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STEP I 1997 question 5 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1997 question 5 solution

(Here, for not-getting-killed-by-LaTeX reasons, I'll use a, b, c, d to denote vectors \mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}. Furthermore, I won't write out the dot product operator symbol, so you should just assume ab means \mathbf a\cdot \mathbf b. And, I'll use the notation a^2 to mean \mathbf a \cdot \mathbf a.)

Let a, b, c, d be the position vectors of points A, B, C, D, respectively. Then position vectors of P, Q, R, S are (a + b)/2, (b + c)/2, (c + d)/2, (d + a)/2, respectively.

To prove: |AB|^2 - |BC|^2 + |CD|^2 - |DA|^2 = 2|QS|^2 - 2|PR|^2

We shall prove the equivalent form

2|AB|^2 + 2|CD|^2 - 4|QS|^2 = 2|BC|^2 + 2|DA|^2 - 4|PR|^2.

The LHS can be written in vector form as

2(a-b)^2+2(c-d)^2 - (b+c-a-d)^2

= 2(a-b)^2 + 2(c-d)^2 - (b-a)^2 - (c-d)^2 - 2(b-a)(c-d)

= a^2 - 2ab + b^2 + c^2 - 2cd + d^2 - 2bc - 2da + 2ac + 2bd

=a^2 + b^2 + c^2 + d^2 + 2ac + 2bd - 2ab - 2bc - 2cd - 2da

The RHS is

2(b-c)^2 + 2(a-d)^2 -(a+b-c-d)^2

= 2(b-c)^2 + 2(a-d)^2 - (a-d)^2 - (b-c)^2 - 2(a-d)(b-c)

= b^2 - 2bc + c^2 + a^2 - 2ad + d^2 - 2ab - 2dc + 2ac + 2bd

= a^2 + b^2 + c^2 + d^2 + 2ac + 2bd - 2ab - 2bc - 2cd - 2da

and so LHS = RHS, and we have proven

|AB|^2 - |BC|^2 + |CD|^2 - |DA|^2 = 2|QS|^2 - 2|PR|^2 = 2|QS|^2 - 2|PR|^2.

Now, since |QS|^2 - |PR|^2 only depends on the lengths of the rods, and the rods are rigid, the expression is going to stay constans as the rods move.

Similarly, we shall prove

2|AC||BD|\cos\theta = |AB|^2 + |CD|^2 - |BC|^2 - |DA|^2

Written in vectors, this is

2(a - c)(b - d) = (b-c)^2 + (d-a)^2 - (a-b)^2 - (c-d)^2

LHS:

2(a-c)(b-d) = 2ab + 2cd - 2bc - 2ad

RHS:

(b-c)^2 + (d-a)^2 - (a-b)^2 - (c-d)^2

= b^2 - 2bc +c^2 + d^2 - 2ad + a^2 - a^2 + 2ab - b^2 - c^2 + 2cd - d^2

= 2ab + 2cd - 2bc - 2ad

and so LHS = RHS, and

2|AC||BD|\cos\theta = |AB|^2 + |CD|^2 - |BC|^2 - |DA|^2

Since |AC||BD|\cos\theta now only depends on the lengths of the rods, it will remain constant. QED.

(Actually, there are two angles \theta formed by AC and BD and depending on which one you choose, you'll end up with different values of |AC||BD|\cos\theta (one of the values is minus the other). But as long as you stick with the same angle, it's going to be the same scalar product, and so the value remains constant.)

Solution by ukgea

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