RaleD wrote: On semicircle, with diameter $|AB|=d$, are given points $C$ and $D$ such that: $|BC|=|CD|=a$ and $|DA|=b$ where $a, b, d$ are different positive integers. Find minimum possible value of $d$ R U sure you post the correct box? According to me, you should post on the geometry box, maybe.

I think this is more NT, when you are solving this you will use relation $d(d-b)=2a^2$ where $d, b, a$ are positive integers; and you need to find minimal $d$ for such equation is possible (in set of positive integers). Anyway, another number theory question wasn't in competition.

We have: + $S_{\bigtriangleup BDC }=\frac{1}{2}CD.CB\sin \widehat{CBD}$ $=\frac{1}{2}a^2. \sin (\frac{\pi}{2}+\widehat{ACD})$ $=\frac{1}{2}a^2. \cos\widehat{ACD}$ $=\frac{1}{2}a^2.\frac{AC^2+a^2-b^2}{2AC.a}$ $=\left (\frac{d^2-b^2}{\sqrt{d^2-a^2}} \right ) .\frac{a}{4} $;$(1)$ and + $S_{\bigtriangleup BDC }=\frac{BD.BC.CB}{4R}$ $=\frac{a^2\sqrt{d^2-b^2}}{2d} ;(2)$ From $(1)$ and $(2)$, easy to obtain: $d^2-bd-2a^2=0$ Hence: $\sqrt{\Delta }=\sqrt{b^2+8a^2}=k \in \mathbb{N}; $ $d=\frac{b+\sqrt{b^2+8a^2} }{2} =\frac{b+k}{2}\in \mathbb{N}; 2|(b+k), $ $\min d \Leftrightarrow \min (b+k)$

tuan119 wrote: We have: + $S_{\bigtriangleup BDC }=\frac{1}{2}CD.CB\sin \widehat{CBD}$ $=\frac{1}{2}a^2. \sin (\frac{\pi}{2}+\widehat{ACD})$ $=\frac{1}{2}a^2. \cos\widehat{ACD}$ $=\frac{1}{2}a^2.\frac{AC^2+a^2-b^2}{2AC.a}$ $=\left (\frac{d^2-b^2}{\sqrt{d^2-a^2}} \right ) .\frac{a}{4} $;$(1)$ and + $S_{\bigtriangleup BDC }=\frac{BD.BC.CB}{4R}$ $=\frac{a^2\sqrt{d^2-b^2}}{2d} ;(2)$ From $(1)$ and $(2)$, easy to obtain: $d^2-bd-2a^2=0$ Hence: $\sqrt{\Delta }=\sqrt{b^2+8a^2}=k \in \mathbb{N}; $ $d=\frac{b+\sqrt{b^2+8a^2} }{2} =\frac{b+k}{2}\in \mathbb{N}; 2|(b+k), $ $\min d \Leftrightarrow \min (b+k)$ You didn't find minimum of $d$. (it is fixed integer)

RaleD wrote: On semicircle, with diameter $|AB|=d$, are given points $C$ and $D$ such that: $|BC|=|CD|=a$ and $|DA|=b$ where $a, b, d$ are different positive integers. Find minimum possible value of $d$ First notice that $A,D,C,B$ are in this order on the semicircle (order A,C,D,B would imply $B=D$ and so $b=d$, impossible). Let $2\theta$ be the angle $BOC$. We get the equations $a=d\sin\theta$ $b=d\sin(\frac{\pi}2-2\theta)=$ $d\cos 2\theta=$ $d(1-2\sin^2\theta)=$ $d(1-2\frac{a^2}{d^2})$ And so the equation $bd=d^2-2a^2$ which may be written $2a^2=d(d-b)$ If $p|b$ and $p|d$, then $p^2|2a^2$ and so $p|a$ and we have a smaller solution when dividing $a,b,d$ by $p$ So $b,d$ are coprime. And so : either $d=2m^2$ and $d-b=n^2$ and $a=mn$ either $d=m^2$ and $d-b=2n^2$ and $a=mn$ In the first case ($d=2m^2$ and $d-b=n^2$ and $a=mn$): $m=1$ implies $d=2$ and $b=a=1$ impossible $m=2$ implies a possible solution $(a,b,d)=(2,7,8)$ In the second case ($d=m^2$ and $d-b=2n^2$ and $a=mn$) : $m=1$ is impossible (no value for $b$) $m=2$ is impossible since then $b=2$ and so $n=1$ and $a=2=b$ $m=3$ gives a solution greater that the one we already found. Hence the answer : $\boxed{d=8}$

Bosnia and Herzegovina 2011? Not originally. This is Euclid 1999 Problem 10b. Official solutions here.

RaleD wrote: You didn't find minimum of $d$. (it is fixed integer) I didn't find minimum of $d$. Because, BigSams had the answer. I Post the answer when BigSams had a plan that answered, that was not very good. May be.