Solutions

311.

Given a square with a side length 1, let P
be a point in the plane such that the sum of the distances from P
to the sides of the square (or their extensions) is equal to 4. Determine the set of
all such points P.
Solution. If the square is bounded by the lines
x = 0, x = 1, y = 0 and y = 1 in the Cartesian plane,
then the required locus is equal the octagon whose vertices
are (0, 2), (1, 2), (2, 1), (2, 0), (1,
1),
(0,
1), (
1, 0), (
1, 1). Any point on the locus must
lie outside of the square, as within the square the sum of
the distances is equal to 2. If, for example, a point
on the locus lies between x = 0 and x = 1, the sum of the
distances to the vertical sides is 1, and it must be
1 unit from the nearer horizontal side and 2 units
from the farther horizontal side.
If, for example, a point on the locus lies to the left of
x = 0 and above y = 1 and has coordinates (u, v), then
u + (1 + u ) + v + (v  1) = 4 

or
u + 1  u + v + v  1 = 4 or v  u = 2 . 

Thus it can be shown that every point on the locus lies on the
octagon, and conversely, it is straightforward to verify that
each point on the octagon lies on the locus.

312.

Given ten arbitrary natural numbers. Consider the
sum, the product, and the absolute value of the difference
calculated for any two of these numbers. At most how many of
all these calculated numbers are odd?
Solution. Suppose that there are k odd numbers and
10
 k even numbers, where 0
£ k
£ 10.
There are k(10
 k) odd sums, k(10
 k) odd differences and
^{1}/
_{2}k(k
 1) odd products (on the presumption that the
numbers chosen are distinct), giving a total of

1
2

(39k  3k^{2}) = 
3
2


é ë


æ è


13
2

2


ö ø

 
æ è


13
2

 k 
ö ø

2


ù û



odd results. This quantity achieves its maximum when k = 13/2,
so the maximum number 63 of calculated numbers occurs when
k = 6 or k = 7.
Comment. If we allow a number to be operated with itself,
then the maximum occurs when k = 7.

313.

The three medians of the triangle ABC partition it
into six triangles. Given that three of these triangles have
equal perimeters, prove that the triangle ABC is equilateral.
Solution. [P. Shi] Let a, b, c be the respective lengths
of the sides BC, CA, AB, and u, v, w the respective lengths
of the medians AP, BQ, CR (P, Q, R the respective midpoints of
BC, CA, AB). If G is the centroid of the triangle ABC,
then
AG:GP = BG:GQ = CG:GR = 2:1 . 

We need three preliminary lemmata.
Lemma 1. 4u^{2} = 2b^{2} + 2c^{2}  a^{2}; 4v^{2} = 2c^{2} + 2a^{2}  b^{2};4w^{2} = 2a^{2} + 2b^{2}  c^{2}.
Proof. This can be established, either by representing the
medians vertorially in terms of the sides and applying the cosine
law to the whole triangle, or by applying the cosine law to pairs
of inner triangles along an edge. ª
Lemma 2. u < v; u = v; u > v according as
a > b; a = b; a < b, with analogous results for other pairs of
medians and sides.
Proof. 4(u^{2}  v^{2}) = 3(b^{2}  a^{2}). ª
Lemma 3. If triangle BCR and BCQ have the same perimeter,
then b = c.
Proof. Equality of the perimeters is equivalent to
BR + RC = BQ + QC, so that Q and R are points on an ellipse
with foci B and C. Since RQ is parallel to the major axis
containing BC, R and Q are reflections of each other in
the minor axis, so that RB = QC. Hence b = c. ª
We now establish conditions under which the triangle must be isosceles.
Lemma 4. Suppose that two adjacent inner triangles along the
same side of triangle ABC have the same perimeter. Then
triangle ABC is isosceles.
Solution. For example, the equality of the perimeters
of BPG and CPG is equivalent to

1
2

a + 
1
3

u + 
2
3

v = 
1
2

a + 
1
3

u + 
2
3

w 

Lemma 5. Suppose that two adjacent inner triangles sharing
a vertex of triangle ABC have the same perimeter. Then triangle
ABC is isosceles.
Proof. For example, suppose that triangle BRG and PGB
have the same perimeter. Produce CR to point T so that
GR = RT. Thus, BR is a median of triangle BGT. Produce
AP to point S so that GP = PS. Thus, CP is a median of
triangle CPS.
Since GP joins midpoints of two sides of triangle CTB,
TB  GP and TB = 2GP = GS. Since triangle PGB and
PSC are congruent (SAS), BG = SC. Also, TG = 2RG = GC.
Hence, triangles TBG and GSC are congruent (SSS).
Let X be the midpoint of GC. A translation that takes T to
G takes triangle TBG to triangle GSC and median BR to median
SX. We have that
Perimeter(SXC) = Perimeter(BRG) = Perimeter(PGB) = Perimeter(PSC) . 

Applying Lemma 3, we deduce that GS = GC, whence u = w and
a = c.
ª
Lemma 6. If two opposite triangles (say, BGP and AQG)
have equal perimeters, then triangle ABC is isosceles.
Proof. The equality of the perimeters of BGP and AQG
implies that

1
2

a + 
1
3

u + 
2
3

v = 
1
2

b + 
1
3

v + 
2
3

u 

By Lemma 2, the latter equation holds if and only if a = b.
ª
Let us return to the problem. There are essentially four different
cases for the three inner triangles with equal perimeters.
Case 1. The three are adjacent (say BRG, BPG, CPG).
Then by Lemmata 4 and 5, a = b = c.
Case 2. Two are adjacent along a side and the third is opposite
(say BPG, CPG, AQG). Then, by Lemmata 4 and 6, a = b = c.
Case 3. Two are adjacent at a vertex and the third is
opposite (say BPG, CQG, AQG.) Then, by Lemmata 5 and 6,
a = b = c.
Case 4. No two are adjacent (say BPG, CQG, ARG).
Then we have

1
2

a + 
1
3

u + 
2
3

v = 
1
2

b + 
1
3

v + 
2
3

w = 
1
2

c + 
1
3

w + 
2
3

v . 

Thus
3(a  b) = 2(w  u) + 2(w  v) . 
 (1) 
Similarly,
3(b  c) = 2(u  v) + 2(u  w) ; 
 (2) 
3(c  a) = 2(v  u) + 2(v  w) . 
 (3) 
Suppose, wolog, that a
³ b
³ c. Then u
£ v
£ w,
so that, by (2), 3(b
 c)
£ 0
Þ b = c
Þ v = w. But then, by (3), 3(c
 a) = 2(v
 u)
³ 0
Þ c = a.
The result follows.

314.

For the real numbers a, b and c, it is known that
and
Find the value of the expression
M = 
1
1 + a + ab

+ 
1
1 + b + bc

+ 
1
1 + c + ca

. 

Solution 1. Putting the first equation over a common
denominator and using the second equation yields that
whence


= 
1
1 + a + (1/c)

+ 
1
1 + b + (1/a)

+ 
1
1 + c + (1/b)


 
 
= 
c
1 + c + ac

+ 
a
1 + a + ab

+ 
b
1 + b + bc


 
 
= 
c
1 + c + (1/b)

+ 
a
1 + a + (1/c)

+ 
b
1 + b + (1/a)


 
 
= 
bc
1 + b + bc

+ 
ac
1 + c + ac

+ 
ab
1 + a + ab

. 

This yields three different expressions for M over denominators
of the form 1 + a + ab, which when added together yield
3M = 3 or M = 1.
Solution 2. [V. Krakovna] It is clear that abc ¹ 0
for the expressions to be defined. As before, abc = 1, and

1
1 + a + ab

= 
1
1 + a + (1/c)

= 
c
c + ca + 1

. 

Hence


= 
1
1 + a + ab

+ 
1
1 + b + bc

+ 
1
1 + c + ca


 
 
= 
c
c + ca + 1

+ 
1
1 + b + bc

+ 
1
1 + c + ca


 
 
= 
c + 1
1 + c + ca

+ 
1
1 + b + bc


 
 
= 
b(c+1)
b + bc + 1

+ 
1
1 + b + bc

= 
bc + b + 1
b + bc + 1

= 1 . 


315.

The natural numbers 3945, 4686 and 5598 have the
same remainder when divided by a natural number x. What is the
sum of the number x and this remainder?
Solution. Observe that 5598
 4686 = 912 = 16 ×57
and 4686
 3945 = 741 = 13 ×57, so that if a divisor leaves
equal remainders for the three numbers, the divisor must also divide
evenly into 57. Since 5598 = 98 ×57 + 12,
4686 = 82 ×57 + 12 and 3945 = 69 ×57 + 12.
the number x must be 1, 3, 19 or 57. The sums of the number and
the remainder are respectively 1, 3, 31 and 69.

316.

Solve the equation
x^{2}  3x + 2 + x^{2} + 2x  3  = 11 . 

Solution. The equation can be rewritten
x  1 [ x  2 + x + 3 ] = 11 . 

When x
£ 3, the equation is equivalent to
neither of whose solutions satisfies x
£ 3.
When
3
£ x
£ 1, the equation is equivalent to
5x + 5 = 11 and we get the solution x =
6/5.
When 1 < x < 2, the equation is equivalent to 5x
 5 = 11
which has no solution with 1 < x < 2.
Finally, when 2
£ x, the equation is equivalent to
2x
^{2}  x
 12 = 0 and we obtain the solution
x =
^{1}/
_{4}(1 +
Ö{97}).
Thus the solutions are x = 6/5, (1 + Ö{97})/4.

317.

Let P(x) be the polynomial
P(x) = x^{15}  2004x^{14} + 2004x^{13}  ¼ 2004x^{2} + 2004x , 

Calculate P(2003).
Solution 1. For each nonnegative integer n, we have that
x^{n+2}  2004x^{n+1} + 2003x^{n} = x^{n}(x  1)(x  2003) . 

Therefore,


= (x^{15}  2004x^{14} + 2003x^{13}) + (x^{13}  2004x^{12}+ x^{11}) + ¼+ (x^{3}  2004x^{2} + 2003x) + x 
 
 
= (x^{13} + x^{11} + ¼+ x)(x  1)(x  2003) + x , 

whereupon P(2003) = 2003.
Solution 2. [R. Tseng]
P(x) = (x^{15}  2003x^{14})  (x^{14}  2003x^{13}) +(x^{13}  2003x^{12})  ¼ (x^{2}  2003x) + x 

whence
P(2003) = 0
 0 + 0
 ¼ 0 + 2003 = 2003.