Geometry with a circle from a Romanian contest

Question

Let $M$ be a point exterior to a circle $\mathcal{C}(O, r)$. Let us consider the tangents $MA$ and $MB$ to this circle ($A$ and $B$ are on the circle). The parallel lines through $A$ and $B$ to the lines $MB$ and $MA$ intersect the circle the second time in $C$ and $D$, respectively, and the lines $MC$ and $MD$ intersect the circle the second time in $E$ and $F$ respectively. Let $\{Q\}=BF \cap MA$ and $\{P\}=AE\cap MB$. If $x\cdot \overrightarrow{AB} + y \cdot \overrightarrow{AC} + z \cdot \overrightarrow{AQ}+ t \cdot \overrightarrow{MP}=\overrightarrow{0}$, where $x, y, z, t\in \mathbb{R}^{*}$, then it is true that:
a) $2x+z=1$
b) $x+z=0$
c) $\frac{z-t}{y}=\frac{AC}{MP}$
d) $\frac{z+t}{y}=\frac{AC}{PB}$
e) $\frac{z-t}{y}=\frac{4AC}{MB}$.

This question appeared in a Romanian contest for high school students approximately a month ago. I have been trying to solve it for quite a while, but I haven't made much progress. I think that I should write the power of $M$ with respect to our circle because this would give me some relation between those lengths. Another thing that I observed, yet I don't know how to prove, is that $PQ \parallel AB$. This together with Thales' Theorem would give us that $\frac{MP}{PB}=\frac{MQ}{QA}$. Apart from these, I haven't found much more.

EDIT: As suggested in the comments, I will add here that the correct answer is c).

Answer 1

First of all, let us prove that $P$ and $Q$ are midpoints of $\overline{MB}$ and $\overline{MA}$.

Proof :

Let $$G=AC\cap BF,\quad H=AC\cap BD,\quad I=AC\cap MD$$

Also, let $$AG=a,\quad HI=b,\quad MB=AH=MA=BH=k,\quad BF=t$$

Since $\triangle{GQA}\sim\triangle{BQM}$, we have $$QA:QM=AG:MB=a:k\implies QA=\frac{ak}{a+k},QM=\frac{k^2}{a+k}$$

Since $\triangle{IDH}\sim\triangle{MDB}$, we have $$HD:BD=IH:MB=b:k\implies BD=\frac{k^2}{b+k}$$

Since $\triangle{MQF}\sim\triangle{DBF}$, we have $$QM:BD=QF:BF\implies \frac{k^2}{a+k}:\frac{k^2}{b+k}=QF:t\implies QF=\frac{(b+k)t}{a+k}$$

Since $QA^2=QF\times QB$, we have $$\bigg(\frac{ak}{a+k}\bigg)^2=\frac{(b+k)t}{a+k}\times\bigg(t+\frac{(b+k)t}{a+k}\bigg)$$ $$\implies t^2=\frac{a^2k^2}{(b+k)(a+b+2k)}\tag1$$

Since $\triangle{DMB}\sim\triangle{BMF}$, we have $$MD:MB=BD:FB\implies MD:k=\frac{k^2}{b+k}:t\implies MD=\frac{k^3}{t(b+k)}$$

Since $\triangle{FDB}\sim\triangle{FMQ}$, we have $$DF:MF=BF:QF=t:\frac{(b+k)t}{a+k}\implies MF=\frac{k^3}{t(a+b+2k)}$$

Since $\triangle{DMB}\sim\triangle{BMF}$, we have $$MB:MD=MF:MB\implies k:\frac{k^3}{t(b+k)}=\frac{k^3}{t(a+b+2k)}:k$$ $$\implies t^2=\frac{k^4}{(b+k)(a+b+2k)}\tag2$$

It follows from $(1)(2)$ that $a=k$, i.e. $AG=MB$ from which $AQ=MQ$ follows.

Similarly, we have $BP=MP$. $\quad\blacksquare$

Let $\vec{AM}=\vec a,\vec{AH}=\vec b$.

Then, we have $$\begin{align}&x\cdot \overrightarrow{AB} + y \cdot \overrightarrow{AC} + z \cdot \overrightarrow{AQ}+ t \cdot \overrightarrow{MP}=\overrightarrow{0} \\\\&\implies x(\vec a+\vec b) + y \cdot \frac{AC}{AH}\cdot \vec b + z \cdot \frac 12\vec a+ t\cdot\frac 12\vec b=\overrightarrow{0} \\\\&\implies \bigg(x+\frac z2\bigg)\vec a+\bigg(x+\frac y2\cdot \frac{AC}{MP}+\frac t2\bigg)\vec b=\vec 0 \\\\&\implies x+\frac z2=0\qquad\text{and}\qquad x+\frac y2\cdot \frac{AC}{MP}+\frac t2=0 \\\\&\implies 2x+z=0\qquad\text{and}\qquad \frac{z-t}{y}=\frac{AC}{MP}\end{align}$$

• Suppose that (a) is correct. Then, we have $1=2x+z=0$.

• Suppose that (b) is correct. Then, we have $x=z=0$.

• Suppose that (d) is correct. Then, we have $\frac{z+t}{y}=\frac{z-t}{y}\implies t=0$.

• Suppose that (e) is correct. Then, we have $\frac{z-t}{2y}=\frac{z-t}{y}\implies z=t\implies AC=0$.

Therefore, the only correct option is $\color{red}{\text{(c)}}$.

Answer 2

We can consider the special case where the tangents make a $60^\circ$ angle at $M$. Then $C$ and $D$ coincide, as do $E$ and $F$; we find $P$ and $Q$ to be the midpoints of $\overline{MB}$ and $\overline{MA}$. (The midpoint property happens to be true in all cases.)

Defining $\vec{p}:=\overrightarrow{MP}$ and $\vec{q}:=\overrightarrow{AQ}$, we clearly have $\overrightarrow{AB}=2(\vec{p}+\vec{q})$ and $\overrightarrow{AC}=2\vec{p}$. Therefore, $$0 = x\,\overrightarrow{AB} + y\,\overrightarrow{AC} + z\,\overrightarrow{AQ} + t\,\overrightarrow{MP} = (2x+2y+t)\vec{p}+(2x+z)\vec{q}$$ Since $\vec{p}$ and $\vec{q}$ are linearly independent, the individual coefficients must vanish. (So, right away, we know that condition (a) doesn't hold.) We can readily deduce $$\frac{z-t}{y} = 2 = \frac{|AC|}{|MP|}$$ which is condition (c).

Further, we can rule-out the other conditions by considering a particular solution, say, $(x,y,z,t) = (1,1,-2,-4)$ and noting that $|AC|/|PB|=2$ and $|AC|/|MB|=1$.

Assuming(!) that at least one condition always holds, that condition must be (c). $\square$

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Addendum. Here's a proof of "the midpoint property" (specifically, $P$ is the midpoint of $\overline{MB}$) in general. Let $O$ be the center of the circle.

• Since $\overline{AC}\parallel\overline{MB}$ (and thus $\overline{AC}\perp\overline{OB}$), we have $\angle AOB=\angle BOC$; let $\theta$ be the common measure.
• Then, inscribed angle $\angle AEC=\frac12(360^\circ-2\theta)$, so that $\angle AEM$ also has measure $\theta$.
• Further, "inscribed" angle $\angle MAC$ (with side $\overline{AM}$ tangent to the circle) also has measure $\theta$.

Now, drop a perpendicular from $M$ to the extension of $\overline{AC}$, and let $A'$ be the reflection of $A$ in that perpendicular.

• That perpendicular necessarily passes through the circumcenter $O'$ of $\triangle MAA'$; moreover, since $\overline{MB}\parallel\overline{AA'}$, we have that $\overline{MB}$ is tangent to $\bigcirc O'$ at $M$.

• As $\angle MEA$ and $\angle MA'A$ are necessarily supplementary, $\square MEAA'$ is a cyclic quadrilateral. In particular, $E$ lies on $\bigcirc O'$.

We have, then, that $\overleftrightarrow{AE}$ is the radical axis of the two circles, and thus necessarily bisects their common tangent segment, $\overline{MB}$, at $P$. $\square$

And ... To show that condition (c) holds in general ... Let's define $s:=r/2$ and $\phi:=\theta/2$ (to avoid fractions).

Take $\vec{p}$ and $\vec{q}$ as before (so that $\overrightarrow{AB}$ is still $2(\vec{p}+\vec{q})$). Note that $|\vec{p}|=|\vec{q}|=s\tan\phi$. Defining $\vec{c}:=\overrightarrow{AC}$, we have $|\vec{c}|=4s\sin2\phi=8s\sin\phi\cos\phi=8s\tan\phi\cos^2\phi$, so that $\vec{c} = 8\vec{p}\,\cos^2\phi$.

Then, the vanishing vector sum condition implies $$2x+8y\cos^2\phi+t \;=\; 2x+z \;=\; 0$$ so that $$\frac{z-t}{y}= 8\cos^2\phi = \frac{|\vec{c}|}{|\vec{p}|}=\frac{|AC|}{|MP|}$$ as desired. $\square$