353. Some problems of the analytical geometry.

by Virgil Nicula, Aug 8, 2012, 6:23 PM

PP1. Consider an ellipse $\mathcal E$ with foci $F_1$ and $F_2$ . For a random point $P\in\mathcal E$ let $\left\{\begin{array}{ccc} F_1E & ; & E\in PF_2\\\\ F_2D & ; & D\in PF_1\end{array}\right|$

be the angle bisectors of $\triangle F_1PF_2$ . Find the area of $\triangle DPE$ in function of the point $P(x,y)\in\mathcal E$ .

Proof. Let $\frac {X^2}{a^2}+\frac {Y^2}{b^2}=1$ be the equation of the ellipse $\mathcal E$ with foci $F_1(c,0)$ , $F_2(-c,0)$ , where $c>0$ and $a^2=b^2+c^2$ . If $P(x,y)$ $\in\mathcal E$ , then denote $S\in F_1F_2$ so that the ray $[PS$

is the angle-bisector of $\widehat{F_1PF_2}$ . Thus, $PF_1+PF_2=2a$ and $F_1F_2=$ $SF_1+SF_2=$ $2c$ . The equation of the tangent line $t=PP$ to $\mathcal E$ is $b^2x\cdot X+a^2y\cdot Y=a^2b^2$ with the slope

$s_t=-\frac {b^2x}{a^2y}$ . Thus, the slope of the normal line $n=PS$ in same point $P\in \mathcal E$ is $s_n=\frac {a^2y}{b^2x}$ and the equation of $n$ is $Y-y=\frac {a^2y}{b^2x}\cdot (X-x)$ . Therefore, $S\in n\cap Ox\implies$

$\boxed{S\left(\frac {c^2x}{a^2},0\right)}$ . Using the theorem of the angle-bisector for $[PS$ obtain that $\frac {PF_1}{PF_2}=$ $\frac {SF_1}{SF_2}=$ $\frac {c-\frac {c^2x}{a^2}}{\frac {c^2x}{a^2}+c}=$ $\frac {a^2-cx}{a^2+cx}$ $\implies$ $\frac {PF_1}{a^2-cx}=$ $\frac {PF_2}{a^2+cx}=$ $\frac {PF_1+PF_2}{2a^2}=$ $\frac {2a}{2a^2}=$ $\frac 1a$ .

In conclusion, the lengths of the sides for $\triangle F_1PF_2$ are $F_1F_2=2c$ and $\boxed{\begin{array}{c} PF_1=a-\frac {cx}{a}\\\\ PF_2=a+\frac {cx}{a}\end{array}}$ . From the well-known relations $\left\{\begin{array}{c} PD=\frac {F_2P\cdot F_1P}{F_2P+F_2F_1}\\\\ PE=\frac {F_1P\cdot F_2P}{F_1P+F_1F_2}\end{array}\right|$ obtain that

$\left\{\begin{array}{c} PD=\frac {\left(a^2+cx\right)\left(a^2-cx\right)}{a\left(a^2+cx+2ac\right)}\\\\ PE=\frac {\left(a^2+cx\right)\left(a^2-cx\right)}{a\left(a^2-cx+2ac\right)}\end{array}\right|$ . So $[DPE]=[F_1PF_2]\cdot \frac {PD}{PF_1}\cdot\frac {PE}{PF_2}=$ $cy\cdot\frac {a^2+cx}{a^2+2ac+cx}\cdot\frac {a^2-cx}{a^2+2ac-cx}\implies$ $\boxed{\ [DPE]=\frac {cy\left(a^4-c^2x^2\right)}{\left(a^2+2ac\right)^2-c^2x^2}\ }$ .

Remark. $\cos\widehat{F_1PF_2}=\frac {PF_1^2+PF_2^2-F_1F_2^2}{2\cdot PF_1\cdot PF_2}=$ $\frac{\left(a-\frac {cx}{a}\right)^2+\left(a+\frac {cx}{a}\right)^2-4c^2}{2\left(a-\frac {cx}{a}\right)\left(a+\frac {cx}{a}\right)}\implies$ $\boxed{\cos\widehat{F_1PF_2}=\frac {a^4+c^2x^2-2a^2c^2}{\left(a^2-cx\right)\left(a^2+cx\right)}}$ . Analogously prove $\boxed{\ \begin{array}{c} \cos\widehat{PF_1F_2}=\frac {a(c-x)}{a^2-cx}\\\\ \cos\widehat{PF_2F_1}=\frac {a(c+x)}{a^2+cx}\end{array}\ }$ .

PP2. Find the minimum of $x^2+y^2$ , where $\{x,y\}\subset\mathbb R$ and $(x+5)^2+(y-12)^2=196$ .

Proof 1. This problem is equivalently (geometrically) with the finding of the point $P(x,y)$ on the circle $w=C(S,14)$ with the radius $R=14$ and the center

is $S(-5,12)$ such that $P$ has minimal distance from the origin. Observe that the origin is inside w.r.t. $w$ and $OS=13$ . In conclusion, the minimum value is equally to $1$ .

Proof 2. For any point $P(x,y)\in w$ exists $\phi\in \left[0,2\pi\right)$ so that $\left\{\begin{array}{c} x=-5+14\cos\phi\\\ y=12+14\sin\phi\end{array}\right\|$ . Thus, our problem begins the finding $\phi$ for which $x^2+y^2=365+28(-5\cos\phi +12\sin\phi )$

is minimum $\iff$ $-5\cos\phi +12\sin\phi$ is minimum. Since $|-5\cos\phi +12\sin\phi|\le 13$ obtain that the minimum of $x^2+y^2$ is $365-13\cdot 28$ , i.e. this minimum is equally to $1$ .

PP3. In an orthogonal cartesian system $xOy$ let a fixed $P (a, b)$ and mobile $M\in Ox\ ,$ $N\in Oy$ so that $PM\perp PN$ . Find the geometrical locus of $L\in MN$ for which $PL\perp MN$ .

Proof. Denote the fixed $A(a,0)\ ,\ B(0,b)$ and the mobile $M(m,0)$ , $N(0,n)$ , $K\in AB\cap MN$ . I"ll show that $PK\perp MN$ , i.e. $K\equiv L$ and the geometrical locus of $L$ is the fixed

$AB$ . Therefore, $PM\perp PN$ $\iff$ $\boxed{b(n-b)=a(a-m)}\ (*)$ and $\left\{\begin{array}{cccc} AB\ : & \frac xa+\frac yb=1 & \iff & bx+ay=ab\\\\ MN\ : & \frac xm+\frac yn=1 & \iff & nx+my=mn\end{array}\right\|\ \cap$ $\implies$ $K\left\{\begin{array}{c} x_K=\frac {am(n-b)}{an-bm}\\\\ y_K=\frac {bn(a-m)}{an-bm}\end{array}\right|$ . Thus, $PK\perp MN$

$\iff$ $\frac {b-y_K}{a-x_K}\cdot\frac nm=1$ $\iff$ $\frac {b-\frac {bn(a-m)}{an-bm}}{a-\frac {am(n-b)}{an-bm}}\cdot\frac nm=1$ $\iff$ $\frac {bm(n-b)}{an(a-m)}\cdot\frac nm=1$ , what is truly from the relation $(*)$ .

Lemma. The projection of $P(x,y)$ on $d(X,Y)\equiv aX+bY+c=0$ is $P_0:\ \boxed{\begin{array}{c} x_0=x-\frac {a}{a^2+b^2}\cdot d(x,y)\\\\ y_0=y-\frac {b}{a^2+b^2}\cdot d(x,y)\end{array}}$ .

Proof 1. $\left\{\begin{array}{ccc} P_0\in d & \iff & ax_0+by_0+c=0\\\\ PP_0\perp d & \iff & \frac {x_0-x}{a}=\frac {y_0-y}{b}\end{array}\right\|$ . Thus, $\frac {a\left(x_0-x\right)}{a^2}=\frac {b\left(y_0-y\right)}{b^2}=\frac {a\left(x_0-x\right)+b\left(y_0-y\right)}{a^2+b^2}=$ $\frac{\left(ax_0+by_0\right)-(ax+by)}{a^2+b^2}=$

$\frac {-c-(ax+by)}{a^2+b^2}=$ $\frac {-\left(ax+by+c\right)}{a^2+b^2}=$ $-\frac {d(x,y)}{a^2+b^2}\implies$ $\frac {x_0-x}{a}=\frac {y_0-y}{b}=-\frac {d(x,y)}{a^2+b^2}$ . In conclusion, the projection of $P(x,y)$ on the line

$d(X,Y)\equiv aX+bY+c=0$ is $P_0:\left\{\begin{array}{c} x_0=x-\frac {a}{a^2+b^2}\cdot d(x,y)\\\\ y_0=y-\frac {b}{a^2+b^2}\cdot d(x,y)\end{array}\right\|$ .[/hide]

Proof 2 (classic). Let the slope $p$ of $PM$ . Thus, the slope of $PN$ is $-\frac 1p$ . Thus, $\left\{\begin{array}{cccc} PM\ : & y-b=p\cdot (x-a) & \implies & M\left(a-\frac bp,0\right)\\\\ PN\ : & y-b=-\frac 1p\cdot (x-a) & \implies & N\left(0, b+\frac ap\right)\end{array}\right\|\implies$ equation of $MN$ is

$\boxed{\frac{x}{a-\frac bp}+\frac {y}{b+\frac ap}=1}\ (*)$ , i.e. $d(x,y)\equiv p(a+bp)x+p(ap-b)y-(ap-b)(a+bp)=0$ . Prove easily that $d(a,b)=ab\left(1+p^2\right)$ and $\left[p(a+bp)\right]^2+\left[p(ap-b)\right]^2=$

$p^2\left(1+p^2\right)\left(a^2+b^2\right)$ . Apply upper lemma. The projection $L$ of $A$ on $d$ is $\left\{\begin{array}{c} x_L=a-\frac {p(a+bp)}{p^2\left(1+p^2\right)\left(a^2+b^2\right)}\cdot ab\left(1+p^2\right)=\frac {a^2(pa-b)}{p\left(a^2+b^2\right)}=\frac {a^3}{a^2+b^2}-\frac {a^2b}{p\left(a^2+b^2\right)}\\\\ y_L=b-\frac {p(ap-b)}{p^2\left(1+p^2\right)\left(a^2+b^2\right)}\cdot ab\left(1+p^2\right)=\frac {b^2(pb+a)}{p\left(a^2+b^2\right)}=\frac {b^3}{a^2+b^2}+\frac {ab^2}{p\left(a^2+b^2\right)}\end{array}\right\|$ $\implies$

$\frac {\frac {a^3}{a^2+b^2}-x}{y-\frac {b^3}{a^2+b^2}}=\frac {a^2b}{ab^2}=\frac ab\implies$ $b\left(\frac {a^3}{a^2+b^2}-x\right)=a\left(y-\frac {b^2}{a^2+b^2}\right)\iff$ $bx+ay=\frac {ba^3+ab^3}{a^2+b^2}\implies$ $bx+ay=ab\implies$ locus of $L$ is $\frac Xa+\frac Yb=1$ , i.e. the line $AB$ .

PP4. Let $ABCD$ be a square so that $AB$ lies along the line $y=x+8$ and $C\ ,\ D$ lie on the parabola $y=x^2$ . Find all values of $AB$ .

Proof. Let $C\left(c,c^2\right)$ and $D\left(d,d^2\right)$ . Thus, $CD\parallel AB\iff$ $\frac {d^2-c^2}{d-c}=1\iff$ $\boxed{s=c+d=1}\ (1)$ , where $\boxed{CD=|c-d|\sqrt 2}\ (*)$ and must that the distance

$\delta_{AB}(C)=CD\iff$ $\frac {\left|c-c^2+8\right|}{\sqrt 2}=|c-d|\sqrt 2\stackrel{(1)}{\iff}$ $|cd+8|=2|c-d|\stackrel{\mathrm{cd=p}}{\iff}$ $(p+8)^2=4(1-4p)\iff$ $p^2+32p+60=0\left|\begin{array}{c} \nearrow\ -2\\\\ \searrow\ -30\end{array}\right|$ . Therefore,

the sum $s=c+d=1$ and the produkt $p=\left\{\begin{array}{ccccc} -2 & \implies & t^2-t-2=0\iff\{c,d\}=\{-1,2\} & \implies & AB\stackrel{(*)}{=}3\sqrt 2\\\\ -30 & \implies & t^2-t-30=0\iff\{c,d\}=\{-5,6\} & \implies & AB\stackrel{(*)}{=}11\sqrt 2\end{array}\right\|$ $\implies$ $\boxed{AB\in\left\{3\sqrt 2,11\sqrt 2\right\}}$ .

PP5. Let a square $ABCD$ and $N\in (BC)$ , $P\in (CD)$ . Denote $O\in AC\cap BD$ and the midpoint $M$ of $[AB]$ . Prove that $m\left(\widehat{NOP}\right)=\frac{\pi}{4}\iff MN\parallel AP\ .$

Proof 1 (analytic). Suppose w.l.o.g. that $\left\{\begin{array}{cc} A(0,0)\ ;\ B(0,2)\ ; & C(2,2)\ ;\ D(2,0)\\\\ N(a,2)\ ,\ a>1\ ; & P(2,b)\ ,\ b>1\end{array}\right\|$ . Observe that $M(0,1)$ , $O(1,1)$ and $MN\parallel AP\iff s_{MN}=s_{AP}\iff$ $\frac 1a=\frac b2\ ,$

where $s_d$ is the slope of the line $d$ . Thus, $\boxed{MN\parallel AP\iff ab=2}\ (1)$ . If denote $\phi =m\left(\widehat{NOP}\right)$ , then $\tan\phi =\frac {s_{ON}-s_{OP} }{1+s_{ON}\cdot s_{OP}}=$ $\frac {\frac 1{a-1}-\frac {b-1}1}{1+\frac {b-1}{a-1}}\implies$ $\tan\phi =\frac {a+b-ab}{a+b-2}\implies$

$\boxed{\phi =\frac {\pi}{4}\iff ab=2}\ (2)$ . In conclusion, using the equivalencies $(1)$ and $(2)$ obtain that $MN\parallel AP\iff ab=$ $2\iff$ $\phi =\frac {\pi}{4}$ , i.e. $m\left(\widehat{NOP}\right)=\frac{\pi}{4}\iff MN\parallel AP$ .

Proof 2. Suppose w.l.o.g. $AB=2$ . Let $\left|\begin{array}{c} BN=1+a\\\\ DP=1+b\end{array}\right|$ , where $\{a,b\}\subset (0,1)$ , midpoints $R$ , $S$ of $[BC]$ , $[CD]$ and $\left|\begin{array}{c} u=m\left(\widehat{NOC}\right)\\\\ v=m\left(\widehat{POS}\right)\end{array}\right|$ . So $\frac {NB}{NC}=\frac {\sin (90^{\circ}-u)}{\sin u}\iff$

$\tan u=\frac {1-a}{1+a}$ and $\tan v=b$ . Therefore, $m\left(\widehat{NOP}\right)=\frac {\pi}{4}\iff u=v$ . In conclusion, $\boxed{m\left(\widehat{NOP}\right)=\frac {\pi}{4}\iff b=\frac {1-a}{1+a}}\ (1)$ . On other hand $MN\parallel AP \iff$

$MBN\sim PAD\iff$ $\frac {BN}{BM}=\frac {AD}{AP}\iff$ $1+a=\frac 2{b+1}\iff$ $(1+a)(b+1)=2\iff$ $b(a+1)=1-a\iff$ $b=\frac {1-a}{1+a}$ , i.e. $\boxed{MN\parallel AP\iff b=\frac {1-a}{1+a}}\ (2)$ .

PP6. $[AB]$ is major axis of elipse $\Sigma$ with center $O$ and let $F$ be one of its foci. For $P\in\Sigma$ let $CD$ be chord through $O$ such that $CD\parallel PP$ and $Q\in PF\cap CD$ . Prove $PQ=OA$ .

Proof. $b^2x^2+a^2y^2=a^2b^2$ isquation of the ellipse $\Sigma$ , where $b^2+c^2=a^2$ and let $F(\epsilon c,0)$ be a focus, where $\epsilon ^2=1$ . Let $P(au,bv)\in \Sigma$ be a mobile point, where

$u^2+v^2=1$ . Then the slope of the tangent $PP$ in the point $P\in \Sigma$ to the ellipse $\Sigma$ is $m_{PP}=-\frac{bu}{av}$ . The equation of the line $CD$ is $y=-\frac{bu}{av}\cdot x$ and the equation of the line $CF$ is

$y=\frac{bv}{au-\epsilon c}\cdot (x-\epsilon c)$ . Thus, for the intersection point $Q\in CD\cap PF$ we have $x_Q=\frac{\epsilon acv^2}{a-\epsilon cu}$ , $y_Q=-\frac{\epsilon bcuv}{a-\epsilon cu}$ and $\left(\frac{a-\epsilon cu}{a}\right)^2\cdot PQ^2=$ $\left[u(a-\epsilon cu)-\epsilon cv^2\right]^2+b^2v^2=$

$(au-\epsilon c)^2+b^2v^2=$ $a^2u^2+b^2(1-u^2)+c^2-2\epsilon acu=$ $(a^2-b^2)u^2+a^2-2\epsilon acu=$ $c^2u^2+a^2-2\epsilon acu=$ $(a-\epsilon cu)^2\Longrightarrow PQ=a$ , i.e. $PQ=OA$ .

PP7. Let the circle $w=C(O,r)$ , the fixed points $\{A,B\}\subset w$ and a fixed line $d$ . For a mobile point $P\in w$ denote the

intersections $M\in PA\cap d$ and $N\in PB\cap d$ . Prove that the circle with the diameter $MN$ is always tangent to a fixed circle.

Proof. Suppose w.l.o.g. $w$ is $x^2+y^2=1$ , $d$ is $y=k$ and $\left\{\begin{array}{ccc} A & (\cos a,\sin a)\\\\ B & (\cos b,\sin b)\\\\ P & (\cos p,\sin p)\end{array}\right\|$ . I"ll find the mobile $M(u,k)$ , $N(v,k)$ which is function of $p\in[0,2\pi ]\ :$

$\blacktriangleright\ \left|\begin{array}{ccc} u & k & 1\\\\ \cos a & \sin a & 1\\\\ \cos p & \sin p & 1\end{array}\right|=0\implies$ $u=\frac {k(\cos a-\cos p)+\sin (a-p)}{\sin a-\sin p}=$ $\frac {k\sin\frac {a+p}2\sin\frac {p-a}2+\sin \frac{a-p}2\cos\frac {a-p}2}{\sin\frac{a-p}2\cos\frac {a+p}2}=$ $\frac {-k\sin\frac {a+p}2+\cos\frac {a-p}2}{\cos\frac {a+p}2}$ $\implies$

$\boxed{u=\frac {-k\left(\tan\frac p2+\tan\frac a2\right)+1+\tan\frac p2\tan\frac a2}{1-\tan\frac p2\tan\frac a2}}\ (1)$ .

$\blacktriangleright\ \left|\begin{array}{ccc} v & k & 1\\\\ \cos b & \sin b & 1\\\\ \cos p & \sin p & 1\end{array}\right|=0\implies$ $v=\frac {k(\cos b-\cos p)+\sin (b-p)}{\sin b-\sin p}=$ $\frac {k\sin\frac {b+p}2\sin\frac {p-b}2+\sin \frac{b-p}2\cos\frac {b-p}2}{\sin\frac{b-p}2\cos\frac {b+p}2}=$ $\frac {-k\sin\frac {b+p}2+\cos\frac {b-p}2}{\cos\frac {b+p}2}$ $\implies$

$\boxed{v=\frac {-k\left(\tan\frac p2+\tan\frac b2\right)+1+\tan\frac p2\tan\frac b2}{1-\tan\frac p2\tan\frac b2}}\ (2)$ .

Let $\left\{\begin{array}{c} \tan \frac a2=m\\\\ \tan\frac b2=n\\\\ \tan\frac p2=l\end{array}\right\|$ . $(1)$ and $(2)$ become $\odot\begin{array}{ccc} \nearrow & u=\frac {-k(l+m)+1+lm}{1-lm} & \searrow\\\\ \searrow & v=\frac {-k(l+n)+1+ln}{1-ln} & \nearrow\end{array}\odot$ . Prove easily $\left\{\begin{array}{ccc} u+v & = & \frac {[k(m+n)-2mn]l^2+2k(mn-1)l+2-(m+n)}{(1-ml)(1-nl)}\\\\ uv & = & \frac {(k-m)(k-n)l^2+\left[(m+n)\left(k^2+1\right)-2k(1+mn)\right]l+k^2mn-k(m+n)+1}{(1-ml)(1-nl)}\end{array}\right\|$ .

Prove easily the circle with the diameter $MN$ has equation $x^2+y^2-(u+v)x-2ky+k^2+uv=0$ a.s.o.

PP8. $k$ and $l$ (intersecting in $A$ ) are tangent to ellipse $\xi$ in $B$ and $C$ . Let $F$ and $G$ be focuses of $\xi$ . Let $D$ be the projection of $A$ on $FG$ . Prove that $\widehat{ADB}\equiv\widehat{ADC}\ .$

Proof. Suppose w.l.o.g. that the equation of $\xi$ is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\ .$ Let mobile $\{A,B\}\subset\xi$ , i.e. $\left\{\begin{array}{ccc}B(am,bn) & , & \boxed{m^{2}+n^{2}=1}\\\\ C(ap,br) & , & \boxed{p^{2}+r^{2}=1}\end{array}\right\|\Longrightarrow$

$\left\{\begin{array}{ccc}\mathrm{the\ tangent\ }BB & : & bmx+any-ab=0\\\\ \mathrm{the\ tangent\ }CC & : & bpx+ary-ab=0\end{array}\right\|$ $\Longrightarrow$ $A\left[\frac{a(r-n)}{mr-np}\ ,\ \frac{b(m-p)}{mr-np}\right]\Longrightarrow$ $\left\{\begin{array}{c}\mathrm{\ the\ slope\ of\ }DB\mathrm{\ is\ }s(DB)=\frac{b}{a}\cdot (mr-np)\cdot \frac{n}{m(mr-np)+n-r}\\\\ \mathrm{the\ slope\ of\ }DC\mathrm{\ is\ }s(DC)=\frac{b}{a}\cdot (mr-np)\cdot \frac{r}{p(mr-np)+n-r}\end{array}\right\|\ .$

Therefore, $s(DB)+s(DC)=$ $\boxed * \cdot [(mr-np)(mr+np)+(n-r)(n+r)]=$ $\boxed * \cdot [r^{2}(m^{2}-1)+n^{2}(1-p^{2})]=$ $\boxed * \cdot (-r^{2}n^{2}+r^{2}n^{2})=0\ ,$

where $\boxed * =\frac{b\cdot(mr-np)}{a\cdot [p(mr-np)+n-r]\cdot [m(mr-np)+n-r]}\ .$ In conclusion, $s(DB)+s(DC)=0\ ,$ i.e. $\widehat{ADB}\equiv\widehat{ADC}\ .$

Remark. You can prove easily if the ellipse $\xi$ is a circle $c=C(O)$ and $D$ is the projection of the mobile $A$ to the fixed diameter $[XY]\ .$ Indeed, $B,C,D$ belong to the circle

with the diameter $OA$ and $\widehat{ABC}\equiv\widehat{ACB}\ ,\ \widehat{ABC}\equiv\widehat{ADC}\ ,\ \widehat{ACB}\equiv\widehat{ADB}$ $\Longrightarrow$ $\widehat{ADB}\equiv\widehat{ADC}\ .$ Then you can project on an any plane with the common line $XY$ .

PP9. Let the parabola $w$ with the equation $y=x^2$ and the line $d$ with the equation $y=2x-3$ . For a mobile point $P(a,b)\in d$ , where $a^2>b$ denote

$\{M,N\}\subset w$ so that $P\in MM\cap NN$ (for $X\in w$ let $XX$ - the tangent line to $w$ ). Find the minimum value of the area $[PMN]$ , letting $P$ moves on $d$ .

Proof. $P\in d\iff \boxed{b=2a-3}\ (*)$ and the equations of $MM$ , $NN$ are $:\ \left\{\begin{array}{ccccc} M\left(m,m^2\right) & \implies & MM\ :\ y-m^2=2m(x-m) & \implies & y=2mx-m^2\\\\ N\left(n,n^2\right) & \implies & NN\ :\ y-n^2=2n(x-n) & \implies & y=2nx-n^2\end{array}\right\|$ ,

where $m\ne n\ .$ Thus, $\left\{\begin{array}{ccc} P\in MM & \implies & b=2ma-m^2\\\\ P\in NN & \implies & b=2na-n^2\end{array}\right\|$ $\implies$ $\left\{\begin{array}{cccccccc} 2ma-m^2=2na-n^2 & \implies & 2a(m-n)=m^2-n^2 & \implies & m+n & = & 2a & (1)\\\\ n\left(b+m^2\right)=m\left(b+n^2\right) & \implies & b(m-n)=mn(m-n) & \implies & mn & = & b & (2)\end{array}\right\|\ .$

The area $S=[PMN]=\frac 12\cdot|\Delta|$ , where $\Delta =\left|\begin{array}{ccc} a & b & 1\\\\ m & m^2 & 1\\\\ n & n^2 & 1\end{array}\right|\ \stackrel{1\wedge 2}{=}\ \left|\begin{array}{ccc} \frac {m+n}2 & mn & 1\\\\ m & m^2 & 1\\\\ n & n^2 & 1\end{array}\right|\ \stackrel{\begin{array}{c} L_2:=L_2-L_1\\\ L_3:=L_3-L_1\end{array}}{=}\ \left|\begin{array}{ccc} \frac {m+n}2 & mn & 1\\\\ \frac{m-n}2 & m(m-n) & 0\\\\ \frac{n-m}2 & n(n-m) & 0\end{array}\right|=$ $\left|\begin{array}{cc} \frac{m-n}2 & m(m-n)\\\\ \frac{n-m}2 & n(n-m)\end{array}\right|=$

$\frac {(m-n)^2}2\cdot \left|\begin{array}{cc} 1 & m\\\\ -1 & -n\end{array}\right|\implies$ $\Delta=\frac {(m-n)^3}2$ and $\boxed{S=\frac 14\cdot |m-n|^3}\ (3)\ .$ In conclusion, $S$ is $\min\iff$ $|m-n|$ is $\min\iff$ $|m-n|^2=(m-n)^2$ is $\min\iff$

$(m+n)^2-4mn$ is $\min\iff$ $4a^2-4b$ is $\min\iff$ $a^2-b$ is $\min\stackrel{*}{\iff} a^2-2a+3$ is $\min\iff$ $\boxed{a=1\ \wedge\ b=-1}$ , i.e. $P(1,-1)$ $\iff$ $m+n=2\ \wedge\ mn=-1\iff$

$|m-n|=2\sqrt 2\ \stackrel{3}{\iff}\ 4S=16\sqrt 2\iff$ $\boxed{S=4\sqrt 2}\ .$ I denoted $L_k$ - the line $k\in \overline{1,3}$ in the determinant $\Delta$ .

PP10. Let $\left\{M\left(\frac {m^2}{2p},m\right)\ ,\ N\left(\frac {n^2}{2p},n\right)\right\}\subset\mathcal P$ with equation $y^2=2px$ and $P(a,b)\in MM\cap NN$ , where $XX$ is the tangent to $\mathcal P$ at $X\in\mathcal P$ . Prove that $\left\{\begin{array}{ccc} 2pa & = & mn\\\\ 2b & = & m+n\end{array}\right\|$

Proof. $2yy'=2p\implies$ $y'=\frac py$ and the tangent $TT$ to $\mathcal P$ , where $T\left(\frac {t^2}{2p},t\right)\in \mathcal P$ is $y-t=\frac pt\left(x-\frac {t^2}{2p}\right)$ $\implies$ $\boxed{y=\frac {px}t+\frac t2}\ (*)$ $\implies$ $\left\{\begin{array}{cccc} MM\ : & y & = & \frac {px}m+\frac m2\\\\ NN\ : & y & = & \frac {px}n+\frac n2\end{array}\right\|$ $\implies$

$\left\{\begin{array}{c} \frac {px}m+\frac m2=\frac {px}n+\frac n2 \implies n\left(2px+m^2\right)=m\left(2px+n^2\right)\implies 2px(m-n)=mn(m-n)\ \stackrel{m\ne n}{\implies}\ 2px=mn\\\\ \frac {y-\frac m2}{y-\frac n2}=\frac {\frac {px}m}{\frac {px}n}\implies\frac {2y-m}{2y-n}=\frac nm\implies m(2y-m)=n(2y-n)\implies 2y(m-n)=m^2-n^2\implies\ m+n=2y\end{array}\right\|$ $\implies$ $\left\{\begin{array}{ccc} 2pa & = & mn\\\\ 2b & = & m+n\end{array}\right\|$

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263. Newton's sums. Applications.

262. IMO Shortlist 2005, Geometry 6th.

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257. Seeking roots of 2th equation with complex coef.

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251. Problems of Analytical Geometry.

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245. An inequality in an acute triangle.

244. Some properties of cyclic orthodiagonal quadrilaterals.

243. Nice ! (a+b) is constant.

242. A general geometrical problems.

241. Areas in a triangle.

240. Some interesting geometrical inequalities.

239. Algebraic marathon.

238. Some metrical problems.

237. Some problems with ... problems.

236. A locus with the metrical relation.

235. Proofs of some geometry problems with complex numbers.

234. Some simple and nice problems from contests.

233. An interesting identity and an easy & nice inequality.

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232. Some geometrical inequalities (own).

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231. A conditioned minimum value.

230. Some usual synthetical properties in a triangle.

229. Concurs online MATEFBC, ed. 5 -subiect 2.

228. OJM - Romania, 2010 problem 3.

227. IMO ShortList 2003 (problem 5).

226. Some inequalities from Calculus (Real Analysis).

225. An interesting problems from Kunny.

224. Some nice and easy inequalities in a triangle.

223. An inequality with complex numbers-RMO shortlist 2010.

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219. A very nice geometrical inequality.

218. Some problems from the Suiss IMO Selection Team 2006.

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216. Nice ! Find the length of hypotenuse (Belarus - 2010)

215. Limit of the unique root for polynominal equations.

214. A range of some functions.

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212. A simple applications of the Casey's theorem.

211. Pretty hard problem !

210. Tangential circle of the triangle ABC.

209. Nice problem, nice proof !

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207. Some interesting problems of Toshio Seimiya, Japan.

206. Four problems with extremum in a quadrilateral.

205. A "slicing" problem with 11 methods.

204. Saraghin 2010 (three geometry problems).

203. An application of the Pascal's permutation theorem.

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202. Concurrent lines in a right triangle.

201. Applications of Desarques' and Pappus' theorems.

200. Some nice applications of the inversion.

199. Some properties of symmedian. Two extensions.

198. Generalized Carnot's lemma.

197. Triangles outside of a triangle and concurrence.

195. Semicircle. Nice application of harmonical division.

194. The Lalescu's sequence. Properties.

193. An inequality with n! .

192. An easy and nice problem of Valentin Vornicu.

191. Six nice problems with the incircle of a triangle.

190. Very nice ! Coculescu's Contest seniors 2010.

189. Pascal's theorem and some nice and easy applications.

188. Some problems from vectorial geometry.

187. Distancies.

186. Without the l'Hospital's rules.

185. The Durrande's relation.

184. An interesting geometry problem from RMO 2010.

183. A remarkable characterization of equilateral triangle.

182. Another two nice problems with complex numbers (own).

181. Two special inequalities from CRUX (own).

180. Own proposed problem from CRUX (with complex numbers).

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179. The range of |z| , |z^2+a|=|bz+c|. Particular cases.

178. Some nice and simple geometry problems.

177. A nice proof of one identity in a triangle.

176. Another nice problems for middle school.

175. A Geometric Proof (without trig.) of Heron's Formula.

174. Some interesting properties of modulus (middle school).

173. An interesting trigonometric equations.

172. The Zelinsky's problem.

171. Another "slicing" proposed problems (middle school).

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170. A nice "slicing" proposed problem for middle school.

169. Some metrical problems in a circle (III).

168. Some properties of the Lemoine's point.

167. Nice problems - OIM, 1995 and ... Toshio Seimyia.

166. Some constant geometrical expressions.

165. Some metrical problems (5) in a circle (II).

164. Some metrical problems (5) in a circle.

163. Applications of Ptolemy's inequality. Ex. - China,1998.

162. Generalization of proposed problem from Mongolia, 2000.

161. A difficult inequality (barycentrical coordinates)

160. A very nice "slicing" problem.

159. Multiple derivatives in a point.

158. Very nice and difficult limits.

157. Steinbart's theorem. Applications.

156. Some problems with perpendicularities.

155. A constant ratio in a triangle ABC.

154. Difficult inequality (without derivatives eventually).

153. Geometrical minimum/maximum.

152. Concurrence, cyclically, metrics, symmedian a.s.o.

151. Miscellaneous problems for the admission at math.

150. Some (5) problems with fixed points.

149. Some (8) problems with collinearity or concurrence.

148. A special class of systems.

147. A problem with two polynominals.

146. Synthetic, metric, trigonometric and analitycal proofs.

145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

143. Assess of a ratio (nice and difficult problem).

142. Inscribed/circumscribed quadrilateral w.r.t. ABC.

141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).

140. A synthetical property <==> linear/angled relation.

139. A caracterization of a right angle in a triangle.

138. Generalized "slicing" problems (own).

137. Stewart's relation.

136. A geometrical interpretation of a system (own).

135. Three strong geometrical inequalities.

September 2010

134. Some problems with projections/perpendiculars (2).

133. Some problems with projections/perpendiculars (1).

132. Asupra unei inegalitati intr-un triunghi.

131. A stronger Form of the Steiner-Lehmus Theorem (own).

130. A triangle and a fixed point.

129. BX=CY and three properties.

128. Value of IG and two inequalities in neighbourhood of 0.

127. Pagina educativa.

126. Routh's theorem.

125. I like very much this Darij Grinberg's message.

124. A nice and easy problem with a right-angled triangle.

123. My extension and its proof.

122. An identity for an equilateral triangle. Extension.

121. Viete's relations with cos A , cos B , cos C.

120. Two equal neighbouring angles/sides in a quadrilateral.

119. Inspired by a Miculita's problem.

118. An usual remarkable geometrical locus.

117. Some applications of harmonic division/quadrilateral.

116. The length of AE (nine methods !).

115. An difficult equation with one radical.

114. A property of tangential triangle for ABC.

113. Range for a function.

112. The equivalence relation ".s.s."

111. Simple, but interesting ...

110. German TST 2004 - Exam V , Problem 1

109. Two equal areas ==> find A.

108. Old, short and nice proof of Euler's relation.

107. XY parallel BC and X,Y are isogonal/izotomic points.

106. OLM - Bucuresti (geometrie).

105. A problem with three circles.

104. Concursul Revistei de Matematica din Galati (RMG).

103. Metrical usual relations in triangle. Applications.

102. Characterization of a metrical relation in a triangle.

101. O problema cu numere complexe.

100. Bobiller's theorem.

99. Some simple and nice geometry problems.

98. Symmedian & median (metrical relations).

97. Problem of butterfly.

August 2010

96. Integrals.

95. Three circles in an angle.

94. Equations of angle bisectors (interior and exterior).

93. The Appolonius' construct problems.

92. CA , CB tangent to parabola - OIMU 2008 Problem 4.

91. JBMO 2010, Problem 3.

90. Nice and difficult inequality in ABC (own).

89. Similarity (parallel/antiparallel).

88. A nice geometric locus conditioned metrically.

87. An interesting vectorial identity.

86. A very nice maxmin.

85. Problem "slicing" : 60-24-12.

84. IRAN national math olympiad - 2010

83. A metrical characterization of concurrence.

82. AA' , BN , CM concur in Gergonne point.

81. Sawayama and Thebault’s theorem.

80. Three concurrent perpendiculars (the Carnot's lemma).

79. A quadrilateral with many perpendicularities.

78. Jakobi's theorem.

77. Two equivalent perpendicularities.

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75. Extended Steinbart Theorem.

74. Inegalitatea Erdos-Mordell.

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70. Colinearity with pole/polar - H\in PQ .

71. An angle questions (synthetic/trigonometric proofs).

69*1. Slicing problem 3.

69. Position of x1, x2 w.r.t. a given real number k.

68. Some nice applications of "pole/polar".

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66. Remarkable perpendicularities in a triangle.

65. Parallel to BC through I .

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63. A nice problem with radical center (livetolove212).

62. Nice and easy problem with a right triangle.

61. IMO Shortlist 2009 (very nice problem and its proof).

60. Mediteranean M.C. 2010 Problem 3.

59. IMO 2010, problem 4.

58. A cevian and two incircles.

57. Opinii si comentarii relativ la inv. rom. II

56. A extremum problem with areas.

55. Two important circles - mixtilinear incircle/excircle.

54. Cinci grade de libertate.

53. An remarkable concurrency.

52. Harmonical division.

51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.

51. A nice problem with perpendicularity from Jaime.

50. IMAC 2010 seniors, the first day.

49. The midpoint of a cevian in an acute triangle.

June 2010

48. An inequality in a triangle : h_a+max{b,c} ...

47. Interesting and difficult identities in a triangle.

46. Outside of a quadrilateral.

45. \cos B+\cos C=1 <==> nice property.

44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.

43. A "slicing" problem with a parameter.

42. IMAC 2010 Problema 2.

41. ONM 2010 Calarasi Problema 3.

May 2010

40. Lema lui Haruki.

39. A nice and remarkable problem with Brocard's point.

38. Minimum area of triangle touching with semicircle.

37. Own extension of the Steiner-Lehmus' problem.

36. ONM (juniori) - 2010 Iasi, problema 4.

35. Nice problem from Arqady (Eduard_Tur).

34. IMO Shortlist 2002 geometry problem 7.

April 2010

32. Linkuri diverse.

31.Some nice metrical problems.

29. Inegalitate integrala de la Kunny.

28. A fost odată ...

27.Problem of Darij Grinberg (I - centroid of triangle DEF).

26. Outside of a triangle.

25. Geometric equations.

24. A fine and cool inequalities.

23. A very nice exponential inequality (own - from CRUX).

22. Some interesting limits (sequence/function).

21 bis. Some problems with complex numbers.

21. Probleme de geometrie (Yetty & I).

20. O ineg.cu suma de AG/AI .

19. Own proposed problems in CRUX Mathematicorum - Canada.

18. O problema a lui Toshio Seimiya.

17. O inegalitate "tare" intr-un triunghi.

16. Length of the (interior) angle bisector (10 proofs).

15. Inequalities stronger than Panaitopol's.

14. Opinii si comentarii relativ la inv. rom.

13. Inegalitatea de la Sorin Borodi.

12. Inegalitatea I. Maftei & S. Radulescu (2).

11. Inegalitatea I. Maftei & S. Radulescu (1).

10. Walker's inequality and its extension.

9. Inegalitatea HO+HA+HB+HC >= 3R .

8. The first problem Shortlist ONM 2010 Romania.

7. Some interesting "slicing" problems.

6. Interesting problems from JBMO.

5. Theoretical preliminary for inequalities.

4. Remarkable geometrical inequalities (2).

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