Japan Mathematical Olympiad

First, let us show that the following lemma holds.
Lemma: Suppose a triangle $T$ contains a triangle $T'$, and that the lengths of the radii of $T$ and $T'$ are the same. Then, we must have $T = T'$.
Proof: It is clear that the incircle of a triangle is uniquely determined by the fact that it is the circle having the largest value for its radius among circles contained in the triangle. So, if we denote by $\Gamma$ and $\Gamma'$ the incircles of the triangles $T$ and $T'$, respectively, then since $\Gamma'$ is contained in $T'$ and $T' \subset T$, we must have that the length of the radius of $\Gamma'$ must be less than or equal to that of $\Gamma$. Since these lengths are equal by assumption, we must have $\Gamma = \Gamma'$ by the uniqueness. Then, $\Gamma$ must be contained in $T'$ so we must have $T = T'$.

Denote by $A, B, C, D$ the 4 points given for the problem. We may assume that the triangle $ABC$ has the largest area among the 4 triangles formed by any 3 of the 4 given points. Denote by $\ell_A$ the line parallel to the side $BC$ and going through the point $A$. If the point $D$ lies on the other side from the points $B, C$ with respect to the line $\ell_A$, then the triangle $DBC$ will have the area bigger than that of the triangle $ABC$, contrary to our assumption. So, $D$ must lie either on the line $\ell_A$ or on the same side as $B, D$ with respect to $\ell_A$.

If we denote by $\ell_B, (\ell_C)$ the line parallel to $AC$ ($AB$) going through the point $B$ ($C$), then we can conclude in the same way as above that the point $D$ lies inside of the triangle (possibly on the boundary) formed by the lines $\ell_A, \ell_B, \ell_C$. Denote by $A', B'$ and $C'$ the points of intersections of the lines $\ell_B, \ell_C$, of the lines $\ell_C, \ell_A$ and of the lines $\ell_A, \ell_B$, respectively.

Now suppose the point $D$ lies in the triangle $ABC$. Then, the triangle $ABC$ contains the triangle $ABD$. But since the radii of the incircles of $ABC$ and $ABD$ have the same length by assumption, we must have $D = C$ by the Lemma proved above. But this contradicts the fact that the 4 given points are distinct. Hence the point $D$ must lie outside of the triangle $ABC$. We may assume without loss

of generality that $D$ lies inside of the triangle $AB'C$. Since the sides $AB, B'C$ are parallel and so are the sides $AB', BC$, the quadrilateral $AB'CB$ is a parallelogram, and so, the triangles $ABC$ and $CB'A$ are congruent and hence the length of the radii of the incircles of these 2 triangles are equal. Since the triangle $CDA$ is contained in the triangle $CB'A$ and the radii of their incircles have the same length, we must have $D = B'$ by the Lemma. Consequently, we conclude that the quadrilateral $ABCD$ is a parallelogram, and therefore, the triangles $ABC$ and $BCD$ have the same area. Since the radii of the incircles of these triangles are also the same, we conclude that the perimeters of these triangles also have the same length (since the area of the triangle is given by $sr$, where $s$ is the length of its perimeter and $r$ is the length of its radius). Thus we get $AB + BC + CA = BC + CD + DB$ and since $AB = CD$ as $ABCD$ is a parallelogram, we see that $CA = DB$ must hold. Thus, the diagonals $AC$ and $BD$ of this parallelogram have the same length, and we conclude that the parallelogram is a rectangle, and all 4 triangles formed by any 3 of the points $A, B, C, D$ are congruent.