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).

Problem:

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.

Solution

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}

Finally,

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
3 years, 1 month ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

There are no comments in this discussion.

×

Problem Loading...

Note Loading...

Set Loading...