Suppose $\bigtriangleup ABC$ is isosceles in $A$. Let $D$ be the midpoint of $AB$. Let the intersection of $\omega$ and $SN$ be $K$. Let $MK$ cut $TN$ at $R$. Pascal's theorem for $MMTKKN$ gives $S-P-R$. So it suffices to show that $R$ $\in$ $BC$. By Desargues theorem on $\bigtriangleup CTM$ and $\bigtriangleup BNK$ its enough to show that $BK$, $CM$ and $AS$ are concurrent. Since $CM$ and $AS$ meet on the centroid of $\bigtriangleup ABC$ its enough to show that $BK$ is a median in $\bigtriangleup ABC$. Or equivalently $B$, $K$ and $T$ collinear. $\angle STB$ $=$ $\angle ABT$ $=$ $\angle ACD$. $\angle TAN$ $=$ $\angle ANM$ $=$ $\angle SDN$ hence $MSND$ isosceles trapezoid from which: $\angle MNS$ $+$ $\angle DMN$ $=$ $\angle CDB$ $=$ $\angle CAD$ $+$ $\angle ACD$ which implies: $\angle MNS$ $=$ $\angle ACD$. To finish notice that: $\angle STK$ $=$ $\angle MNS$ $=$ $\angle ACD$ $=$ $\angle STB$ or $B$, $K$ and $T$ collinear. The reverse proof is very similar.

Let $R=AS\cap (AMT), U=AM\cap TT$ (the tangent to $(AMT) $ at $T$ ) First $A(T,R;M,N)=(T,S,M,\infty _{ST})=-1 \implies TR $ goes through $P$ more Pascal's with $TTRAMM$ yields $P,S,U $ are collinear so $P\in BC\iff U\in BC $ but $UT$ is tangent to $(ATM) \iff \frac{TM^2}{TA^2}=\frac{UM}{UA} (1)$ . In the other hand $ABC$ isoceles $\iff TM=\frac12 TA \iff \frac{TM^2}{TA^2}=\frac14 (2)$ ; but if $AM$ cuts $BC$ at $U'$ by applying Menelaus 's in the triangle $TSB$ with $A-M-U'$ and in the triangle $SMU'$ with $A,T,C$ we get $AM=\frac34 AU'$ or $\frac{U'M}{U'A}=\frac14$ so $U\in BC\iff U=U' \iff \frac{UM}{UA}=\frac{1}{4} $ therefore by $(1)$ and $(2)$ we conclude.

Let $D=AS\cap(ATM)$. Note that \[-1=(ST;M\infty)\stackrel{A}{=}(DT;MN),\]so $T,D,P$ collinear. Thus, we have $P=TD\cap MM$. By Pascal on $TTMMAD$, we have \[TT\cap MA,S,P\]collinear. Thus, $P\in BC$ if and only if $TT\cap MA\in BC$. Letting $E=AM\cap SC$, we then have \[P\in BC\iff ET^2=EM\cdot EA.\]We now use barycentric coordinates on $\triangle ASC$ to finish. Here $A=(1,0,0)$, $S=(0,1,0)$, $C=(0,0,1)$, $a=SC$, $b=AC$, and $c=AS$. It is easy to see that $M=(1/4,1/2,1/4)$, $T=(1/2,0,1/2)$, and $E=(0,2/3,1/3)$. By barycentric distance formula, we have \begin{align*} ET^2 &= -a^2(2/3)(-1/6)-b^2(-1/2)(-1/6)-c^2(2/3)(-1/2)=a^2/9-b^2/12+c^2/3 \\ EM^2 &= -a^2(1/6)(1/12)-b^2(-1/4)(1/12)-c^2(1/6)(-1/4)=-a^2/72+b^2/48+c^2/24 \\ EA^2 &= -a^2(2/3)(1/3)-b^2(-1)(1/3)-c^2(-1)(2/3)=-2a^2/9+b^2/3+2c^2/3. \end{align*}It is not hard to see that \[ET^4-EM^2\cdot EA^2 = \frac{1}{108}(a^2+9c^2)(a^2+c^2-b^2).\]This expression is $0$ if and only if $a^2+c^2=b^2$, or that $\angle ASC=\pi/2$, or that $ABC$ is isosceles with apex $A$. This finishes the proof.

This seems to be a straightforward application of complex numbers. Set $\omega$ to be the unit circle with $A=a$ ; $T=1$ and $M=m$ . So $c=2-a$ . Note that$n=\frac ma$ . So we have , $p= \frac {2m}{a+1}$ . So we need to show that $$ \frac {p-s}{s-c} = \frac {a+1-2ma}{(a+1)(2m+a-3)} \in \mathbb R$$ After taking its conjugate and perform a 10-min calculation, it boils down to having := $$ a^2m+m+6am = 4(am^2+a)$$$$ \implies a+\frac 1a +6 = 4(m +\frac 1m)$$$$ \implies (a-1)(\frac 1a -1) = 4 (m-1)(\frac 1m -1)$$$$ \implies |a-1| =2|m-1| \implies AT = 2MT$$ However this obviously implies $AB=AC$ ,so we're done. $\blacksquare$

why did you post it in this thread make a new thread also re your problem: if p is an odd prime then p^p + 1 is even and larger than 2, so thus not a prime. so p = 2 => p^p + 1 = 5 is the only solution.

Apparently I forgot how to complex, forgot that I could WLOG one of the complex coordinates as $1$ like what @2above did. Complex Solution. Take $\omega$ as the unit circle. Since $AN \equiv AB \parallel TS$, we know that $ANTM$ is a trapezoid, and therefore we can let $A(a), T(b), M(c)$ and $N \left( \frac{bc}{a} \right)$. Now, $T$ is the midpoint of $AC$, $M$ is the midpoint of $ST$, and $S$ is the midpoint of $BC$, which gives us \[ C(2b - a), S(2c - b), B (4c + a - 4b) \]Simple calculation gives us $P \left( \frac{2bc}{a + b} \right)$ as well. Now we want to prove that $|AB| = |AC| \Leftrightarrow P \in ST$. Now, the condition of $|AB| = |AC|$ is equivalent to \[ |4c + a - 4b - a | = |2b - 2a | \Leftrightarrow (2c - 2b ) \left( \frac{2}{c} - \frac{2}{b} \right) = (b - a) \left( \frac{2}{b} - \frac{2}{a} \right) \Leftrightarrow c(a^2 + b^2 + 6ab) = 4a(b^2 + c^2)\]The condition of $P \in ST$ is equivalent to \begin{align*} \frac{\frac{2bc}{a + b} - 2b + a}{3b - a - 2c} &= \frac{\frac{2a}{(a + b)c} - \frac{2}{b} + \frac{1}{a}}{\frac{3}{b} - \frac{1}{a} - \frac{2}{c}} \\ \left( \frac{2bc}{a + b} - 2b + a \right)(3ac - bc - 2ab) &= \left( \frac{2a^2 b}{a + b} - 2ac + bc \right)(3b - a - 2c ) \\ \frac{a - b}{a + b} ( c(a^2 + b^2) + 6abc - 4ab^2 - 4ac^2) &= 0 \end{align*}Since $a \not= b$, the conclusion follows.