Advanced book on Mathematics Olympiad

Real Analysis 565

which is equivalent tocoscosBPDAP C=vv^12. Snell’s law follows once we note that the angles of
incidence and refraction are, respectively, the complements of∠AP Cand∠BPD.

509.LetD, E, Fbe the projections of the incenter onto the sidesBC,AC, andAB,
respectively. If we setx=AF,y=BD, andz=CE, then

cot

A

2

=

x

r

, cot

B

2

=

y

r

, cot

C

2

=

z

r

.

The lengthsx, y, zsatisfy

x+y+z=s,

x^2 + 4 y^2 + 9 z^2 =

(

6 s

7

) 2

.

We first determine the triangle similar to the one in question that has semiperimeter equal
to 1. The problem asks us to show that the triangle is unique, but this happens only
if the planex+y+z=1 and the ellipsoidx^2 + 4 y^2 + 9 z^2 =^3649 are tangent. The
tangency point must be at an extremum off (x, y, z)=x+y+zwith the constraint
g(x, y, z)=x^2 + 4 y^2 + 9 z^2 =^3649.
We determine the extrema offwith the given constraint using Lagrange multipliers.
The equation∇f=λ∇gbecomes

1 = 2 λx,

1 = 8 λy ,

1 = 18 λz.

We deduce thatx = 21 λ,y = 81 λ, andz= 181 λ, which combined with the constraint
g(x, y, z)=^3649 yieldsλ=^4972. Hencex=^3649 ,y= 499 , andz= 494 , and sof (x, y, z)=1.
This proves that, indeed, the plane and the ellipsoid are tangent. It follows that the
triangle with semiperimeter 1 satisfying the condition from the statement has sides equal
tox+y=^4349 ,x+z=^4549 , andy+z=^1349.
Consequently, the unique triangle whose sides are integers with common divisor
equal to 1 and that satisfies the condition from the statement is 45, 43, 13.
(USA Mathematical Olympiad, 2002, proposed by T. Andreescu)

510.Leta, b, c, dbe the sides of the quadrilateral in this order, and letxandybe the
cosines of the angles formed by the sidesaandb, respectively,candd. The condition
that the triangle formed byaandbshares a side with the triangle formed bycandd
translates, via the law of cosines, into the relation

a^2 +b^2 − 2 abx=c^2 +d^2 − 2 cdy.