Two tangents meet on BC

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#1 h Feb 21, 2019, 11:13 AM • 4 Y

Y by FISHMJ25, itslumi, Adventure10, Mango247

Let $\triangle ABC$ be a triangle, and let $S$ and $T$ be the midpoints of the sides $BC$ and $CA$ , respectively. Suppose $M$ is the midpoint of the segment $ST$ and the circle $\omega$ through $A, M$ and $T$ meets the line $AB$ again at $N$ . The tangents of $\omega$ at $M$ and $N$ meet at $P$ . Prove that $P$ lies on $BC$ if and only if the triangle $ABC$ is isosceles with apex at $A$ .

Proposed by Reza Kumara, Indonesia

This post has been edited 1 time. Last edited by MNJ2357, Feb 21, 2019, 11:24 AM

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#2 h Feb 21, 2019, 2:39 PM • 3 Y

Y by CyclicISLscelesTrapezoid, Adventure10, soryn

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.

This post has been edited 1 time. Last edited by aqwxderf, Feb 21, 2019, 2:40 PM
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PROF65

#3 h Feb 21, 2019, 7:33 PM • 1 Y

Y by Adventure10

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.

This post has been edited 1 time. Last edited by PROF65, Feb 21, 2019, 7:37 PM

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yayups

#4 h Mar 27, 2020, 3:40 AM

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https://lh3.googleusercontent.com/-5WjJtAcS5KM/XnsMV0QHi8I/AAAAAAAAFoQ/8V14DyqqBBod0b_JAy0xT6ogecQhAQbMgCK8BGAsYHg/s0/2020-03-25.png

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.

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#5 h Jul 12, 2020, 10:04 AM • 2 Y

Y by SenatorPauline, AlastorMoody

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. $\blacksquare$

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khina

#7 h Jul 12, 2020, 5:18 PM • 4 Y

Y by stroller, Inconsistent, Aryan-23, ChineseHacker

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.

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#9 h Nov 29, 2020, 1:42 PM • 3 Y

Y by GorgonMathDota, Aritra12, Aryan-23

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.

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#10 h May 11, 2021, 12:22 AM

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