Solutions and Comments
 37.

Let ABC be a triangle with sides a, b, c,
inradius r and circumradius R (using the conventional
notation). Prove that

r 2R

£ 
abc
 ______________________ Ö2(a^{2} + b^{2})(b^{2} + c^{2})(c^{2} + a^{2}) 

. 

When does equality hold?
Solution.


 (b^{2} + c^{2})(1  cosA) = b^{2} + c^{2}  2bc cosA  (b^{2} + c^{2}) + (b^{2} + c^{2})cosA 
 

 

Þ a^{2} ³ (b^{2} + c^{2})(1  cosA) = 2(b^{2} + c^{2})sin^{2} (A/2) . 

With similar inequalities for b and c, we find that
a^{2} b^{2} c^{2} ³ 8(a^{2} + b^{2})(b^{2} + c^{2})(c^{2} + a^{2})sin^{2} (A/2) sin^{2} (B/2) sin^{2} (C/2) . 

Since r = 4Rsin(A/2) sin(B/2) sin(C/2), the desired
result follows. Equality holds if and only if the triangle
is equilateral.
Comment. The identity in the solution can be obtained as
follows. Let 2s = a + b + c. Then
while
a = 2R sinA = 4R sin 
A 2

cos 
A 2

. 

Hence

ar s  a

= 4R sin^{2} 
A 2

. 

Using similar identities for the other sides, we find that

abcr^{3} (s  a)(s  b)(s  c)

= 64R^{3} sin^{2} 
A 2

sin^{2} 
B 2

sin^{2} 
C 2

. (*) 

Note that the area
D of the triangle is given by
D = rs = 
abc 4R

= 
 ________________ Ös(s  a)(s  b)(s  c)

, 

so that the left side of (*) becomes
4R
Dr
^{2}(rs)
D^{2} = 4Rr
^{2} .
Substituting this in, dividing by 4R and taking the square root yields
r = 4R sin 
A 2

sin 
B 2

sin 
C 2

. 

 38.

Let us say that a set S of nonnegative real
numbers if hunkydory if and only if, for all x and
y in S, either x + y or x  y  is in
S. For instance, if r is positive and n is a natural
number, then S(n, r) = { 0, r, 2r, ¼, nr }
is hunkydory. Show that every hunkydory set is
{ 0 }, is of the form S(n, r) or has exactly four
elements.
Solution 1. { 0 } and sets of the form
{ 0, r } are clearly hunkydory. Let S be a
nontrivial hunkydory set with largest postive
element z. Then 2z \not
Î S, so 0 = z  z
Î S. Thus, every hunkydory set contains 0.
Suppose that S has at least three elements, with
least positive element a.
Suppose, if possible, that S contains an element that
is not a positive integer multiple of a. Let b be the
least nonmultiple of a. Then 0 < b  a < b. Since
b  a cannot be a multiple of a (why?), we must have
b  a \not Î S and b + a Î S. Since z is the largest
element of S, z  a and z  b belong to S. However,
(z  a)  (z  b) = b  a does not belong to S, so
2z  (a + b) = (z  a) + (z  b) Î S. Therefore,
2z  (a + b) £ z, whence z £ a + b, so that
z = a + b. Thus, S contains { 0, a, b, a+b },
with a + b the largest element. This subset is already
hunkydory. But suppose, if possible, S contains more
elements. Let c be the smallest such element.
Then 0 < (a + b)  c Î S Þa £ (a + b)  c Þ c < b Þc = ma for some positive integer m ³ 2.
Since b + ma > b + a, b  ma must belong to S, and
so be a multiple of a. This yields a contradiction.
Hence, S must be equal to { 0, a, b, a+b }.
The only remaining case is that S consists solely of
nonnegative multiples of some element a. Let na be the
largest such multiple. If n = 2, then S = S(2, a).
Suppose that n > 2. Then (n  1)a Î S, so S contains
{ 0, a, (n1)a, na }, which is hunkydory.
Suppose S contains a further multiple ma with
2 £ m £ n2. Since a Î S and na + a > na,
(n1)a Î S, so that
n  (m + 1)a = (n1)a  ma Î S Þ(m+1)a = n  [n  (m+1)a] Î S. By induction, it can
be shown that ka Î S for m £ k £ n. In particular,
(n  2)a Î S so that 2a = na  (n2)a Î S. But then
3a, 4a, ¼, na are in S and so S = S(n, a). The
desired result follows,
Solution 2. [S. Niu] Let S = { a_{0}, a_{1}, ¼, a_{n} },
with a_{0} < a_{1} < ¼ < a_{n}. The elements a_{n}  a_{n},
a_{n}  a_{n1}, ¼, a_{n}  a_{0} are n+1 distinct elements
of S listed in increasing order, and so a_{0} = 0, and for each
i with 0 £ i £ n, we must have that a_{n}  a_{i} = a_{ni}. Let i £ ^{n}/_{2}. Then i £ n  i and
so a_{i} £ a_{ni}; thus, a_{i} £ (a_{n})/2 £ a_{ni}.
Thus, if j > k ³ n/2, a_{j}  a_{k} Î S.
Since 0 < a_{n1}  a_{n2} < a_{n}  a_{n2} = a_{2}, it follows that a_{n1} a_{n2} = a_{1}. Also, 0 < a_{n1}  a_{n2} = a_{1} < a_{n1}  a_{n3} < a_{n}  a_{n3} = a_{3}, so that a_{n1}  a_{n3} = a_{2}.
Continuing on in this way, we find that, for i ³ n/2,
0 < a_{n1}  a_{n2} < a_{n1}  a_{n3} < ¼ < a_{n1}  a_{n  i} < a_{n}  a_{n  i} = a_{i}, 

whence a
_{n1}  a
_{nj} = a
_{j1} for 1
£ j
£ (n/2).
Now 0 < a_{n2}  a_{n3} < a_{n1}  a_{n3} = a_{2} so
a_{n2}  a_{n3} = a_{1}. We can proceed in this fashion to
obtain that, for j ³ n/2, a_{j+1}  a_{j} = a_{1}. Hence, for
i £ (n/2)  1, a_{i+1}  a_{i} = (a_{n}  a_{ni1})  (a_{n} a_{ni}) = a_{ni}  a_{ni1} = a_{1}.
Let n = 2m. Then a_{i} = ia_{1} and a_{ni} = a_{n}  ia_{1} for
1 £ i £ m, so that a_{m} = ma_{1} and a_{n} = a_{m} + ma_{1} = na_{1}. It follows that a_{k} = ka_{1} for 1 £ k £ n and
S = S(n, a_{1}).
Let n = 2m+ 1. If m = 0, then S = { 0, a_{1} } = S(1, a_{1}).
If m = 1, then S = { 0, a_{1}, a_{3}  a_{1}, a_{3} } = { 0, a_{1}, a_{2}, a_{1} + a_{2} } is a 4element hunkydory set.
Let m ³ 2. Then, for 1 £ i £ m, a_{i} = ia_{1} and
a_{ni} = a_{n}  ia_{i}. Now a_{m+1} = a_{n}  a_{m} > a_{n1}  a_{m} > ¼ ³ a_{m+2}  a_{m} > a_{m+1}  a_{m} ³ a_{1}.
Since { a_{n}  a_{m} = a_{m + (m+1)}  a_{m}, a_{n1}  a_{m}, ¼, a_{m+2}  a_{m} , a_{m+1}  a_{m} } contains m+1 elements,
we must have a_{m+j}  a_{m} = a_{j} for 1 £ j £ n  m = m + 1.
Therefore, a_{i} = ia_{i} for 1 £ i £ n. (Why does this last
statement fail to follow when m = 1?)
 39.

(a) ABCDEF is a convex hexagon, each of
whose diagonals AD, BE and CF pass through a common
point. Must each of these diagonals bisect the area?
(b) ABCDEF is a convex hexagon, each of whose diagonals
AD, BE and CF bisects the area (so that half the area of
the hexagon lies on either side of the diagonal). Must the
three diagonals pass through a common point?
Solution 1. (a) No, they need not bisect the area. Let
the vertices of the hexagon have coordinates (1, 0),
(1, 1), (1, 1), (1, 0), (t, t), (t, t)
with t > 0 but t ¹ 1. The diagonals with equations
y = 0, y = x and y = x intersect in the origin but
do not bisect the area of the hexagon.
(b) Let the hexagon be ABCDEF and suppose that the
intersection of the diagonals AD and BE is on the same
side of CF as the side AB. Thus, AB, CD and
EF border on triangles whose third vertices form a
triangle at the centre of the hexagon (we will show this
triangle to be degenerate). Let a, b, c, d, e,
f be the lengths of the rays from the respective
vertices A, B, C, D, E, F to the vertices of
the central triangle, whose sides are x, y, z
so that the lengths of AD, BE and CF are respectively
a + x + d, b + y + e, c + z + f. All lowercase
variables represent nonnegative real numbers.
Let the areas of the bordering on FA, AB, BC,
CD, DE, EF be respectively a,
b, g, d, e, f, and
let the area of the central triangle be l.
Then, since each diagonal bisects the area of the
hexagon, we have that
>From the first two equations, we find that
d =
a+
l. Similarly,
f =
g+
l and
b =
e+
l.
Using the fact that the area of a trangle is half the
product of adjacent sides and the sine of the angle
between them, and the equality of opposite angles,
we find that
1 = 
a+ l d

= 
(a + x)(f + z) cd



1 = 
g+ l f

= 
(b + y)(c + z) ef



1 = 
e+ l b

= 
(d + x)(e + y) ab

. 

Multiplying these three equations together yields that
abcdef = (a + x)(b + y)(c + z)(d + x)(e + y)c + z) , 

whence x = y = z = 0. Thus, the central triangle
degenerates and the three diagonals intersect in a common
point.
Solution 2. (a) No. Let ABCDEF be a regular hexagon.
The diagonals AD, BE, CF intersect and each diagonal
does bisect the area. Let X be any point other than
F on the diagonal CF for which ABCDEX is still
a convex hexagon. The diagonals of this hexagon are the
same as those of the regular hexagon, and so have a common
point of intersection. However, the diagonals AD and
BE no longer intersect the area of the hexagon.
(b) [X. Li] Let
ABCDEF be a given convex hexagon, each of whose diagonals
bisect its area. Suppose that the diagonals AD and
CF intersect at G. As in Solution 1, we can determine
that the areas of triangles AGF and DGC are
equal, whence AG ·GF = CG ·GD,
or AG/GD = CG/GF. Therefore,
DAGC ~ DDGF (SAS). It follows that
AC/DF = AG/GD = CG/GF, ÐCAG = ÐFDG, and so
AC  DF. In a similar way, we find that
BF  CE and AE  BD, so that DACE ~ DDFB and AC/DF = CE/FB = EA/BD.
Suppose diagonals AD and BE intersect at H. Then, as
above, we find that AG/GD = AC/DF = EA/BD = EH/HB = AH/HD,
so that H = G. Hence, the three diagonals have the point
G in common.
 40.

Determine all solutions in integer pairs (x, y)
to the diophantine equation x^{2} = 1 + 4y^{3}(y + 2).
Solution 1. Clearly, (x, y) = (
±1, 0), (
±1, 2) are
solutions. When y = 1, the right side is negative and there
is no solution. Suppose that y
³ 1; then
(2y^{2} + 2y)^{2} = 4y^{4} + 8y^{3} + 4y^{2} > 4y^{4} + 8y^{3} + 1 

and
(2y^{2} + 2y  1)^{2} = 4y^{4} + 8y^{3}  4y + 1 < 4y^{4} + 8y^{3} + 1 

so that the right side is between two consecutive squares,
and hence itself cannot be square.
Suppose that y £ 3. We first observe that for a given
product p of two positive integers, the sum
of these positive integers has a minium
value of 2Öp (why?) and a maximum value of
1 + p. This follows from the fact that, for integers
u with 1 £ u £ p,
(1 + p)  (u + p/u) = (u  1)[(p/u)  1] ³ 0 . 

We have that


[(2y^{2} + 2y  1)  x] = (2y^{2} + 2y  1)^{2}  x^{2} 
 
= (4y^{4} + 8y^{3}  4y + 1)  (4y^{4} + 8y^{3} + 1) 
 

 

Since 2y
^{2} + 2y  1 = y
^{2} + (y + 1)
^{2}  2 is positive,
at least one of the factors on the left is positive.
Since the product is positive, both factors are positive.
By our observation on the sum of the factors, we find that
4y^{2} + 4y  2 £ 1  4y , 

which is equivalent to
However, this does not hold when y
£ 3.
Therefore, the only solutions are the four that we identified
at the outset.
Solution 2. Since x must be odd, we can let
x = 2z + 1 for some integer z, so that the equation
becomes z(z+1) = y^{4} + 2y^{3}. We can deal with the
cases that y = 0, 1, 2 directly to obtain the solutions
(x, y, z) = (1, 0, 0), (1, 0, 1), (1, 2, 0), (1, 2, 1).
Henceforth, suppose that y ³ 1 or y £ 3,
so that y^{4} + 2y^{3} is positive. Let
f(t) = t(t + 1). Then f(t) is increasing for
t ³ 0 and f(t) = f(t  1) for every integer
t; thus, we need check only that y^{4} + 2y^{3} does not
coincide with a value taken by f(t) for nonnegative
values of t.
Now
f(y^{2} + y) = y^{4} + 2y^{3} + 2y^{2} + y > y^{4} + 2y^{3} ; 

f(y^{2} + y  1) = y^{4} + 2y^{3}  y ¹ y^{4} + 2y^{3} ; 



= y^{4} + 2y^{3}  2y^{2}  3y + 2 
 
= y^{4} + 2y^{3}  (2y + 1)(y + 1) + 3 < y^{4} + 2y^{3} . 

 

It follows that
f(t) can never assume the value y
^{4} + 2y
^{3}
for any positive t, and hence for any t. Thus, the
solutions already listed comprise the complete solution set.
 41.

Determine the least positive number p for which
there exists a positive number q such that

 ____ Ö1 + x

+ 
 ____ Ö1  x

£ 2  
x^{p} q



for 0
£ x
£ 1. For this least value of p, what
is the smallest value of q for which the inequality is
satisfied for 0
£ x
£ 1?
Comments. Recall the binomial expansion
(1 + x)^{n} = 1 + nx + 
n(n1) 2!

x^{2} + 
n(n1)(n2) 3!

x^{3} + ¼+ 
n(n1)¼(nr+1) r!

x^{r} + ¼ . 

When n is not a nonnegative integer, this is an infinite
series that converges when 0
£ x
 < 1 to
(1 + x)
^{n}. The partial sums constitute a close approximation.
When n =
^{1}/
_{2}, we have that
(1 ±x)^{1/2} = 1 ± 
1 2

x  
1 8

x^{2} ± 
1 16

x^{3}  
5 128

x^{4}±¼ 

so that
(1 + x)^{1/2} + (1  x)^{1/2} ~ 2  
x^{2} 4

 
x^{4} 8

£ 2  
x^{4} 4

. 

This suggests that we are looking for (p, q) = (2, 4). However, the
approximation approach is not sufficiently rigorous, and we need
to find an argument in finite terms that will work.
Solution 1. Observe that, for 0 £ x £ 1,
 Ö

1 ±x

£ 1 ± 
1 2

x  
1 8

x^{2}± 
1 16

x^{3} 

Û1 ±x £ 1 ±x + 
5 64

x^{4} ± 
1 64

x^{5}+ 
1 256

x^{6} 

The last inequality clearly holds, so the first must as well.
Hence

 ____ Ö1 + x

+ 
 ____ Ö1  x

£ 2(1  
1 8

x^{2}) = 2  
1 4

x^{2} 

so the pair (p, q) = (2, 4) works for all x
Î [0, 1].
Suppose, for some constants p and c with 0 < p < 2 and
c > 0,

 ____ Ö1 + x

+ 
 ____ Ö1  x

£ 2  2cx^{p} 

for 0
£ x
£ 1. For this range of x, this is equivalent to
2 + 2 
 _____ Ö1  x^{2}

£ 4  8cx^{p} + 4c^{2} x^{2p} 

Û 
 _____ Ö1  x^{2}

£ 1  4cx^{p} + 2c^{2} x^{2p} 

Û 1  x^{2} £ 1  8cx^{p} + 20c^{2} x^{2p}  16c^{3} x^{3p} + 4c^{4} x^{4p} 

8c £ x^{2p} + 4c^{2} x^{p}(5  4cx^{2p} + c^{3} x^{3p}) . 

However, for x sufficiently small, the right side can be made
less than 8c, yielding a contradiction. Hence, when
0 < p < 2, there is no value that yields the desired
inequality.
Now we look at the situation when p = 2 and q > 0.
For 0 £ x £ 1,

 ____ Ö1 + x

+ 
 ____ Ö1  x

£ 2  
x^{2} q



Û 2(1 + 
 _____ Ö1  x^{2}

) £ 4  
4x^{2} q

+ 
x^{4} q^{2}



Û 
 _____ Ö1  x^{2}

£ 1  
2x^{2} q

+ 
x^{4} 2q^{2}



Û 1  x^{2} £ 1  
4x^{2} q

+ 
5x^{4} q^{2}

 
2x^{6} q^{3}

+ 
x^{8} 4q^{4}



Û 0 £ x^{2} 
é ê
ë


æ ç
è

1  
4 q


ö ÷
ø

+ 
x^{2} q^{2}


æ ç
è

5  
2x^{2} q

+ 
x^{4} 4q^{2}


ö ÷
ø


ù ú
û

. 

If q < 4, then the quantity in square brackets is negative for small
values of x. Hence, for the inequality to hold for all x
in the interval [0, 1], we must have q
³ 4. Hence, p must be at
least 2, and for p = 2, q must be at least 4.
Solution 2. [R. Furmaniak] The given inequality is
equivalent to


³ 
x^{p}
2  
 ____ Ö1 + x

 
 ____ Ö1  x



 
= 
x^{p}(2 + 
 ____ Ö1 + x

+ 
 ____ Ö1  x

) 
4  (2 + 2 
 _____ Ö1  x^{2}

) 


 
= 
x^{p}(2 + 
 ____ Ö1 + x

+ 
 ____ Ö1  x

) 


 
= 
x^{p} (2 + 
 ____ Ö1 + x

+ 
 ____ Ö1  x

)(1 + 
 _____ Ö1  x^{2}

) 
2x^{2}

. 

 

If p < 2, then the right side becomes arbitrarily large
as x gets close to zero, so the inequality becomes
unsustainable for any real q. Hence, for the inequality
to be viable, we require p
³ 2. When p = 2, we
can cancel x
^{2} and see
by taking x = 0 that q
³ 4. It remains to verify the
inequality when (p, q) = (2, 4). We have the following
chain of logically equivalent statements, where
y =
Ö[(1  x
^{2})] (note that 0
£ x
£ 1):

 ____ Ö1 + x

+ 
 ____ Ö1  x

£ 2  
x^{2} 4



Û 2 + 2 
 _____ Ö1  x^{2}

£ 4  x^{2} + 
x^{4} 16



Û 32 
 _____ Ö1  x^{2}

£ x^{4}  16x^{2} + 32 

Û 32y £ 1  2y^{2} + y^{4}  16 + 16y^{2} + 32 

0 £ y^{4} + 14y^{2}  32y + 17 = (y  1)^{2}(y^{2} + 2y + 17) = (y  1)^{2}[(y + 1)^{2} + 16] . 

Since the last inequality is clearly true, the first holds and the
result follows.
 42.

G is a connected graph; that is, it consists of
a number of vertices, some pairs of which are joined by edges,
and, for any two vertices, one can travel from one to another
along a chain of edges. We call two vertices adjacent
if and only if they are endpoints of the same edge. Suppose
there is associated with each vertex v a nonnegative integer
f(v) such that all of the following hold:
(1) If v and w are adjacent, then
f(v)  f(w)
 £ 1.
(2) If f(v) > 0, then v is adjacent to at least one vertex w
such that f(w) < f(v).
(3) There is exactly one vertex u such that f(u) = 0.
Prove that f(v) is the number of edges in the chain with the
fewest edges connecting u and v.
Solution. We prove by induction that f(x) = n if and only
if the shortest chain from u to x has n members. This
is true for n = 0 (and for n = 1). Suppose that this
holds for 0 £ n £ k.
Let f(x) = k + 1. There exists a vertex y adjacent to
x for which h = f(y) < k + 1. By the induction hupothesis,
y can be connected to u by a chain of h edges, so
x can be connected to u by a chain of h + 1 edges.
Hence, h + 1 ³ k + 1. From these two inequalities,
we must have h = k, so x can be connected to u by
a chain of k + 1 edges. There cannot be a shorter
chain, as, by the induction hypothesis, this would mean that
f(x) would have to be less than k + 1.
Let the shortest chain connecting x to u have k + 1 edges.
Following along this chain, we can find an element z adjacent
to x connected to u by k edges. This must be one of the
shortest chains between u and z, so that f(z) = k.
By hypothesis (1), f(x) must take one of the values k  1
and k + 1. The first is not admissible, since there is no
chain with k  1 edges connecting u and x. Hence
f(x) = k + 1.