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

by Virgil Nicula, Aug 28, 2010, 3:22 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=364018

$\blacktriangleright$ PP1. $A,B$ of $\triangle ABC$ belong to parabola $\mathbb P$ with equation $y=ax^2 + bx + c$ , $a>0$ and $CA$ , $CB$ are tangent to $\mathbb P$ . Find $\frac{S^2}{m_c^3}$ , where $m_c$ is length of $C$-median.

Preliminary. We know that $\boxed {(2ax+b)^2=4ay+\Delta}\ (*)$ for any point $P(x,y)$ of parabola $\mathbb P$ , where $\Delta=b^2-4ac$ . For $\mathbb P$ is well-known the vertex $V\left(-\frac {b}{2a}\ ,\ -\frac {\Delta}{4a}\right)$ , the focus

$F\left(-\frac {b}{2a},-\frac {\Delta}{4a}+\frac p2\right)$ , i.e. $FV=\frac p2\ ,$ $(a>0\iff p>0)$ and the director $d$ with equation $y=\left(-\frac {b}{2a},-\frac {\Delta}{4a}-\frac p2\right)$ so that for any point $P(x,y)\in\mathbb P$ there is the definition relation

$\delta_d(P)=PF$ . Thus, $\left(y+\frac {\Delta}{4a}+\frac p2\right)^2=$ $\left(x+\frac {b}{2a}\right)^2+\left(y+\frac {\Delta}{4a}-\frac p2\right)^2$ $\iff$ $\left(x+\frac {b}{2a}\right)^2=p\left(y+\frac {\Delta}{4a}\right)$ $\iff$ $(2ax+b)^2=ap(4ay+\Delta)$ $\stackrel{(*)}{\iff}$

$4ay+\Delta=ap(4ay+\Delta)\ ,\ \forall\ P\in \mathbb P$ , i.e. $\forall\ y\ge -\frac {\Delta}{4a}$ $\iff$ $\boxed{\ p=\frac 1a\ }\ (**)$ .

Proof 1 (analytic). Suppose w.l.o.g. that equation of parabola $\mathbb P$ is $y^2=2px$ , where vertex is $V(0,0)$ , focus is $F\left(\frac p2,0\right)$ and equation of director $d$ is $x=-\frac p2$ . Suppose that

$A\left(\frac {m^2}{2p},m\right)\ ,$ $B\left(\frac {n^2}{2p},n\right)$ . $C(x,y)$ is intersection of tangents in $A$ , $B$ to parabola $\mathbb P$ and which have equations $\left\|\begin{array}{ccc}
AA & \implies  & my=px+\frac {m^2}{2}\\\\
BB & \implies & ny=px+\frac {n^2}{2}\end{array}\right\|$ . Thus, $\boxed{\begin{array}{c}
m+n=2y\\\\
mn=2px\end{array}}$ . The

midpoint of $[AB]$ is $M\left(\frac {m^2+n^2}{4p},\frac {m+n}{2}\right)$ and $CM\parallel Ox$ . Thus, $C\left(\frac {mn}{2p},\frac {m+n}{2}\right)$ and $m_c=CM=x_M-x_C=$ $\frac {m^2+n^2}{4p}-\frac {mn}{2p}$ $\iff$ $\boxed {4p\cdot m_c=(m-n)^2}$ . The area

$S=[ABC]$ is $S=\frac 12\cdot$ $\mod\left|\begin{array}{ccc}
\frac {m^2}{2p} & m & 1\\\\
\frac {n^2}{2p} & n & 1\\\\
\frac {mn}{2p} & \frac {m+n}{2} & 1\end{array}\right|\iff$ $16p^3S=\mod\left|\begin{array}{ccc}
m^2 & 2pm & 2p\\\\
n^2 & 2pn & 2p\\\\
mn & p(m+n) & 2p\end{array}\right|\iff$ $8pS=\mod\left|\begin{array}{ccc}
m^2 & 2m & 1\\\\
n^2 & 2n & 1\\\\
mn & (m+n) & 1\end{array}\right|$ . With invariant transformations

$\left\|\begin{array}{c}
L_1:=L_1-L_3\\\\
L_2:=L_2-L_3\end{array}\right\|$ get $8pS=\mod\left|\begin{array}{ccc}
m(m-n) & m-n & 0\\\\
n(n-m) & n-m & 0\\\\
mn & m+n & 1\end{array}\right|\implies$ $\boxed {8pS=|m-n|^3}$ $\iff\frac{S^2}{m_c^3}=p$ $\stackrel{(**)}{\ \iff\ }$ $\boxed {\frac{S^2}{m_c^3}=\frac 1a}$ .

Proof 2 (shorter, synthetic-analytic). $S=\frac {c\cdot m_c}{2}\cdot \sin\left(\widehat{AB,Ox}\right)$ $\iff$ $\boxed {S=\frac {m_c}{2}\cdot \mathrm{pr}_{Oy}AB}$ . Observe that $\mathrm{pr}_{Oy}AB=|m-n|$

and $CM\parallel Ox$ $\iff$ $m_c=|x_M-x_C|=\frac {m^2+n^2}{4p}-\frac {mn}{2p}$ , i.e. $m_c=\frac {(m-n)^2}{4p}$ . Therefore, the area $S$ of $\triangle ABC$ is $S=\frac {|m-n|^3}{8p}$ a.s.o.

Propose four similar problems which use the remarkable relations $\boxed{\begin{array}{c}
m+n=2y\\\\
mn=2px\end{array}}$ between the points $A\left(\frac {m^2}{2p},m\right)\in\mathbb P\ ,\ B\left(\frac {n^2}{2p},n\right)\in\mathbb P$ and $C(x,y)\in AA\cap BB$ .
[/size]

AP1 $\blacktriangleright$ Let $\mathbb P$ be the parabola with equation $y^2=2px$ , where $p>0$ . Consider a fixed point $D\left(x_0,y_0\right)$ and two mobile points $\{M,N\}\subset\mathbb P$

so that $D\in MN$ . Find the locus of $L\in MM\cap NN$ , where mean by $XX$ - the tangent in the point $X\in\mathbb P$ to the parabola $\mathbb P$ . Study some particular cases.

AP2 $\blacktriangleright$ Let $\mathbb P$ be the parabola with equation $y^2=2px$ , where $p>0$ . Consider a fixed circle $w=C(A,r)$ , where $A(r,0)$ and two mobile points $\{M,N\}\subset\mathbb P$

so that $AB$ is tangent to $w$ . Find the locus of $L\in MM\cap NN$ , where mean by $XX$ - the tangent in the point $X\in\mathbb P$ to the parabola $\mathbb P$ . Study some

particular cases. What is the locus of $L$ if instead of the circle $w$ have the circle $w'=C(O,r)$ ? What happen in the general case of some circle ?

AP3 $\blacktriangleright$ Let $\mathbb P_k$ , $\overline{1,3}$ be three parabolas with equations $y^2=2p_kx$ , $k\in\overline{1,3}$ where $0<p_1<p_2<p_3$ and $p_2^2=p_1p_3$ . Consider two mobile

points
$\{M,N\}\subset \mathbb P_2$ so that the line $MN$ is tangent to $\mathbb P_1$ . Prove that the intersection of tangents to $\mathbb P_2$ in the points $M$ , $N$ belongs to $\mathbb P_3$ .

AP4 $\blacktriangleright$ Given are the parabola $ y^{2}= 2px$ and the circle $ x^{2}+y^{2}= px$, where $ p > 0$ . The tangent in a mobile point $ M$ which belongs to the circle cut the

parabola in the points $ X$, $ Y$ . Denote the intersection $ L$ of the tangents in the points $ X$, $ Y$ to the parabola. Ascertain the geometrical locus of the point $ L$ .



$\blacktriangleright$ PP2. Let $f:\mathbb R\rightarrow\mathbb R$ be a function, where $f(x)=mx^3+nx^2+px+r$ and $m\ne 0$ . Let $d$ be a line so that $d\cap \mathbb G_{f}=\left\{P_1,P_2,P_3\right\}$ .

For any $k\in\overline{1,3}$ the tangent to $\mathbb G_f$ in the point $P_k\in\mathbb G_f$ cut again $\mathbb G_f$ in the point $R_k$ . Prove that the points $R_k\ ,\ k\in\overline{1,3}$ are collinearly.


Proof. Let $\left\{\begin{array}{ccc}
P_k\left(x_k, y_k\right) & ; & y_k=f\left(x_k\right)\\\\
R_k\left(r_k,z_k\right) & ; & z_k=f\left(r_k\right)\end{array}\right\|$ where $k\in\overline{1,3}$ . Thus, $P_1\in P_2P_3 \iff \left|\begin{array}{ccc}
x_1 & f\left(x_1\right) & 1\\\\
x_2 & f\left(x_2\right) & 1\\\\
x_3 & f\left(x_3\right) & 1\end{array}\right|=0\iff$ $\left|\begin{array}{ccc}
x_1 & f\left(x_1\right) & 1\\\\
x_2-x_1 & f\left(x_2\right)-f\left(x_1\right) & 0\\\\
x_3-x_1 & f\left(x_3\right)-f\left(x_1\right) & 0\end{array}\right|=0\iff$

$\left(x_2-x_1\right)\left(x_3-x_1\right)\left(x_3-x_2\right)\left[m\left(x_1+x_2+x_3\right)+n\right]=0\iff$ $\boxed{x_1+x_2+x_3=-\frac nm}$ . For any $k\in\overline{1,3}\ ,\ R_k\in P_kP_k$ $\implies$ for any $k\in\overline{1,3}\ ,$ $2x_k+r_k=-\frac nm\implies$

$\sum_{k=1}^3\left(2x_k+r_k\right)=-\frac {3n}{m}\iff$ $2\sum_{k=1}^3x_k+\sum_{k=1}^3r_k=-\frac {3n}{m}\iff$ $-\frac {2n}{m}+\sum_{k=1}^3r_k=-\frac {3n}{m}\iff$ $\sum_{k=1}^3r_k=-\frac nm\iff$ $R_k\ ,\ k\in\overline{1,3}$ are collinearly.
This post has been edited 63 times. Last edited by Virgil Nicula, Nov 27, 2015, 6:50 AM

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