Advanced book on Mathematics Olympiad

Geometry and Trigonometry 615

Let us rephrase this in geometric terms. We are required to include a segmenty=px+q,
0 ≤x≤1, between two circular arcs.
The arcs are parts of two circles of radius 1 and of centersO 1 ( 0 ,

√ 2 − 1
2 ) and
O 2 ( 0 ,−

√ 2 − 1
2 ). By examining Figure 76 we will conclude that there is just one such
segment. On the first circle, consider the pointsA( 1 ,

√

2 − 1

2 )andB(^0 ,

√
2 + 1
2 ). The dis-
tance fromBtoO 2 is

√

2, which is equal to the length of the segmentAB. In the isosceles
triangleBO 2 A, the altitudes fromO 2 andAmust be equal. The altitude fromAis equal
to the distance fromAto they-axis, hence is 1. Thus the distance fromO 2 toABis 1
as well. This shows that the segmentABis tangent to the circle centered atO 2. This
segment lies between the two arcs, and above the entire interval[ 0 , 1 ]. Being inscribed
in one arc and tangent to the other, it is the only segment with this property.
This answers the problem, by showing that the only possibility isp=−1,q=

√

2 + 1

2.

M

A

B

O 2

O 1

Figure 76

(Romanian Team Selection Test for the International Mathematical Olympiad, 1983)

597.The fact that the points(xi,x^1 i)lie on a circle means that there exist numbersA,B,
andCsuch that

xi^2 +

1

xi^2

+ 2 xiA+ 2

1

xi

B+C= 0 , fori= 1 , 2 , 3 , 4.

View this as a system in the unknowns 2A, 2 B, C. The system admits a solution only if
the determinant of the extended matrix is zero. This determinant is equal to

∣∣

∣∣

∣∣

∣

∣∣

∣

x^21 +x^12

1

x (^1) x^111
x^22 +x^12
1
x (^2) x^121
x^23 +x^12
3
x (^3) x^131
x^24 +x^12
4
x (^4) x^141

∣∣

∣∣

∣∣

∣

∣∣

∣

=

∣

∣∣

∣∣

∣∣

∣∣

x 12 x (^1) x^111
x 22 x (^2) x^121
x 32 x (^3) x^131
x 42 x (^4) x^141

∣

∣∣

∣∣

∣∣

∣∣

+

∣∣

∣∣

∣∣

∣

∣∣

∣

1

x 12 x^1

1

x 11

1

x 22 x^2

1

x 21

1

x 32 x^3

1

x 31

1

x 42 x^4

1

x 41

∣∣

∣∣

∣∣

∣

∣∣

∣