608 Geometry and Trigonometry
and the analogous formulas for−→v 2 ,−→v 3 , and−→v 4. Since the rational multiple of a vector and
the sum of two vectors can be constructed with straightedge and compass, we can construct
the vectors−→vi,i = 1 , 2 , 3 ,4. Then we take the vectors
−−→
A′B =−−→v 1 ,−−→
B′C=−−→v 2 ,
−−→
C′D=−−→v 3 , and
−−→
D′A=−−→v 4 from the pointsA′,B′,C′, andD′to recover the vertices
B,C,D, andA.
Remark.Maybe we should elaborate more on how one effectively does these construc-
tions. The sum of two vectors is obtained by constructing the parallelogram they form.
Parallelograms can also be used to translate vectors. An integer multiple of a vector can
be constructed by drawing its line of support and then measuring several lengths of the
vector with the compass. This construction enables us to obtain segments divided into
an arbitrary number of equal parts. In order to divide a given segment into equal parts,
form a triangle with it and an already divided segment, then draw lines parallel to the
third side and use Thales’ theorem.
584.LetObe the intersection of the perpendicular bisectors ofA 1 A 2 andB 1 B 2 .We
want to show thatOis on the perpendicular bisector ofC 1 C 2. This happens if and only
if(
−−→
OC 1 +
−−→
OC 2 )·
−−→
C 1 C 2 =0.
Set−→
OA=
−→
l,−→
OB=−→m,−→
OC=−→n,−−→
AA 2 =−→a,−−→
BB 2 =
−→
b,−−→
CC 2 =−→c. That the
perpendicular bisectors ofA 1 A 2 andB 1 B 2 pass throughOcan be written algebraically as
( 2−→
l +−→a +−→c)·(−→c −−→a)=0 and ( 2 −→m+−→a +−→
b)·(−→a −−→
b)= 0.The orthogonality of the sides of the rectangles translates into formulas as
(−→m−−→
l)·−→a = 0 ,(−→m−−→n)·−→
b = 0 ,(−→n−−→
l)·−→c = 0.We are required to prove that( 2 −→n +
−→
b +−→c)·(−→
b −−→c)=0.And indeed,( 2 −→n+−→
b +−→c)·(−→c −−→
b)= 2 −→n ·−→c − 2 −→n·−→
b +−→c^2 −−→
b^2
= 2 (−→m−−→
l)·−→a + 2−→
l ·−→c − 2 −→m·−→
b +−→c^2 −−→
b^2
= 2 −→m·−→a − 2 −→m·−→
b +−→a^2 −−→
b^2 + 2−→
l ·−→c − 2−→
l ·−→a −−→a^2 +−→c^2 = 0.Hence the conclusion.
585.LetH′be the orthocenter of triangleACD. The quadrilateralsHPBQandHCH′A
satisfyHC⊥BP,H′C⊥HP,H′A⊥HQ,AH⊥BQ,AC⊥HB(see Figure 73). The
conclusion follows from a more general result.
Lemma.LetMNPQandM′N′P′Q′ be two quadrilaterals such thatMN⊥N′P′,
NP⊥M′N′,PQ⊥Q′M′,QM⊥P′Q′, andMP⊥N′Q′. ThenNQ⊥M′P′.