Solutions and comments.
Notes. A rectangular hyperbola is an hyperbola
whose asymmptotes are at right angles.

97.

A triangle has its three vertices on a rectangular
hyperbola. Prove that its orthocentre also lies on the
hyperbola.
Solution 1. A rectangular hyperbola can be represented as
the locus of the equation xy = 1. Let the three vertices of
the triangle be at (a, 1/a), (b, 1/b), (c, 1/c). The altitude
to the points (c, 1/c) has slope
(a
 b)/(a
^{1}  b
^{1}) = ab and its equation is y = abx + (1/c)
 abc. The altitude to
the point (a, 1/a) has equation y = bcx + (1/a)
 abc. These
two lines intersect in the point (
1/abc,
abc) and the result
follows.
Solution 2. [R. Barrington Leigh] Suppose that the equation
of the rectangular hyperbola is xy = 1. Let the three vertices
be at (x_{i}, y_{i}) (i = 1, 2, 3), and let the orthocentre be
at (x_{0}, y_{0}). Then
(x_{1}  x_{2})(x_{0}  x_{3}) = (y_{1}  y_{2})(y_{0}  y_{3}) 

and
(x_{1}  x_{3})(x_{0}  x_{2}) = (y_{1}  y_{3})(y_{0}  y_{2}) . 

Crossmultiplying these equations yields that
(x_{1}  x_{2})(y_{1}  y_{3})(x_{0}  x_{3})(y_{0}  y_{2}) = (x_{1}  x_{3})(y_{1}  y_{2})(x_{0}  x_{2})(y_{0}  y_{3}) , 

whence
(1  x_{1}y_{3}  x_{2}y_{1} + x_{2}y_{3})(x_{0}y_{0}  x_{0}y_{2}  x_{3}y_{0} + x_{3}y_{2}) = (1  x_{1}y_{2}  x_{3}y_{1} + x_{3}y_{2})(x_{0}y_{0}  x_{0}y_{3}  x_{2}y_{0} + x_{2}y_{3}) . 

Collecting up the terms in x
_{0}y
_{0}, x
_{0}, y
_{0}, and the rest,
and simplifying, yields that x
_{0}y
_{0} = 1, as desired.

98.

Let a_{1}, a_{2}, ¼, a_{n+1}, b_{1}, b_{2}, ¼, b_{n} be nonnegative real numbers for which
(i) a
_{1} ³ a
_{2} ³ ¼ ³ a
_{n+1} = 0,
(ii) 0 £ b_{k} £ 1 for k = 1, 2, ¼, n.


Suppose that m = ëb_{1} + b_{2} + ¼+ b_{n} û+ 1. Prove that

n å
k=1

a_{k} b_{k} £ 
m å
k=1

a_{k} . 

Solution. Note that m
1
£ b
_{1} + b
_{2} +
¼+ b
_{m} < m.
We have that
a_{1}b_{1} + a_{2}b_{2} + ¼ 

+ a_{m} b_{m}+ a_{m+1} b_{m+1} + ¼+ a_{n} b_{n} 
 
£ a_{1} b_{1} + a_{2} b_{2} + ¼+ a_{m} b_{m} + a_{m} (b_{m+1} + b_{m+2} + ¼+ b_{n}) 
 
< a_{1} b_{1} + a_{2} b_{2} + ¼+ a_{m} b_{m}+ a_{m} (m  b_{1}  b_{2}  ¼ b_{m}) 
 
= a_{1} b_{1} + a_{2} b_{2} + ¼+ a_{m} b_{m} + a_{m} (1  b_{1})+ a_{m} (1  b_{2}) + ¼+ a_{m} (1  b_{m}) 
 
£ a_{1} b_{1} + a_{2} b_{2} + ¼+ a_{m} b_{m} +a_{1} (1  b_{1}) + a_{2} (1  b_{2}) + ¼+ a_{m} (1  b_{m}) 
 
= a_{1} + a_{2} + ¼+ a_{m} . 



99.

Let E and F be respective points on sides
AB and BC of a triangle ABC for which AE = CF. The
circle passing through the points B, C, E and the circle
passing through the points A, B, F intersect at B and
D. Prove that BD is the bisector of angle ABC.
Solution 1. Because of the concyclic quadrilaterals,
ÐDEA = 180
^{°}  ÐBED =
ÐDCF and
ÐDFC = 180
^{°}  ÐDFB =
ÐDAB .
Since, also, AE = CF,
DDAE
º DDFC (ASA)
so that AD = DF. In the circle through ABFD, the
equal chords AD and DF subtend equal angles ABD and
FBD at the circumference. The result follows.
Solution 2. ÐCDF = ÐCDE  ÐFDE = 180^{°}  ÐABC  ÐFDE = ÐFDA  ÐFDE = ÐEDA and ÐAED = 180^{°}  ÐBED = ÐBCD = ÐFCD. Since AE = CF, DEAD º DCFD (ASA). The altitude from D to AE is equal to
the altitude from D to FC, and so D must be on the bisector of
ÐABC.
Solution 3. Let B be the point (0, 1) and D the point
(0, 1). The centres of both circles are on the right
bisector of BD, namely the xaxis.
Let the two circles have equations
(x  a)^{2} + y^{2} = a^{2} + 1 and (x  b)^{2} + y^{2} = b^{2} + 1.
Suppose that y = mx  1 is a line
through B; this line intersects the circle of equation
(x  a)^{2} + y^{2} = a^{2} + 1 in the point

æ ç
è


2(m+a) m^{2} + 1

, 
m^{2} + 2am  1 m^{2} + 1


ö ÷
ø



and the circle of equation (x
 b)
^{2} + y
^{2} = b
^{2} + 1
in the point

æ ç
è


2(m+b) m^{2} + 1

, 
m^{2} + 2bm  1 m^{2} + 1


ö ÷
ø



The distance between these two points is the square root of

é ê
ë


2(ab) m^{2} + 1


ù ú
û

2

+ 
é ê
ë


2m(ab) m^{2} + 1


ù ú
û

2

= 
4(ab)^{2}(1 + m^{2}) (m^{2} + 1)^{2}

= 
4(ab)^{2} m^{2} + 1

. 

Now suppose that the side AB of the triangle has equation
y = m_{1} x  1 and the side BC the equation y = m_{2} x  1,
so that (A, E) and (C, F) are the pairs of points where
the lines intersect the circles.
Then, from the foregoing paragraph, we must have
m_{1}^{2} + 1 = m_{2}^{2} + 1 or 0 = (m_{1}  m_{2})(m_{1} + m_{2}).
Since the sides are distinct, it follows that m_{1} = m_{2}
and so BD bisects ÐABC.

100.

If 10 equally spaced points around a circle
are joined consecutively, a convex regular inscribed decagon
P is obtained; if every third point is joined, a
selfintersecting regular decagon Q is formed. Prove that
the difference between the length of a side of Q and
the length of a side of P is equal to the radius of the
circle. [With thanks to Ross Honsberger.]
Solution 1. Let the decagon be ABCDEFGHIJ. Let
BE and DI intersect at K and let AF and DI
intersect at L. Observe that
AB
 DI
 EH and BE
 AF
 HI, so that ABKL and
KIHE are parallelograms. Now AB is a side of P and
HE is a side of Q, and the length of the segment
IL is the difference of the lengths of EH = IK and AB = KL.
Since L, being the intersection of the diameters AF and DI,
is the centre of the circle, the result follows.
Solution 2. [R. Barrington Leigh]
Use the same notation as in Solution 1. Let
O be the centre of P. Now, AB is an edge of P,
AD is an edge of Q, DO is a radius of the circle and
BG a diameter. Let AD and BO intersect at U. Identify
in turn the angles ÐDOU = 72^{°}, ÐDAB = 36^{°}, ÐABU = 72^{°}, ÐDUO = ÐBUA = 72^{°}, whence AU = AB, DU = DO and
AD  AB = AD  AX = DX = DO, as desired.
Solution 3. Label the vertices of P as in Solution
1. Let O be the centre of P, and V be
a point on EB for which EV = OE. We have that
ÐAOB = 36^{°}, ÐDOB = ÐOBA = 72^{°},
ÐBOE = 108^{°} and ÐOEB = ÐOBE = 36^{°}.
Also, ÐEOV = ÐEVO = 72^{°} and OE = EV = OA = OB. Hence, DDAB = DEVO (SAS), so that OV = AB.
Since ÐBVO = 108^{°} and ÐBOV = 36^{°},
ÐOBV = 36^{°}, and so BV = OV = AB.
Hence BE  AB = EV + BV  AB = EV = OE, the radius.
Solution 4. Let the circumcircle of P and Q have radius
1. A side of P is the base of an isosceles triangle with
equal sides 1 and apex angle 36^{°}, so its length is
2sin18^{°}. Likewise, the length of a side of Q is
2sin54^{°}. The difference between these is
2sin54^{°}  2sin18^{°} = 2cos36^{°} 2cos72^{°} = 2t  2(2t^{2}  1) = 2 + 2t  4t^{2} 

where t = cos36
^{°}.
Now

= cos36^{°} =  cos144^{°} = 1  2cos^{2} 72^{°} 
 
= 1  2(2t^{2}  1)^{2} = 8t^{4} + 8t^{2}  1 , 


so that

= 8t^{4}  8t^{2} + t + 1 = (t + 1)(8t^{3}  8t^{2} + 1) 
 
= (t + 1)(2t  1)(4t^{2}  2t  1) . 


Since t is equal to neither
1 nor
^{1}/
_{2}, we must have
that 4t
^{2}  2t = 1. Hence
2sin54^{°}  2sin18^{°} = 2  (4t^{2}  2t) = 1 , 

the radius of the circle.

101.

Let a, b, u, v be nonnegative. Suppose that
a^{5} + b^{5} £ 1 and u^{5} + v^{5} £ 1. Prove that
a^{2}u^{3} + b^{2}v^{3} £ 1 . 

[With thanks to Ross
Honsberger.]
Solution. By the arithmeticgeometric means inequality,
we have that

2a^{5} + 3u^{5} 5

= 
a^{5} + a^{5} + u^{5} + u^{5} + u^{5} 5

³ Ö[5 ]a^{10}u^{15} = a^{2}u^{3} 

and, similarly,

2b^{5} + 3v^{5} 5

³ b^{2} v^{3} . 

Adding these two inequalities yields the result.

102.

Prove that there exists a tetrahedron ABCD, all
of whose faces are similar right triangles, each face having
acute angles at A and B. Determine which of the edges of
the tetrahedron is largest and which is smallest, and find the
ratio of their lengths.
Solution 1.
Begin with AB, a side of length 1. Now construct a
rectangle ACBD with diagonal AB, so that
AC
 =
BD
 = s < t =
AD
 =
BC
.
The requisite values of s and t will be determined in due
course. We want to show that we can fold up D and C
from the plane in which AB lies (like folding up the wings of
a butterfly) in such a way that we can obtain the desired
tetrahedron.
When the triangles ADB and ACB lie flat, we see that
C and D are distance 1 apart. Suppose that, when we have
folded up C and D to get the required tetrahedron, they
are distance r apart. Then ACD should be a right triangle
similar to ABC. The hypotenuse of DACD cannot be
AC as AC < AD. Nor can it be CD, for then, we would have
AD = BC, AC = AC, and CD would have to have length 1,
possible only when ABCD is coplanar. So the hypotenuse must
be AD. The similarity of DADC and DABC
would require that
where r =
CD
. Thus, 1/t = t/s or s = t
^{2}
and t/s = s/r or r = s
^{2}/t = t
^{3}. So we must fold C and
D until they are distance t
^{3} apart.
Is this possible? Since DACB is right,
1 = t^{2} + s^{2} = t^{2} + t^{4}, whence s = t^{2} = ^{1}/_{2}(1 + Ö5) < 1. Hence r < 1.
To arrange that we can make the distance between C and
D equal to r, we must show that r exceeds the minimum
possible distance between C and D, which occurs when
DADB is folded flat partially covering DACB.
Suppose this has been done, with ABCD coplanar and C, D
both on the same side of AB. Let P and Q be the respective
feet of the perpendiculars to AB from C and D. Then
CP  = DQ  = t^{3} , AP  = QB  = t^{4} , AQ  = PB  = t^{2} , 

and
CD  = PQ  = t^{2}  t^{4} = (t^{4} + t^{6})  t^{4} = t^{6} < t^{3} . 

When C and D are located, we have AB  = 1,
AD  = BC  = t, AC  = BD  = t^{2} and CD  = t^{3}. Since all
faces of the tetrahedron ABCD have sides in the ratio
1 : t : t^{2}, all are similar right triangles and
AB : CD = 1 : t^{3}.
Solution 2. Let a = ÐCAB and AB  = 1.
By the condition on the acute angles of triangles ACB and
ACD, ÐACB = ÐADB = 90^{°}, so that the
triangles DACD and DADB, being similar and
sharing a hypotenuse, are congruent.
Suppose, if possible, that ÐBAD = a. Then
AC = AD and so DACD must be isosceles with its
right angle at A, contrary to hypothesis.
So, ÐABD = a and BD  = AC  = cosa, AD  = BC  = sina.
Consider DACD. Suppose that ÐACD = 90^{°}.
If ÐDAC = a, then DABC º DADC and 1 = AB  = AD  = sina,
yielding a contradiction. Hence ÐADC = a,
AD  = AC /sina = cosa/sina and CD  = AC cota = cos^{2} a/sina. Hence, looking at AD ,
we have that

cosa sina

= sinaÞ 0 = cosa sin^{2} a = cos^{2} a+ cosa 1 . 

Therefore,
cos
a =
^{1}/
_{2}(
Ö5
 1) and
sin
^{2} a = cos
a.
Observe that BC sina = sin^{2} a = cosa = BD  and
BC cosa = sinacosa = cos^{2} a/sina = CD ,
so that triangle BCD is right with ÐCDB = 90^{°}
and similar to the other three faces.
We need to check that this setup is feasible. Using spatial
coordinates, take
C ~ (0, 0, 0) A ~ (0, cosa, 0) B ~ (sina, 0, 0) . 

Since
ÐACD = 90
^{°}, D lies in the plane y = 0
and so has coordinates of the form (x, 0, z). Since
ÐCDB = 90
^{°}, CD
^DB, so that
0 = (x, 0, z)·(x  sina, 0, z) x^{2} + z^{2}  xsina , 

Now
CD
 = cos
asin
a forces
cos
^{2} asin
^{2} a = x
^{2} + z
^{2}. Hence
xsina = cos^{2} asin^{2} aÞ x = cos^{2} asina . 

Therefore
z^{2} = (cos^{2} a cos^{4} a)sin^{2} a = cos^{2} asin^{4} aÞz = cosasin^{2} a , 

Hence D
~ (cos
^{2} asin
a, 0, cos
asin
^{2} a).
Thus, letting sina = t = ^{1}/_{2}(Ö5  1),
we have A ~ (0, t^{2}, 0), B ~ (t, 0, 0),
C ~ (0, 0, 0), D ~ (t^{5}, 0, t^{4}) with t^{4} + t^{2}  1 = 0, and AB  = 1, AD  = BC  = t, BD  = AC  = t^{2} and
CD  = t^{3}. [Exercise: Check that the
coordinates give the required distances and similar right
triangles.] The ratio of largest to smallest edges is
1 : t^{3} = 1 : [^{1}/_{2}(Ö5  1)]^{3/2} = 1 : Ö{2 + Ö5}.
We need to dispose of the other possibilities for DACD.
By the given condition, ÐDAC ¹ 90^{°}.
If ÐADC = 90^{°}, then we have essentially the
same situation as before with the roles of a and
its complement, and of C and D switched.
Comment. Another way in that was used by several solvers
was to note that there are four right angles involved among the
four sides, and that at most three angles can occur at a given
vertex of the tetrahedron. It is straightforward to argue that
it is not possible to have three of the right angles at either
C or D. Since all right angles occur at these two vertices,
then there must be two at each. As an exercise, you might want
to complete the argument from this beginning.