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A rectangular field has width $a$ and length $a+1$. We cut it into 3 triangles that all have integer side lengths. If all triangles have a different area, then what’s the minimum value of $a$? Please don’t use computers.


3 Answers 3


The minimum value for $a$ is:



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First, let the rectangle be ABDC where A is the top left and C the bottom left vertices. Let E be the point on AC defining the three triangles. Let F be the projection of E onto BD. Without loss of generality, assume BF > FD.

We note that $a$ = EF must be larger than both BF and FD since BF + FD = $a+1$ and FD cannot be 1 (since there is no Pythagorean triple with 1). So $a$ must be the middle term (not the smallest nor the largest) in the two triangles BFE and FDE.

Now we list all Pythagorean triples where the middle term is less than or equal to 24, which are:
3, 4, 5
6, 8, 10
9, 12, 15
12 16, 20
15, 20, 25
18, 24, 30
5, 12, 13
10, 24, 26
7, 24, 25
8, 15, 17
20, 21, 29

Let each triple be $(b, a, c)$, then we need to find two triples $(b_1, a, d_1)$ and $(b_2, a, d_2)$ where $b_1 + b_2 = a+1$.

First, we note that the triples must be different since if the two triples are the same, it will make the area of ABE and EDC the same, violating the constraint of all triangles having different areas. Now, these two different triples need to have the same $a$, so we have either $a=12$ or $a=24$. For $a=12$, there is only one possibility: $(9, 12, 15)$ and $(5, 12, 13)$, however $5+9 \neq 12+1$, so $a=12$ does not work.

The next smallest $a$ is then $a=24$, which is shown in my answer above with $(18, 24, 30)$ and $(7, 24, 25)$ since $18+7 = 25 = 24+1$.

  • 1
    $\begingroup$ +1! Did you enumerate the Pythagorean triples by hand to strictly respect the no computer tag? ;) $\endgroup$
    – JKHA
    Jun 26, 2020 at 5:09
  • 3
    $\begingroup$ Yes, of course, hehe. And apparently I missed one, according to Wikipedia. I just added it. I used triples in the form (2k+1, 2k^2+2k, 2k^2+2k+1) as I believed all small triples are of this form. So I missed (20, 21, 29). $\endgroup$
    – justhalf
    Jun 26, 2020 at 6:02
  • $\begingroup$ @justhalf: that generator expression might best be restated as (2k+1, 2k(k+1), 2k(k+1)+1). $\endgroup$
    – smci
    Jun 26, 2020 at 19:17
  • 1
    $\begingroup$ 1. C & D in your proof are swapped relative to your diagram. 2. Your enumeration of triads omits (8, 15, 17). $\endgroup$
    – PM 2Ring
    Jul 28, 2020 at 17:19
  • 1
    $\begingroup$ Thanks PM 2Ring for catching the errors! I have updated the Pythagorean triples, but I cannot update the image now, so I updated the text instead. $\endgroup$
    – justhalf
    Jul 29, 2020 at 1:40

The general formulas for creating right angle triangles with integer sides are A=m^2+n^2, B=2mn, c=m^2-n^2 if m=n+1 then two sides of the triangle are consecutive. In your problem the smallest values for m,n are m=4 and n=3. So a+1=25 and a=24.

First triangle has sides 7,24,25

Second triangle has sides 18,24,30

Third triangle has sides 25,25,30.

  • 1
    $\begingroup$ Vassilis: that's only the formula for a primitive PT. But it's not established here that all the PTs are primitive, hence you have to use the more general $k(2mn, m^2-n^2, m^2+n^2)$. The multiplier $k$ introduces more complexity. But we're thinking along the same lines, see my answer. $\endgroup$
    – smci
    Jun 26, 2020 at 21:33
  • $\begingroup$ When two sides of a Pythagorian triangle differ by 1, the triangle is primitive. The question states A=a+1 and B=a. $\endgroup$ Jun 26, 2020 at 22:24
  • $\begingroup$ No it doesn't. It states a and a+1 are sides of different PTs; a is itself a side of two distinct PTs and a+1 of a third. Not one PT. You can even see that from the solution (spoiler), one PT has the multiplier k=6, the other has k=5, the bottom one is primitive (k=1). They are not the same PT. They are three different non-primitive PTs. $\endgroup$
    – smci
    Jun 27, 2020 at 2:42

I thought of approaching this mainly in a number-theoretic way, consider divisibility, to heavily constrain the heuristic search so it can be done by hand:

  • $(a+1)$ is a member of >=1 PT
  • $a$ is a member of >=2 distinct PTs
    • moreover, considering the left-hand side of the square:
    • call the LHS side $b$, $b$ is in one of those PTs involving $a$, WLOG call this the 'bottom' PT. Clearly $b < a$, since $b < (a+1)/2 < a$
      • so $(a, b)$ are the legs of a very small PT, with $b <= a/2$, i.e. $a >= 2b$
    • then ($a+1) - b$ is in the other PT involving $a$.
      • so $(a, (a+1-b))$ are the legs of another (distinct) small PT, with $(a+1-b) > a/2$
    • we want to find minimal $a$ for which this works. That implies $b$ is (not necessarily minimal but) very small and a leg of a PT (possibly the prime leg), so $b>=3, a>=6$ and we only need to check candidates for $b = 3,4,5,6,7...$
  • exactly one of $a, a+1$ must be even; which one? (intuition suggests it's far more likely to be $a$ (since it's a member of >=2 PTs, hence likely highly composite), but can't see how to prove that without piecewise considering both cases (a-even/odd) and showing that a-even gives minimal solution)
    • (Hunch: if we could prove $(a+1)$ is odd (but unlikely to be prime). Hence $a, (a+1)$ would likely be a tuple of composite numbers, with $a$ even and highly composite, and $(a+1)$ a composite product of odd primes. Hence $(a+1)$ would have to be divisible by at least one of 3,5,7..., to give minimal $a$. Hence we consider the case $3 | (a+1)$, then try 5 and show $(a+1) = 5^2$ works and thus achieves a minimum)
    • this is essentially piecewise-considering divisibility-by-2 and -by-3 of $a, (a+1)$
    • Note also: if the leg of a PT is prime, that implies it was a primitive PT (k=1), and moreover we must have k(m^2-n^2) = 1(m-n)(m+n) = 1(1)(m+n) = (some odd prime p):
      • This adds the constraint m=n+1 for the primitive PT.
      • Further, when a prime p is the prime leg of a primitive PT, then coprimeness says the other sides can't be divisible by p.
      • So, we could prune our search down to primitive PTs where the prime leg is p=3, 5, 7..., thus {(3,4,5), (5,12,13), (7,24,25), (11,60,61)...}
  • only at most one of $a, a+1$ might be divisible-by-3 and once we determine that's a=24, then (a+1) has to be an odd composite not divisible by 3, hence must be divisible by one of 5,7,... and for minimality it's probably divisible by 5.
  • consider also that a is a member of two distinct PTs, one of them primitive (the 'bottom' one containing (a,b)), then the 'top' PT containing (a,a+1-b) is likely non-primitive. It seems plausible that to minimize a, a has to be even (or else we just brute-force and consider even a first). So, for candidates for a we want to consider small even numbers which are members of >=2 PTs, one of which is definitely primitive, i.e. for the definitely-primitive PTs we only need to consider {(3,4,5), (5,12,13), (8,15,17), (7,24,25), (9,40,41), (11,60,61), ... } and thus we only need to consider a = 12,24,40,60.... (Oh and that suggests it might be easy to show a is divisible by 3)
  • haven't even used the all-triangles-different-area constraint (/ Heron's Law). I don't think we even need that since many of the sidelengths here are coprime, which will guarantee the areas distinct.
  • anyway these constraints set up a heavily bounded search over the PTs generated by: $k((m-n)(m+n), 2mn, m^2 + n^2)$

So this sets up a heuristic search which finds (a=24, (a+1)=5^2=25, b=7) and the other side $(a+1)-b = 25-7 = 18$ Since we knew a >= 2b

We only needed to consider the candidates a = 12,24,40,60,... and b= prime legs p = 3,5,7,11... where p doesn't also divide a, and also a >= 2b.

You could turn that into one or two simple Python generators.

(If anyone can help me firm up the intuitive parts of the above, please leave constructive comment. Even if you ignore all the non-essential intuitive bits, this still heavily constrains the heuristic search, i.e. satifisies no-computers and v small number of candidates to check)


Hunch, unproven: Ultimately, minimizing a or (a+1) comes down to setting up that one of these is an odd square, or odd composite. Say we prove piecewise or by divisibility-by-3 arguments that 3^2 and 3*7 aren't possible, then 5^2 must be the next smallest choice. Thus I suspect (a+1) = 5^2 has the property that it's the smallest such number which can be partitioned into 7 (a small odd prime leg of a primitive PT also containing a=24), and 18 (a (highly composite) number that occurs in a different non-primitive PT that contains a=24).


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