Find all integer sided triangles whose area and peremeter are equal

Number Theory Level 5

Determine all triangles whose sides are integers, and the areas and perimeters are (numerically) equal. What is the sum of the longest sides of each of the triangles?

Extra Credit for proving that those are the only solutions. (Put your proof in the solutions section)

Explicit Example: If the triangles are (7, 8, 9 ) and ( 13 , 5, 9). Then enter your solution as 9+13 = 22

The answer is 89.

4 solutions

Agnishom Chattopadhyay
Jun 10, 2014

These people can explain the solution better than I can.

Nice Prob. Pic is from PE 91 right? Enjoyed that one..

Thaddeus Abiy - 7 years ago

yep, that's correct.

Agnishom Chattopadhyay - 7 years ago

Gosh dang it I missed one! :O

Finn Hulse - 7 years ago

Yes god damn it i solved this in my first attempt after spending 10 mins on it

Aayush Mani - 7 years ago

I done for 88...just 1 less

Kevin Patel - 6 years, 12 months ago

May I ask how you came up with 88? Yeah, it's one less numerically speaking, but that's a pretty far-off error considering that 88 must have been a sum of lengths from the set of triangles that satisfy the conditions of the problem. What was your source of error here?

13 less, from missing triangle 5-12-13, or 10 less, from missing triangle 6-8-10, might have made more sense (for me at least).

Michael Raphael Panganiban - 6 years, 12 months ago

Notice that this problem is equivalent to find all triangles with integer sides whose inradius is $2$ . It's because: $r=\dfrac{2A}{P}$ , where $r$ is the inradius, $A$ is the area and $P$ is the perimeter. So, when $r=2 \Rightarrow A=P$ .

Alan Enrique Ontiveros Salazar - 6 years, 11 months ago

This is crucial.

Lu Chee Ket - 6 years, 10 months ago

Any new ideas with this approach?

Sal Gard - 4 years, 11 months ago

[b + 3 + 20/ (b - 5), b, 5 + 20/ (b - 5)] for b = 10, 15 and 25; {9, 7 and 6 produce redundant worse order.}

[b + 8/ (b - 4), b, 4 + 8/ (b - 4)] for b = 8 and 12; {6 and 5 produce redundant worse order.}

These are the solutions. b and c can interchange. No need to search further.

Lu Chee Ket - 6 years, 9 months ago

The graphics is a disaster. So any triangle is allowed not just rectangular ones. Crazy!

Mark Macqueen - 7 years ago

What is a rectangular triangle, eh?

Agnishom Chattopadhyay - 7 years ago

Those with a right angle. The graphics are misleading. However, I guessed you meant turned triangles are considered as 1 only, as you did not draw all of them. The question made people who have obtained some answers to think for a while by its ambiguity.

Lu Chee Ket - 6 years, 10 months ago

I agree with you.

Lu Chee Ket - 6 years, 10 months ago

Lu Chee Ket
Aug 15, 2014

Featuring proof for number theory

Area to perimeter is proportional to dimensional length; the larger the size of an object, the greater the ratio of numerical value of area to numerical value of perimeter. Numerical area can hardly equal to numerical perimeter when an object is getting greater and greater.

With s = 0.5 (a + b + c) and for perimeter = area: 2 s = sqrt[s (s - a) (s - b) (s - c)] => 4 s = (s - a) (s - b) (s - c); (10, 8, 6), (13, 12, 5), (17, 10, 9), (20, 15, 7) and (29, 25, 6) are obtainable as the 5 triangles of even numerical value of areas and also perimeters.

4 s = (s - a) (s - b) (s - c) simplified into 4 s = (s - b) (s - c) for ultimate case. When s = 0, s = a = b = c = 0. Area makes perimeter while perimeter may make no area.

4 x 12 = 2 x 4 x 6:
4 s = (12 - 10) (s - b) (s - c); s^2 - 16 s + 48 = 0; (s - 4) (s -12) = 0

4 x 15 = 2 x 3 x 10:
4 s = (15 - 13) (s - b) (s - c); s^2 - 19 s + 60 = 0; (s - 4) (s -15) = 0

Such that 4 s = 2 (s - b) (s - c); s^2 - (b + c + 2) s + b c = 0; b + c = 2 + b c/ 4

4 x 18 = 1 x 8 x 9:
4 s = (18 - 17) (s - b) (s - c); s^2 - 23 s + 90 = 0; (s - 5) (s -18) = 0

4 x 21 = 1 x 6 x 14:
4 s = (21 - 20) (s - b) (s - c); s^2 - 26 s + 105 = 0; (s - 5) (s - 21) = 0

4 x 30 = 1 x 5 x 24:
4 s = (30 - 29) (s - b) (s - c); s^2 - 35 s + 150 = 0; (s - 5) (s - 30) = 0

Proceeding: 4 s = 1 (s - b) (s - c); s^2 - (b + c + 4) s + b c = 0; b + c = 1 + b c/ 5

At ultimate stage,

b + c = 1 + b c/ 5 where c = (5 b - 5)/ (b - 5) or else b = (5 c - 5)/ (c - 5).

b + c = 1 + b c/ 5 is having solutions of (17, 10, 9), (20, 15, 7) and (29, 25, 6) only, whereby

b + c = 2 + b c/ 4 is having solutions of (10, 8, 6) and (13, 12, 5) only. c = (4 b - 8)/ (b - 4) or else b = (4 c - 8)/ (c - 4) similarly.

Based on c = (5 b - 5)/ (b - 5) for ultimate case, c = 6 or 7 or 9 only. There can be other (b, c)'s which can satisfy this equation but they cannot be integers-only. Consequently, there is no more integer solution after (a, b, c) = (29, 25, 6) with 'a' taking a lead. For a = 30 ahead, (s - a) (s - b) (s - c) is greater than 4 s more occasionally while no equal can occur; when this happens, it is like a growing of sizes and hence area is tending to always greater than perimeter in their respective numerical value. Area/ Perimeter is proportional to a length.

For circle of radius = 2 units rolling on a 30 units line segment with (a, b, c) of longest side as base,

a, b, c, A, P, h, s, (s - a),

C, A, B:

30; 30; 4.297541; 64.29754; 64.29754; 4.286503; 32.14877; 2.14877;

8.214732; 85.89263; 85.89263;

30; 29.68858; 4.311422; 64; 64; 4.266667; 32; 2;

8.262822; 90; 81.73718;

30; 26.04988; 5.950124; 62; 62; 4.133333; 31; 1;

9.129706; 126.8699; 44.0004;

30; 15.54299; 15.54299; 61.08597; 61.08597; 4.072398; 30.54299; 0.542986;

15.18929; 149.6214; 15.18929;

Highest figures at side edges and lowest figures at the middle, integer (s - a) i.e. k of 2 and 1 make perimeters and areas integers. (s - a) = 1 is believed to have the tendency for even more integers and tending to side edges, 2 extra units to perimeter; (s - a) = 2 makes 4 extra units to perimeter. k = 1 and k = 2 are meant for all integers at radius = 2. k >= 3 shall be found not for radius = 2.

In fact, 30 <= a <= 3000 had been found invalid with such triangle using computer. These are written in an attempt to deny validity at extreme such as infinity, actually. Meet with valid cases is main reason of found integers, however, there is a need to be certain of invalid case at extreme, not unable to draw but no integers for all.

4 s = (s - a) (s - b) (s - c)

=> 4 s = k (s - b) (s - c)
{(s - a) is replaced with k of 1, 2, 3, 4 or 5.}

=> s^2 - (b + c + 4/ k) s + b c = 0
{s: One common and one 0.5 of perimeter.}

k = 1 with s = 5: c = (5 b - 5)/ (b - 5)

k = 2 with s = 4: c = (4 b - 8)/ (b - 4)

k = 3 with s = 3: c = (3 b - 5)/ (b - 3)

k = 4 with s = 2: c = (2 b - 2)/ (b - 2)

k = 5 with s = 1: c = (5 b - 1)/ (5 b - 5)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

k = 1 with s = 5: (17, 10, 9), (20, 15, 7) and (29, 25, 6);

k = 2 with s = 4: (10, 8, 6) and (13, 12, 5);

k = 3 with s = 3: (4, 1, 1), (- 4/ 3, 1, 1), (23/ 3, 7, 4), (-5, 7, 4), (20/ 3, 5, 5) and (- 4, 5, 5);

k = 4 with s = 2: (5, 4, 3) and (-3, 4, 3);

k = 5 with s = 1: NIL

Completing the task by c = f (b) is simple compared to further computer search. Nevertheless, initial results found by computer were first taken, only then we proceed to confirm for all other possibilities. (4, 1, 1) and (5, 4, 3) cannot be answers. Not valid with other answer.

Lu Chee Ket - 6 years, 10 months ago

Completed proof:

(1, 1, 1) for s = 1.5 is a minimum but not taken as common minimum s.

For k = 1 only, such that s - a = 1, it happens that radius = 2 allows only possible formation tending to corner than middle. Area enclosed with maximum base length ought to time with only little height.

s^2 - (b + c + 4) s + b c = 0

c = (s b + 4 s - s^2)/ (b - s) and a = Perimeter - (b + c);

With s = 7, c = (7 b -21)/ (b - 7): (19, 14, 11), (24, 21, 9), (37, 35, 8); {Not equal}

With s = 6, c = (6 b - 12)/ (b - 6): (18, 12, 10), (19, 14, 9), (22, 18, 8), (33, 30, 7); {Not equal}

With s = 5, c = (5 b - 5)/ (b - 5): (17, 10, 9), (20, 15, 7), (29, 25, 6); {Wanted}

With s = 4, c = (4 b)/ (b - 4): (16, 8, 8), (14, 12, 6), (25, 20, 5); {Zero area or not equal}

With s = 3, c = (3 b + 3)/ (b - 3): (15, 7, 6), (16, 9, 5), (21, 15, 4); {Not formed}

With s = 2, c = (2 b + 4)/ (b - 2): (14, 6, 4), (17, 10, 3); {Not formed}

With s = 1, c = (b + 3)/ (b - 1): (12, 3, 3), (13, 5, 2); {Not formed}

With s = 0, c = 0: NIL; {Invalid}

Hence, k = s - a = 1 with s = 5 substituted to s^2 - (b + c + 4/ k) s + b c = 0 has resembled whole search for extreme base of a >>= 30, with a < (b + c) <= (a + 2) hopefully, but found invalid. k >= 2 for a >>= 30 may only form integer triangles with radius > 2 units, as the base is becoming huge for not a small area. Therefore, the 5 answers are the only answers. Sum = 89.

Lu Chee Ket - 6 years, 10 months ago

KEY: s^2 - (b + c + 4/ k) s + b c = 0 with k = 1.

Lu Chee Ket - 6 years, 10 months ago

All answers for (a, b, c) from s^2 - (b + c + 4/ k) s + b c = 0

a = (b + c) - 4 = (1/ 2) b c - (b + c) from k = 2:

(10, 8, 6) and (13, 12, 5);

a = (b + c) - 2 = (2/ 5) b c - (b + c) from k = 1:

(17, 10, 9), (20, 15, 7) and (29, 25, 6); potential to have further possibility is here but invalid.

Nothing to tell whether k = 2 or k = 1 can have more answers. The fact is k = 1 should be ultimate but found only 3 answers.

Lu Chee Ket - 6 years, 10 months ago

c^3 - 60 c^2 + 16 c + 960 = 0

z^3 - 1184 z - 14720 = 0

c = z + 20

(30, 30, c); {Isosceles}

(30, 17 + sqrt(161), 17 - sqrt(161)); {Right angled (one more)}

(30, 16 + sqrt(101), 16 - sqrt(101)); {Obtuse (s - a) = 1}

(30, 15 + 120/ 221, 15 + 120/ 221); {Middle}

Lu Chee Ket - 6 years, 9 months ago

(30, 17 + sqrt(161), 17 - sqrt(161)) is obtained via z^3 - 38 z^2 - 884 z + 34680 = 0 with z = (b + c) in a way with no presumed initial. The cubic equation is not a genuine cubic equation as it is a case in between some critical change. Finally, it is found to be (s - a) = k = 2. With simple presumed sides of (30, b, 34 - b), it is obtainable easily.

(30, 16 + 2 sqrt(33), 16 - 2 sqrt(33)) is hence found for (s - a) = k = 1 as (30, b, 32 - b). For s = 3, (30, 18 + 8 sqrt(3), 18 - 8 sqrt(3)) seem to hold but actually not. For s = 4, (30, 34, 4) is also a failure. For, s >= 5, all are not answer.

A missing one is however not answer of all integers. Nevertheless, it can be seen that only (s - a) = k <= 2 can have valid answers to right order.

Lu Chee Ket - 6 years, 9 months ago

The merit finally is we search for (a, b, a - b + 2) and (a, b, a - b + 4) only particularly for a > 3000 which had not been considered for example, and relations of the 5 found are clearer.

{[(b - k)^2 + 4]/ [(b - k) - 4/ k], b, [(b - k)^2 + 4]/ [(b - k) - 4/ k] - b + 2 k} is to become 1 dimensional search with varying b when only k = 1 and k = 2 are to be considered.

I utilized only (1 + 1/2) minutes to finish b = 1 to 10,000,000 search for k = 1, 2, 3, 4 and 5 via 32-bits LONGINT using Turbo Pascal and confirmed to have no other than the 5 answers. MOD and DIV are applied with b, p, p1, p2, q, q1, q2, r, r1, r2, s, s1, s2, t, t1 and t2 as variables, (a, b, a - b + 2 k) actually. Only k = 1 and k = 2 arise with answers to preserve the first as longest side without added routines.

Lu Chee Ket - 6 years, 9 months ago

[(b - k)^2 + 4] MOD [(b - k) - 4/ k] = 0 becomes the core or the key for reason meeting each other. Reducing the rank simplify the search with a risk of fault introduced. Therefore, we study these carefully so that nothing is omitted to the simplification. Once we finish to an extreme to resemble infinity, divergence shall tell the certainty for only 5 answers. After b = 1 to 25 with k = 1 and k = 2, display using Excel appeared to be the limit at b = 25.

Lu Chee Ket - 6 years, 9 months ago

[c + 3 + 20/ (c - 5), 5 + 20/ (c - 5), c] for c = 9, 7 and 6; {10, 15 and 25 produce redundant worse order.}

[c + 8/ (c - 4), 4 + 8/ (c - 4), c] for c = 6 and 5; {8 and 12 produce redundant worse order.}

b and c can interchange. No need to search further.

Lu Chee Ket - 6 years, 9 months ago

Márcio Nascimento
Jun 26, 2014

I runned a little program to discover the candidats! 100 was just a fun limit. If you take off the "if", you will see a clearly divergence after 30, between areas and perimeters!

main()
{

    int a, b, c;
    double p;

    for (a = 1 ; a < 100 ; a++)
        for (b = a ; b > (a - b) ; b--)
            for (c = b ; c > (a - b) ; c--)
            {
                p = (a+b+c)/2.0;
                if (p+p==sqrt(p*(p-a)*(p-b)*(p-c)))
                     printf ("%d %d %d\n", a, b, c);
            }
}

Just 4 p == (p-a) (p-b) (p-c) could be better.

Lu Chee Ket - 6 years, 10 months ago

I just tried to be "didactic"! Ty.

Márcio Nascimento - 6 years, 9 months ago

All right.

Lu Chee Ket - 6 years, 9 months ago

[b + 3 + 20/ (b - 5), b, 5 + 20/ (b - 5)] for b = 10, 15 and 25; {9, 7 and 6 produce redundant worse order.}

[b + 8/ (b - 4), b, 4 + 8/ (b - 4)] for b = 8 and 12; {6 and 5 produce redundant worse order.}

b and c can interchange. No need to search further.

Lu Chee Ket - 6 years, 9 months ago

I doidnt understand!!! Explain more!!!

Márcio Nascimento - 6 years, 9 months ago

I appreciate the for-loops which are very good. The thing is they are raw without searching for simplification for final. Rank of 3 dimensional search cannot go far to complete for a proof. After investigation, we can find that only certain groups are critical for answers required. Your search can be resolved into obvious search which can stop quite soon without further doubt. Example 8/ (b - 4), not many integers can satisfy and hence no other possibility. The answer must be completed or otherwise this question should not be answered.

Lu Chee Ket - 6 years, 9 months ago

Ajit Athle
Jun 12, 2014

I believe (5, 12, 13), (6, 8, 10), (6, 25, 29), (7, 15, 20) and (9, 10, 17) are the only integral solutions that can be obtained by using Heron's formula for area equating it to the perimeter.

[c + 3 + 20/ (c - 5), 5 + 20/ (c - 5), c] for c = 9, 7 and 6; {10, 15 and 25 produce redundant worse order.}

[c + 8/ (c - 4), 4 + 8/ (c - 4), c] for c = 6 and 5; {8 and 12 produce redundant worse order.}

b and c can interchange. No need to search further. Not believe only but we can be certain.

Lu Chee Ket - 6 years, 9 months ago

Wait, any triangle is allowed? I thought it was just the ones in the picture...

Thomas Darkmoon - 7 years ago

None of those triangles could ever satisfy the integer condition.

Harsh Sheth - 7 years ago

Find all integer sided triangles whose area and peremeter are equal

The answer is 89.

4 solutions

0 pending reports