STEP I 1990 question 6 solution

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STEP I 1990 question 6 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1990 question 6 solution

Let $\mathbf a, \mathbf b, \mathbf c, \mathbf d$ be the position vectors of A, B, C, D respectively. Without loss of generality, let $\mathbf b = \mathbf 0$ . Then $\mathbf d = \mathbf a + \mathbf c$ .

(i) The statement is equivalent to

$\mathbf a^2 + \mathbf c^2 + (\mathbf a + \mathbf c - \mathbf c)^2 + (\mathbf a + \mathbf c - \mathbf c)^2 = (\mathbf a - \mathbf c)^2 + (\mathbf a + \mathbf c)^2$

(where squaring a vector denotes taking the dot product of that vector with itself) This is in turn equivalent to

$\mathbf a^2 + \mathbf c^2 + \mathbf a^2 + \mathbf c^2 = \mathbf a^2 - 2\mathbf a\cdot \mathbf c + \mathbf a + 2\mathbf a \cdot \mathbf c + \mathbf c^2$

which is evidently true. Thus the statement is proved.

(ii) |AC^2 - BD^2|

$= | (\mathbf a - \mathbf c)^2 - (\mathbf a + \mathbf c)^2|$

$= | \mathbf a^2 - 2\mathbf a \cdot \mathbf c + \mathbf c^2 - \mathbf a^2 - 2\mathbf a \cdot \mathbf c - \mathbf c^2|$

$= |-4\mathbf a \cdot \mathbf c|$

$= |4 \mathbf a \cdot \mathbf c|$

$= |4 |\mathbf a| |\mathbf c| \cos \theta |$

where $\theta$ is either the angle ABC or the angle BCD (it doesn't matter which, since the cosine of these two angles are negatives of each other and this distinction disappears when taking the modulus anyway.)

Now, the expression $|\mathbf c| |\cos \theta|$ is the length of the projection of BC on AB. By symmetry, this is also the length of the projection of AD onto CD, and it is also the projection of BC on CD and AD on AB. Note that $|\mathbf a|$ in this case represents the length of AB or CD, and thus the whole expression can be seen as four times the area of the rectangle whose sides are either of BC, AD projected on either of AB, CD, together with AB or CD.

Instead, if we consider $|\mathbf a| |\cos \theta|$ and $|\mathbf c|$ we can similarly see that the expression is four times the area of the rectangle whose sides are either of AB, CD projected on BC, AD, together with BC or AD. And thus, the expression is four times the area of the rectangle formed by any side and the projection of any adjacent side onto that side, as required.

For the next case, the result we shall prove is:

|AB^2 - AD^2| is the area of the rectangle whose sides are one diagonal and the projection of the other diagonal onto the first diagonal.

This is because

|AB^2 - AD^2|

$= |\mathbf a^2 - \mathbf (a + c - a)^2|$

$= |\mathbf a^2 - \mathbf c^2|$

$= |(\mathbf a + \mathbf c)\cdot(\mathbf a - \mathbf c)|$

$= |BD| |AC| |\cos \theta|$

Where $\theta$ this time is the angle between the diagonals. (Again, it doesn't matter if we take the obtuse or the acute angle, because of the modulus sign).

Now since $|AC| |\cos \theta|$ and $|BD| |\cos \theta|$ are the length of the projections of AC on BD and BD on AC respectively, we can see the above expression as either BD multiplied by length of projection of AC on BD, or AC multiplied by the length of the projection of BD onto AC. This thus proves the statement.