Advanced book on Mathematics Olympiad

Geometry and Trigonometry 609

A

B

C

D

H

P

Q

H

Figure 73

Proof.Let

−−→

MN=−→v 1 ,

−→

NP =−→v 2 ,

−→

PQ=−→v 3 ,

−−→

QM =−→v 4 , and

−−−→

M′N′ =−→w 1 ,

−−→

N′P′=−→w 2 ,

−−→

P′Q′=−→w 3 ,

−−−→

Q′M′=−→w 4. The conditions from the statement can be

written in vector form as

−→v 1 ·−→w 2 =−→v 2 ·−→w 1 =−→v 3 ·−→w 4 =−→v 4 ·−→w 3 = 0 ,

−→v 1 +−→v 2 +−→v 3 +−→v 4 =−→w 1 +−→w 2 +−→w 3 +−→w 4 =−→ 0 ,

(−→v 1 +−→v 2 )·(−→w 2 +−→w 3 )= 0.

We are to show that

(−→v 2 +−→v 3 )·(−→w 1 +−→w 2 )= 0.

First, note that

0 =(−→v 1 +−→v 2 )·(−→w 2 +−→w 3 )=−→v 1 ·−→w 2 +−→v 1 ·−→w 3 +−→v 2 ·−→w 2 +−→v 2 ·−→w 3

=−→v 1 ·−→w 3 +−→v 2 ·−→w 2 +−→v 2 ·−→w 3.

Also, the dot product that we are supposed to show is zero is equal to

(−→v 2 +−→v 3 )·(−→w 1 +−→w 2 )=−→v 2 ·−→w 1 +−→v 2 ·−→w 2 +−→v 3 ·−→w 1 +−→v 3 ·−→w 2
=−→v 2 ·−→w 2 +−→v 3 ·−→w 1 +−→v 3 ·−→w 2.
This would indeed equal zero if we showed that−→v 1 ·−→w 3 +−→v 2 ·−→w 3 =−→v 3 ·−→w 1 +−→v 3 ·−→w 2.
And indeed,
−→v
1 ·
−→w
3 +
−→v
2 ·
−→w
3 =(
−→v
1 +
−→v
2 )·
−→w
3
=−(−→v 3 +−→v 4 )·−→w 3 =−−→v 3 ·−→w 3 −−→v 4 ·−→w 3 =−−→v 3 ·−→w 3
=−−→v 3 ·−→w 3 −−→v 3 ·−→w 4 =−−→v 3 ·(−→w 3 +−→w 4 )
=−→v 3 ·(−→w 1 +−→w 2 )=−→v 3 ·−→w 1 +−→v 3 ·−→w 2.