Iranian Mathematical Olympiad

Denote by $C(O,R)$ the circle with center $O$ and radius $R$. If ${\omega }_1=C(O_1,R_1)$ and ${\omega }_2=C(O_2,R_2)$, then $$d({\omega }_1,{\omega }_2{)}^2=O_1{O}_2^2-(R_1-R_2{)}^2,$$ and $d({\omega }_1,{\omega }_2)$ is defined if and only if the right hand side is nonnegative. This is equivalent to the existence of the common external tangents of these circles.

a) Let $\omega =C(O,R)$ and ${\omega }_i=C(O_i,R_i)$ for $1\le i\le n$. We have $$\frac{1}{n}\sum _{i=1}^{n}d(\omega ,{\omega }_i{)}^2=\frac{1}{n}\sum _{i=1}^{n}O{O}_i^2+\frac{1}{n}\sum _{i=1}^n(R-R_i{)}^2.$$ Suppose that $\bar{R}=\frac{1}{n}(R_1+R_2+\cdots +R_n)$. Then $$\frac{1}{n}\sum _{i=1}^n(R-R_i{)}^2=R^2-2R\bar{R}+\frac{1}{n}\sum _{i=1}^n{R}_i^2=(R-\bar{R}{)}^2+\frac{1}{n}\sum _{i=1}^n(\bar{R}-R_i{)}^2.$$ Similarly, let $\bar{O}$ be the centroid of all the $O_i$'s ($1\le i\le n$). Then $$\frac{1}{n}\sum _{i=1}^{n}O{O}_i^2=O{\bar{O}}^2+\frac{1}{n}\sum _{i=1}^n\bar{O}{O}_i^2.$$ Therefore, the circle $C(\bar{O},\bar{R})$ satisfies the problems condition. Now, if two circles ${\bar{\omega }}_1=C(P_1,r_1)$ and ${\bar{\omega }}_2=C(P_2,r_2)$ both satisfy the problems condition, for each circle $\omega$ we can write (C is a constant value here): $$\begin{array}{rl}d(\omega ,{\bar{\omega }}_1{)}^2-d(\omega ,{\bar{\omega }}_2{)}^2& =C\\ \Rightarrow & O{P}_1^2-O{P}_2^2+(R-r_1{)}^2-(R-r_2{)}^2=C\\ \Rightarrow & O{P}_1^2-O{P}_2^2-2R(r_1-r_2)=C\end{array}$$ By fixing $O$, we have $r_1=r_2$ and hence $O{P}_1^2-O{P}_2^2$ is a constant value. Thus $P_1=P_2$ and ${\bar{\omega }}_1={\bar{\omega }}_2$. Therefore, $\bar{\omega }=C(\bar{O},\bar{R})$ is the unique circle satisfying the problems condition.

b) We can write $x_3=(1-\alpha )x_1+\alpha x_2$ for some $\alpha \in \mathbb{R}$. Since the distances of $O_1$ and $O_2$ from the common tangent are $R_1$ and $R_2$, respectively, we conclude that the distance of $O_3$ from this line is $|(1-\alpha )R_1+\alpha R_2|$ (If the number in the absolute value sign is negative, it means that $O_1$ and $O_2$ are in the two sides of that line). So $$R_3=|(1-\alpha )R_1+\alpha R_2|.$$ Furthermore, the centroid of ${\omega }_1$ and ${\omega }_2$ is $\bar{\omega }=C(\frac{x_1+x_2}{2},\frac{R_1+R_2}{2})$. Therefore, putting $\alpha =\frac{1}{2}$ in the definition of ${\omega }_3$ implies $$\begin{array}{rl}d(\omega ,{\omega }_1)& =d(\omega ,{\omega }_2)\\ & \Rightarrow O{O}_1^2-(R-R_1{)}^2=O{O}_2^2-(R-R_2{)}^2\\ & \Rightarrow O{O}_1^2-O{O}_2^2=R_1^2-R_2^2-2R(R_1-R_2)\\ & \Rightarrow (x-x_1{)}^2-(x-x_2{)}^2=R_1^2-R_2^2-2R(R_1-R_2)\\ & \Rightarrow 2x(x_1-x_2)-2R(R_1-R_2)=x_1^2-x_2^2-R_1^2+R_2^2.\end{array}$$ We have $$\begin{array}{rl}d(\omega ,{\omega }_3{)}^2& =(x-x_3{)}^2+y^2-(R-R_3{)}^2\\ & =(x-((1-\alpha )x_1+\alpha x_2){)}^2+y^2-(R-((1-\alpha )R_1+\alpha R_2){)}^2\\ & =((x-x_1)+\alpha (x_1-x_2){)}^2+y^2-((R-R_1)+\alpha (R_1-R_2){)}^2.\end{array}$$ Suppose that circles $\omega$, ${\omega }_1$ and ${\omega }_2$ are fixed and $\alpha$ varies. We can write $d(\omega ,{\omega }_3{)}^2$ as a polynomial of $\alpha$ such as $p_2{\alpha }^2+p_1\alpha +p_0$. We know $p_2=(x_1-x_2{)}^2-(R_1-R_2{)}^2$ is nonnegative because $d({\omega }_1,{\omega }_2)$ is defined. Furthermore, the coefficient of $\alpha$ is $$p_1=2(x_1-x_2)(x-x_1)-2(R_1-R_2)(R-R_1).$$ Because of the relation between $X$ and $R$, $$x_1^2-R_1^2-x_2^2+R_2^2-2x_1(x_1-x_2)+2R_1(R_1-R_2)=-(x_1-x_2{)}^2+(R_1-R_2{)}^2.$$ If $p_1=p_2=0$, $d(\omega ,{\omega }_3)$ is independent of $\alpha$. Otherwise, the minimum value of $d(\omega ,{\omega }_3)$ is for $\alpha =\frac{1}{2}$ and the assertion is proved.

c) First, note that the radical center of three circles ${\omega }_1,{\omega }_2$ and ${\omega }_3$, as a circle with radius zero is equidistant from these circles (if this point is outside of all the circles). Therefore, we have Lemma 1. If two circles $C_1$ and $C_2$ are both equidistant from ${\omega }_1$ and ${\omega }_2$, then the direct homothetic center of $C_1$ and $C_2$ lies on the radical axis of ${\omega }_1$ and ${\omega }_2$. If two circles $C_1$ and $C_2$ have equal radius then the direct homothetic center of them is not defined. In this case, the line passing through the centers of $C_1$ and $C_2$ is parallel to the radical axis of ${\omega }_1$ and ${\omega }_2$. Let ${\omega }_i=C(O_i,R_i)$ and $C_i=C(P_i,r_i)$. We can assume that $O_i=(x_i,0)$ and $P_i=(a_i,b_i)$. If $(x,0)$ lies on the radical axis of ${\omega }_1$ and ${\omega }_2$, then $$(x-x_1{)}^2-R_1^2=(x-x_2{)}^2-R_2^2\Rightarrow x=\frac{x_1^2-x_2^2-R_1^2+R_2^2}{2(x_1-x_2)}.$$ Denote this value by $c$. We proved in part b that $$2a_i(x_1-x_2)-2r_i(R_1-R_2)=x_1^2-R_1^2-x_2^2+R_2^2\Rightarrow a_i=r_i\frac{R_1-R_2}{x_1-x_2}+c.$$ The direct homothetic center of $C_1$ and $C_2$ lies on $P_1{P}_2$ and hence it can be shown that $$S=\frac{r_2}{r_2-r_1}{P}_1-\frac{r_1}{r_2-r_1}{P}_2\text{ (whenever }{r}_1\ne r_2),$$ so the first coordinate of $S$ is $$\frac{r_2{a}_1-r_1{a}_2}{r_2-r_1}=\frac{1}{r_2-r_1}\left[r_2\left(r_1\frac{R_1-R_2}{x_1-x_2}+c\right)-r_1\left(r_2\frac{R_1-R_2}{x_1-x_2}+c\right)\right]=c.$$ Hence $S$ lies on the radical axis of ${\omega }_1$ and ${\omega }_2$. Otherwise, if $r_1=r_2$, by the above equations we get $a_1=a_2$ and thus the line passing through the centers of $C_1$ and $C_2$ is parallel to the radical axis of ${\omega }_1$ and ${\omega }_2$.

d) The answer is no. Let ${\omega }_i=C(O_i,R_i)$ and $d_{ij}=|O_i-O_j|$. We assume that $R_1\ge R_2\ge R_3\ge R_4$. According to the problems assumption we must have $$d_{ij}^2-(R_i-R_j{)}^2=1.$$ Note that these equations are invariant under addition of a constant number to $R_i$'s ($1\le i\le n$). Therefore, we can suppose that $R_4=0$ and $O_4$ lies on the radical axis of ${\omega }_1$ and ${\omega }_2$. According to the following figure we have: $$\begin{array}{rl}{a}^2+x^2-R_1^2& =1,\\ a^2+y^2-R_2^2& =1,\\ (x+y{)}^2-(R_1-R_2{)}^2& =1.\end{array}$$



Subtracting the sum of the first two equations from the third one implies: $$2xy+2R_1{R}_2+2-2a^2=1\Rightarrow a=\sqrt{\frac{1}{2}+xy+R_1{R}_2}.$$ On the other hand, by subtraction of the first equation from the second one we have: $$\{\begin{array}{l}{x}^2-y^2=R_1^2-R_2^2\\ x+y=d_{12}\end{array}\Rightarrow \begin{array}{rl}x-y& =\frac{R_1^2-R_2^2}{d_{12}}\\ \Rightarrow \{x,y\}=\frac{d_{12}}{2}\pm \frac{R_1^2-R_2^2}{2d_{12}}& \\ \Rightarrow xy=\frac{d_{12}^2}{4}-{\left(\frac{R_1^2-R_2^2}{d_{12}}\right)}^2.& \end{array}$$ Hence $$a=\sqrt{\frac{1}{2}+\frac{d_{12}^2}{4}-\frac{(R_1^2-R_2^2{)}^2}{4d_{12}^2}+R_1{R}_2}.$$ Now, replacing $R_1$, $R_2$ and $R_3$ by $R_1-R_3$, $R_2-R_3$ and 0 yields: $$b=\sqrt{\frac{1}{2}+\frac{d_{12}^2}{4}-\frac{((R_1-R_3{)}^2-(R_2-R_3{)}^2{)}^2}{4d_{12}^2}+(R_1-R_3)(R_2-R_3)}.$$

There are two possibilities. Case 1. $O_3$ and $O_4$ are not on the same side of $O_1{O}_2$. In this case we have $$O_3{O}_4^2-R_3^2=1\Rightarrow 1+R_3^2=O_3{O}_4^2\ge (a+b{)}^2\ge a^2+b^2.$$ On the other hand, $$a^2>\frac{1}{2}+R_1{R}_2,$$ $$b^2>\frac{1}{2}+(R_1-R_3)(R_2-R_3),$$ so $$R_3^2>R_1{R}_2+(R_1-R_3)(R_2-R_3)\Rightarrow (R_1+R_2)R_3>2R_1{R}_2.$$ But this is in contradiction with the assumption that $R_1\ge R_2\ge R_3$.

Case 2. $O_3$ and $O_4$ are on the same side of $O_1{O}_2$. In this case we have $$\begin{array}{rl}1+R_3^2& =(a-b{)}^2+(x-z{)}^2=(a^2+x^2)+(b^2+z^2)-2ab-2xz\\ & =1+R_1^2+1+(R_1-R_3{)}^2-2ab-2xz\\ \Rightarrow 0& =1+2R_1^2-2R_1{R}_3-2ab-2xz\\ \Rightarrow 4ab+4xz& =2+4R_1(R_1-R_3).\end{array}$$ Substituting $d_{12}^2=1+(R_1-R_2{)}^2$ implies $$a=\sqrt{\frac{1}{2}+\frac{d_{12}^2}{4}-\frac{(R_1^2-R_2^2{)}^2}{4d_{12}^2}+R_1{R}_2}=\sqrt{\frac{3}{4}+\frac{(R_1+R_2{)}^2}{4}-\frac{(R_1-R_2{)}^2(R_1+R_2{)}^2}{4(1+(R_1-R_2{)}^2)}}$$ Similarly, $$b=\sqrt{\frac{3}{4}+\frac{(R_1+R_2-2R_3{)}^2}{4}-\frac{(R_1-R_2{)}^2(R_1+R_2-2R_3{)}^2}{4(1+(R_1-R_2{)}^2)}}$$ Defining $$\begin{array}{rl}t& :=R_1-R_2,\\ s& :=R_1+R_2-2R_3,\\ r& :=R_1+R_2,\end{array}$$ we have $$\begin{array}{rl}a& =\sqrt{\frac{3}{4}+\frac{r^2}{4}-\frac{r^2{t}^2}{4(1+t^2)}}=\sqrt{\frac{3}{4}+\frac{r^2}{4(1+t^2)}},\\ b& =\sqrt{\frac{3}{4}+\frac{s^2}{4(1+t^2)}}.\end{array}$$ Also, $$xz=\left(\frac{d_{12}}{2}+\frac{R_1^2-R_2^2}{2d_{12}}\right)\left(\frac{d_{12}}{2}+\frac{(R_1-R_3{)}^2-(R_2-R_3{)}^2}{2d_{12}}\right)=\frac{1+t^2}{4}+\frac{rt}{4}+\frac{st}{4}+\frac{rs{t}^2}{4(1+t^2)}.$$ Now, we have: $$\begin{array}{rl}4ab+4xz& =2+4R_1(R_1-R_3)\\ \Rightarrow \sqrt{\left(3+\frac{r^2}{1+t^2}\right)\left(3+\frac{s^2}{1+t^2}\right)}+1+t^2+rt+st+\frac{rs{t}^2}{1+t^2}& =2+(r+t)(s+t).\end{array}$$ Multiplication by $1+t^2$ and some computations yields: $$\sqrt{(3+3t^2+r^2)(3+3t^2+s^2)}=1+t^2+rs,$$ which is in contradiction with the Cauchy-Schwarz inequality.