Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A question from my vector calculus assignment. Geometry, anything visual, is by far my weakest area. I've been literally staring at this question for hours in frustrations and I give up (and I do mean hours). I don't even now where to start... not feeling good over here.

Question:

In the diagram below $ABCD$ is a parallelogram with $P$ and $Q$ the midpoints of the the sides $BC$ and $CD$, respectively. Prove $AP$ and $AQ$ trisect $BD$ at the points $E$ and $F$ using vector methods.

Image: enter image description here

Hints: Let $a = OA$, $b = OB$, etc. You must show $ e = \frac{2}{3}b + \frac{1}{3}d$, etc.


I figured as much without the hints. Also I made D the origin and simplified to $f = td$ for some $t$. And $f = a + s(q - a)$ for some $s$, and $q = \frac{c}{2}$ and so on... but I'm just going in circles. I have no idea what I'm doing. There are too many variables... I am truly frustrated and feeling dumb right now.

Any help is welcome. I'm going to go watch Dexter and forget how dumb I'm feeling.

share|improve this question
2  
Depending on which "Dexter" show you intend to watch that might not make you feel any better. –  anon Oct 4 '11 at 1:15
    
The one that kills people. I will make me feel better lol –  iDontKnowBetter Oct 4 '11 at 1:17
    
I'm not sure how you can adapt this approach to your "vector" requirement, but: letting $D:(0,0)$, $C:(c,0)$, $A:(a,b)$, and $B:(a+c,b)$ (why?), use the two-point form of the equation of a line to get the equations of the lines $\overline{AQ}$ and $\overline{AP}$, where e.g. $Q:(c/2,0)$ (why?). Find the intersection points of those lines with $\overline{DB}$. Check that those two intersection points are in fact trisection points for $\overline{BD}$. –  J. M. Oct 4 '11 at 1:31
    
(I couldn't resist; here's a Mathematica "proof": (({x, y} /. First@Solve[{y == InterpolatingPolynomial[{{0, 0}, {a + c, b}}, x], y == InterpolatingPolynomial[{{a, b}, #}, x]}, {x, y}]) & /@ {{c/2, 0}, {a + 2 c, b}/2}) === Map[{a + c, b} # &, {1, 2}/3]) –  J. M. Oct 4 '11 at 1:42
    
@J.M I already tried this and I keep getting anything but the correct answer. I get point of intersection (-4c, 0), which makes no sense. This is the most frustrating thing ever. I'm just going to leave this question blank. I've never been so frustrated by something. — And there's no way I could come up with any of the answers below. I barely understand them. –  iDontKnowBetter Oct 4 '11 at 4:23

4 Answers 4

Note that EBP and EDA are similar triangles. Since 2BP=AD, it follows that 2EB=ED, and thus 3EB=BD. Which is to say, AP trisects BD.

share|improve this answer
    
Slick, but doesn't use vectors. –  anon Oct 4 '11 at 2:23

There are as usual many approaches. We deal with $F$ only. Basically the same method will work for $E$. Actually, we don't need to do anything for $E$, just a little reflection, and twisting our necks around. Or else we can simply rename our points: interchange $B$ and $D$. Recycling is a good thing.

Let $u$ be the vector $DC$, and $v$ the vector $DA$. Then $DB$ is the vector $u+v$.

We want to show that $DF=(1/3)(u+v)$.

To do this, it is enough to show that with this choice of $F$, the vector $AF$ is parallel to the vector $FQ$. Compute. For our choice of $F$, we have $$AF=(1/3)(u+v)-v=(1/3)u -(2/3)v.$$ Also, $$FQ=(1/2)u-(1/3)(u+v)=(1/6)u -(1/3)v.$$ The parallelism is obvious, the vector $AF$ is twice the vector $FQ$. As a little bonus we get therefore an additional geometric result, that $AF=2FQ$.

Comment: Since we were given the answer, we were able to save a few steps. Clearly $DF=\lambda(u+v)$ for some $\lambda$. We verified that $\lambda=1/3$. But what if we had not been given the answer? And what if $Q$ was not the midpoint of $DC$, but a division point of $DC$ in some ratio other than $1:1$? We sketch how to do the same problem, without being supplied the $1/3$. Minor modification will take care of other division ratios.

The idea is much the same as before. We have $DF=\lambda(u+v)$ for some now unknown $\lambda$. And $AF=\kappa FQ$ for some unknown constant $\kappa$. Thus we have $$AF=\lambda(u+v)-v=\lambda u +(\lambda-1)v,$$ $$FQ=(1/2)u-\lambda(u+v)=(1/2-\lambda)u -\lambda v.$$ The condition $AF=\kappa FQ$ comes down to the two equations $$\lambda =\kappa(1/2-\lambda) \text { and } \lambda-1=-\kappa\lambda.$$ To solve, substitute $\lambda-1$ for $-\kappa\lambda$ in the first equation. Quickly we get $\kappa=2$, and then $\lambda=1/3$.

share|improve this answer

The statement is invariant under affine transformations of the plane, so we can assume that $A=(0,1)$, $B=(1,1)$, $C=(1,0)$ and $D=(0,0)$. Then of course $Q=(1/2,0)$ and $P=(1,1/2)$, and we can compute. The line through $A$ and $P$ is $$x+2y=2;$$ the line through $A$ and $Q$ is $$2x+y=1;$$ finally the line through $B$ and $D$ is of course $$x-y=0.$$ Computing the intersection of $AP$ and $BD$, we get the point $E=(2/3,2/3)$, and the intersection of $AQ$ and $BD$ is $F=(1/3,1/3)$. It is clear, then, that $E$ and $F$ trisect the segment $BD$.

 

The thing to takehome from this is that sometimes symmetry allows to reduce a theorem to a computation.

share|improve this answer
    
Doesn't that map a square? How do you get $x + 2y = 2$ and so on? — I'm giving up on this question. I've no understanding of geometry. –  iDontKnowBetter Oct 4 '11 at 4:25
    
The line whose equation is $x+2y=2$ passes through $A$ and $P$, so it must the the unique line through $A$ and $P$. Etc. –  Mariano Suárez-Alvarez Oct 4 '11 at 4:51

Since we know what exactly we want to prove, I will first give a "cheating solution" that makes a neat homework answer but provides little insight. After all, such tricks are also useful sometimes. :)

I will prove that $E$ divides $BD$ in the ratio $2:1$. It seems convenient to fix the origin at $C$. Then the vertices $B$, $D$ and $A$ are respectively at $\mathbf b$ and $\mathbf d$ and $\mathbf b + \mathbf d$. Now, let $X$ be the point $\frac{2}{3} \mathbf b + \frac{1}{3} \mathbf d$. (This is the part that is most unsatisfactory.)

We will show that $A$, $X$ and $C$ are collinear. To verify this, just compute: $$ \overline{XA} = (\mathbf b +\mathbf d) - (\frac{2}{3} \mathbf b + \frac{1}{3} \mathbf d) = \frac{1}{3} \mathbf b + \frac{2}{3} \mathbf d = \frac{1}{3} (\mathbf b + 2\mathbf d), $$ and $$ \overline{PA} = (\mathbf b +\mathbf d) - \frac{1}{2} \mathbf b = \frac{1}{2} (\mathbf b + 2\mathbf d). $$ Now is it evident that $\overline{PA}$ is parallel to the vector $\overline{XA}$? What does this mean? How does this fact help you?


To make this into a more systematic proof, we will proceed roughly as J.M. suggests, but using the vector notation. Let $E$ be the point of intersection of $AP$ with $BD$. We want to find $E$.

Since $E$ lies inside the segment $BD$, $E$ can be written as $\alpha \mathbf b + (1-\alpha) \mathbf d$ for some $\alpha \in [0,1]$ (the exact range in which $\alpha$ lies is not that important). Then, $E$ will lie in $AP$ iff $\overline{EA}$ is parallel to $\overline{PA}$. But we can compute these vectors: $$ \overline {EA} = (\mathbf b +\mathbf d) - (\alpha \mathbf b + (1-\alpha) \mathbf d) = (1-\alpha) \mathbf b + \alpha \mathbf d, $$ and $$ \overline{PA} = (\mathbf b +\mathbf d) - \frac{1}{2} \mathbf b = \frac{1}{2} \mathbf b + \mathbf d. $$ Now under what condition will these two vectors be parallel to each other?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.