i hope i didn't make any mistakes any mistakes, here's my solution: we have $MN = AP \sin{A}$, therefore $T=MN+BP+CP = AP \sin{A}+BP+CP = a ( \frac{AP}{2R}+\frac{\sin{m}+\sin{n}}{\sin{(m+n)}})$ (by the sin law) , where $m,n$ are respectively the angles $\angle{PBC}$ and $\angle{PCB}$. therefore $T$ is minimum when $\frac{AP}{2R}+\frac{\sin{m}+\sin{n}}{\sin{(m+n)}}$ is minimum, but $\frac{\sin{m}+\sin{n}}{\sin{(m+n)}}= \frac{BP+CP}{BC}\geq 1$, with equality when $P \in BC$, therefore $T$ is minimum when $AP$ is minimum, therefore $P$ is minimum when $P \equiv A$. PS: i think there is a particulary case when $P$ is on the perpendicular bisector of $BC$, but that is trivial.

pohoatza wrote: 1) we have $MN = AP \sin{A}$ 2) therefore $T$ is minimum when $\frac{AP}{2R}+\frac{\sin{m}+\sin{n}}{\sin{(m+n)}}$ is minimum, but $\frac{\sin{m}+\sin{n}}{\sin{(m+n)}}= \frac{BP+CP}{BC}\geq 1$, with equality when $P \in BC$, therefore $T$ is minimum when $AP$ is minimum 3) with equality when $P \in BC$, [...] therefore $T$ is minimum when $P \equiv A$. 1) Why? We have $AM = AN = AP$, so by the sines law, $MN = AM \sin{A}$ $\Leftrightarrow$ $MN = \sin{ANM}\cdot \frac{MN}{\sin{A}}\cdot \sin{A}$ $\Leftrightarrow$ $\angle ANM = 90^{\circle}$. But $ABC$ is isosceles! 2) $AP$, $BP$ and $CP$ are not independent! 3) So $T$ is minimum when $P \in BC$ and $P \equiv A$. How come?!?

at 1) you answered yourself $AM=AP$, and $MN=2AM \sin{A}$, therefore $MN=2AP \sin{A}$ as like i said. 2)-3) i don't understand what you mean, i meant $\frac{\sin{m}+\sin{n}}{\sin{(m+n)}}= \frac{BP+CP}{BC}$ is minimum when $P \in BC$, and it's minimum is $1$, therefore now for $T$ to realise his minimun, it's necesary that $P \equiv A \in BC$, because $AP$ is the only thing that varies in the expresion.

pohoatza wrote: at 1) you answered yourself $AM=AP$, and $MN=AM \sin{A}$, therefore $MN=AP \sin{A}$ as like i said. (...) You didn't read after that! There's an '$\Leftrightarrow$' (equivalence) sign there. pohoatza wrote: 2)-3) i don't understand what you mean. pohoatza wrote: [...] $P \equiv A \in BC$[...] In an isosceles triangle, $A \in BC$ ?!? That's a bit weird... I think 3) is very clear. Obviously it's a mistake to state that when $T$ is minimum, both $P \in BC$ and $P \equiv A$ hold.

oh, now i see at the first part, i typed badly the first time, we have $MN = 2AP \sin{A}$ , not simply $AP$., but nothing changes in the rationament., having at the denominator in $E_{1}$ , $R$, which is also constant for your second observations, let's take it otherwise to understand, $E_{2}$ is minimum when it's $1$, and obvious $AP \geq 0$, and because there exists such a $P$ for which the minimum $1$ is realizable, the answer is good, $P \equiv A \in BC$., and if you insist with the fact that $ABC$ is isosceles, that means $P \equiv A$, where $A$ is the midpoint of $BC$, therefore this is an degenarate "isoscel triangle", otherwise the observation that $ABC$ is isoscel is more likely there to separate a case in which the whole perpendicular bisector of $BC$, may contain the point $P$ with the minimum desired.

pohoatza wrote: oh, now i see at the first part, i typed badly the first time, we have $MN = 2AP \sin{A}$ , not simply $AP$., but nothing changes in the rationament., having at the denominator in $E_{1}$ , $R$, which is also constant And also nothing changes in my "rationament" (how did you come to this word? it doesn't even exist - just check here), too! I get from your post that $\sin{ANM}= \frac{1}{2}$, which is usually false! pohoatza wrote: [...] the answer is good, $P \equiv A \in BC$. $ABC$ is a triangle; why should $A \in BC$? $A,B,C$ are fixed points, only $P$ is varying.

in a circle $w$, let's consider a chord $MN$, thus we have $MN = 2R \sin{\widehat{MN}}$, just replace it here., and $A \in BC$ just in the equality case of the triangle ineq., because the triangle inequality is most likely strict, i will stop arquing with you about nothing. yes, when $A$ is on $BC$ is the most particulary case when the "best" minimum is realized. for a "general" minimum, we must simply study the minimum realised in $E_{1}$ , respective $E_{2}$, and we observe that $P \equiv A$ is the "best", hope now it's clear.

That position is not the one where the minimum occurs.. think more!

i hope the following is correct

edit : small writing errors.

PhilAndrew wrote: Let $ABC$ be an isosceles triangle ($AB=AC$) and let $P$ be a mobile point which belongs to its interior. Denote the intersection points $M$, $N$ of the circle $\mathcal{C}(A, AP)$ with the sides $AB$ and $AC$ respectively. Find the position of the point $P$ when the sum $MN+BP+CP$ is minimum. Lemma. For any $x\in\mathcal R$, $a\sin x+b\cos x\le \sqrt{a^{2}+b^{2}}$ and for any $x\in\left[0,\frac{\pi}{2}\right]$, $a>0$, $b>0$ we have $\min\{a,b\}\le a\sin x+b\cos x\le\sqrt{a^{2}+b^{2}}$. Particularly, for any $x>0\ ,\ x\sin\frac{A}{2}+\cos \frac{A}{2}\le \sqrt{1+x^{2}}$, i.e. $\boxed{\sqrt{1+x^{2}}-x\sin\frac{A}{2}\ge \cos\frac{A}{2}}$ with equality iff $x=\tan\frac{A}{2}$. Proof of the proposed problem. Denote the middlepoint $D$ of the side $[BC]$. I"ll suppose w.l.o.g. that $DB=DC=1$, $AD=a>0$. I n the minimum position of the mobile point $P$ we have obviously $P\in (AD)$ (prove easily !). Denote $PD=x\in (0,a)$. Thus, $AP=a-x$, $MN=2AP\sin \frac{A}{2}=2(a-x)\sin\frac{A}{2}$, $PB=PC=\sqrt{1+x^{2}}$, $\tan\widehat{PBC}=x$. Therefore, $MN+PB+PC$ is minimum $\Longleftrightarrow$ $(a-x)\sin\frac{A}{2}+\sqrt{1+x^{2}}$ is minimum $\Longleftrightarrow$ $\sqrt{1+x^{2}}-x\sin\frac{A}{2}$ is minimum $\Longleftrightarrow$ (from the above lemma) $x=\tan \frac{A}{2}$ $\Longleftrightarrow$ $\tan\widehat{PBC}=\tan\frac{A}{2}$ $\Longleftrightarrow$ $m(\widehat{PBC})=\frac{A}{2}$ $\Longleftrightarrow$ $PB\perp AC$ $\Longleftrightarrow$ the point $P$ is the orthocenter of $\triangle ABC$.