628 Geometry and Trigonometry
The substitutiont=tanu 2 changes this into
2
a+d∫
dt
1 +aa−+ddt^2.
Ifa=dthe answer to the problem is^1 atanx− 2 α+C.Ifaa−+dd>0, the answer is
2
√
a^2 −d^2arctan(√
a−d
a+dtan
x−α
2+C
)
,
while ifaa−+dd<0, the answer is
1
√
d^2 −a^2ln∣
∣∣
∣∣
∣
1 +
√
d−a
d+atanx−α
2
1 −√
d−a
d+atanx−α
2∣
∣∣
∣∣
∣
+C.
615.The first equation is linear, so it is natural to just solve for one of the variables, say
u, and substitute in the second equation. We obtain
2 xy=z(x+y−z),or
z^2 −xz−yz+ 2 xy= 0.This is a homogeneous equation. Instead of looking for its integer solutions, we can
divide through by one of the variables, and then search for the rational solutions of the
newly obtained equation. In fancy language, we switch from a projective curve to an
affine curve. Dividing byy^2 gives
(
z
y
) 2
−
(
z
y)(
x
y)
−
(
z
y)
+ 2
(
x
y)
= 0.
The new equation is
Z^2 −ZX−Z+ 2 X= 0 ,which defines a hyperbola in theXZ-plane. Let us translate the original problem into
a problem about this hyperbola. The conditionsx≥yandm≤xybecomeX≥1 and
X≥m. We are asked to find the largestmsuch that any point(X, Z)with rational
coordinates lying on the hyperbola and in the half-planeX≥1 hasX≥m.
There is a standard way to see that the points of rational coordinates are dense in
the hyperbola, which comes from the fact that the hyperbola isrational. Substituting
Z=tX, we obtain