[asy][asy] import geometry; import olympiad; size(150); pair a, b, c, d, o, m, n; a = (-9, 10); b = (11,10); c = -a; d = -b; o = extension(a, c, b, d); m = 0.5(o+b); n = 0.5(c+d); draw(a--n--m--a--b--c--d--a--c); draw(o--b); label("$A$",a,dir(a)); label("$B$",b,dir(b)); label("$C$",c,dir(c)); label("$D$",d,dir(d)); label("$M$",m,dir(-15)); label("$N$",n,dir(n)); label("$O$",o,dir(85)); draw(circumcircle(b, c, d), dotted); draw(o--d, dashed); [/asy][/asy] If $\triangle ABC\sim\triangle AMN$, then it follows that $A$ is the center of the spiral similarity mapping $BC$ to $MN$. It is well known that the center of the spiral similarity mapping $BC$ to $MN$ is $(BCX)\cap (MNX)\neq X$, where $X=BM\cap CN$. In particular, since $BM\cap CN=D$, we must have $A$ lie on $(BCD)$. The only way for a parallelogram to be cyclic is if it is a rectangle, so we now know that $ABCD$ is a rectangle. In order to show that $ABCD$ must also be a square, we use barycentric coordinates (!). Set reference triangle $\triangle ABC$, and it is straightforward to compute that $D=(1, -1, 1), O = \left(\frac{1}{2}, 0, \frac{1}{2}\right), M = \left(\frac{1}{4}, \frac{1}{2}, \frac{1}{4}\right), N = \left(\frac{1}{2}, -\frac{1}{2}, 1\right)$. Let the circumcircle of $\triangle ADN$ be $-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)$, and we wish to show that if $M$ lies on this as well, then $a\sqrt{2} = c\sqrt{2}=b$. Since $A$ lies on the circle, $u = 0$. By plugging in $D$, we find $$a^2-b^2+c^2-v+w = 0\implies v = w,$$since $a^2+c^2=b^2$. Similarly, by plugging in $N$, $$\frac{a^2}{2}-\frac{b^2}{2}+\frac{c^2}{4}+\frac{v}{2}=0$$$$\implies \frac{a^2+c^2-b^2}{2}+\frac{v}{2}=\frac{c^2}{4}$$$$\implies v =w= \frac{c^2}{2}.$$Thus, the circumcircle of $\triangle ADN$ is $-a^2yz-b^2zx-c^2xy+(x+y+z)\left(\frac{c^2(y+z)}{2}\right)=0$. Now, plugging in $M$, $$-\frac{a^2}{8}-\frac{b^2}{16}-\frac{c^2}{8}+\frac{3c^2}{8}=0$$$$\implies -2a^2-b^2+4c^2=0=a^2+c^2-b^2$$$$\implies -3a^2+3c^2=0\implies a = c\implies ABCD \text{ is a square.}$$

parmenides51 wrote: Let $ABCD$ be a parallelogram of center $O$. Points $M$ and $N$ are the midpoints of $BO$ and $CD$, respectively. Prove that if the triangles $ABC$ and $AMN$ are similar, then $ABCD$ is a square. Lemma: For any two similar triangles $ABC,A_0B_0C_0$ with the same orient and any real number $k$, define $A_1$ on $AA_0$ such that $\vec{A_1A_0}=k\vec{A_0A}$. Define $B_1,C_1$ similarly. Then the triangle $A_1B_1C_1$ is similar to $ABC$ in the same orient. The proof is not trivial but left as an exercise to the reader. Apply the lemma on triangles $ABC$ and $AMN$ and real number $k=-\frac{1}{2}$ to get $AOD$ is similar to $ABC$ hence $\angle DAO=\angle OAB$ and consequently $AD=AB$ so we get $\angle AOD=90$. From the similarity we get $\angle ABC=90$ and since $ABCD$ is a rhombus the conclusion follows. The following problems are nicely solvable by the above lemma. 2007 Olympic Revenge Iran 2015

draw a parallel from $M$ to $BC$ to intersect $DC$ at $P$ and extend $AM$ to intersect $DC$ at $T$ we have: $MP=\frac{3}{4} .AD \implies \frac{TP}{TD}=\frac{3}{4} \implies TD=3DC$ claim:$AMND$ cyclic. proof by angle chasing. Let $DC=a,AM=b$ so compute the power of $T$: $3a.\frac{5}{2}a=4b.3b \implies 5a^2=8b^2$ note: you can find length of median by $\sqrt{(2x^2+2y^2-z^2)/4}$ and then with little computing we are done

matinyousefi wrote: parmenides51 wrote: Let $ABCD$ be a parallelogram of center $O$. Points $M$ and $N$ are the midpoints of $BO$ and $CD$, respectively. Prove that if the triangles $ABC$ and $AMN$ are similar, then $ABCD$ is a square. Lemma: For any two similar triangles $ABC,A_0B_0C_0$ with the same orient and any real number $k$, define $A_1$ on $AA_0$ such that $\vec{A_1A_0}=k\vec{A_0A}$. Define $B_1,C_1$ similarly. Then the triangle $A_1B_1C_1$ is similar to $ABC$ in the same orient. The proof is not trivial but left as an exercise to the reader. Apply the lemma on triangles $ABC$ and $AMN$ and real number $k=-\frac{1}{2}$ to get $AOD$ is similar to $ABC$ hence $\angle DAO=\angle OAB$ and consequently $AD=AB$ so we get $\angle AOD=90$. From the similarity we get $\angle ABC=90$ and since $ABCD$ is a rhombus the conclusion follows. The following problems are nicely solvable by the above lemma. 2007 Olympic Revenge Iran 2015 This lemma was appeared in this pdf with many examples. mean geometry