A Geometry Problem from “The Monthly”, March 1998

A Geometry Problem from "The Monthly". March 1998

In a review in The American Mathematical Monthly, March 1998, the following problem was stated. Prove that in a figure like

$$ A B + B F = A D + D F \Rightarrow A C + C F = A E + E F $$

After a while I found myself considering the figures with the points AC,D fixed but the line BFE moving parallel to itself, with the dotted lines through D and C its extreme positions:

Below we give the positions of $B, F, E$ when the moving line is in its leftmost position - i.e. coinciding with the dotted line through $D$ - and later when the moving line is in its rightmost position.

First we are interested in the values of $\alpha$ and $\gamma$ when the moving line is in this leftmost position, and where $\alpha$ and $\gamma$ - corresponding to antecedent and consequent respectively - are given by $\alpha = {AB} + {BF} - {AD} - {DF}$ and $\gamma = {AC} + {CF} - {AE} - {EF}$ Denoting these leftmost values by ${\alpha 0}$ and ${\gamma 0}$ ,we read from the above figure

(0) ${\alpha 0} = p + r - q,\;{\gamma 0} = p + s - q$

Next we are interested in the values of $\alpha$ and $\gamma$ , when the moving line is

in its rightmost position. The points B, F, E are then located as shown in the following figure

all i $\gamma \times \gamma + \alpha = 0$ and $\gamma - 1$ respectively,we read from the

$\alpha 1 = P - q - s,\;\gamma 1 = P - Q - R.$

Our proof obligation is $\alpha = 0 \Rightarrow \gamma = 0$ . Actually, and not surprisingly, we shall show $\alpha = 0 \equiv \gamma = 0$ . We can do so because all components of $\alpha$ and $\gamma$ , and hence $\alpha$ and $\gamma$ themselves are linear functions of the position of the moving line (measured in distance from, say, the leftmost dotted line). Because ${\alpha 0} > 0$ and ${\alpha 1} < 0$ ,these two values determine by linear interpolation the unique position at which $\alpha = 0$ ,and similarly for $\gamma$ . We don't need to

determine these positions, we only need to show that they are the same, which follows from

$$ \alpha 0 \cdot \gamma 1 = \alpha 1 \cdot \gamma 0 $$

Thanks to (0) and (1), the latter proof

$$ (p - q + r) \cdot (P - Q - R) = (P - q - s) \cdot (P - q + s) $$

Because (in the 2nd figure) the dotted lines are parallel, we have

(2) $p \cdot Q = q \cdot P\;q \cdot R = r \cdot Q\;r \cdot P = p \cdot R$

and the above proof obligation can be simplified to

(3) $P \cdot P + q \cdot Q - r \cdot R = {P}^{2} + {q}^{2} - {s}^{2}$

We are getting on more and more familiar ground. Using the Cosine Rule - i.e. $c^{2} = a^{2} + b^{2} - 2 \cdot a \cdot b \cdot \cos \gamma$ - we deduce in the 2nd figure

true

$\equiv \left{ \begin{array}{l} \text{Cosine Rule in } \bigtriangleup {ACD}\text{ and }\bigtriangleup {CHD}} \ \end{array}\right.$

$$ p ^ {2} = q ^ {2} + s ^ {2} + 2 \cdot q \cdot s \cdot \cos \delta \quad \wedge $$

$$ R ^ {2} = (Q - q) ^ {2} + s ^ {2} - 2 \cdot (Q - q) \cdot s \cdot \cos \delta $$

$$ \Rightarrow \quad \left{e l i m i n a t e \quad c o s. \delta \right} $$

$$ \begin{array}{l} (Q - q) \cdot P ^ {2} + q \cdot R ^ {2} = q \cdot Q \cdot (Q - q) + Q \cdot s ^ {2} \ \equiv {\text {d i s t r i b u t i o n a n d r e a r r a n g i n g} } \ Q \cdot P ^ {2} + Q \cdot q ^ {2} - Q \cdot s ^ {2} = q \cdot P ^ {2} + q \cdot Q ^ {2} - q \cdot R ^ {2} \ = {(2) a n d Q \neq 0 } \ P ^ {2} + q ^ {2} - s ^ {2} = p \cdot P + q \cdot Q - r \cdot R \ \equiv {\mathrm {d e f . o f} (3) } \ (3) \ \end{array} $$

and this completes my proof.

Note Introduction and elimination of cos. $\delta$ as in the last calculation is a standard device. (End of Note.)

The recognition of the linear dependence on the displacement of the moving line was crucial. The separate naming of $p,q,r$ and of $P,Q,R$ freed us from mentioning their constant ratio where it did not matter.

Austin

Good Friday, 10 April 1998

prof. d. Edsger W. Dgksta

Department of Computer Sciences

The University of Texas at Austin

Austin, TX 78712 - 1188

USA