shobber wrote: Find the least positive integer $n$ ($n\geq 3$), such that among any $n$ points (no three are collinear) in the plane, there exist three points which are the vertices of a non-isoscele triangle. It seems that ti suffices to consider the diameter of the set and then work some cases, am I right? Daniel

shobber wrote: Find the least positive integer $n$ ($n\geq 3$), such that among any $n$ points (no three are collinear) in the plane, there exist three points which are the vertices of a non-isosceles triangle. Proof. I'll prove that $n = 7$ fits. To show this, consider the following Lemma. Assume that the set $M$ consists of $5$ points in the plane such that every triangle composed of three points of $M$ is isosceles. Then there exists a set $N \subset M$ consisting of $4$ points $A$, $B$, $C$, $D$ such that $\overline{AB}=\overline{AC}=\overline{CD}$ and $\overline{AD}=\overline{BC}=\overline{BD}$. Moreover, the fifth point $E \in M \setminus N$ lies on the perpendicular bisectors of the sides $\overline{AC}$ and $\overline{BD}$ (since $AC \parallel BD$). Proof of the Lemma. Let $M={A}$, $B$, $C$, $D$, ${E}$ be a given set with the property that every triangle composed of three points of $M$ is isosceles and let $G$ be its graph. Since no three points are collinear, every edge of $G$ appears at least twice. Let without loss of generality $\overline{AB}=\overline{AC}$ and $\overline{BC}=\overline{BD}$. Suppose that $\overline{AB}=\overline{BD}$. Consider triangle $\triangle ACD$. If $\overline{AC}=\overline{AD}$, then $D$ is the intersection of circles $(A)$ and $(B)$ with diameters $\overline{AB}$. If $\overline{AC}=\overline{CD}$, then $D$ is the intersection of circles $(C)$ and $(B)$ with diameters $\overline{AB}$. In both cases, $ADBC$ is a rhombus composed of two equilateral triangles $\triangle ADB$ and $\triangle ABC$. Now, $E$ does certainly not lie inside $ADBC$ (if so, $E$ must be the circumcenter of $\triangle ADB$ or $\triangle ABC$, a contradiction). Suppose now wlog that $\overline{BE}=\overline{BC}$ or $\overline{CE}=\overline{BC}$ (since $E$ must not lie on every perpendicular bisector). Then $\overline{AE} \ne \overline{BC}$ and $\overline{DE} \ne \overline{BC}$. Hence $E$ lies on the perpendicular bisector of $\overline{AD}$. Therefore, $BC \perp DE$. But then triangle $\triangle AEC$ is not isosceles, a contradiction. If $\overline{AD}=\overline{CD}$, then $D$ is the intersection of the perpendicular bisector of $\overline{AC}$ and circle $(B)$ with diameter $\overline{AB}$. Suppose that $AEBC$ is no rhombus composed of two equilateral triangles, since we already dealt with this case (just switch the roles of $D$ and $E$) and consider triangle $\triangle AEC$. Suppose wlog that $\overline{AE}=\overline{AC}$, since $E$ must not lie on the perpendicular bisector of $\overline{AC}$. Therefore, $\overline{EC}=\overline{EB}$ (triangles $\triangle ACB$ is isosceles), from which we conclude that $E$ is the intersection of the perpendicular bisector of $\overline{BC}$ and the circle $(A)$ with diameter $\overline{AB}$. Now there are just $4$ possibilities for $(D$, $E)$. It's easy to show that each of them leads to a contradiction. Therefore, we may suppose that $\overline{AB}=\overline{AC}$ and $\overline{BC}=\overline{BD}$, but $\overline{AB} \ne \overline{BD}$. Consider two cases: 1. Suppose that $\overline{AD}=\overline{BD}$. Since $\triangle ADC$ is isosceles, we either have $\overline{AC}=\overline{CD}$ or $\overline{AD}=\overline{DC}$. In the latter case, switching $(A$, $B$, $C$, $D) \to (D$, $C$, $B$, $A)$ leads directly to the contradiction obtained one paragraph above. Hence $\overline{AC}=\overline{CD}$ and thus $\overline{AB}=\overline{AC}=\overline{CD}$ and $\overline{AD}=\overline{BC}=\overline{BD}$. A quick angle chase shows that $AC\perp BD$ and moreover $\angle CBD=36^{\circ}$ (this actually determines every angle in the configuration $ADBC$). Suppose that $E$ does not lie on the perpendicular bisector of $BD$ and let wlog $\overline{DE}=\overline{DB}$. If $\overline{EC}=\overline{ED}$, then triangles $\triangle BDC$ and $\triangle EDC$ are concurrent, which means that $E$ is the image of $B$ when beeing reflected on $DC$. But then $A$, $C$, $E$ are collinear (this follows from the angles), which is a contradiction. Hence $\overline{EC}=\overline{DC}$, which means that triangles $\triangle EDC$ and $\triangle ADC$ are concurrent. Consequenty, $B$, $C$, $E$ are concurrent (an angle chase again), which is a contradiction again. Hence $E$ lies on the perpendicular of $\overline{BD}$ which proves the lemma. 2. Suppose that $\overline{AD}=\overline{AB}$. Hence $D$ is the intersection of circles $(A)$ with diameter $\overline{AB}$ and $(B)$ with diameter $\overline{BC}$. Wlog suppose that $E$ does not lie on the perpendicular bisector of $\overline{AD}$ (since $AB$ is an axis of symmetry). If $\overline{AE}=\overline{AD}$, i.e. $E$ belongs to the circumcircle of $\triangle BCD$, we get $\overline{BE}=\overline{DE}$. Let $\angle DAB= \angle BAC= 2\alpha$. An angle chase now shows that triangle $\triangle BEC$ has angles $\angle BEC=\alpha$, $\angle ECB=1/2\alpha$ and $\angle CBE= 180-3/2\alpha$. Since $\triangle BEC$ is isosceles and $\alpha \ne 90^{\circ}$, we get $\alpha =72^{\circ}$. Thus, if we switch $(A$, $B$, $C$, $D$, $E) \to (E$, $D$, $B$, $A$, $C)$, the conditions of the lemma are fulfilled. Suppose now $\overline{DE}=\overline{AD}$. If $\overline{BE}=\overline{ED}$, $ABED$ is a rhombus. Therefore, $\overline{EC}=\overline{BC}$, since $E \not \equiv A$. This yields $\overline{DC}=\overline{AD}$, since $C$, $A$, $E$ are not collinear, which is a contradiction. If $\overline{BE}=\overline{BD}$, we get $\overline{AE}=\overline{BE}$ as well, since $E \not \equiv C$, $D$. This finally proves the lemma, since these are the only two cases we have to consider ($\triangle ABD$ is isosceles and $AB \ne BD$). $\square$ Suppose now we've got a set $M$ conisting of $7$ points, such that every triangle composed of three points of $M$ is isosceles. By the lemma, there exist four points $A$, $B$, $C$, $D$ of $M$ such that they form an equilateral trapezoid and $\overline{AB}=\overline{AC}=\overline{CD}$ and $\overline{AD}=\overline{BC}=\overline{BD}$. Moreover, the remaining three points must all lie on the perpendicular bisector of $\overline{AC}$, clearly a contradiciton. To see that $n=6$ does not fit, take a regular $5$-gon and its circumcenter. Therefore, $n=7$, as required. $\square$

i don't understand. The problem is really easy! Let $S(n)$ is number of isosceles triangles in $n$ points. Then $S(n)\le n(n-1)$ and we can find $n<8$ when there isn't a nonisoscele triangle.

The answer stays the same when the condition of points being colinear is removed. The answer is n=7. To see why n=6 is not sufficient, take a regular pentagon along with its center, and this set of 6 points clearly satisfies the condition of a good set. Given 2 points A,B in a set S that satisfies problem's condition, all other points in S are either on the circle (A,B) (meaning, with center A, passing through B), on (B,A) or on [A,B] (meaning the perpendicular bisector of A,B). Assume A,B,C are coliniar in S. Then, wlog B is the midpoint of segment AC. (A,B) is tangent to (C,B) and [A,B] is parallel to [CB] and therefore every other point in S has to lie on (B,A). But (A,C),(C,A) are tangent to (B,A) at C,A so every other point of S has to lie on [A,C]. Buy [A,C] intersects (B,A) at 2 points that complete the configuration to a square and its center, which are only 5 points. So we may assume that there are no triples of colliniar points in S.