Advanced book on Mathematics Olympiad

208 4 Geometry and Trigonometry

The condition that these three points be collinear translates to

1

2 (ac−bd)

∣

∣∣

∣∣

∣

ac 1

bd 1

ab(c−d) cd(a−b) ac−bd

∣

∣∣

∣∣

∣

= 0.

This is verified by direct computation. 

Example.In a circle are inscribed a trapezoid with one side as diameter and a triangle
with sides parallel to the sides of the trapezoid. Prove that the two have the same area.

Solution.We refer everything to Figure 26. Assume that the circle has radius 1, and the
trapezoid has vertices( 1 , 0 ),(a, b),(−a, b)and(− 1 , 0 ).

b,−a

a,b

(0,1)

(1,0)

(

()

)

Figure 26

The triangle is isosceles and has one vertex at( 0 , 1 ). We need to determine the
coordinates of the other two vertices. One of them lies where the parallel through( 0 , 1 )
to the line determined by( 1 , 0 )and(a, b)intersects the circle. The equation of the line is

y=

b

a− 1

x+ 1.

The relationa^2 +b^2 =1 yieldsb^2 =( 1 −a)( 1 +a),or 1 −ba=^1 +ba. So the equation of
the line can be rewritten as

y=−

1 +a

b

x+ 1.

Now it is easy to guess that the intersection of this line with the circle is(b,−a)(note that
this point satisfies the equation of the circle). The other vertex of the triangle is(−b,−a),
so the area is^12 ( 2 b)( 1 +a)=b+ab. And the area of the trapezoid is^12 ( 2 a+ 2 )b=b+ab,
the same number. 

589.Prove that the midpoints of the sides of a quadrilateral form a parallelogram.