
120-209-241

Mathematics and mathematicians love to generalise. No sooner have we got a result than we start to wonder whether it is part of a broader pattern or not.

So are there more such "multi-positional numbers" like 221 (is there a better term than that?).

I have now found 8 more such and thus 9 in all

(1) 325 is such a number:

as hypoteneuse: 36-323-325 or 204-253-325
as longer leg: 228-325-397
as shorter leg: 325-52812-52813

(2) 425 is such a number:

as hypoteneuse: 87-416-425 or 297-304-425
as longer leg: 168-425-457
as shorter leg: 425-90312-90313

(3) 493 is such a number:

as hypoteneuse: 132-475-493 or 155-468-493
as longer leg: 276 493 565
as shorter leg: 493-121524-121525

(4) 725 is such a number:

as hypoteneuse: 333-644-725 or 364-627-725
as longer leg: 108-725-733
as shorter leg: 725-262812-262813

(5) 1025 is such a number:

as hypoteneuse: 64-1023-1025 or 496-897-1025
as longer leg: 528-1025-1153
as shorter leg: 1025-525312-525313

(6) 1189 is such a number:

as hypoteneuse: 611-1020-1189 or 660-989-1189
as longer leg: 420-1189-1261
as shorter leg: 1189-706860-706861

(7) 1517 is such a number:

as hypoteneuse: 165-1508-1517 or 795-1292-1517
as longer leg: 156-1517-1525
as shorter leg: 1517-1150644-1150645

(8) 1537 is such a number:

as hypoteneuse: 385-1488-1537 or 312-1505-1537
as longer leg: 984-1537-1825
as shorter leg: 1537-1181184-1181185

I may well have missed some, combing through lists of triples manually. This sounds like a job for specialist maths software like Mathematica.

OBSERVATIONS

1) All nine have a choice of two triangles where the number acts as a hypoteneuse. Is this a necessary prerequisite for them to also have this property as well?

2) None of the numbers is divisible by 3 or 4 even though in every triple one of the numbers is always divisible by 3, one by 4 and one by 5.

Answer 2

Tuncay U's answer is not correct.
I can find 2 triples with 13: 5,12,13
and 13,84,85. I can prove there no 3rd one.
Proof: Suppose a² + 13² = c².
Then (c+a)(c-a) = 169.
So c+a and c-a must be divisors of 169.
But the only divisors of 169 are 1,13 and 169.
So either
c+a = 13
c -a = 13
which yields c = 13
or c + a = 169
c -a = 1
which yields c = 85.
So there is no third triple.
The same type of argument works for p = any
prime number which equals 1(mod 4).
For p = 3(mod 4), p can never be the
largest member of a Pythagorean triple.
The correct answer is 15.
The 3 triples are
15,112,113 (or 15,20,25 or 15,36,39)
8,15,17
and
9 12 15.
Bruce Birchall's answer of 221 for primitive
triples is indeed correct.
Let's find the possible values of c < 221.
Then we will attempt to find primitive triples
with a value of c equal to one of the legs.
First, recall that for primitive triples a,b,c,
a = u²-v², b = 2uv, c = u²+v².
So we can rule out:
a). all even values of c because that would mean
a was also even and a,b,c would not be a
primitive triple.
b) all primes p with p = 3(mod 4). By Fermat's
2 squares theorem any prime factor p of c such
that p = 3(mod 4) must occur in c to an even power.
c) all primes p with p = 1(mod 4), by the argument above.
d) the square of any prime p = 3(mod 4). The
only way to represent p² as the sum of 2 squares
is p²+0², so c cannot equal p².
Thus the only possible values of c less than 221
are
c = 25,65,85,145,169,185 and 205.
For 25 and 169 we find only the following primitive triples,
using a divisor argument similar to the one above.
7,24,25
25,312,313
119,120,169
169,14280,14281.
Here are the only other primitive triples found:
16,63,65
33,56,65
65,2112,2113
65,72,97
13,84,85
36,77,85
85,132,157
85,3612,3613
17,144,145
24,143,145
145,408,433
145, 10512,10513
57,176,185
104,153,185
185,672,697
185,17112,17113
84,187,205
133,156,205
205,21012,21013.
We see that none of these triples has an
allowable value of c as the longer leg,
so 221 is indeed the smallest possible
value of c for primitive triples.

Answer 3

If the fast leg has a length of x feet, the lengthy leg is x + 3 and the hypotenuse is x + 6 (x + 6)^2 = x^2 + (x + 3)^2 = x^2 + x^2 + 6x + 9 x^2 + 12x + 36 = 2x^2 + 6x + 9 6x = x^2 - 27 x^2 - 6x - 27 = 0 x^2 - 9x + 3x - 27 = 0 x(x - 9) + 3(x - 9) = 0 (x + 3)(x - 9) = 0 which potential x is comparable to the two -3 or +9 subsequently the lengths of the three facets of the backyard of Alex are 9 feet, 12 feet and 15 feet.

Answer 4

BruceBirchall's latest three numbers to possess this property (325, 425 and 1025) are all prime multiples of a square number (13 x 25, 17 x 25 and 41 x 25)

Is there a conjecture to be devised from that observation?

Answer 5

ZERO!

The value of all three is 0 if your triangle is a point.

Answer 6

Hello There isn't any the number keep getting smaller as you asubject one from the previous one There is no SMallest number for any thing!!!!!!!!!!!!!!!!!!

Answer 7

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