shobber wrote: Cyclic quadrilateral $ABCD$ has positive integer side lengths $AB$, $BC$, $CA$, $AD$. It is known that $AD=2005$, $\angle{ABC}=\angle{ADC} = 90^{\circ}$, and $\max \{ AB,BC,CD \} < 2005$. Determine the maximum and minimum possible values for the perimeter of $ABCD$. Proof. Let $\overline{AB}=a$, $\overline{BC}=b$, $\overline{CD}=c$ and $\overline{DA}=d=2005$. Since $\angle{ABC}=\angle{ADC} = 90^{\circ}$, we have by the Pythagorean theorem $a^2+b^2-c^2=2005^2$ $(*)$. Hence if we find a solution $(a$, $b$, $c)$ of $(*)$ satisfying $\max \{a$, $b$, $c\} < 2005$, then $a$, $b$, $c$, $d$ are the sidelengths of a cyclic quadrilateral. Let us first determine the minimum value of the perimeter of $ABCD$, i.e. the minimum value of $a+b+c$, since $d=2005$ is constant. Let wlog $a \ge b$ and thus $a\ge 1000$. The triple $(2004$, $45$, $4)$ satisfies $(*)$ and gives $a+b+c=2053$. I'll now prove that $2053$ is indeed the minimum. We evidently have $(b+c)^2>(b+c)(b-c) = (2005-a)(2005+a)$ and thus $a+b+c>a+\sqrt{(2005-a)(2005+a)}=:f(a)$. $f$ is convex. Moreover, we have $f(1000)>2053$, $f(2003)>2053$ and $f(2004)<2053$. This, together with the convexity, implies $f(a)>2053$ on $[1000$, $2003]$. Hence the minimum is attained for $a=2004$. A quick computation shows that $b+c \ge 49$ and thus the minimum of $a+b+c+d$ is $4058$. Next, I'll show that the maximum value of $a+b+c+d$ is $7772$ and thus that the maximum value of $a+b+c$ is $5767$. Lemma. If $pq=rs$ for some positive integers $p$, $q$, $r$, $s$ satisfying $r>p>q>s$, then $p+q<r+s$. Proof of the Lemma. Assume $p+q \ge r+s$ and thus $pq+q^2 \ge qr+qs \Leftrightarrow rs+q^2\ge qr+qs \Leftrightarrow (r-q)(s-q) \ge 0$, which is clearly a contradiction, since $r>q$ and $q>s$. Therefore, the Lemma is proved. Back to the problem. If $b+c>2005+a$, then we have by the lemma that $2b=(b+c)+(b-c) > (2005+a)+(2005-a)=4010$ and thus $b>2005$, clearly a contradiction. Hence $b+c < 2005 +a$ and $b-c>2005-a$. Let $b-c=2005-a+x$ and $b+c=2005+a-y$. Consequently, \[y= \frac{x(2005+a)}{2005-a+x}. \quad \quad (**)\] Since $2005+a \equiv 2005-a \mod 2$, $x$ is even. Assume that $a+b+c=5767$ is not the maximum. Then $2005+2a-y=a+b+c >5767 \Leftrightarrow 2a-3762>y$. Suppose that $x \ne 2$ and thus $x \ge 4$. From $(**)$ we get $y \ge 4(2005+a)/(2009-a)$. Therefore, $(a-1881)(2009-a) >2(2005+a) \Rightarrow 0>(a-63)^2$, which is a contradiction. Therefore, $x=2$. A similar estimation gives $(a-1881)(2007-a) > 2005+a$. Let $a=1881+a'$. Hence $(a-1881)(2007-a)>2005+a \Leftrightarrow a'(126-a') > 3886+a'$ and therefore $881>(2a'-125)^2 \Leftrightarrow 19 > |2a'-125|$. Thus, we get $53\le a' \le 72$ and consequently $1934 \le a \le 1953$. Hence $2007-a \in [54$, $73]$. Since $y$ is an integer, $2007-a | 2(2005+a) \Leftrightarrow 2007-a | 8024=2^3 \cdot 17 \cdot 59$. Thus, $2007-a=59$ or $2007-a=68$. Both values of $a$ lead to $a+b+c=5767$ and therefore the maximum. Thus, the minimum value of the perimeter is $4058$, the maximum value is $7772$.

I think there is a slight mistake in the above for the minimum value, since I believe $(2004, 45, 4)$ doesn't work. The below solution should give the correct value of the minimum. I think that the minimum value of the perimeter of $ABCD$ is $4160$, attained at $(1999, 155, 1).$ We will now show that this is minimal. Let $a, b, c$ be as $\mathbf{FelixD}$ defined above, also assuming that $a \ge b$. As $\mathbf{FelixD}$ did above, we can observe that $a+b+c > 4160$ whenever $a \le 1998$ (plug them into $f$), and so it hence only suffices to find the minimal values of $a+b+c$ when $a \in \{1999, 2000, 2001, 2002, 2003, 2004\}$. Since we have that $2005^2 - a^2 = b^2 - c^2$, from $b+c | 2005^2-a^2$ and $b+c > b-c$, we can easily finish with case-checking when noting that $2005^2-2004^2 = 19 \cdot 211, 2005^2 - 2003^2 = 48 \cdot 167, 2005^2-2002^2 = 3 \cdot 4007, 2005^2 - 2001^2 = 8 \cdot 2003, 2005^2 - 2000^2 = 89 \cdot 225,$ and $2005^2 - 1999^2 = 154 \cdot 156,$ where each factor pair is the one which minimizes the larger factor. Since all of them other than $156$ are greater than $160$, and $a \ge 2000$ except for the $156$, it's readily seen that $4160$ is the minimum here.