251. Problems of Analytical Geometry.

by Virgil Nicula, Mar 11, 2011, 7:03 AM

PP1. Let $w=C(O,r)$ be a circle and let $d$ be a line which is tangent to $w$ in the fixed point $A\in w$ . Consider two mobile

points $\{M,N\}\subset d$ so that $\overline{AM}\cdot\overline {AN}=k\in \mathbb R$ , where $k$ is constant and $k\not\in \{\pm r\}$ . Denote the intersection $L$

of the tangents (different from $d$) from $M$ , $N$ to $w$ . Prove that the geometrical locus of $L$ is a line $l$ which is parallelly with $d$ .


Proof. Choose one system of coordinates $xOy$ so that the equation of the circle $w$ is $x^2+y^2=1$ and the equation of the line $d$ is $y=1$ , i.e. $A(0,1)$ .

Consider two mobile points $M(m,1)$ , $N(n,1)$ for which $mn=k\in \mathbb R$ (constant) and $k\not\in\{\pm 1\}$ . Show easily that the slopes $s_m$ , $s_n$ of the tangents

$t_m$ , $t_n$ from $M$ , $N$ respectively (different from $d$) to $w$ are $\left\|\begin{array}{ccc}
\delta_{t_m}(O)=1 & \implies & s_m=\frac {2m}{m^2-1}\\\\
\delta_{t_n}(O)=1 & \implies & s_n=\frac {2n}{n^2-1}\end{array}\right\|$ and the equations of the tangents $t_m$ , $t_n$

are $\left\|\begin{array}{ccccc}
t_m & : & y-m=\frac {2m}{m^2-1}\cdot (x-m) & \implies & 2mx+\left(1-m^2\right)y-\left(1+m^2\right)=0\\\\
t_n & : & y-n=\frac {2n}{n^2-1}\cdot (x-n) & \implies & 2nx+\left(1-n^2\right)y-\left(1+n^2\right)=0\end{array}\right\|$ $\implies$ $L\left(\frac {m+n}{1+mn}, \frac {1-mn}{1+mn}\right)\in t_m\cap t_n$ .

In conclusion, $y_L=\frac {1-k}{1+k}$ , i.e. the geometrical locus of the mobile point $L$ is a line $l$ for which $l\parallel d$ . See second and third problems from
here.

See
here and here another problems of analytical geometry.


PP2. Ascertain $\min_{\{a,b\}\subset\mathbb R}\left(a^2+b^2\right)$ for which the equation $x^4+ax^3+bx^2+ax+1=0\ (*)$ has at least a real root.

Proof. For one fixed $x$ the relation $(*)$ is the equation $x\left(x^2+1\right)\cdot \underline a+x^2\cdot \underline b+\left(x^4+1\right)=0$ of a line $d$ in the analytic system $aOb$ . The distance $\sqrt {a^2+b^2}$ from

origin $O$ to $P(a,b)\in d$ must be equally to the distance $\frac {x^4+1}{\sqrt{x^2\left(x^2+1\right)^2+\left(x^2\right)^2}}$ from the origin to the line $d$ . Thus, $\sqrt {a^2+b^2}=\frac {x^4+1}{\sqrt{x^2\left(x^2+1\right)^2+\left(x^2\right)^2}}$ .

The minimum value of the function $f(x)=\frac {x^4+1}{\sqrt{x^2\left(x^2+1\right)^2+\left(x^2\right)^2}}$ , $x\in\mathbb R$ is $f(1)=\frac {2}{\sqrt 5}$ . In conclusion, $\min_{\{a,b\}\subset\mathbb R}\left(a^2+b^2\right)=\frac 45$ .



PP3. Show that if the equation $x^4 + ax^3 + 2x^2 + bx + 1=0$ has at least a real root, then $a^2 + b^2\ge 8$ .

Proof. Consider the equation $x^3\cdot a+x\cdot b+\left(x^2+1\right)^2=0$ of the line $d$ in the analytical system $aOb$ , where $x\in \mathbb R$ . The distance $\sqrt {a^2+b^2}$ from the origin $O$

to $M(a,b)\in d$ is equally to the distance $\delta =\frac {\left(x^2+1\right)^2}{\sqrt {x^2\left(x^4+1\right)}}$ from the origin to the line $d$ . Thus, $\sqrt {a^2+b^2}=\delta \ge 2\sqrt 2$ . In conclusion, $a^2+b^2\ge 8$ .

Remark. Clearly $x\ne 0$. Thus, $x^2+ax+2+\frac {b}{x}+\frac{1}{x^2}=0\iff$ $\frac{a^2+b^2}{4}=\left(x+\frac{a}{2}\right)^2+\left(\frac {1}{x}+\frac{b}{2}\right)^2+2\ge 2\implies $ $a^2+b^2\ge 8$ .



PP4. Let $\mathbb E$ be an ellipse with foci $F$ and $F'$ . Consider for a point $P\in \mathbb E$ the tangent line $t$ and the normal line $n$

through the point $P$ to the given ellipse. Prove that normal line $n$ is the angle bisector of the angle $\widehat{FPF'}$ .


Proof. Let $\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$ be the equation of the ellipse $\mathbb E$ with the foci $F(c,0)$ , $F'(-c,0)$ , where $a^2=b^2+c^2$ . Let $P(au,bv)\in\mathbb E$ , where $u^2+v^2=1$ .

Observe that $\frac {2x}{a^2}+\frac {2yy'}{b^2}=0$ $\implies$ $y'(x,y)=-\frac {b^2x}{a^2y}$ $\implies$ $y'(P)=y'(au,bv)=$ $-\frac {b^2au}{a^2bv}\implies$ the slope of the tangent line to the ellipse in the point $P$ is

$s_t(P)=-\frac {bu}{av}$ and the slope of the normal line to the ellipse in the point $P$ is $s_n(P)=\frac {av}{bu}$ . Therefore, the equation of the tangent line $t$ is $y-bv=-\frac {bu}{av}\cdot (x-au)$

and the equation of the normal line $n$ is $y-bv=\frac {av}{bu}\cdot (x-au)$ . If denote $T\in Ox\cap t$ and $N\in Ox\cap n$ , then prove easily that $T\left(\frac au,0\right)$ and $N\left(\frac {c^2u}{a},0\right)$ .

Observe that $\left(x_F+x_{F'}\right)\cdot\left(x_T+x_N\right)=2\cdot\left(x_F\cdot x_{F'}+x_T\cdot x_N\right)\iff$ $0=2\cdot\left(-c^2+\frac au\cdot\frac {c^2u}{a}\right)$ , what is truly.

In conclusion, the division $(F',N,F,T)$ is harmonically and in this case $PT\perp PN\iff$ $\widehat{NPF'}\equiv\widehat{NPF}$ .



PP5. Let $ABCD$ be a square and let $O$ be its center. If $P$ is a point lying on the segment $[AD]$ , prove that there exists the points $M$ ,

$N$ lying on the segments $[AB]$ and $[BC]$ respectively such that $O$ is the orthocenter of $\triangle\, MNP$ if and only if $3\cdot AP\ge 2\cdot AD$ .


Proof. Choose $A(-1,-1)$ , $B(-1,1)$ , $C(1,1)$ , $D(1,-1)$ and $M(-1,m)$ , $N(n,1)$ , $P(p,-1)$ , where $|m|\le 1$ , $|n|\le 1$ , $|p|\le 1$ .

Therefore, $O$ is the orthocenter of $\triangle MNP\iff$ $\left\{\begin{array}{ccc}
PO\perp MN & \implies & \underline m+p\cdot \underline n=1-p\\\\
MO\perp PN & \implies & 2\cdot \underline m-\underline n=-p\end{array}\right|\iff$ $\left\{\begin{array}{c}
m=-\frac {p^2+p-1}{2p+1}\\\\
n=\frac {2-p}{1+2p}\end{array}\right|$ .

Since $AD=2$ and $AP=p+1$ , from $|m|\le 1$ , $|n|\le 1$ obtain that $p\ge \frac 13\iff$ $3(p+1)\ge 2\cdot 2\iff$ $3\cdot AP\ge 2\cdot AD$ .

Remark. $\left\{\begin{array}{ccccc}
|m|\le 1 & \iff & |p^2+p-1|\le |2p+1| & \iff & (p^2+3p)(p^2-p-2)\le 0\\\\
|n|\le 1 & \iff & |p-2|\le |2p+1| & \iff & (3p-1)(p+3)\ge 0\end{array}\right|$ $\implies$ $p\in\left[\frac 13,1\right]$ .



PP6. Find the maximum area of the triangle formed by the intersection points between an ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ (a>b>0)$ with the line passing through one of the foci and the other focus.


Proof. Given the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ with foci $F'(-c,0)$ and $F(c,0)$ . Let $y=\lambda\cdot (x-c)$ be the equation of a line $d$ through $F$ . Can suppose w.l.o.g. by

symmetry w.r.t. the $xx'$-axis that $\lambda >0$ and the line $d$ cuts the ellipse in $M$ and $N$ , where $y_n<0<y_m$ . Solving the system $\left\{\begin{array}{c}
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\\\\
y=\lambda\cdot (x-c)\end{array}\right\|$ obtain that

$b^2\left(c+\frac y\lambda\right)^2+a^2y^2=$ $a^2b^2\iff$ $b^2\left(y+\lambda c\right)^2+a^2\lambda ^2y^2=$ $a^2b^2\lambda ^2\iff$ $\left(a^2\lambda ^2+b^2\right)y^2+2b^2c\lambda y+\lambda ^2b^2\left(c^2-a^2\right)=0$ . Since

$\boxed{a^2=b^2+c^2}$ obtain that $\boxed{\left(a^2\lambda ^2+b^2\right)y^2+2b^2c\lambda y-\lambda ^2b^4=0}$ . Since $\Delta '=$ $b^4c^2\lambda ^2+\lambda ^2b^4\left(a^2\lambda ^2+b^2\right)=$ $\lambda ^2b^4\left(c^2+a^2\lambda ^2+b^2\right)$

$=\lambda ^2b^4\left(a^2+a^2\lambda ^2\right)=$ $a^2b^4\lambda ^2\left(\lambda ^2+1\right)$ obtain that $\left\{y_m,y_n\right\}=\frac{\lambda b^{2}(-c\pm a\sqrt{1+\lambda ^{2}})}{a^{2}\lambda ^{2}+b^{2}}$ . Since $F'F=2c$ (constant) $[AMN]=c\cdot \left(y_m-y_n\right)$

is $\mathrm{max.}\iff$ $y_m-y_n=\frac {2ab^2\lambda \sqrt{\lambda ^2+1}}{a^2\lambda ^2+b^2}$ is $\mathrm{max.}\iff$ $f(\lambda )=\frac {\lambda \sqrt{\lambda ^2+1}}{a^2\lambda ^2+b^2}$ is $\mathrm{max.}$ and in this case $[AMN]=2acb^2\cdot f(\lambda )$ . Prove easily that

$f'(\lambda )\ .s.s.^{(*)}\ \lambda ^2\cdot \left(2b^2-a^2\right)+b^2$ . In conclusion, $\left|\begin{array}{c}
b<a\le b\sqrt 2\ \implies\ f(\lambda )\le f(\infty )=\frac {1}{a^2}\ \implies\ [AMN]\le\frac {2cb^2}{a}\\\\
b\sqrt 2<a\ \implies\ f(\lambda )\le f\left(\frac {b}{\sqrt{a^2-2b^2}}\right)=\frac {1}{2bc}\ \implies\ [AMN]\le ab\end{array}\right|$ $\implies [AMN]\le ab$ .

$(*)$ Remark. I"ll use the notation $X\ .s.s.\ Y\ \iff\ X=Y=0\ \vee\ XY>0\ \iff\ \mathrm{sign} (X)=\mathrm{sign} (Y)$ , i.e. $X$ and $Y$ have Same Sign.


PP7. If $(x-3)^2+(y-3)^2 = 4$ , then find range of $xy$ .

Proof 1. Using the geometric interpretation, it means finding the tangency points between the circle centred at $(3,3)$ of radius $2$ and the family of equilateral

hyperbolae $xy = k$. Because of the symmetry of the configuration, these occur at points on the first bisectrix $x=y$ . Thus at $x=y=3\pm \sqrt{2}$,

therefore $(3-\sqrt{2})^2 \leq xy \leq (3+\sqrt{2})^2$ .

Proof 2. Using the substitutions $\left\{\begin{array}{c}
x=3+2\cos \phi\\\
y=3+2\sin\phi\end{array}\right\|$ , where $\phi\in\left[0,2\pi\right]$ obtain that $xy=2\sin 2\phi +6(\cos\phi +\sin\phi )+9$ , i.e. $xy=f(t)$ , where

$f(t)=2t^2+6t+7$ and $t=\cos\phi +\sin\phi\ ,\ |t|\le\sqrt 2$ . Since $-\frac 32<-\sqrt 2<\sqrt 2$ prove easily that the range of the function $f:\left[-\sqrt 2,\sqrt 2\right]\rightarrow \mathbb R$

is $\left[f\left(-\sqrt 2\right), f\left(\sqrt 2\right)\right]$ , i.e. $\left[11-6\sqrt 2, 11+6\sqrt 2\right]$ , i.e. $\left|xy-11\right|\le 6\sqrt 2$ .
This post has been edited 101 times. Last edited by Virgil Nicula, Nov 22, 2015, 11:07 AM

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