The smaller circle can only cover less than the bigger circle, because otherwise the diameter of the smaller circle (the biggest distance it can cover) would be greater than or equal to the greater diameter, but this would be a contradiction. Now, since the smaller circles covers strictly less than half of the greater circumference, it follows that we can pick a semicircumference which covers all the points.

A proof by induction: Let the radius of the $\omega$ be $r$. The assumption means that there is a point(For example:$P$) that has a distance less than $r$ to all $n$ points. When $n=1$, it's obvious. Now suppose the result holds for less than $n$ points. First we will prove some lemmas:

Choose a given point on the $\omega$ and name it $N$; Now we know that there is a diameter(call it $t$) in which all $n-1$ points remain on one side(WLOG Above) of the diameter. If $N$ lies on that side, it will hold the statement; Now assume $N$ lies on the other side. consider $P$, we have the following cases:

Name objects as shown in diagram:

One can see if $\angle{AOB}<180$, there will be a diameter that holds the condition.

). $$\angle{AOB}\geq180\to \text{at least one of} \angle{O'OA},\angle{O'OB}\geq90 $$WLOG$\angle{O'OB}\geq90\to\angle{OO'B}<90\to\angle{O'OB}>\angle{OO'B}$ On the other hand: $$ \overline{OB}=r>\overline{O'B}=r' $$Wich is contradiction in $\triangle{O'OB}$!

Here's another solution: *In this solution the "smaller circle" means "the circle with smaller radius" Main claim . In the non-obtuse triangle $\triangle ABC$ , the smallest circle which contains all $A ,B,C$ or passes through them , is $(ABC)$ . Proof . Name that circle $\omega_{min}$ . Assume that such circle passes through none of them . clearly $(ABC)$ is smaller than $\omega_{min}$ (because it's inside of $\omega_{min}$) . If $\omega_{min}$ passes through exactly one of $A,B,C$ , WLOG $A$ , draw two circles $\omega_{B}$ , $\omega_{C}$ which are tangent to $\omega_{min}$ at $A$ and pass through $B,C$ respectively ; this is possible by PLC , but clearly one of $\omega_{B}$ , $\omega_{C}$ is inside the another(or they're equal) . Absurd again . So $\omega_{min}$ passes through at least two of $A,B,C$ , WLOG $B,C$ . But then $O_{min}$ is on the perpendicular bisector of $BC$ and $AO_{min} \leq BO_{min}$ . But $\triangle ABC$ was an non-obtuse triangle so $O_{min} = O_{\triangle ABC}$ . So the claim holds. Back to the main Problem ,name that circle $\gamma$, we have 2 cases : Case 1 : $n=2$ . in this case , if $AB$ was a diagram of $\omega$ ,we'll do the same stuffs as we do in cases 1,2 of the claim : if $\gamma$ passes through none of $A,B$ , then $\omega$ is smaller because it's inside of $\gamma$ and if $\gamma$ passes through one of $A,B$ , WLOG $A$ , we draw a tangent circle to $\gamma$ at $A$ which passes through $B$; this is possible by PLC . this circle is clearly smaller than $\gamma$. Absurd . So $AB$ is not a diagram . Now , Let $A'$ be the antipode of $A$ in $\omega$ and let $A''$ be a point on $\omega$ such that $A'' , B $ are on different sides of $AA'$ and $\overarc {AA''} < \overarc {BA'}$ . Now , if $A''' = A''O \cap \omega$ , $A''A'''$ have the desired property and we're done in this case. Case 2:$3 \leq n$ . $n=3$ is another version of the claim . Let $4 \leq n$ and $AB$ be the largest segment between $A_{i}A_{j}$'s . Now , if there was two points $X,Y$ of $A_{i}$'s on the different sides of $AB$ , at lease one of $\angle AXB , \angle AYB$ is less than $90$ (by case 1 , none of them is equal to $90$) . WLOG $\angle AYB \leq 90$ . But in the triangle $\triangle AYB$ , $AB$ is the largest side and $\angle AYB$ is less than $90$ . so $\triangle AYB$ is an acute triangle . Contradiction by the claim . So all of $n$ points are in one side of $AB$ . Doing the similar stuffs as we do in the last part of the case $1$ , we're done .

Let's Call the circle with smaller radius Q. O1 and O2 are centers of W and Q. then let's assume there's no such diameter. A is one of those n-points so if AB is a diameter then there exists a point A' such that arc ABA' is more than 180. O1 and O2 both are on perpendicular bisector of AA'. M is the midpoint of AA' 1) O1 lies between M and O2 then by Trigonometry O2A > O1A and it's a contradiction. 2) O2 lies between M and O1 then if O1O2 meets W at P we have O2P > O1P and it's a contradiction. so such diameter exists.

Second solution : let O1 and O2 be the centers of W and the smaller circle. perpendicular at O1 to O1O2 meets W at P and Q. we'll show PQ is our diameter. let's assume it's not so there exists a K at the other side of PQ w.r.t O2 then angle KO1O2 is more than 90 so KO2 > KO1 and it's a contradiction so there's no point at the other side of PQ.

alinazarboland wrote: $\omega_{C}$ which are tangent to $\omega_{min}$ at $A$ and pass through $B,C$ respectively ; this is possible by PLC , but clearly one of $\omega_{B}$ , $\omega_{C}$ is inside the another(or they're equal) . . Hello, What is "PLC"?