Obviously, triangle DEF is the pedal triangle of isodynamic point. So, these line are concurrent at Fermat's point (which point satisfying $ \angle APB=\angle BPC=\angle CPA=120^o$)

mr.danh wrote: Obviously, triangle DEF is the pedal triangle of isodynamic point. What is this point, and why?

Let $ \triangle A'B'C'$ be an arbitrary circumscribed equilateral triangle in $ \triangle ABC.$ Circumcircles $ (X),$ $(Y),$ $(Z)$ of $ \triangle A'CB,$ $\triangle B'AC,$ $\triangle C'BA$ concur at its Miquel point WRT $ \triangle A'B'C',$ i.e. the 1st Fermat point $ F$ of $ \triangle ABC.$ Lines $ FA,FB,FC$ are pairwise radical axes of the circles $ (X),(Y),(Z).$ Then $ \triangle XYZ$ is equilateral an its sidelines are the perpendicular bisectors of $ FA,FB,FC.$ Let $ M,N$ be the projections of $ Z,Y$ on $ B'C'$ and $ Z'$ the projection of $ Y$ on $ ZM.$ If $ Z \not\equiv Z',$ then $ YZ \ge YZ' = MN = \frac {_1}{^2}B'C'$ and we get similar expressions cyclically. Thus, $ \triangle A'B'C'$ attains its maximum area (perimeter) if and only if $ \triangle A'B'C'$ is centrally similar to $ \triangle XYZ,$ that is, if $ \triangle A'B'C'$ is centrally similar to the pedal triangle $\triangle F_1F_2F_3$ of the 1st Isodynamic point $ F'$ of $\triangle ABC.$ For each inscribed equilateral triangle $ \triangle DEF$ in $ \triangle ABC$ there exists another equilateral triangle $ \triangle D'E'F'$ circumscribed in $ \triangle ABC$ and centrally similar to $ \triangle DEF.$ By Gergonne-Ann theorem we have $ [ABC]^2 = [DEF] \cdot [D'E'F']$ $\Longrightarrow$ $ [DEF]$ attains its minimum iff $ [D'E'F']$ attains its maximum, in other words, if $ \triangle D'E'F'$ is centrally similar to the pedal triangle of $ F'. $ Then $ \triangle DEF$ is identical with $ \triangle F_1F_2F_3.$

By Miquel's Theorem, the circumcircles of AQR, BPR, CPQ intersect at a point M. Claim: $ \angle MPC = \angle MQA = \angle MRB = 90^{\circ}$ Sketch of proof: Suppose otherwise. Let P', Q'. R' be feet of perpendiculars from M to BC, CA, AB. From $ \Delta MPP' \cong \Delta MQQ' \cong \Delta MRR'$ we get $ \Delta P'Q'R' \cong \Delta PQR$, with $ \Delta P'Q'R'$ smaller, a contradiction. So the perpendiculars through P, Q, R to BC, CA, and AB intersect. The triangles ABC, PQR are orthologic, hence the perpendiculars through A, B, C to QR, RP, PQ intersect (This can be proved with Carnot's Theorem and Pythagorean Theorem) P.S. It seems that the proof works even if "PQR is equilateral" is replaced by "PQR is similar to a given triangle"

Place $a,b,c$ on the unit circle. WLOG $\triangle{PQR}$ (which has the same orientation as $\triangle{ABC}$) is oriented such that $p+q\omega+r\omega^2=0$, where $\omega=e^{2\pi i/3}$ satisfies $\omega^2+\omega+1=0$. Then $|p-q|=|q-r|=|r-p|$ is the side length of $\triangle{PQR}$. Taking the conjugate of this equation and using $P\in BC$, etc. \[\frac{b+c-p}{bc}+\frac{c+a-q}{ca}\omega^2+\frac{a+b-r}{ab}\omega=0,\]so plugging in $p=-q\omega-r\omega^2$, we have \[q-r=\frac{bc+ca\omega+ab\omega^2+r(a+b\omega^2+c\omega)}{\omega(a-b\omega)}.\]Let $r=ta+(1-t)b$ for some $t\in\mathbb{R}$. Then \begin{align*} |q-r| &= \left|\frac{bc+ca\omega+ab\omega^2+t(a-b)(a+b\omega^2+c\omega)+b(a+b\omega^2+c\omega)}{\omega(a-b\omega)}\right| \\ &= \left|\frac{(c-b)(a-b\omega)\omega+t(a-b)(a+b\omega^2+c\omega)}{\omega(a-b\omega)}\right|. \end{align*}Thus to minimize $|q-r|$ (which clearly also minimizes $|r-p|=|p-q|=|q-r|$!), we must have \begin{align*} t=\Re\left(\frac{\omega(b-c)(a-b\omega)}{(a-b)(a+b\omega^2+c\omega)}\right)\implies r &= b+(a-b)\Re\left(\frac{\omega(b-c)(a-b\omega)}{(a-b)(a+b\omega^2+c\omega)}\right) \\ &= b+\frac{\omega(b-c)(a-b\omega)(bc-a^2)}{2(a+b\omega^2+c\omega)(bc+ca\omega+ab\omega^2)}. \end{align*}Thus \begin{align*} q-r &= \frac{bc+ca\omega+ab\omega^2+r(a+b\omega^2+c\omega)}{\omega(a-b\omega)} \\ &= \frac{bc+ca\omega+ab\omega^2}{\omega(a-b\omega)} + \frac{b(a+b\omega^2+c\omega)}{\omega(a-b\omega)} + \frac{(b-c)(bc-a^2)}{2(bc+ca\omega+ab\omega^2)} \\ &= (c-b)+\frac{(b-c)(bc-a^2)}{2(bc+ca\omega+ab\omega^2)} \end{align*}and if $AX\perp QR$ and $BX\perp RP$, then \begin{align*} -\frac{x-a}{\overline{x}-\frac{1}{a}} = \frac{q-r}{\overline{q-r}} &= \frac{-(b-c)+\frac{(b-c)(bc-a^2)}{2(bc+ca\omega+ab\omega^2)}}{\frac{b-c}{bc}+\frac{(b-c)(bc-a^2)}{2abc(a+b\omega^2+c\omega)}} \\ &= \frac{abc(a+b\omega^2+c\omega)}{bc+ca\omega+ab\omega^2}\frac{bc-a^2-2(bc+ca\omega+ab\omega^2)}{bc-a^2+2a(a+b\omega^2+c\omega)} \\ &= -\frac{abc(a+b\omega^2+c\omega)}{bc+ca\omega+ab\omega^2} \end{align*}and similarly, \[\frac{x-b}{\overline{x}-\frac{1}{b}} = \frac{bca(b+c\omega^2+a\omega)}{ca+ab\omega+bc\omega^2} = \omega^2\frac{abc(a+b\omega^2+c\omega)}{bc+ca\omega+ab\omega^2}.\]Solving for $x$ (or $\overline{x}$), we get an equation invariant under the cyclic rotation $a\mapsto b\mapsto c\mapsto a$, as desired. Edit: $\text{\LaTeX}$ error

mr.danh wrote: Obviously, triangle DEF is the pedal triangle of isodynamic point. So, these line are concurrent at Fermat's point (which point satisfying $ \angle APB=\angle BPC=\angle CPA=120^o$) Page 6 of the following document proves this. (Theorem 2.5) http://awesomemath.org/wp-content/uploads/reflections/2010_6/Isodynamic_moon_c.pdf

Dear Mathlinkers, accepting that the point in question is the first isodynamic point of ABC, the first Napoleon triangle is homothetic to PQR see : http://perso.orange.fr/jl.ayme vol. 2 the Lester's circle p. 14 This Napoleon triangle being orthologic to ABC, we are done Sincerely Jean-Louis

Generalization: Fix a triangle $\triangle A_1B_1C_1.$ Let $\triangle ABC$ be a triangle and let $A_2, B_2, C_2$ lie on $BC, CA, AB$ so that $\triangle A_2B_2C_2 \sim \triangle A_1B_1C_1$, and $[A_2B_2C_2]$ is minimal among all such triangles. Then $\triangle A_2B_2C_2$ and $\triangle ABC$ are orthologic. Proof: Let $X, Y, Z$ vary on $BC, CA, AB$, respectively, so that $\triangle XYZ \sim \triangle A_1B_1C_1.$ We claim that as $X, Y, Z$ vary, the Miquel Point $M$ of $\{X, Y, Z\}$ WRT $\triangle ABC$ is fixed. Indeed, from cyclic quadrilaterals, we deduce that \[\measuredangle ZXY = \measuredangle ZXM + \measuredangle MXY = \measuredangle ZBM + \measuredangle MCY = -\measuredangle BMC - \measuredangle CAB,\]where the last step follows from examining quadrilateral $MBAC.$ Thus, since $\measuredangle ZXY$ and $\measuredangle CAB$ are fixed, it follows that $\measuredangle BMC$ is fixed as well. Similar arguments show that $\measuredangle CMA$ and $\measuredangle AMB$ are fixed, and it follows that $M$ is fixed. Now, if $R_A$ is the circumradius of $\triangle AYZ$, the Law of Sines yields $YZ = 2R_A\sin\angle YAZ.$ But since $\angle YAZ$ is fixed, it follows that $YZ$ is minimal when $R_A$ is minimal. Because $\overline{AM}$ is a fixed chord on $\odot(AYZ)$, we see that $R_A$ is minimized when $\overline{AM}$ is a diameter of $\odot(AYZ)$, in which case $\angle AYM = \angle AZM = 90^{\circ}.$ Thus, $Y, Z$ are just the projections of $M$ onto $AC, AB$, respectively. Meanwhile, $\measuredangle MXB = \measuredangle MYB = 90^{\circ}$, implying that $X$ is the projection of $M$ onto $BC.$ Hence, $\triangle A_2B_2C_2$ is the pedal triangle of $M$ WRT $\triangle ABC$, and it follows that $\triangle A_2B_2C_2$ and $\triangle ABC$ are orthologic, as desired. $\square$ Remark: It follows from a well-known property of isogonal conjugates that the perpendiculars from $A, B, C$ to $B_2C_2, C_2A_2, A_2B_2$, respectively, concur at the isogonal conjugate of $M$ WRT $\triangle ABC.$

Let $M$ be the Miquel point of triangle $XYZ$ wrt triangle $ABC$. We claim that $M$ is in fact the first isodynamic point. Indeed, $\angle BMC=\angle BMX+\angle CMX=\angle BZX+\angle CYX=\angle BAC+60^{\circ}$ and so it follows that the claim is true. Suppose that $UVW$ is the pedal triangle of $M$ wrt $ABC$. Clearly, $XY/UV=MX/MU \ge 1$ and similar relations hold (using spiral similarity about $M$). Thus, $XY+YZ+ZX \ge UV+VW+WU$. Now, it is clear that $UVW$ is the triangle with minimal perimeter satisfying our conditions, and perpendiculars from $A,B,C$ to $VW,WU,UV$ are concurrent at point $F$, the first Fermat point of triangle $ABC$. Comment In particular, any three points $X,Y,Z$ on $BC,CA,AB$ which move in a way that $\triangle XYZ$ has a fixed shape satisfy the condition that if area of $XYZ$ is minimal then $ABC,XYZ$ are orthologic. The proof is similar, in particular dynamic ideas motivate both

Bump because AIME

Miquel's Theorem implies that $(AQR), (BRP), (CPQ)$ meet at some point $X$. The Extended LoS yields $$\frac{XB}{XC} = \frac{\sin{XPB} \cdot \frac{RP}{\sin {RBP}}}{\sin{XPC} \cdot \frac{PQ}{\sin{PCQ}}} = \frac{\sin{C}}{\sin{B}}.$$Analogously, we find $\frac{XA}{XB} = \frac{\sin{B}}{\sin{A}}$, so cross multiplying gives $$XA \cdot \sin{A} = XB \cdot \sin{B} = XC \cdot \sin{C}.$$Now, let $DEF$ be the pedal triangle of $X$. Because $BFXD$ and $CDXE$ are cyclic with diameters $XB$ and $XC$ respectively, we have $$FD = XB \cdot \sin{B} = XC \cdot \sin{C} = DE$$from the Extended LoS. Similarly, we deduce $EF = FD$, so $DEF$ is indeed equilateral. Now, FTSOC, assume $PQR$ and $DEF$ don't coincide. In addition, define $R_A$ as the radius of $(AQXR)$. Then, since $XA$ is not a diameter of $(AQXR)$, $$QR = 2R_A \cdot \sin QAR > XA \cdot \sin A = EF$$which contradicts the condition that $PQR$ is minimal. It follows that $PQR$ coincides with $DEF$, so orthology finishes. $\blacksquare$ Remarks: We can use Spiral Similarities to prove $DEF$ is equilateral. If $DEF$ coincides with $PQR$, then we're done. If it doesn't, then the Spiral Center Lemma implies $XDE \overset{+}{\sim} XPQ$ and $XEF \overset{+}{\sim} XQR$, so $$XDP \overset{+}{\sim} XEQ \overset{+}{\sim} XFR$$which means $XDEF \overset{+}{\sim} XPQR$, as required.