Mediterranean Mathematical Competition PETER O' HALLORAN MEMORIAL

Consider the auxiliary triangle $DEF$ with $D=(-12,5)$, $E=(12,5)$, and $F=(0,13)$. Since the triangle $DEF$ is symmetric with respect to the $y$-axis, condition (i) yields that its Mediterranean point $M$ lies on the $y$-axis, and hence has coordinates $(0,m)$ with $5\le m\le 13$. Define a non-negative real number $a$ by $a^2+25=m^2$, and note that $0\le a\le 12$. Then the point $A=(a,5)$ lies on the side $DE$ of the triangle $DEF$, and points $A$ and $M$ both are at distance $m$ from the origin. Then $A$, $M$, and the midpoint $C=(12a,12(m+5))$ of $AM$ form a (degenerate) triangle that is symmetric with respect to the line running through the origin and $C$. By (i) the Mediterranean point of $AMC$ lies on this line and hence must be $C$. On the other hand since $DEF$ contains $AMC$, and since $AMC$ contains the Mediterranean point $M$ of $DEF$, condition (ii) yields that the Mediterranean point of $AMC$ must be $M$. This implies that $M$ and $C$ coincide, which in turn implies $a=0$ and $m=5$. Hence $M=(0,5)$ is the Mediterranean point of triangle $DEF$.

Now let us return to the original triangle with vertices $(-3,5)$, $(12,5)$, $(3,11)$ in the problem statement: $DEF$ contains this triangle, and the triangle contains the Mediterranean point $M=(0,5)$ of $DEF$. Then, (ii) yields that its Mediterranean point is $(0,5)$.

---

Alternative solution.

A more careful analysis reveals that the two stated conditions uniquely determine the Mediterranean point of every triangle: It is the point in the triangle that is closest to the origin. This can be seen as follows: (1) First consider a triangle $ABC$ that contains the origin $(0,0)$. Construct an equilateral triangle $DEF$ that contains $ABC$ and that has $(0,0)$ as its center of gravity. Since $DEF$ is symmetric with respect to three lines through the origin, condition (i) implies that its Mediterranean point is the origin. Since $DEF$ contains $ABC$ and the origin is in $ABC$, condition (ii) yields that the Mediterranean point of $ABC$ is the origin.

(2) Next consider a triangle $ABC$ that does not contain the origin $(0,0)$, whose circumcenter is in $(0,0)$, and that furthermore satisfies $AC=BC$. Consider the circle $k_0$ with center in $(0,0)$ that goes through the Mediterranean point $M$ of $ABC$. Suppose for the sake of contradiction that $M$ is not the midpoint of side $AB$. Then $k_0$ also contains some point $P\ne M$ in triangle $ABC$. Let $Q$ denote the midpoint of $MP$. Then condition (i) yields that $Q$ is the Mediterranean point of the degenerate triangle $MPQ$, while condition (ii) implies that the Mediterranean point of $MPQ$ is point $M$. That's a contradiction. Therefore the Mediterranean point of $ABC$ is $M$, and this is indeed its closest point to the origin.

(3) Finally consider an arbitrary triangle $ABC$ that does not contain the origin. Let $M$ denote the point in the triangle $ABC$ that is closest to the origin. Construct a circle $k_1$ that has its center in $(0,0)$ and that touches $ABC$ from outside in point $M$. Construct another circle $k_2$ with center in $(0,0)$ that satisfies the following: The tangent of $k_1$ at point $M$ intersects circle $k_2$ in the points $D$ and $E$. The half-line that starts in $(0,0)$ and goes through point $M$ intersects circle $k_2$ in the point $F$. The radius of $k_2$ is chosen sufficiently large, so that $DEF$ entirely contains $ABC$. Then (2) yields that the Mediterranean point of $DEF$ is the point $M$, and condition (ii) yields that the Mediterranean point of $ABC$ is also $M$. Once again, $M$ is the point of $ABC$ that is closest to the origin.

(4) It remains to observe that for the triangle with vertices in $(-3,5)$, $(12,5)$ and $(3,11)$, the closest point to the origin is point $(0,5)$; hence the desired Mediterranean point is $(0,5)$.