Solutions and Comments
 31.

Let x, y, z be positive real numbers
for which x^{2} + y^{2} + z^{2} = 1. Find the minimum
value of
Solution 1. [S. Niu] Let a = yz/x, b = zx/y and
c = xy/z. Then a, b, c are positive, and the problem
becomes to minimize S = a + b + c subject to
ab + bc + ca = 1. Since
2(a^{2} + b^{2} + c^{2}  ab  bc  ca) = (ab)^{2} + (bc)^{2} + (ca)^{2} ³ 0, 

we have that
a
^{2} + b
^{2} + c
^{2} ³ ab + bc + ca. Thus,


= ab + bc + ca £ a^{2} + b^{2} + c^{2} 
 
= (a + b + c)^{2}  2(ab + bc + ca) = (a + b + c)^{2}  2 = S^{2}  2 

 

so S
³ Ö3
with equality if and only if a = b = c, or
x = y = z = 1/
Ö3. The desired result follows.
Solution 2. We have that


= 
x^{2}y^{2} z^{2}

+ 
y^{2}z^{2} x^{2}

+ 
z^{2}x^{2} y^{2}

+ 2(x^{2} + y^{2} + z^{2}) 
 
= 
1 2


é ê
ë


æ ç
è


x^{2}y^{2} z^{2}

+ 
z^{2}x^{2} y^{2}


ö ÷
ø

+ 
æ ç
è


x^{2}y^{2} z^{2}

+ 
y^{2}z^{2} x^{2}


ö ÷
ø

+ 
æ ç
è


y^{2}z^{2} x^{2}

+ 
z^{2}x^{2} y^{2}


ö ÷
ø


ù ú
û

+ 2 
 
³ x^{2} + y^{2} + z^{2} + 2 = 3 

 

by the ArithmeticGeometric Means Inequality. Equality holds
if and only if xy/z = yz/x = zx/y, which is equivalent to
x = y = z. Hence
S
³ Ö3 if and only if x = y = z = 1/
Ö3.
 32.

The segments BE and CF are altitudes of the
acute triangle ABC, where E and F are points on the
segments AC and AB, respectively. ABC is
inscribed in the circle Q with centre O. Denote the
orthocentre of ABC by H, and the midpoints of BC and
AH be M and K, respectively. Let ÐCAB = 45^{°}.


(a) Prove, that the quadrilateral MEKF is a
square.


(b) Prove that the midpoint of both diagonals
of MEKF is also the midpoint of the segment OH.


(c) Find the length of EF, if the radius of
Q has length 1 unit.
Solution 1. (a) Since AH is the hypotenuse of
right triangles AFH and AHE, KF = KH = KA = KE. Since
BC is the hypotenuse of each of the right triangles
BCF and BCE, we have that MB = MF = ME = MC.
Since
ÐBAC = 45
^{°}, triangles ABE,
HFB and
ACF are isosceles right triangles, so
ÐACF =
ÐABE =
ÐFBH =
ÐFHB = 45
^{°} and
FA = FC, FH = FB.
Consider a 90^{°} rotation with centre F that takes
H ® B. Then FA ® FC, FH ® FB,
so DFHA ® DFBC and K ® M.
Hence FK = FM and ÐKFM = 90^{°}.
But FK = KE and FM = ME, so MEKF is an equilateral
quadrilateral with one right angle, and hence is a square.
(b) Consider a 180^{°} rotation (halfturn) about the
centre of the square. It takes K « M,
F « E and H « H¢. By part
(a), DFHA º DFBC and AH ^BC.
Since KH  MH¢ (by the halfturn), MH¢^BC.
Since AH = BC, BM = ^{1}/_{2}BC = ^{1}/_{2}AH = KH = MH¢, so that BMH¢ is a right isosceles triangle and
ÐCH¢M = ÐBH¢M = 45^{°}. Thus,
ÐBH¢C = 90^{°}. Since ÐBAC = 45^{°},
H¢ must be the centre of the circle through ABC. Hence
H¢ = O. Since O is the image of H by a halfturn about the
centre of the square, this centre is the midpoint of
OH¢ as well as of the diagonals.
(c) EF  = Ö2 FM  = Ö2 BM  = OB  = 1.
Solution 2. [M. Holmes] (a) Consider a Cartesian plane
with origin F (0, 0) and xaxis along the line AB.
Let the vertices of the triangle be A (1, 0),
C (0, 1), B (b, 0). Since the triangle is acute,
0 < b < 1. The point E is at the intersection of the
line AC (y = x + 1) and a line through B with
slope 1, so that E = ( ^{1}/_{2}(b  1),^{1}/_{2}(b + 1). H is the intersection point of the
lines BE and CF, so H is at (0, b); K is the
midpoint of AH, so K is at (^{1}/_{2}, ^{b}/_{2});
M is the midpoint of BC, so M is at (^{b}/_{2},^{1}/_{2}). It can be checked that the midpoints of
EF and KM are both at (^{1}/_{4}(b1), ^{1}/_{4}(b+1)).
The slope of EF is (b+1)/(b1) and that of KM is the
negative reciprocal of this, so that EF ^KM.
It is straightforward to check that the lengths of
EF and KM are equal, and we deduce that EKFM is a
square.
(b) O is the intersection point of the right bisectors of
AB, AC and BC. The line x + y = 0 is the right bisector
of AC and the abscissae of points on the right bisector of
BC are all ^{1}/_{2}(b  1).
Hence O is at
(^{1}/_{2}(b1), ^{1}/_{2}(1  b)). It can be checked that
the midpoint of OH agress with the joint midpoint of
EF and KM.
(c) This can be checked by using the coordinates of points
already identified.
Comment. One of the most interesting theorems in
triangle geometry states that for each triangle there
exists a circle that passes through the following
nine special points: the three midpoints of the
sides; the three intersections of sides and
altitudes (pedal points); and the three midpoints
of the segments connecting the vertices to the
orthocentre. This circle is called the
ninepoint circle. If H is the orthocentre and
O is the circumcentre, then the centre of the
ninepoint circle is the midpoint of OH. Note that
in this problem, the points M, E, F, K belong to the
ninepoint circle.
 33.

Prove the inequality a^{2} + b^{2} + c^{2} + 2abc < 2,
if the numbers a, b, c are the lengths of the sides
of a triangle with perimeter 2.
Solution 1. Let u = b + c  a, v = c + a  b and
w = a + b  c, so that 2a = v + w, 2b = u + w and
2c = u + v. Then u, v, w are all positive and
u + v + w = 2. The difference of the right and left sides
multiplied by 4 is equal to


 
= 8  (v + w)^{2}  (u + w)^{2}  (u + v)^{2} (v + w)(u + w)(u + v)] 
 
= 8  (u + v + w)^{2}  (u^{2} + v^{2} + w^{2}) (2  u)(2  v)(2  w) 
 
= 8  4  (u^{2} + v^{2} + w^{2})  8+ 4(u + v + w)  2(vw + uw + uv) + uvw 
 
= 4  (u + v + w)^{2}  8 + 4×2 + uvw 
 
= 4  4  8 + 8 + uvw = uvw > 0 

 

as desired.
Solution 2. [L. Hong] The perimeter of the triangle
is a + b + c = 2. We have that

a^{2} + b^{2} + c^{2} + 2abc 

= (a + b + c)^{2} + 2(a  1)(b  1)(c  1)+ 2  2(a + b + c) 
 
= 4 + 2(a  1)(b  1)(c  1) + 2  4 = 2(a  1)(b  1)(c  1) + 2 . 

 

Since a < b + c, b < c + a and c < a + b,
it follows that a < 1, b < 1, c < 1, from which
the result follows.
 34.

Each of the edges of a cube is 1 unit in length,
and is divided by two points into three equal parts.
Denote by K the solid with vertices at these points.


(a) Find the volume of K.


(b) Every pair of vertices of K is
connected by a segment. Some of the segments are
coloured. Prove that it is always possible to find
two vertices which are endpoints of the same number
of coloured segments.
Solution. (a) The solid figure is obtained by slicing
off from each corner a small tetrahedron, three of whose
faces are pairwise mutually perpendicular at one vertex;
the edges emanating from that vertex all have length
1/3, and so the volume of each tetrahedron removed is
1/3(1/2·1/3 ·1/3 )(1/3) = 1/162.
Since there are eight such tetrahedra removed, the
volume of the resulting solid is 1  4/81 = 77/81.
(The numbers of vertices, edges and faces of the solid
are respectively 24, 36 and 14.)
(b) The polyhedron has 3 ×8 = 24 vertices.
Each edge from a given vertex is joined to 23 vertices. The possible number
of coloured segments emanating from a vertex is one of the
twentyfour numbers, 0, 1, 2, ¼, 23. But it is not
possible for one vertex to be joined to all 23 others and
another vertex to be joined to no other vertex. So there
are in effect only 23 options for the number of coloured
segments emanating from each of the 24 vertices. By the
Pigeonhole Principle, there must be two vertices with the
same number of coloured segments emanating from it.
 35.

There are n points on a circle whose
radius is 1 unit. What is the greatest number of
segments between two of them, whose length exceeds
Ö3?
Solution. [O.Bormashenko] The side of the equilateral
triangle inscribed in a circle of unit radius is
Ö3. So the segment with length
Ö3 is a
chord subtending an angle of 120
^{°} at the centre.
Therefore, there is no triangle with three vertices on
the circle each of whose sides are longer than
Ö3.
Consider the graph whose vertices are all n given points
and whose arcs all have segments longer than
Ö3.
This graph contains no triangles.
Recall Turan's theorem (see the solution of problem 23
in Olymon 1:4: Let G be a graph with
n vertices. Denote by l(G) the number of its edges
and t(G) the number of triangles contained in the graph.
If t(G) = 0, then l(G) £ ën^{2}/4 û.
>From this theorem, it follows that the number of
segments with chords exceeding Ö3 is at most
ën^{2}/4 û.
To show that this maximum number can be obtained, first
construct points A, B, C, D on the circle, so that
the disjoint arcs
AB and CD subtend angles of 120^{°} at the
centre. If n = 2k+1 is odd, place k points on the
arc BC and k+1 points on the arc DA. Any segment
containing a point in BC to a point in DA must
subtend an angle exceeding 120^{°}, so its
length exceeds Ö3. There are exactly
k(k+1) = ë(2k+1)^{2}/4 û such segments.
If n = 2k is even, place k points in each of the
arcs BC and CA, so that there are exactly
k^{2} = ë(2k)^{2}/4 û such segments.
In either case, the maximum number of segments
whose length exceeds Ö3 is ën^{2}/4û.
 36.

Prove that there are not three rational
numbers x, y, z such that
x^{2} + y^{2} + z^{2} + 3(x + y + z) + 5 = 0 . 

Solution. Suppose the x = u/m, y = v/m and
z = w/m, where m is the least common multiple of
the denominators of x, y and z. Then,
multiplying the given equation by m^{2} yields
u^{2} + v^{2} + w^{2} + 3(um + vm + wm) + 5m^{2} = 0 . 

A further multiplication by 4 and a rearrangement of
terms yields
(2u + 3m)^{2} + (2v + 3m)^{2} + (2w + 3m)^{2} = 7m^{2} . 

This is of the form
p^{2} + q^{2} + r^{2} = 7n^{2} (*) 

for integers p, q, r, n.
Suppose that n is even. Then, considering the equation
modulo 4, we deduce that p, q, r must also be even.
We can divide the equation by 4 to obtain another
equation of the form (*) with smaller numbers.
We can continue to do this as long as the resulting
n turns out to be even. Eventually, we
arrive at an equation for which n is odd, so that
the right side is congruent to 7 modulo 8. But there
is not combination of integers p, q and r for which
p
^{2} + q
^{2} + r
^{2} º 7 (mod 8), so that the equation is
impossible.