I almost forgot how fun challenging math was until my plane ride last week where I solved IMO problem 1. Then I fell right back in love with math again. Anyway, let's begin. (Please excuse my notations and terminology, I'm a bit rusty on that).

Triangle \(BCF\) has a right angle at \(B\). Let \(A\) be the point on line \(CF\) such that \(FA = FB\) and \(F\) lies between \(A\) and \(C\). Point \(D\) is chosen so that \(DA = DC\) and \(AC\) is the bisector of \(\angle{DAB}\). Point \(E\) is chosen so that \(EA = ED\) and \(AD\) is the bisector of \(\angle{EAC}\). Let \(M\) be the midpoint of \(CF\). Let \(X\) be the point such that \(AMXE\) is a parallelogram. Prove that \(BD\), \(FX\), and \(ME\) are concurrent.

Let \(\overline{BC}=a,~\text{and}~\overline{FB}=bi\)

We are given

\[\boxed{\begin{align} C &= \langle a,0 \rangle\\ B &= \langle 0,0 \rangle\\ M &= \langle \frac{a}{2},\frac{b}{2} \rangle \\ F &= \langle 0,b \rangle \end{align}}\]

Now let angle \(\angle \text{JAB}=\theta\) and the rest are labeled in the picture for you. (IGNORE POINT G). Also, define \(\overline{CF}=\sqrt{a^2+b^2}=r\).

Now define \(P_x\) as the \(x\)-coordinate of point \(P\) and define \(P_y\) similarly. Segments \(\overline{BF}\) and \(\overline{FA}\) are both length \(b\). Thus \(C_x\cdot \dfrac{b}{r}=A_x\) and \(F_y\cdot \dfrac{b+r}{r}=A_y\).

Some arithmetic yields \(\boxed{A= \langle -\frac{ab}{r},\frac{b^2}{r}+b \rangle}\)

Next, it is obvious that since \(\overline{CA}=b+r\), segment \(\overline{DC}=\overline{AD}=\dfrac{b+r}{2\cos(\theta)}\).

And it follows

\(D=\vec{C}+\frac{b+r}{2\cos(\theta)}\cdot \text{cis}\left(\dfrac{\pi}{2}\right)\)

\(D= \langle a-\frac{b+r}{2}\tan(\theta), \frac{b+r}{2} \rangle\)

Similar triangles tells us that \(\overline{JB}=\frac{ab}{r}\) and point \(A\) tells us that \(\overline{JA}=\frac{b^2}{r}+b\).

Thus the picture shows us that \(\tan{\theta}=\frac{a}{b+r}\)

Amazingly, this means that point \(D\) is directly over point \(M\) with coordinates \(\boxed{D= \langle \frac{a}{2}, \frac{b+r}{2} \rangle}\)

Next, it is apparent that \(\overline{AE}=\overline{ED}=\dfrac{\overline{AD}}{2\cos(\theta)}=\dfrac{b+r}{4\cos^2(\theta)}\).

Pythagorean theorem tells us that \(\overline{AB}=b\sqrt{2+\frac{2b}{r}}\). Unbelievably, this is pretty much the only root we have to deal with besides \(r\) in all of this problem. And this root goes away immediately. Props to the maker of this beautiful problem.

From \(\triangle ABJ\) we know what \(\cos^2(\theta)\) is and thus \(\overline{AE}=\dfrac{b+r}{4\cos^2(\theta)}=\dfrac{r}{2}\).

Thus \(E=\vec{A}+\text{cis}\left(\frac{3\pi}{2}+\frac{r}{2}\right)\)

Not surprisingly, everything cancels once again and we are left with

\(\boxed{E= \langle 0,b+\frac{r}{2} \rangle}\)

Finally,

\(\vec{X} =\vec{M}-\vec{A}+\vec{E}\)

\(\boxed{X =\langle \frac{a}{2}+\frac{ab}{r}, \frac{b+r}{2} -\frac{b^2}{r} \rangle}\)

We now have

\[\boxed{\begin{align} C &= \langle a,0 \rangle\\ B &= \langle 0,0 \rangle\\ M &= \langle \frac{a}{2},\frac{b}{2} \rangle \\ F &= \langle 0,b \rangle \\ A &= \langle -\frac{ab}{r},\frac{b^2}{r}+b \rangle \\ D &= \langle \frac{a}{2}, \frac{b+r}{2} \rangle \\ E &= \langle 0,b+\frac{r}{2} \rangle \\ X &=\langle \frac{a}{2}+\frac{ab}{r}, \frac{b+r}{2} -\frac{b^2}{r} \rangle \end{align}}\]

I didn't take the time to actually work out the lines in variable form. But plugging in values for a and b show that the three lines are concurrent.

I unfortunately lost the paper that I had the coordinates of intersection on. But the coordinates are sooo pretty, no radicals, just simple, elegant fractions.

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