Almost Isosceles

We'll try to make the triangle as close to isosceles as possible by looking for right-angled triangles whose shorter sides differ by $1$ . Clearly the sides of such a triangle will not share any factors, so we're looking for primitive Pythagorean triples $a,b,c$ and we can put $a=x^2-y^2$ and $b=2xy$ (and $c=x^2+y^2$ ). There are two cases to consider; either $a-b=1$ or $b-a=1$ ; but, as we'll see, they lead to the same equation to solve.

Start with $a-b=1$ : $x^2-y^2-2xy=1$

Writing $u=x-y$ and $v=y$ , this is Pell's equation : $u^2-2v^2=1$

Now consider $b-a=1$ : $2xy-x^2+y^2=1$

This time, put $u=x+y$ and $v=x$ . Once again, we get $u^2-2v^2=1$

so this is the only equation we have to solve.

To find solutions to Pell's equation, we need a fundamental solution $(u_1,v_1)$ . It's easy to see that $(3,2)$ works. All further solutions (see the linked article on Pell above) are then found using $u_{n+1} = u_n u_1 + 2v_n v_1 \quad \quad v_{n+1} = u_n v_1 + u_1 v_n$

and these can be used to generate the triangle's sides. Per the two cases outlined above, each pair $(u_n,v_n)$ generates two solution triangles. For example, from $(u_1,v_1)=(3,2)$ we get

Case $a-b=1$ : $x-y=3,y=2$ so that $x=5$ ; $(a,b,c)=(21,20,29)$

Case $b-a=1$ : $x+y=3,x=2$ so that $y=1$ ; $(a,b,c)=(3,4,5)$

We then calculate $(u_2,v_2)=(17,12)$ giving the triples $(697,696,985$ and $(119,120,169)$ , and so on, until we find the triangle with the longest sides possible less than $10^{10}$ .

This is the triple $(5406093003,5406093004,7645370045)$ which has perimeter $\boxed{18457556052}$ .

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At the beginning, we assumed the best approach was to look for triangles whose shorter sides differed by just $1$ . But could there be a better solution whose shorter sides differed by some other integer? Well, in this case, no: say there were a better solution (ie a triangle with angles closer to $45^{\circ}$ with sides differing by $d>1$ . Then, considering the tangent of this angle, we would need $\frac{a'+d}{a'}-1<\frac{a+1}{a}-1$ where $a=5406093003$ from above and $a'$ is the shortest side of the new triangle. Simplifying, $\frac{d}{a'}<\frac{1}{a}$ But now note that since $a'<10^{10}$ and $a>\frac12 \cdot 10^{10}$ , this is impossible.

A right triangle with all integer sides can’t have an angle that is exactly $45°$ , because then its legs $a$ and $b$ would be equal, and by the Pythagorean Theorem its hypotenuse would be $c = \sqrt{2}a$ , a non-integer.

The next best scenario would be if the legs had a difference of $1$ and with $a$ is as large as possible, so that $b = a + 1$ . Then by the Pythagorean Theorem, $a^2 + (a + 1)^2 = c^2$ , which rearranges $(2a + 1)^2 - 2c^2 = -1$ , a negative Pell equation where $c$ can be expressed recursively by the equation $c_n = 6c_{n - 1} - c_{n - 2}$ , with $c_0 = 1$ and $c_1 = 5$ , and whose next few terms can be calculated as $29$ , $169$ , $985$ , $5741$ , $33461$ , $195025$ , $1136689$ , $6625109$ , $38613965$ , $225058681$ , $1311738121$ , and $7645370045$ for $c_n < 10^{10}$ (see sequence A001653 ). Using the largest $c_n$ value that is less than $10^{10}$ , $c = 7645370045$ , and $a$ and $b$ calculate to $a = 5406093003$ and $b = 5406093004$ .

Finally, we need to show that the ratio of legs $\frac{a}{b} = \frac{5406093003}{5406093004}$ is the closest to $1$ for any other integer leg difference $k \geq 2$ . Let’s say there is an integer leg difference $k \geq 2$ such that $\frac{a'}{a' + k} > \frac{5406093003}{5406093004}$ . Then after some rearranging, $a' > k \cdot 5406093003 > 10^{10}$ , which is not allowed.

Therefore, the perimeter of the right triangle with all integer sides less than $10^{10}$ with an angle closest to $45°$ is $a + b + c = \boxed{18457556052}$ .