Ponder This Challenge - September 2024 - Sibling triangles

This riddle was suggested by Latchezar Christov - thanks Latchezar!

We will call two triangles "sibling pairs" if they are two non-identical triangles with integer side lengths respectively $(a,b,c_1)$ and $(a,b,c_2)$, where $c_1 and c_2$ are the only side of each triangle with a different length from the other triangle, since there are two common side lengths $a,b$ **and** both triangles must have a common nonzero area.

An example is the pair of triangles with side lengths $(5,5,6)$ and $(5,5,8)$. The area of both is 12. Another example is $(7,4,7)$ and $(7,4,9)$ with area $\sqrt{180}$.

For some $a$ and $b$, there may not be any sibling pair at all (e.g., $a=3$ and $b=2$) and for some, there may be more than one. For example, if $a=24$ and $b=23$, three sibling pairs exist with the following side lengths:

• $(24,23,19)$ and $(24,23,43)$, area = $\sqrt{41580}$
• $(24,23,23)$ and $(24,23,41)$, area = $\sqrt{55440}$
• $(24,23,29)$ and $(24,23,37)$, area = $\sqrt{71820}$

Your goal: Find two integers $a,b$ such that $a>b$ and there are **exactly** 50 sibling pairs with common side lengths $a$ and $b$.

**A bonus** "*" will be given for solving the above problem with the additional constraint that at least two of those 50 sibling pairs must have **integer** areas. The areas of the remaining pairs may be irrational.

Solution

• Click here to view the solution

  September 2024 Solution:

  A solution to the main problem is $a,b=65909,57052$

  A solution to the bonus is $a,b = 11102305, 3338835$

  The area of a triangle with sides a,b,c is given by Heron's formula: denote $s=\frac{a+b+c}{2}$, and the area $A$ is $A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\frac{1}{4}\sqrt{4a^{2}b^{2}-\left(a^{2}+b^{2}-c^{2}\right)^{2}}$. Thus, if we want to find $c_{1},c_{2}$ such that the area of the triangles $\left(a,b,c_{1}\right)$ and $\left(a,b,c_{2}\right)$ is equal, it sufficies to satisfy the equation $\left(a^{2}+b^{2}-c_{1}^{2}\right)=\pm\left(a^{2}+b^{2}-c_{2}^{2}\right)$. Assuming $c_{1}\ne c_{2}$ this simplifies to $2\left(a^{2}+b^{2}\right)=c_{1}^{2}+c_{2}^{2}$, which is related to a classical problem in number theory - the representation of an integer N as the sum of two squares.

  Given N, finding such representation can be hard unless one has the factorization of N, and then one case use a recursive algorithm for finding all such representations thanks to the Brahmagupta–Fibonacci identity

  $\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=\left(ac-bd\right)^{2}+\left(ad+bc\right)^{2}$

  Which shows how to obtain a representation as sums of squares of a number $N=N_{1}N_{2}$ from such representations of $N_{1},N_{2}$.

  We are looking for solutions to to the equation $2N=c_{1}^{2}+c_{2}^{2}$ where $N=a^{2}+b^{2}$. However, not every $\left(c_{1},c_{2}\right)$ which satisfy the equation lead to a triangle, since the area might turn out to be zero or negative: we need to have $a-b$, assuming $a>b$. We proceed as follows, given some $N=p_{1}^{r_{1}}\cdots p_{k}^{r_{k}}$:

  1. Compute the list of all $\left(a,b\right)$ such that $N=a^{2}+b^{2}$
  2. Compute the list of all $\left(c_{1},c_{2}\right)$ such that $2N=c_{1}^{2}+c_{2}^{2}$
  3. Check whether there are $\left(a,b\right)$ such that exactly 50 pairs $\left(c_{1},c_{2}\right)$ satisfy $a-b<_{1}<a+b$(for the bonus: also check whether we have at least two integer areas).

  To solve the problem, one iterates on various values of $N=p_{1}^{r_{1}}\cdots p_{k}^{r_{k}}$. It is best to look at the first few primes congruent to 1 modulo 4, since product of such primes is guaranteed to be representable as sum of squares. The relatively small solution $\left(a,b\right)=\left(433316,384137\right)$ can be found using $N=5^{5}\cdot13^{5}\cdot17^{2}$ (where 5,13,17 are the first primes congruent to 1 modulo 4). The even smaller solution $\left(65909,57052\right)$ is obtained by considering two more primes: $N=5\cdot13^{2}\cdot17^{2}\cdot29^{2}\cdot37$.

  Finding solution to the basic problem is extremely quick given efficient implementations for steps 1-3, but finding a solution to the bonus can still be difficult.

Solvers

• Lazar Ilic (29/8/2024 3:42 PM IDT)
• *King Pig (30/8/2024 2:17 AM IDT)
• Daniel Chong Jyh Tar (30/8/2024 2:26 AM IDT)
• *Bertram Felgenhauer (30/8/2024 3:44 AM IDT)
• Alper Halbutogullari (30/8/2024 5:03 AM IDT)
• Lorenz Reichel (30/8/2024 1:27 PM IDT)
• Alex Fleischer (30/8/2024 3:34 PM IDT)
• *Sanandan Swaminathan (30/8/2024 8:32 PM IDT)
• John Tromp (30/8/2024 10:07 PM IDT)
• *Bert Dobbelaere (31/8/2024 12:55 AM IDT)
• Julian Ma (31/8/2024 4:03 AM IDT)
• Dan Dima (31/8/2024 9:17 AM IDT)
• Loukas Sidiropoulos (31/8/2024 11:21 AM IDT)
• Yi Jiang (31/8/2024 1:36 PM IDT)
• Viplov Jain (31/8/2024 4:27 PM IDT)
• Paul Lupascu (31/8/2024 5:39 PM IDT)
• John Spurgeon (31/8/2024 7:21 PM IDT)
• *Victor Chang (31/8/2024 8:09 PM IDT)
• Kipp Johnson (1/9/2024 3:02 PM IDT)
• Harald Bögeholz (1/9/2024 3:16 PM IDT)
• Gary M. Gerken (1/9/2024 5:15 PM IDT)
• *Yan-Wu He (1/9/2024 6:56 PM IDT)
• *Hugo Pfoertner (2/9/2024 12:35 AM IDT)
• Marco Bellocchi (2/9/2024 1:41 AM IDT)
• Shirish Chinchalkar (2/9/2024 2:32 AM IDT)
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• Richard Gosiorovsky (2/9/2024 9:14 AM IDT)
• Andreas Knüpfer (2/9/2024 9:28 AM IDT)
• Jiri Spitalsky (2/9/2024 11:34 AM IDT)
• *Guangxi Liu (2/9/2024 1:10 PM IDT)
• Reiner Martin (2/9/2024 2:37 PM IDT)
• Dan Ismailescu (2/9/2024 6:54 PM IDT)
• *Daniel Bitin (2/9/2024 11:49 PM IDT)
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• Sri Mallikarjun J (3/9/2024 4:20 PM IDT)
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• Adrian Neacsu (5/9/2024 1:07 AM IDT)
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• Jesse Rearick (6/9/2024 9:22 AM IDT)
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• *Vladimir Volevich (7/9/2024 12:24 AM IDT)
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• *Mark Beyleveld (7/9/2024 4:40 PM IDT)
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• Reda Kebbaj (8/9/2024 4:33 AM IDT)
• *Radu-Alexandru Todor (8/9/2024 2:06 PM IDT)
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• Ashfaque Shaikh (9/9/2024 5:54 PM IDT)
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• *Martin Thorne (9/9/2024 9:49 PM IDT)
• Gyozo Nagy (9/9/2024 10:08 PM IDT)
• Jared Kittleson (9/9/2024 10:16 PM IDT)
• *Karl-Heinz Hofmann (10/9/2024 11:09 AM IDT)
• Mathias Schenker (10/9/2024 1:53 PM IDT)
• Joaquim Carrapa (11/9/2024 1:30 AM IDT)
• Karl D'Souza (11/9/2024 2:40 AM IDT)
• *Dieter Beckerle (13/9/2024 11:18 AM IDT)
• Nadir S. (13/9/2024 12:09 PM IDT)
• Rudy Cortembert (14/9/2024 12:48 AM IDT)
• Fakih Karademir (14/9/2024 2:15 AM IDT)
• *Stephen Ebert and Matthew Lee (14/9/2024 8:28 AM IDT)
• Kang Jin Cho (14/9/2024 11:41 AM IDT)
• *Vincent Picaud (15/9/2024 3:21 PM IDT)
• *Sanjay Rao (16/9/2024 1:06 PM IDT)
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• Hakan Summakoğlu (21/9/2024 3:21 PM IDT)
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• *David F.H. Dunkley (23/9/2024 1:22 AM IDT)
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• *David Greer (23/9/2024 11:43 PM IDT)
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• *Andreas Stiller (26/9/2024 1:42 AM IDT)
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• Pierre Parquier (3/11/2024 10:33 AM IDT)