A beautiful geometry problem

Question

Let $PP'$ and $QQ'$ be two parallel lines tangent to a circle of center $C$ and radius $r$ in the points $P$ and $Q$, respectively. $P'Q'$ cuts de circle in $M$ and $N$. Let $Y$ and $X$ be the points in which $Q'Q$ is cut by $PN$ and $PM$, respectively. Given the lengths $PP'= p$, $QQ'= q$ and $2r = d$, find the lengths $QY = y$ and $QX = x$.

I've been struggling with this problem for a couple of days, so a hint or a solution would be welcome.

Now, what makes this problem beautiful is the fact that if you let $p=-\dfrac{2a}b$ and $q =-\dfrac{c}{2b}$ then the lengths $y$ and $x$ will be the real roots of the equation $ax^2 + 2bx + c = 0$.

Answer 1

HINTS:

Direct ( BFI :) ) method

Let radius of circle be $ r$

Transversal through circle center

$$ \frac{y-r}{x-p} = \frac{2r}{p+q} $$

Central circle Equation

$$ x^2+y^2 =r^2$$

Points of intersection $$ ( N,M)= (x_1,y_1), (x_2,y_2) $$

Form Left slant line Equation L in which put $ \rightarrow y=0 $

Form Right slant line Equation R in which put$ \rightarrow y=0 $

(Next verify the problem beauty etc. in hindsight.)

Answer 2

$MX=x*\dfrac x{\sqrt{d^2+x^2}}$ ,$MQ=\sqrt{x^2-\dfrac{x^4}{d^2+x^2}}=\dfrac{xd}{\sqrt{d^2+x^2}}$ $PM=\sqrt{d^2-MQ^2}=d\sqrt{1-\dfrac{x^2}{d^2+x^2}}=\dfrac{d^2}{\sqrt{d^2+x^2}}$

Since $PMP'\sim MXQ'$

$p:q+x=PM:MX=d^2/\sqrt{d^2+x^2}:x^2/\sqrt{d^2+x^2}=d^2:x^2$

$⇔d^2(q+x)=px^2$

$⇔px^2-d^2x-d^2q=0$

$x=(d^2+d\sqrt{d^2+4pq})/2p $

Similarly, since $PNP'\sim NQ'Y$

$p:q-y=PN:NY=d^2:y^2 $ $py^2+d^2y-d^2q=0 $

$y=(-d^2+d\sqrt{d^2+4pq})/2p$

Then $(X-x)(X-y)=X^2-\sqrt{d^2+4pq}/p*X+qd^2/p$.

but $(X-x)(X+y)=X^2+\dfrac{d^2}pX-\dfrac{qd^2}p$. x and y are solutions of this equation.

Answer 3

I hadn't seen @Takahiro's argument, but mine follows the same basic approach (streamlined slightly with the help of some trig):

$$\begin{align} \triangle PP^\prime M\sim \triangle XQ^\prime M &\implies \frac{|\overline{XQ^\prime}|}{|\overline{PP^\prime}|} = \frac{d\sin^2\theta}{d\cos^2\theta} \quad \left( = \tan^2\theta \right)\\[4pt] &\implies \frac{q+x}{p} = \frac{x^2}{d^2} \end{align}$$ where $d$ is the diameter of the circle. Likewise, we have (but don't show) $$\frac{q-y}{p} = \frac{y^2}{d^2}$$ Thus, $x$ and $-y$ (note ---as Takahiro did--- the sign change!) are roots of $$ z^2 p - d^2 z - d^2 q = 0 \tag{$\star$}$$

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Unfortunately, the substitutions $p = -\frac{2a}{b}$ and $q = -\frac{c}{2b}$ transform $(\star)$ into $$4 a z^2 + 2 b d^2 z - c d^2 = 0 \tag{$\star\star$}$$ which is not the "beautiful" relation promised. Perhaps OP intended $p = -\frac{2 a r^2}{b} = -\frac{ad^2}{2b}$ and $q = \frac{c}{2b}$. (I'm not sure I see what's so "beautiful" about those rather ad hoc assignments, however.)