IMO Day 1 Problem 1: Vector/trig bash

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 BCFBCF has a right angle at BB. Let AA be the point on line CFCF such that FA=FBFA = FB and FF lies between AA and CC. Point DD is chosen so that DA=DCDA = DC and ACAC is the bisector of DAB\angle{DAB}. Point EE is chosen so that EA=EDEA = ED and ADAD is the bisector of EAC\angle{EAC}. Let MM be the midpoint of CFCF. Let XX be the point such that AMXEAMXE is a parallelogram. Prove that BDBD, FXFX, and MEME are concurrent.


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

We are given

C=a,0B=0,0M=a2,b2F=0,b\boxed{\begin{aligned} 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{aligned}}

Now let angle JAB=θ\angle \text{JAB}=\theta and the rest are labeled in the picture for you. (IGNORE POINT G). Also, define CF=a2+b2=r\overline{CF}=\sqrt{a^2+b^2}=r.

Now define PxP_x as the xx-coordinate of point PP and define PyP_y similarly. Segments BF\overline{BF} and FA\overline{FA} are both length bb. Thus Cxbr=AxC_x\cdot \dfrac{b}{r}=A_x and Fyb+rr=AyF_y\cdot \dfrac{b+r}{r}=A_y.

Some arithmetic yields A=abr,b2r+b\boxed{A= \langle -\frac{ab}{r},\frac{b^2}{r}+b \rangle}

Next, it is obvious that since CA=b+r\overline{CA}=b+r, segment DC=AD=b+r2cos(θ)\overline{DC}=\overline{AD}=\dfrac{b+r}{2\cos(\theta)}.

And it follows

D=C+b+r2cos(θ)cis(π2)D=\vec{C}+\frac{b+r}{2\cos(\theta)}\cdot \text{cis}\left(\dfrac{\pi}{2}\right)

D=ab+r2tan(θ),b+r2D= \langle a-\frac{b+r}{2}\tan(\theta), \frac{b+r}{2} \rangle

Similar triangles tells us that JB=abr\overline{JB}=\frac{ab}{r} and point AA tells us that JA=b2r+b\overline{JA}=\frac{b^2}{r}+b.

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

Amazingly, this means that point DD is directly over point MM with coordinates D=a2,b+r2\boxed{D= \langle \frac{a}{2}, \frac{b+r}{2} \rangle}

Next, it is apparent that AE=ED=AD2cos(θ)=b+r4cos2(θ)\overline{AE}=\overline{ED}=\dfrac{\overline{AD}}{2\cos(\theta)}=\dfrac{b+r}{4\cos^2(\theta)}.

Pythagorean theorem tells us that AB=b2+2br\overline{AB}=b\sqrt{2+\frac{2b}{r}}. Unbelievably, this is pretty much the only root we have to deal with besides rr in all of this problem. And this root goes away immediately. Props to the maker of this beautiful problem.

From ABJ\triangle ABJ we know what cos2(θ)\cos^2(\theta) is and thus AE=b+r4cos2(θ)=r2\overline{AE}=\dfrac{b+r}{4\cos^2(\theta)}=\dfrac{r}{2}.

Thus E=A+cis(3π2+r2)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

E=0,b+r2\boxed{E= \langle 0,b+\frac{r}{2} \rangle}


X=MA+E\vec{X} =\vec{M}-\vec{A}+\vec{E}

X=a2+abr,b+r2b2r\boxed{X =\langle \frac{a}{2}+\frac{ab}{r}, \frac{b+r}{2} -\frac{b^2}{r} \rangle}

We now have

C=a,0B=0,0M=a2,b2F=0,bA=abr,b2r+bD=a2,b+r2E=0,b+r2X=a2+abr,b+r2b2r\boxed{\begin{aligned} 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{aligned}}

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.

Note by Trevor Arashiro
4 years, 12 months ago

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