Proving Diagonal Trisection in a Parallelogram using Vector Calculus

[link2001]

the problem goes:

ABCD is a parallelogram in which points P and Q are the midpoints of sides BC and CD, respectively. Use vector calculus to prove that AP and AQ trisect the diagonal BD at the points E and F.


...A _________B
.../...F.../
.../...E.../P
D/________/C
...Q
(Imagine lines from D->B, A->Q and A->P, where AQ passes through E and AP passes through F)


I set the Problem Up like this:
DE = x(DA + AB)
EF = y(DA + AB)
FB = z(DA + AB)

After this I get stuck running around in circles trying to make subsitutions to show that x, y & z = 1/3.

Ideas anyone?

 

[Nate810]

We can use the fact that P and Q are the midpoints of sides BC and CD to solve this problem. Start by noting that BD = DA + AB, since these are the two sides of a parallelogram. Then, we can say that DE = x(BD), EF = y(BD), and FB = z(BD). Now, since P is the midpoint of BC, we can say that BP = CP. By the same logic, AQ = AD. This means that DE + EF = BP and EF + FB = AQ. Combining our equations, we get x(BD) + y(BD) = BP and y(BD) + z(BD) = AQ. Since BP = CP and AQ = AD, we can substitute these in for the right hand side of the equations. We then have: x(BD) + y(BD) = CP, and y(BD) + z(BD) = AD. Now, we can add the two equations together to get (x + y + z)(BD) = CP + AD. Since BD = DA + AB, this becomes (x + y + z)(DA + AB) = CP + AD. Finally, we can divide both sides of the equation by (DA + AB), which gives us x + y + z = (CP + AD)/(DA + AB). Since CP = AD, we can simplify this to x + y + z = 1/2. Since x, y, and z must all be positive numbers, the only way for their sum to equal 1/2 is if they are each equal to 1/3. Thus, DE trisects BD at E, and EF trisects BD at F.

 

[quicknote]

I would approach this problem by breaking it down into smaller, more manageable steps. First, I would define the properties of a parallelogram and the concept of midpoints. Then, I would introduce the concept of vector calculus and explain how it can be used to solve geometric problems.

Next, I would define the points E and F as the intersection points of the diagonals BD and AP/AQ. I would also introduce the concept of trisection, which means dividing a line into three equal parts.

To prove that AP and AQ trisect the diagonal BD, I would use the properties of parallelograms and midpoints to show that the vectors AP and AQ are equal in length and direction to the vectors BE and BF. This can be done using vector addition and subtraction.

Then, I would use the properties of vector calculus, such as the dot product and cross product, to show that the vectors AP and AQ are perpendicular to the diagonal BD. This would demonstrate that the points E and F are indeed the trisection points of BD, as the perpendicular bisector of a line divides it into two equal parts.

Finally, I would use the properties of parallelograms again to show that the lengths of the segments BE and BF are equal to one-third of the length of the diagonal BD. This would prove that AP and AQ trisect the diagonal BD at the points E and F.

Overall, the key to solving this problem using vector calculus is to break it down into smaller steps and use the properties of parallelograms and midpoints to connect the points and vectors in the problem. With careful calculations and substitutions, it is possible to prove that AP and AQ trisect the diagonal BD in a parallelogram.