336. Some problems of the analytical geometry.

by Virgil Nicula, Jan 31, 2012, 12:20 PM

PP1. Let $f(x)=ax^2+bx+c\ ,\ x\in\mathbb R$ . Prove that exists $F$ (focus) and $d$ (directrix) so that $d\parallel Ox$ and $(\forall ) M\in\mathrm G_f$ - graph of $f$ , we have $MF=\delta_d(M)$ - distance of $M$ to $d$ .

Proof. Let $F(m,n)$ , equation $y-p=0$ of $d$ and $M(x,y)\in\mathbb G_f$ , where $y=ax^2+bx+c$ . Thus, $MF=\delta_d(M)$ $(\forall ) M\in\mathbb G_f\iff$ $(x-m)^2+(y-n)^2=(y-p)^2$

$(\forall ) x\in\mathbb R\iff$ $x^2-2mx+m^2-2ny+n^2=-2py+p^2\iff$ $x^2-2mx+2(p-n)\left(ax^2+bx+c\right)+m^2+n^2-p^2=0\iff$

$\left\{\begin{array}{c} 2a(p-n)+1=0\\\\ b(n-p)+m=0\\\\ 2(p-n)c+m^2+n^2-p^2=0\end{array}\right\|\iff$ $\left\{\begin{array}{c} m=\frac {-b}{2a}\\\\ n-p=\frac {1}{2a}\\\\ \Delta +2a(n+p)=0\end{array}\right\|\iff$ $\left\{\begin{array}{c} m=\frac {-b}{2a}\\\\ n=\frac {1-\Delta}{4a}\\\\ p=-\frac {1+\Delta}{4a}\end{array}\right\|$ . Thus, $F\left(\frac{-b}{2a},\frac {1-\Delta}{4a}\right)$ - the focus of parabola $\mathrm G_f$ and which belongs

to the symmetry axis of $\mathrm G_f$ . Equation of the directrix is $y=\frac {-1-\Delta}{4a}$ . So the vertex of $\mathrm G_f$ is $V\left(\frac {-b}{2a}, \frac {-\Delta}{4a}\right)$ and if $P=\mathrm{pr}_d(F)$ , then the vertex $V$ is the midpoint of $[FP]$ .

PP2. Consider the line $d\ :\ 2y=x-1$ and the parabola $p\ :\ y^2=4x$ . Denote

$\{A,B\}=d\cap p$ and the points $C\in p$ so that $CA\perp CB$ . Find the points $C$ .

Proof. Exist $\{a,b,c\}\subset \mathbb R^*$ so that $A\left(a^2,2a\right)$ , $B\left(b^2,2b\right)$ and $C\left(c^2,2c\right)$ , where $a\ne b$ . The slope of the line $d\equiv AB$ is $\frac 12=\frac {2(a-b)}{a^2-b^2}\iff$

$\boxed{a+b=4}\ (1)$ . Observe that the focus of $p$ is $F(1,0)\in d$ , i.e. $F\in AB\iff$ $\left|\begin{array}{ccc} 1 & 0 & 1\\\\ a^2 & 2a & 1\\\\ b^2 & 2b & 1\end{array}\right|=0\iff$ $\left|\begin{array}{ccc} 1 & 0 & 1\\\\ a^2 & a & 1\\\\ b^2 & b & 1\end{array}\right|=0$ .

Thus, $a+a^2b=b+ab^2$ $\iff$ $\boxed{ab=-1}\ (2)$ . Therefore, $CA\perp CB\iff$ $s_{CA}\cdot s_{CB}=-1\iff$

$\frac {2(a-c)}{a^2-c^2}\cdot\frac {2(b-c)}{b^2-c^2}=-1\iff$ $(a+c)(a+b)=-4\iff$ $c^2+(a+b) c+ab+4=0\stackrel{(1)\wedge(2)}{\ \iff\ }$

$c^2+4c+3=0\iff$ $c\in\{-1,-3\}\iff$ $\boxed{C_1(1,-2)\ \wedge\ C_2(9,-6)}$ .

Generalization. Consider the parabola $\mathbb P\ :\ y^2=2px\ ,\ p>0$ with the focus $F$ and the line $d$ with the slope $\lambda$

such that $F\in d$ . Denote $\{A,B\}=P\cap d$ and the points $C\in p$ so that $CA\perp CB$ . Find the points $C$ .

Proof. Exist $\{a,b,c\}\subset \mathbb R^*$ so that $A\left(\frac p2a^2,pa\right)$ , $B\left(\frac p2 b^2,pb\right)$ and $C\left(\frac p2c^2,pc\right)$ , where $a\ne b$ . The slope of the line $d\equiv AB$ is $\lambda =\frac {p(a-b)}{\frac p2\left(a^2-b^2\right)}\iff$

$\boxed{a+b=\frac {2}{\lambda}}\ (1)$ . Observe that the focus of $P$ is $F\left(\frac p2,0\right)\in d$ , i.e. $F\in AB\iff$ $\left|\begin{array}{ccc} \frac p2 & 0 & 1\\\\ \frac p2 a^2 & pa & 1\\\\ \frac p2 b^2 & pb & 1\end{array}\right|=0\iff$ $\left|\begin{array}{ccc} 1 & 0 & 1\\\\ a^2 & a & 1\\\\ b^2 & b & 1\end{array}\right|=0$ .

Thus, $a+a^2b=b+ab^2$ $\iff$ $\boxed{ab=-1}\ (2)$ . Therefore, $CA\perp CB\iff$ $s_{CA}\cdot s_{CB}=-1\iff$ $\frac {p(a-c)}{\frac p2\left(a^2-c^2\right)}\cdot\frac {p(b-c)}{\frac p2\left(b^2-c^2\right)}=-1\iff$

$(a+c)(a+b)+4=0\iff$ $c^2+(a+b) c+ab+4=0\stackrel{(1)\wedge(2)}{\ \iff\ }$ $c^2+\frac {2}{\lambda}c+3=0\iff$ $\boxed{\lambda c^2+2c+3\lambda =0}\ (3)$ .

Observe that the reduced discriminant $\Delta' (\lambda )=1-3\lambda^2$ . Distinguish three cases :

$1\blacktriangleright\ \lambda^2 >\frac 13\iff$ $|\lambda |>\frac {\sqrt 3}{3}\iff$ $\lambda\in\left(-\infty,-\frac {\sqrt 3}{3}\right)\cup\left(\frac {\sqrt 3}{3},\infty\right)\implies$ $c\in\emptyset\iff C\in \emptyset\ .$

$2\blacktriangleright\ \lambda^2=\frac 13\iff$ $|\lambda |=\frac {\sqrt 3}{3}\iff$ $\left\{\begin{array}{ccccc} \lambda=-\frac {\sqrt 3}{3} & \implies & c=\sqrt 3 & \iff & C\left(\frac {3p}{2},p\sqrt 3\right)\\\\ \lambda=\frac {\sqrt 3}{3} & \implies & c=-\sqrt 3 & \iff & C\left(\frac {3p}{2},-p\sqrt 3\right)\end{array}\right\|\ .$

In this case the measure of the angle between the line $d$ and the $x$ -axis is $30^{\circ}$ or $150^{\circ}$ .

$3\blacktriangleright\ \lambda^2<\frac 13\iff$ $|\lambda |<\frac {\sqrt 3}{3}\iff$ $\lambda\in\left(-\frac {\sqrt 3}{3},\frac {\sqrt 3}{3}\right)\iff$ $c_{1,2}=\frac {-1\pm\sqrt{1-3\lambda^2}}{\lambda}\iff$

$C_{1,2}\left[\frac {-p\left(3\lambda^2-2\pm 2\sqrt{1-3\lambda^2}\right)}{2\lambda^2},\frac {p\left(-1\pm\sqrt{1-3\lambda^2}\right)}{\lambda}\right]\ .$

PP3. Consider a variable parabola $\mathbb P_a$ with the equation $y^2=4ax$ , where $a\in \mathbb R^*$ is a parameter and the curve $\mathbb C$ with the equation $y^2+x-y-2=0$ .

Denote $\{M,N\}=\mathbb P_a\cap \mathbb C$ . Ascertain the geometrical locus of the point of the intersection $L\in MM\cap NN$ of the tangents to $\mathbb P$ at the points $M$ and $N$ .

Proof. $\left\{\begin{array}{ccc} \mathbb P_a & : & y^2=4ax\\\\ \mathbb C & : & y^2+x-y-2=0\end{array}\right\|\iff$ $\left\{\begin{array}{ccc} \mathbb P_a & : & y^2=4ax\\\\ \mathbb d & : & y=(4a+1)x-2\end{array}\right\|\ .$ Exist $\{u,v\}\subset \mathbb R^*$ so that $M\left(au^2,2au\right)$ and $N\left(av^2,2av\right)$ .

Let $L(x,y)$ . Thus, $\left\{\begin{array}{ccc} L\in MM & : & y\cdot 2au=2a\left(x+au^2\right)\\\\ L\in NN & : & y\cdot 2av=2a\left(x+av^2\right)\end{array}\right\|\iff$ $\left\{\begin{array}{c} uy=x+au^2\\\\ vy=x+av^2\end{array}\right\|\iff$ $\left\{\begin{array}{c} uv=\frac xa\\\\ u+v=\frac ya\end{array}\right\|\ (*)$ . Therefore, $d=MN\iff$

$(0,-2)\in MN\ \wedge\ s_d=s_{MN}\iff$ $\left|\begin{array}{ccc} 0 & -2 & 1\\\\ au^2 & 2au & 1\\\\ av^2 & 2av & 1\end{array}\right|$ $=0\ \wedge\ \frac {2a(u-v)}{a\left(u^2-v^2\right)}=4a+1\iff$ $\left|\begin{array}{ccc} 0 & -1 & 1\\\\ u^2 & au & 1\\\\ v^2 & av & 1\end{array}\right|$ $=0\ \wedge\ u+v=\frac {2}{4a+1}\iff$

$auv+(u+v)=0\ \wedge\ u+v=\frac {2}{4a+1}\stackrel{(*)}{\iff}$ $\left\{\begin{array}{c} ax+y=0\\\\ (4a+1)y=2a\end{array}\right\|\implies$ $\boxed{4y=x+2}$ . By $s_d$ denoted the slope of the line $d$ .

PP4. Let a parabola $\mathbb P\ :\ y^2=2px\ ,\ p>0$ and a mobile $M\in\mathbb P$ . Let $\{N,P\}\subset [Ox$ - symmetry axis so that $\left\{\begin{array}{c} MP\perp OP\\\ MN\perp OM\end{array}\right\|$ . Prove that $NP$ is constant.

Proof. Exist $m\in\mathbb R^*$ such that $M\left(2pm^2,2pm\right)$ . Therefore, $\boxed{OP=2pm^2}$ and for $N(n,0)$ have $MN\perp MO\iff$ $s_{MN}\cdot s_{MO}=-1\iff$

$\frac {2pm}{2pm^2-n}\cdot \frac 1m=-1\iff$ $n=2p\left(m^2+1\right)\iff$ $\boxed{ON=2p\left(m^2+1\right)}$ . Hence $NP=ON-OP\iff$ $\boxed{NP=2p}$ (constant).

Remark. By $s_d$ denoted the slope of the line $d$ . For $p:=8$ the focus is $F(4,0)$ and obtain the proposed problem from here.

PP5. Show that a parabolic reflector works, as in, any line coming in will be reflected to the focus.

Proof. Let $\mathbb P$ be the parabola with the equation $y^2=4ax$ , i.e. its focus is $F(a,0)$ and its directrix has the equation $y=-a$ . For a mobile point $M\in\mathbb P$ exists

$m\in\mathbb R^*$ so that $M\left(am^2,2am\right)$ . Suppose w.l.o.g. that $m>0$ . The equation of the tangent $MM$ in the point $M$ to the parabola $\mathbb P$ is $2amy=2a(x+2am)$ ,

i.e. $my=x+2am$ with the slope $\boxed{s_{MM}=\frac 1m}$ .Observe that the slope of the line $FM$ is $\boxed{s_{FM}=\frac {2m}{m^2-1}}$ . Denote $L\in MM\cap X$ -axis , $\phi =m(\widehat{MLF})$

and $K\in FL$ so that $F\in LK$ . Thus, $\tan\phi=s_{mm}=\frac 1m$ and $\tan\widehat{MFK}=s_{FM}=\frac {2m}{m^2-1}=$ $\frac {2\cdot\frac 1m}{1-\left(\frac 1m\right)^2}=$ $\frac {2\tan\phi}{1-\tan^2\phi}=\tan 2\phi\implies$

$m(\widehat{MFK})=2\phi\implies m(\widehat{MLF})=m(\widehat{LMF})=\phi$ , i.e. the property "parabolic reflector" from the conclusion of the proposed problem.

PP6. Find $m\equiv \min_{x^2+4xy+3y^2=4}\left(x^2+y^2\right)$ .

Proof 1. Observe that $x^2+4xy+3y^2=4\iff (x+y)(x+3y)=4$ . Denote $\left\{\begin{array}{c} x+y=u\\\ x+3y=v\end{array}\right\|$ . Thus, $\left\{\begin{array}{c} x=\frac {3u-v}{2}\\\\ y=\frac {v-u}{2}\end{array}\right\|$ and our

problem becomes $m=\min_{uv=4}\left(\frac {5u^2+v^2}{2}-8\right)$ . Since $5u^2+v^2\ge 2|uv|\sqrt 5$ , i.e. $5u^2+v^2\ge 8\sqrt 5$ obtain that $\boxed{m=4\left(\sqrt 5-2\right)}$ .

Proof 2. Let $x = r \cos \theta$ , $y = r \sin \theta$ . The extremum of $x^2 + y^2$ , i.e. the extremum of $r^2$ subject to the constraint $x^2 + 3y^2 + 4xy=4$ , i.e. $r^2 (2 \sin 2\theta - \cos 2\theta + 2)=4$ .

To simplify this last expression, let $\tan \alpha = 2$ so that $\cos \alpha = 1/\sqrt{5}$ , giving $4 = r^2\left( \frac{\sin\alpha \sin 2 \theta - \cos\alpha\cos 2\theta}{\cos\alpha} + 2 \right)=$ $r^2\left[2 - \sqrt{5} \cos(\alpha + 2\theta)\right]$ . Thus, a minimum is attained when

$\cos(\alpha + 2\theta) = -1$ or $r^2 = \frac{4}{2+\sqrt{5}} = 4(\sqrt{5} - 2)$ . There is no maximum because as $\theta$ approaches a value such that $\cos(\alpha + 2\theta) \to \frac{2}{\sqrt{5}}$ , $r^2 \to +\infty$ .

Proof 3 (thugzmath10). Let $x^2+y^2=p$ (extremum) and I"ll solve the homogenous system $\left\{\begin{array}{c} x^2+4xy+3y^2=4\\\\ x^2+y^2=p\end{array}\right\|$ . With the substitution $\frac xy=t$ obtain that $\frac {t^2+4t+3}{t^2+1}=\frac 4p\iff$

$(4-p)t^2-4pt+(4-3p)=0$ . Since $x,y$ are real, then and $t\in\mathrm R$ , i.e. the discriminant of the quadratic equation at $t$ is at least $0$ , i.e. $16p^2-4(3p^2-16p+16)\geq 0\iff$

$p^2+16p-16\geq 0\rightarrow (p+8)^2\geq 80\iff$ $p\leq -4\left(2+\sqrt{5}\right)\ ,\ p\geq 4\left(\sqrt{5}-2\right)$ . It's easy to check that the minimum is $4\left(\sqrt{5}-2\right)$ and no maximum occurs.

PP7. Consider three parabolas $\left\{\begin{array}{cc} \mathrm P_a\ : & y=ax^2\\\ \mathrm P_b\ : & y=bx^2\\\ \mathrm P_c\ : & y=cx^2\end{array}\right\|$ , where $0<c<b<a$ and $b^2=ac$ . Prove that for any $\{M,N\}\subset \mathrm P_b$

such that $MN$ is tangent to the parabola $\mathrm P_a$ , the intersection of the tangents in the points $M\ ,\ N$ to the parabola $\mathrm P_b$ belongs to the parabola $\mathrm P_c$ .

Proof. Denote $\left\{\begin{array}{c} M\left(m,bm^2\right)\\\\ N\left(n,bn^2\right)\end{array}\right\|$ and let $T\left(t,at^2\right)$ be a mobile point of $\mathrm P_a$ . The equation of the tangent in the point $T$ to the parabola $\mathrm P_a$ is $y=2atx-at^2$ .Thus,

$\{M,N\}=TT\cap\mathrm P_b$ and $\left\{\begin{array}{c} y=2atx-at^2\\\\ y=bx^2\end{array}\right\|$ $\implies$ $bx^2-2atx+at^2=0$ $\begin{array}{c} \nearrow\ m\\\\ \searrow\ n\end{array}$ . Thus, $\left\{\begin{array}{c} m+n=\frac {2at}{b}\\\\ mn=\frac {at^2}{b}\end{array}\right\|\ (*)$ . Denote $L\in MM\cap NN$ .

Observe that $\left\{\begin{array}{cc} MM\: & y=2bmx-bm^2\\\\ NN\ : & y=2bnx-bn^2\end{array}\right\|\implies$ $\left\{\begin{array}{c} m+n=2x\\\ bmn=y\end{array}\right\|$ , i.e. $L\left(\frac {m+n}{2},bmn\right)$ . Using the relations $(*)$ obtain that $L\left(\frac {at}{b},at^2\right)$ ,

i.e. $\left\{\begin{array}{c} x_L=\frac {at}{b}\\\\ y_L=at^2\end{array}\right\|$ $\implies$ $y_L=a\cdot\left(\frac {bx_L}{a}\right)^2=$ $\frac {b^2}{a}\cdot x_L^2=$ $cx_L^2\implies$ $y_L=cx_L^2$ , i.e. $L\in\mathrm P_c$ . Very nice problem !

PP8. Let $P$ be any point on the parabola $y^2=4ax$ between its vertex and the positive end of the latus rectum (the chord of the parabola passing

through the focus and parallel to the directrix). Let $M$ be the foot of perpendicular from the focus $F$ on the tangent at $P$ . Find the area of $\triangle PMF$ .

Proof. Denote $R$ of the directrix for which $PR\parallel Ox$ . Then $\triangle SPR$ is $P$ -isosceles, the point $M\in Oy$ and it is the midpoint of $[FR]$ . Thus, $F(a,0)$ and for $P\left(\frac {m^2}{4a},m\right)$ , where

$m\in [0,2a]$ have $M\left(0,\frac m2\right)\ ,\ R\left(-a,\frac {m^2}{4a}\right)$ . In conclusion, $[PMF]=\frac 12\cdot [PRF]=$ $\frac 14\cdot PR\cdot y_P=$ $\frac 14\cdot \left(\frac {m^2}{4a}+a\right)m$ $\implies$ $\boxed{[PMF]=\frac {m\left(m^2+4a^2\right)}{16a}}$ .

PP9. Find the equation of tangents to the hyperbola $xy = 1$ which goes through $P(-1,1)$ .

Proof 1. Let $y=mx+n$ be required equation of tangent $t$ . Thus, $P\in t\iff \boxed{n=m+1}\ (1)$ and $\left\{\begin{array}{c} xy=1\\\ y=mx+n\end{array}\right\|$ $\implies$ $x(mx+n)=1$ has a double root, i.e.

$mx^2+nx-1=0$ has $\Delta =0$ , i.e. $\boxed{n^2+4m=0}\ (2)$ . From the relations $(1)$ and $(2)$ obtain that $\left\{\begin{array}{c} m=-3\pm 2\sqrt 2\\\ n=-2\pm 2\sqrt 2\end{array}\right\|$ $\iff$ $\boxed{y=\left(-3\pm2\sqrt 2\right)x-2\pm 2\sqrt 2}\stackrel{\mathrm{(joke)}}{\iff}$

$y+3x+2=\pm2\sqrt 2(x+1)\iff$ $(y+3x+2)^2=8(x+1)^2\iff$ $x^2+y^2+6xy-4x+4y-4=0\iff$ $\boxed{(x+y)^2=4(x+1)(y-1)}$ .

Proof 2 (with the theory). Let $P(-1,1)\not\in\mathbb H$ , where the equation of of the hiperbola $\mathbb H$ is $xy-1=0$ .

$\left[\frac 12\cdot (x_0y+xy_0)-1\right]^2=(xy-1)(x_0y_0-1)\iff$ $\left[\frac 12\cdot (-y+x)-1\right]^2=(xy-1)(-2)\iff$

$(y-x+2)^2+8(xy-1)=0\iff$ $x^2+y^2+6xy-4-4x+4y=0\iff$ $(x+y)^2+4(x+1)(y-1)=0$ .

Proof 3 (with a property). Suppose that the tangent $t$ in the point $T(x,y)\in\mathbb H$ cut the axis $Ox$ in the point $A$ and the axis $Oy$ in the point $B$ . From the well-known

property $TA\stackrel{(*)}{=}TB$ obtain that $A(2x,0)$ , $B\left(0,\frac 2x\right)$ and $P\in AB\iff$ $\left|\begin{array}{ccc} -1 & 1 & 1\\\\ 2x & 0 & 1\\\\ 0 & \frac 2x & 1\end{array}\right|=0$ $\iff$ $x^2-2x-1=0\iff$

the tangent points are $\left\{\begin{array}{c} T_1\left(1+\sqrt 2,-1+\sqrt 2\right)\\\\ T_2\left(1-\sqrt 2,-1-\sqrt 2\right)\end{array}\right\|$ and the equations of the tangent lines are $\left\{\begin{array}{c} \left(-1+\sqrt 2\right)x+\left(1+\sqrt 2\right)y=2\\\\ \left(-1-\sqrt 2\right)x+\left(1-\sqrt 2\right)y=2\end{array}\right\|$ .

$(*)\blacktriangleright$ The equation of the tangent line $t$ in the point $T$ is $Y-\frac 1x=-\frac {1}{x^2}\cdot (X-x)\iff$ $\boxed{X+x^2\cdot Y-2x=0}$ . Thus, $\left\{\begin{array}{ccccc} Y:=0 & \implies & X=2x & \implies & A(2x,0)\\\\ X:=0 & \implies & Y=\frac 2x & \implies & B\left(0,\frac 2x\right)\end{array}\right\|$ .

Remark. Let $\boxed{ax^2+2bxy+cy^2+2mx+2ny+p=0}$ be the equation of a conic $\mathbb C$ and a point $T\left(x_0,y_0\right)$ . If $T\in\mathbb C$ , then the equation of the tangent line from $T$ to

the conic $\mathbb C$ is $\boxed{axx_0+b\left(x_0y+xy_0\right)+cyy_0+m\left(x+x_0\right)+n\left(y+y_0\right)+p=0}$ . If $T\not\in\mathbb C$ (exterior), then the equations of the tangent lines from $T$ to the conic $\mathbb C$ are

$\boxed{\left[axx_0+b\left(x_0y+xy_0\right)+cyy_0+m\left(x+x_0\right)+n\left(y+y_0\right)+p\right]^2}=$ $\boxed{\left(ax^2+2bxy+cy^2+2mx+2ny+p\right)\cdot \left(ax_0^2+2bx_0y_0+cy_0^2+2mx_0+2ny_0+p\right)}$ .

Example. Let $\frac{x^2}{a^2}-\frac {y^2}{b^2}=1$ be the equation of the hyperbola $\mathbb H$ with the asymptotes $y=\pm\frac {bx}{a}$ . Let $T\left(x_0,y_0\right)\in\mathbb H$ , i.e. $b^2x_0^2-a^2y_0^2=a^2b^2\ (1)$ . Then the equation of the tangent

line $t$ in $T\left(x_0,y_0\right)\in\mathbb H$ is $\frac{xx_0}{a^2}-\frac {yy_0}{b^2}=1$ . Denote the intersections $M\left(x_1,y_1\right)$ and $N\left(x_2,y_2\right)$ of $t$ with the asymptotes $ay=bx$ and $ay=-bx$ respectively. Obtain that

$\left\{\begin{array}{cccccc} M\ : & \left|\begin{array}{c} b^2xx_0-a^2yy_0=a^2b^2\\\\ ay=bx\end{array}\right| & \implies & bxx_0-axy_0=a^2b & \implies & x_M=\frac {a^2b}{bx_0-ay_0}\\\\ N\ : & \left|\begin{array}{c} b^2xx_0-a^2yy_0=a^2b^2\\\\ ay=-bx\end{array}\right| & \implies & bxx_0+axy_0=a^2b & \implies & x_N=\frac {a^2b}{bx_0+ay_0}\end{array}\right\|$ . Observe that $\frac {x_M+x_N}{2}=\frac {a^2b^2}{b^2x_0^2-a^2y_0^2}\cdot x_0\stackrel{(1)}{\implies}$

$\boxed{x_M+x_N=2x_0}$ , i.e. the tangent point $T\in\mathbb H$ is the midpoint of the segment $MN$ , where $M$ and $N$ are the intersections of the tangent line $TT$ with the asymptotes of $\mathbb H$ .

PP10.

$\blacktriangleright$ Find the range of $a$ for which there exist two common tangent lines of the curve $f(x)=\frac{8}{27}x^3$ and the parabola $g(x)=(x+a)^2$ other than the $x$ axis.

$\blacktriangleright$ For the range of $a$ found in the previous question, express the area bounded by the two tangent lines and the parabola $y=(x+a)^2$ in terms of $a$ .

Proof. $\blacktriangleright$ The tangent line at $(t,f(t))\ (t\ne 0)$ is $y=f(t)+f'(t)(x-t)$ and touch the parabola $y=g(x)$ $\Longleftrightarrow$ the equation 2th degree

$g(x)=f(t)+f'(t)(x-t)$ , i.e. $g(x)-f'(t)\cdot x-f(t)+tf'(t)=0$ has a double root. Thus, $(x+a)^2-\frac {8t^2}{9}\cdot x-\frac {8t^3}{27}+\frac {8t^3}{9}=0$

$\iff$ $x^2+2\left(a-\frac {4t^2}{9}\right)\cdot x+a^2+\frac {16t^3}{27}=0$ yielding $\Delta=\left(a-\frac {4t^2}{9}\right)^2-\left(a^2+\frac{16t^3}{27}\right)=0$ $\Longleftrightarrow 2t^2-6t-9a=0\ (1)$ .

That should have distinct two real solutions except 0, we have $(-3)^2-2(-9a)>0$ and $-9a\neq 0$ , yielding the desired range of $a$ is $-\frac 12<a<0$ or $a>0$ .

$\blacktriangleright$ Let $\alpha ,\ \beta \ (\alpha <\beta)$ be the roots of $(1)$ , we have $\alpha =\frac{3-3\sqrt{1+2a}}{2},\ \beta =\frac{3+3\sqrt{2a+1}}{2}$ . Denote by $x_1,\ x_2\ (x_1<x_2)$ the $x$ coordinates

of the point of tangency, $x_1=-a+\frac 49\alpha ^2,\ x_2=-a+\frac 49\beta ^ 2$ . $\therefore S=\int_{x_1}^{\frac{x_1+x_2}{2}} (x-x_1)^2dx+\int_{\frac{x_1+x_2}{2}}^{x_2} (x-x_2)^2dx$ $=\frac{1}{12}(x_2-x_1)^3$

$=\frac{1}{12}\left\{\frac 49(\beta ^2-\alpha ^2)\right\}^3$ $=\frac{1}{12}\left\{\frac 49(\beta +\alpha)(\beta -\alpha)\right\}^3$ $=\frac{1}{12}\left(\frac 49\cdot 3\cdot 3\sqrt{1+2a}\right)^3$ $=\frac{16}{3}(2a+1)^{\frac 32}$ .

PP11. Let $w=C(O,1)$ be a circle and $A\in w$ be a fixed point. Let $c=[BC]$ be a fixed chord so that $c\parallel OA$ and the distance $d=\delta_{c}(O)=\frac {\sqrt 2}{3}$ .

For a mobile point $M\in [BC]$ denote $\{A,N\}=AM\cap w$ . Find the maximum length of the segment $[MN]$ and in this case find the $\cos\widehat {OAM}$ .

Proof. $x=m\left(\widehat {OAM}\right)\ \implies\ f(x)\equiv$ $MN=AN-AM=$ $2R\cdot \cos x-\frac {d}{\sin x}$ , i.e. $f(x)=2\cos x-\frac {\sqrt 2}{3\sin x}$ , where

$x\in I\equiv \left\{\arctan\frac {\sqrt 2\left(3-\sqrt 7\right)}{2}\ ,\ \arctan\frac {\sqrt 2\left(3+\sqrt 7\right)}{2}\right\}$ . Thus, $f'(x)=-2\sin x+\frac {\sqrt 2\cdot\cos x}{3\sin^2x}$ and $f'(x)=0\iff$

$2\tan x=\frac {\sqrt 2}{3\sin^2x}\iff$ $3\sqrt 2\cdot \tan^3x =\tan^2x+1\stackrel{(t=\tan x)}{\iff}$ $3t^3\sqrt 2-t^2-1=0\iff$ $\left(t\sqrt 2-1\right)\cdot \left(3t^2+t\sqrt 2+1\right)=0\iff$

$t=\tan x=\frac {\sqrt 2}{2}$ . Since $f'(x)\ .s.s.\ \left(1-\sqrt 2\cdot \tan x\right)$ obtain that $\max_{x\in I}f(x)=f\left(\arctan\frac {\sqrt 2}{2}\right)=\frac {\sqrt 6}{3}$ .

Remark. Prove easily that if $M_0$ is the projection of the point $M$ on the line $OA$ , then in the extremum case we have $\frac {AM_0}{AO}=\frac 23$ , i.e. $AM_0=\frac 23$ .

PP12. Let $ABCD$ be a square with $AB=2$ . For two mobile points $E\in (AB)\ ,\ F\in (CD)$ denote $G\in DE\cap AF$ and $H\in EC\cap BF$ . Prove that $GH\ge 1$ .

Proof (analytic). Consider the square $ABCD$ and given points $\{E,F\}$ so that $\left\{\begin{array}{c} B(0,2)\ ,\ C(2,2)\\\\ E(0,e)\ ,\ F(2,f)\\\\ A(0,0)\ ,\ D(2,0)\end{array}\right\|$ , where $\{e,f\}\subset [0,2]$ . Therefore,

$\left\{\begin{array}{cccc} H\ : & \left|\begin{array}{cc} EC\ : & y-e=\frac {2-e}{2}\cdot x\\\\ BF\ : & y-2=\frac {f-2}{2}\cdot x\end{array}\right| & \implies & \left|\begin{array}{c} x_h=\frac {2(2-e)}{4-(e+f)}\\\\ y_h=\frac {4-ef}{4-(e+f)}\end{array}\right|\\\\ G\ : & \left|\begin{array}{cc} AF\ : & y=\frac f2\cdot x\\\\ ED\ : & y=\frac e2\cdot (2-x)\end{array}\right| & \implies & \left|\begin{array}{c} x_g=\frac {2e}{e+f}\\\\ y_g=\frac {ef}{e+f}\end{array}\right|\end{array}\right\|$ . Thus, $GH^2=\left[\frac {2(2-e)}{4-(e+f)}-\frac {2e}{e+f}\right]^2+\left[\frac {4-ef}{4-(e+f)}-\frac {ef}{e+f}\right]^2=$

$\frac {4\cdot \left[(2-e)(e+f)-e(4-(e+f)\right]^2+\left[(4-ef)(e+f)-ef(4-(e+f))\right]^2} {(e+f)^2[4-(e+f)]^2}=$ $16\cdot \frac {(e-f)^2+(e+f-ef)^2}{(e+f)^2[4-(e+f)]^2}$ .

In conclusion, $GH\ge 1\iff$ $4\sqrt {(e-f)^2+(e+f-ef)^2}\ge (e+f)[4-(e+f)]$ .

PP13. In $\triangle{ABC}$ , $M$ is the midpoint of $BC$ . If $A=(2,1)$ , the orthocenter $H=(-6,3)$ and $M=(2,2)$ , find the coordinates of $B$ and $C$ .

Proof. Since $N(-2,2)$ is the midpoint of $[AH]$ obtain that the midpoint of $[MN]$ is $E(0,2)$ - the center of the Euler's circle for $\triangle ABC$ . Since $E$ is the midpoint of $[HO]$ obtain that

$O\left(6,1\right)$ and $R=4$ , where $O$ is the circumcenter and $R$ is the length of the circumradius for $\triangle ABC$ . Thus, the circumcircle $w=C(O,R)$ has the equation $(x-6)^2+(y-1)^2=16$

The slope $s_{OM}$ is $-\frac 14$ and the equation of $BC$ is $y-2=4(x-2)$ , i.e. $y=4x-6$ . Thus, $\{B,C\}=BC\cap w$ , i.e. $\left\{\begin{array}{c} (x-6)^2+(y-1)^2=16\\\\ y=4x-6\end{array}\right\|$ .

Therefore, $(x-6)^2+(4x-7)^2=16\iff x\in\emptyset$ (proposed problem is absurd).

PP14. Let $M\left(am^2 , 2am \right)$ be a mobile point on the parabola $\mathbb P$ with thw equation $y^2 = 4ax$ , where $a>0$ and $m>0$ . Prove that exist $\{R,S\}\subset\mathbb P$

such that the normals at $R$ and at $S$ intersect at $M$ iff $m^2>8$ . Deduce that : the chord $RS$ pass through a fixed point ; $OM$ and $RS$ meet on a fixed line.

Proof. Let $T\left(at^2,2at\right)\in\mathbb P$ for which $M$ belongs to normal at $T$ . Since $2yy'=4a\iff y'=\frac {2a}{y}$ obtain $y'(T)=\frac {2a}{y_T}=\frac {2a}{2at}=\frac 1t$ . Thus, the equation of the normal $\eta$ at $T$ is

$y-2at=-t\left(x-at^2\right)$ and $M\in \eta\iff$ $2a(m-t)=-at\left(m^2-t^2\right)$ , i.e. $t^2+mt+2=0$ . Thus $m^2>8\iff \Delta>0\iff$ exist $R\left(ar^2,2ar\right)$ and $S\left(as^2,2as\right)$ on $\mathbb P$ so

that the normals at $R$ and $S$ intersect at $M$ and in this case $r+s=-m\ ,\ rs=2$ . The equation of the chord $RS$ is $\left|\begin{array}{ccc} x & y & 1\\\\ ar^2 & 2ar & 1\\\\ as^2 & 2as & 1\end{array}\right|=0\iff$

$2a(r-s)x-a\left(r^2-s^2\right)y+ars=0\iff$ $2x-(r+s)y+ars=0\iff$ $\boxed{2x+my+2a=0}$ . Focus of $\mathbb P$ is $F(a,0)$ , its directrix $\delta$ is $x=-a$

Hence $(-a,0)\in RS\cap \delta$ , i.e. the chord $RS$ pass through a fixed point $D(-a,0)$ . Eliminate parameter $m$ between $\left\{\begin{array}{cc} OM\ : & 2x-my=0\\\\ RS\ : & 2x+my+2a=0\end{array}\right\|$ .

Obtain $x=-\frac a2$ , what is the equation of a fixed $f\parallel \delta$ . Thus, $OM$ and $RS$ meet on a fixed $f$ .

PP15. Let $w$ with the equation $x^2+y^2-2x-2\sqrt{3}y=0$ and $d$ with the equation $d(x,y)\equiv x\cos\phi +y\sin\phi +1=0$ . Find all values for $\phi$ such that $d$ can be a tangent for $w$ .

Proof. Observe that $w$ has center $I\left(1,\sqrt 3\right)$ and its radius has length $2$ , i.e. its equivalent equation is $(x-1)^2+\left(y-\sqrt 3\right)^2=4$ . $d$ is tangent to $w$ $\iff$ the distance of $I$ to $d$ is

equally to $2$ , i.e. $\delta_d(I)=2$ $\iff$ $\frac {\left|d\left(1,\sqrt 3\right)\right|}{\sqrt {\cos^2\phi +\sin^2\phi} }=2\iff$ $\left|\cos\phi +\sqrt 3\sin\phi +1\right|=2\iff$ $\cos\phi +\sqrt 3\sin\phi\in\left\{-3,1\right\}$ . Since $\left|\cos\phi +\sqrt 3 \sin\phi\right|\le 2$ obtain that

$\cos\phi +\sqrt 3\sin\phi =1\iff$ $\frac 12\cdot \cos\phi+\frac {\sqrt 3}{2}\cdot\sin\phi =\frac 12\iff$ $\cos\left(\phi -\frac {\pi}{3}\right)=\cos\frac {\pi}{3}\iff$ $\left(\phi -\frac {\pi}{3}\right)\in \left\{\left| 2k\pi\pm\frac {\pi}{3}\right|k\in\mathbb Z\right\}\iff$ $\phi\in \left\{\left | \ 2k\pi\ ;\ 2k\pi +\frac {2\pi}{3}\ \right|\ k\in\mathbb Z\ \right\}$

Remark.. We observe that $x \cos \phi + y \sin \phi + 1 = 0$ is always tangent to the unit circle at $(x,y) = (-\cos \phi, -\sin \phi)$ since the point obviously satisfies the equation of the line, and the

slope of the line is $m_\phi = -\cot \phi$ , where as the radius to the point of tangency is $\frac{-\sin \phi}{-\cos \phi} = \tan \phi =-\frac {1}{m_\phi}$ and hence is perpendicular. Next, we note that $x^2 + y^2 - 2x - 2\sqrt{3}y = 0$

is equivalent to $(x-1)^2 + (y-\sqrt{3})^2 = 4$ , so the given $w$ is centered at $I\left(1,\sqrt{3}\right)$ with radius $r = 2$ . Thus, the line is tangent to $w$ for $\phi$ corresponding to a common tangency to $w$ and

the unit circle and there are exactly two such lines. These tangents intersect at $Q$ on $\overline{OI}$ which satisfies the relationship $\frac{OQ + OI}{r} = \frac{OQ}{1}$ . Hence $OQ = 2$ and $Q = -I = (-1, -\sqrt{3})$

Thus one tangent line is obviously $x = -1$ and the other is the reflection of this line about $\overline{OI}$ ; i.e., $y + \sqrt{3} = \frac{1}{\sqrt{3}} (x + 1)$ or $\frac{\sqrt{3}}{2}y - \frac{1}{2}x + 1 = 0$ . In the first case, $\boxed{\phi= 0}$ and in the

second, $\boxed{\phi = \frac {2\pi}{3}}$ (not counting multiples of $2\pi$ ).

PP16. Find the maximum and minimum of $a+b$ such that $a+b=\sqrt{2a-1}+\sqrt{4b+3}$ .

Proof. Let $\left\{\begin{array}{ccccc} \sqrt{2a-1}=x & \iff & x\ge 0 & \mathrm{and} & a=\frac {x^2+1}{2}\\\\ \sqrt {4b+3}=y & \iff & y\ge 0 & \mathrm{and} & b=\frac {y^2-3}{4}\end{array}\right\|$ . Our problem becomes "find the maximum/minimum of $x+y$ , where $x\ge 0$ , $y\ge 0$ and

$\frac {x^2+1}{2}+\frac {y^2-3}{4}=x+y$ , i.e. $\boxed{2x^2+y^2-4x-4y-1=0}\ (*)\iff$ $\frac {(x-1)^2}{\frac 72}+\frac {(y-2)^2}{7}=1$ , what is the equation of the ellipse-arc which belongs to the first quadrant. Prove easily that the graph of this arc is above $[AB]$ , where $A\left(1+\sqrt {\frac 32},0\right)$ and $B\left(0,2+\sqrt 3\right)$ . The minimum of $x+y$ is $\boxed{1+\sqrt {\frac 32}}$ and the maximum is touched

when the tangent to the ellipse-arc is parallel with $x+y=0$ . It is possibly iff the maximum is $\lambda\ge 0$ for which the equation $2x^2+(\lambda -x)^2-4\lambda -1=0$ , i.e. the equation

$3x^2-2\lambda x+\lambda^2-4\lambda-1=0$ has $\Delta ' (\lambda ) =0$ , i.e. $2\lambda^2-12\lambda -3=0$ and $\lambda\ge 0$ . Hence the maximum of $x+y$ is $\boxed{3+\sqrt {\frac {21}{2}}}$ .

PP17. A line $d$ passing through $A(7,3)$ meets the circle $w$ with the equation $x^2+y^2-2x+2y-23=0$ at the points $\{B,C\}$ such that $AB=3AC$ . Find the equation of the line $d$ .

Proof. Thus, the equation of $w$ becomes $(x-1)^2+(y+1)^2=25$ , i.e. $w$ has center $I(1,-1)$ and its radius has length $r=5$ . Since $AI=2\sqrt {13}>5$ obtain that $A$ belongs to outside

of $w$ . Using the power of $A$ w.r.t. $w$ obtain that $AB\cdot AC=AI^2-r^2=27$ . From the condition of hypothesis $AB=3AC$ obtain that $AB=9$ , $AC=3$ , $BC=6$ and the midpoint

of $BC$ is $M$ so that $IM=r^2-\left(\frac {BC}{2}\right)^2=25-9=16$ , i.e. $IM=4$ . Hence $d$ is tangent to $w'$ with the center $I$ and the length $4$ of its radius, i.e. with the equation

$(x-1)^2+(y+1)^2=16\iff$ $x^2+y^2-2x+2y-14=0$ . Using the mentioned upper theoretical remark from the problem PP9 obtain that the equations of $d$ is

$[7x+3y-(x+7)+(y+3)-14]^2=$ $\left(7^2+3^2-2\cdot 7+2\cdot 3-14\right)\left(x^2+y^2-2x+2y-14\right)\iff$ $4(3x+2y-9)^2=36\left(x^2+y^2-2x+2y-14\right)\iff$

$(3x+2y-9)^2=9(x^2+y^2-2x+2y-14)\iff$ $5y^2-12xy+36x+54y-207=0\iff$ $5y^2-6(2x-9)y+9(4x-23)=0$ . Observe that

$\Delta '=9(2x-9)^2-45(4x-23)=9(4x^2-56x+196)$ , i.e. $\Delta '=[6(x-7)]^2$ . Thus, $y\in\left\{\frac {3(2x-9)\pm6(x-7)}{5}\right\}=\left\{3,\frac {3(4x-23)}{5}\right\}$ , i.e.

the equations of the tangents from the exterior point $A$ to the circle $w'$ are $:\begin{array}{cccc} \nearrow & y-3 & = & 0\\\\ \searrow & 12x-5y-69 & = & 0\end{array}$ .

PP18. How do you find the circumcircle of $\triangle ABC$ , where $A(3,19)$ , $B (-2,18)$ and $C (-10,6)$ .

Proof. Let circumcircle $w=C(O,R)$ of $\triangle ABC$ , the midpoints $M$ , $N$ of $[BC]$ , $[CA]$ . Prove easily $M(-6,12)$ , $N\left(-\frac 72,\frac {25}{2}\right)$ and slopes $s_{BC}=\frac 32$ , $s_{CA}=1$ . Thus,

equations of bisectors for $[BC]$ , $[CA]$ are $\left\{\begin{array}{cc} OM\ : & y-12=-\frac 23\cdot (x+6)\\\\ ON\ : & y-\frac {25}{2}=-\left(x+\frac 72\right)\end{array}\right\|$ . Circumcenter $O$ is the intersection of $\left\{\begin{array}{c} 2x+3y=24\\\ x+y=9\end{array}\right\|$ , i.e. $\boxed{O(3,6)}$ and length

of the circumradius is $R=OA=19-6$ . i.e. $\boxed{R=13}$ . Equation of circumcircle $w$ is $(x-3)^2+(y-6)^2=13^2$ , i.e. $\boxed{x^2+y^2-6x-12y-124=0}$ .

Remark. $OA\perp Ox$ and $OC\perp Oy$ . Prove easily $a=4\sqrt {13}$ , $b=13\sqrt 2$ , $c=\sqrt {26}$ and $\cos B=-\frac {\sqrt 2}{2}$ , i.e. $B=135^{\circ}$ . Thus, $OA\perp OC$ . From the relation $abc=4RS$

obtain that $4\cdot 13\cdot 13\cdot 2=4\cdot 13\cdot S$ , i.e. $S=26$ . Indeed, $S=\frac 12\cdot\mod\left|\begin{array}{ccc} 3 & 19 & 1\\\\ -2 & 18 & 1\\\\ -10 & 6 & 1\end{array}\right|=26$ . Generally, the circumcircle of $\triangle ABC$ has the equation

$\left|\begin{array}{cccc} x^2+y^2 & x & y & 1\\\\ x^2_1+y^2_1 & x_1 & y_1 & 1\\\\ x^2_2+y^2_2 & x_2 & y_2 & 1\\\\ x^2_3+y^2_3 & x_3 & y_3 & 1\end{array}\right|=0$ . In our case $\left|\begin{array}{cccc} x^2+y^2 & x & y & 1\\\\ 370 & 3 & 19 & 1\\\\ 328 &-2 & 18 & 1\\\\ 136 & -10 & 6 & 1\end{array}\right|=0\iff$ $\left(x^2+y^2\right)\left|\begin{array}{ccc} 3 & 19 & 1\\\\ -2 & 18 & 1\\\\ -10 & 6 & 1\end{array}\right|-$ $x\left|\begin{array}{ccc} 370 & 19 & 1\\\\ 328 & 18 & 1\\\\ 136 & 6 & 1\end{array}\right|+$ $y\left|\begin{array}{ccc} 370 & 3 & 1\\\\ 328 &-2 & 1\\\\ 136 & -10 & 1\end{array}\right|-$

$\left|\begin{array}{ccc} 370 & 3 & 19\\\\ 328 &-2 & 18\\\\ 136 & -10 & 6\end{array}\right|=0\iff$ $52\cdot\left(x^2+y^2\right)-2\cdot12\cdot 13\cdot x-8\cdot 13\cdot 6\cdot y-52\cdot 124=0\iff$ $x^2+y^2-6x-12y-124=0$ .

PP19. $y=f(x)$ is the graph of a one-to-one continuous function $f$ . At each point $P$ on the graph of $y=2x^2$ assume that the areas

$OAP$ and $OBP$ (curve & line) are equal . Here $[PA]$ , $[PB]$ are the horizontal and vertical segments . Determine the function $f$ .

Proof. Let $A$ , $P$ and $B$ be $A(t,f(t))$ , $P\left(\sqrt \frac{f(t)}{2},f(t)\right)$ and $B\left(\sqrt \frac{f(t)}{2}, \frac{f(t)}{2}\right)$ . Thus, $[OAP]=[OBP]\iff$ $[OAT]+[APST]-[OPS]=[OPS]-[OBS]\iff$

$[OAT]+[APST]=2\cdot [OPS]-[OBS]\iff$ $\int^t_0 f(x)\ \mathrm{dx}+f(t)\left(\sqrt \frac{f(t)}{2}-t\right)=2\cdot \int^{\sqrt \frac{f(t)}{2}}_0 2x^2\ \mathrm{dx} - \int^{\sqrt \frac{f(t)}{2}}_0 x^2\ \mathrm{dx}\iff$

$\left|\int^t_0 f(x)\ \mathrm{dx} + f(t)\cdot \left(\sqrt \frac{f(t)}{2} - t\right) - \frac{f(t)}{2}\cdot\sqrt{\frac{f(t)}{2}}=0\ \right|^{\prime}\iff$ $f(t)+\frac{3}{4\sqrt{2}}\cdot f'(t)\sqrt{f(t)} - f(t) - tf'(t)=0\iff$ $\sqrt{f(t)}=\frac{4t\sqrt{2}}{3}$ Hence $\boxed{f(x)=\frac{32}{9}\cdot x^2}$ .

PP20. There is a parabola $\mathbb P$ and a triangle $ABC$ such that $\{A,B,C\}\cap\mathbb P=\emptyset$ and $AB\ ,\ BC\ ,\ AC$ are tangent to $\mathbb P$ . Prove that the focus of $\mathbb P$ belongs to the circumcircle of $\triangle ABC$ .

Proof. Let the equation $y^2=2kx$ of the parabola $\mathbb P$ with the focus $F\left(\frac k2,0\right)$ . Let the tangent points $M\left(2km^2,2km\right)\in BC\cap \mathbb P$ , $N\left(2kn^2,2kn\right)\in CA\cap \mathbb P$ ,

$P\left(2kp^2,2kp\right)\in AB\cap\mathbb P$ . Prove easily that $\left\{\begin{array}{ccc} A\in NN\cap PP & \implies & A\left(2knp,kn+kp\right)\\\\ B\in PP\cap MM & \implies & B\left(2kpm,kp+km\right)\\\\ C\in MM\cap NN & \implies & C\left(2kmn,km+kn\right)\end{array}\right\|$ . For example, $A\in NN\cap PP\implies$ $A\in\left\{\begin{array}{ccc} 2ny & = & x+2kn^2\\\\ 2py & = & x+2kp^2\end{array}\right\|\implies$

$\left\{\begin{array}{ccc} x_A\ & = & 2knp\\\\ y_A & = & k(n+p)\end{array}\right\|$ . The focus of $\mathbb P$ belongs to the circumcircle of $\triangle ABC\iff$ $\left|\begin{array}{cccc} 4k^2n^2p^2+k^2(n+p)^2 & 2knp & k(n+p) & 1\\\\ 4k^2p^2m^2+k^2(p+m)^2 & 2kpm & k(p+m) & 1\\\\ 4k^2m^2n^2+k^2(m+n)^2 & 2kmn & k(m+n) & 1\\\\ k^2 & 2k & 0 & 4\end{array}\right|=0\iff$

$\left|\begin{array}{cccc} 4n^2p^2+(n+p)^2 & np & n+p & 1\\\\ 4p^2m^2+(p+m)^2 & pm & p+m & 1\\\\ 4m^2n^2+(m+n)^2 & mn & m+n & 1\\\\ 1 & 1 & 0 & 4\end{array}\right|=0\stackrel{\left(\begin{array}{c} \mathrm L_2:=\mathrm L_2-\mathrm L_1\\\ \mathrm L_3:=\mathrm L_3-\mathrm L_1\end{array}\right)}{\ \iff\ }$ $\left|\begin{array}{cccc} 4n^2p^2+(n+p)^2 & np & n+p & 1\\\\ 4p^2(m+n)+2p+m+n & p & 1 & 0\\\\ 4n^2(m+p)+2n+m+p & n & 1 & 0\\\\ 1 & 1 & 0 & 4\end{array}\right|=0\stackrel{\left(\mathrm L_3:=\mathrm L_3-\mathrm L_2\right)}{\ \iff\ }$

$\left|\begin{array}{cccc} 4n^2p^2+(n+p)^2 & np & n+p & 1\\\\ 4p^2(m+n)+2p+m+n & p & 1 & 0\\\\ 4(mn+np+pm)+1 & 1 & 0 & 0\\\\ 1 & 1 & 0 & 4\end{array}\right|=0\stackrel{\left(\begin{array}{c} \mathrm L_1:=\mathrm L_1-\mathrm n\mathrm L_2\\\ \mathrm L_2:=\mathrm L_2-\mathrm p\mathrm L_3\\\ \mathrm L_3:=\mathrm L_3-\mathrm L_4\end{array}\right)}{\ \iff\ }$ $\left|\begin{array}{cccc} p^2-4mnp^2-mn & 0 & p & 1\\\\ m+n+p-4mnp & 0 & 1 & 0\\\\ mn+np+pm & 0 & 0 & -1\\\\ 1 & \boxed{1} & 0 & 4\end{array}\right|=0\iff$ $\left|\begin{array}{ccc} p^2-4mnp^2-mn & p & 1\\\\ m+n+p-4mnp & 1 & 0\\\\ mn+np+pm & 0 & -1\end{array}\right|=0$ ,

what is true because $L_1+L_3=pL_2$ . In conclusion, the focus of $\mathbb P$ belongs to the circumcircle of $\triangle ABC$ .

This post has been edited 290 times. Last edited by Virgil Nicula, Nov 19, 2015, 8:46 AM

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