Solutions

262.

Let ABC be an acute triangle. Suppose that
P and U are points on the side BC so that P lies
between B and U, that Q and V are points on the
side CA so that Q lies between C and V, and that
R and W are points on the side AB so that R lies
between A and W. Suppose also that
ÐAPU = ÐAUP = ÐBQV = ÐBVQ = ÐCRW = ÐCWR . 

The lines AP, BQ and CR bound a triangle T_{1}
and the lines AU, BV and CW bound a triangle T_{2}.
Prove that all six vertices of the triangles T_{1} and
T_{2} lie on a common circle.
Solution 1. Note that the configuration requires the feet
of the altitudes to be on the interior of the sides of the triangle
and the orthocentre to be within the triangle. Let
q
be the common angle referred to in the problem. Let XYZ be
that triangle with sides parallel to the sides of triangle ABC
and A on YZ, B on ZX and C on XY. Then A, B, C
are the respective midpoints of YZ, ZX, XY and the orthocentre
H or triangle ABC is the circumcentre of triangle XYZ.
[Why?] Let
r be the common length of HX, HY, HZ.
Let K be the intersection of AP and BQ (a vertex of T_{1}).
Since ÐKPC + ÐKQC = 180^{°}, CQKP is concyclic.
Hence ÐAKB + ÐAZB = ÐPKQ + ÐPCQ = 180^{°}
and AKBZ is concyclic. Since AH ^BC and BH ^AC, the
angle between AH and BH is equal to ÐACB = ÐXZY,
so that AHBZ is also concyclic. Thus, A, H, K, B, Z lie on a
common circle, so that ÐHKZ = ÐHBC = 90^{°}.
Now q = ÐAPC = ÐZAK = ÐZHK, so that, in
the right triangle HKZ, HK  = rcosq.
Similarly, it can be shown that the distance from each vertex of
triangle T_{1} and T_{2} from H is rcosq and the
result follows.
Solution 2. [R. Dan] Let H be the orthocentre of
triangle ABC, let AP and BQ intersect at D, and
let AU and BQ intersect at E. Triangle APU is
isosceles with AP = AU, and AH a bisector of ÐPAU
and a right bisector of PU. Suppose X = AH ÇBC, Y = BH ÇAC and Z = CH ÇAB.
Since triangles APU and BQV are similar, ÐPAH = ÐQBH, so that ABDH is concyclic and ÐADH = ÐABH. Similarly, ACEH is concyclic and ÐAEH = ÐACH. Since the quadrilateral BZYC has right
angles at Z and Y, BZYC is concyclic and
ÐABH = ÐABY = ÐZBY = ÐZCY = ÐZCA = ÐACH . 

Therefore,
ÐADH =
ÐABH =
ÐACH =
ÐAEH.
Since AH is common, ÐADH = ÐAEH and ÐDAH = ÐEAH, triangles ADH and AEH are congruent (ASA)
and HD = HE. Thus, H is equidistant from the intersections
of BQ with both AP and AU. Similarly, H is equidistant from
the intersection of AP and both BQ and BV. Following around,
we can show that H is equidistant from all the vertices of triangle
T_{1} and T_{2}, and the result follows.

263.

The ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
are each used exactly once altogether to form three
positive integers for which the largest is the sum of the
other two. What are the largest and the smallest possible
values of the sum?
Solution 1. Since the sum has at least as many digits as
either of the summands, the sum must have at least four digits.
However, the number of digits of the sum cannot exceed one more
than the number of digits of the larger summand. Hence, the sum
can have at most five digits. However, a fivedigit sum must
arise from the sum of a fourdigit number which is at most
9876 and a singledigit number which is at most 9. Since this
means that the sum cannot exceed 9885, we see that a fivedigit
sum is impossible.
A fourdigit sum can arise either as the sum of two threedigit
numbers or as the sum of a fourdigit and a twodigit number.
In the former case, the sum must exceed 1000 and
be less than 2000 and, in the
latter case, it must be at least 2000.
Thus, the smallest possible sum must be obtained by adding
two threedigit numbers to get a fourdigit sum. Since the
digits of the sum are all distinct, the smallest possible
sum is at least 1023. Since 589 + 437 = 1026, the smallest
sum is at most 1026. We may assume that each digit in the first
summand exceeds the corresponding digit in the second summand.
The only possibilities for a lower sum are
5pq + 4rs = 1023 , 6pq + 3rs = 1024 , 6pq + 3rs = 1025 , 

for digits p, q, r, s. One can check that none of these works.
For the largest sum, let the first summand have four digits and
the second two. The hundreds digit of the first summand is 9
and the thousands digit of the sum exceeds the thousands digit of
the first summand by 1. Since 5987 + 34 = 6021, the largest sum
is at least 6021. The only possibilities to consider for a larger
sum are
79ab + cd = 80ef , 69ab + cd = 70ef , 59ab + cd = 60ef , 

for digits a, b, c, d, e, f. It can be checked that none of these
works.
Thus, the smallest sum is 1026 and the largest is 6021.
Solution 2. [C. Shen] As in Solution 1, we eliminate the
possibility of a fivedigit sum. Suppose that we have
with digits a, b, c, d, e, f, g, h and f = a + 1. There must
be a carry from adding the tens digits and we have two possibilities:
c + e = h , b + d = 10 + g ; 
 (1) 
c + e = 10 + h , b + d = 9 + g . 
 (2) 
In case (1), we have that


= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = a + b + c + d + e + f + g + h 
 
 
= a + 10 + 2g + 2h + f = 2(5 + g + h + a) + 1 , 

which is impossible, as the two sides have different parities.
In case (2), we have that
36 = a + 9 + 2g + 10 + 2h + f = 2(10 + a + g + h) , 

so that a + g + h = 8. Since a, g, h are all positive integers,
a
£ 5 and we have the case 59bc + de = 60gh with c + e
³ 11. The only possibilities for (c, e) are (8, 3),
(8, 4), (7, 4), and these lead to
5978 + 43 = 6021 , 5978 + 34 = 6012 , 5987 + 34 = 6021 . 

The largest sum is 6021.
The smallest sum is at least 1023 and at most 1026 = 589 + 437.
Suppose that
with 3
£ x
£ 6. Since r + w
³ 3 + 4 = 7 and
q + v
³ 7, we have
r + w = 10 + x , q + v = 11 , p + u = 9 . 

Hence


= 3 + 4 + 5 + 6 + 7 + 8 + 9 = p + q + r + u + v + w + x 
 
 
= 9 + 11 + 2x + 10 = 30 + 2x , 

so that x = 6 and 1026 is the smallest sum.

264.

For the real parameter a, solve for real x
the equation
A complete answer will discuss the circumstances under which
a solution is feasible.
Solution 1. Suppose that y =
Ö{a + x}. Note that
x and y are both nonnegative. Then
x
^{2}  a = y and y
^{2}  a = x, whence
0 = (x^{2}  y^{2}) + (x  y) = (x  y)(x + y + 1) . 

Since x + y + 1
³ 1, it follows that y = x and so
0 = x^{2}  x  a = (x  (1/2))^{2}  ((1/4) + a) . 

For a real solution, we require that a
³ 1/4.
For
1/4
£ a
£ 0, both the sum and the product of
the solutions are nonnegative and we get the candidates
When a > 0, the equation has a positive and a negative solution,
and only the positive solution
is up for consideration.
We check that these solutions work. When a ³ 1/4,
x = ^{1}/_{2}(1 + Ö{1 + 4a}) and
so that
a +  Ö

a + x

= 
2

= 
æ è


2


ö ø

2

= x^{2} . 

When
1/4
£ a
£ 0, x =
^{1}/
_{2}(1
 Ö{1 + 4a}),
so that
a +  Ö

a + x

= a + 
æ è


2


ö ø

= 
æ è


2


ö ø

2

= x^{2} . 

(Note that, when a > 0,
and we get an extraneous solution.)
Solution 2. x = Ö{a + Ö{a + x}} Þ x^{2}  a = Ö{a + x}
Þ 0 = x^{4}  2ax^{2}  x + a^{2}  a = (x^{2}  x  a)(x^{2} + x  a + 1) . 

We analyze the possibilities from x
^{2}  x
 a = 0 as
in Solution 1. If, on the other hand,
x
^{2} + x
 (a
 1) = 0, then x =
^{1}/
_{2}(
1
±Ö{4a
 3}),
which is real when a
³ 3/4. The possibility
x =
^{1}/
_{2}(
Ö{4a
 3}
 1) leads to
and
a +  Ö

a + x

= 
2

¹ 
æ è


2


ö ø

2

. 

Thus, x =
^{1}/
_{2}(
Ö{4a
 3}
 1) is extraneous.
Since
^{1}/
_{2}(
1
 Ö{4a
 3}) < 0,
x =
^{1}/
_{2}(
1
 Ö{4a
 3}) < 0 is also extraneous,
Solution 3. For a solution, we require that x ³ 0.
By squaring twice, we are led to the equation
0 = x^{4}  2ax^{2}  x + a^{2}  a = a^{2}  (2x^{2} + 1)a + (x^{4}  x) . 

Solving for a yields


= 
(2x^{2} + 1) +  Ö 
(2x^{2} + 1)^{2}  4(x^{4}  x)

2

= 
(2x^{2} + 1) +  Ö 
4x^{2} + 4x + 1

2


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

= x^{2} + x + 1 , 

or
a = 
(2x^{2} + 1)  (2x + 1)
2

= x^{2}  x . 

(Note that the proper square root has been extracted since
x
³ 1/2.) In the first case

æ Ö


= 
æ Ö

x^{2} + x + 1 +  Ö 
x^{2} + 2x + 1


=  Ö

x^{2} + 2x + 2

> x . 

In the second case,
Thus, only the case, a = x
^{2}  x leads to a valid solution.
Note that a = x
^{2}  x = (x
 ^{1}/
_{2})
^{2}  ^{1}/
_{4},
so that a
³ ^{1}/
_{4} for a solution to work.
Since we require x
³ 0 and a = x(x
 1), we see from the
graph of this equation that there are two valid values of x
when
^{1}/
_{4} £ a
£ 0 and one valid value of x when
0 < a.
Solution 4. [J. Zhao] For any real x, one of the
following must hold:
x >  Ö

a + x

; x <  Ö

a + x

; x =  Ö

a + x

. 

In case of the first,
so that such x does not satisfy the equation. Similarly, we can
reject any x satisfying the second condition as a solution of
the equation. Hence for every solution of the given equation,
we must have that x =
Ö{a + x} or x
^{2}  x
 a = 0.
We now finish off as in the previous solutions.
Comment. Many solvers did not pay attention to the feasibility
of the solutions. Solution 3 was particularly insidious, because it
was easy to skip the analysis that only one of the values of x
gave a solution when a > 0. Surd equations is a dandy topic for
students to lose points they should gain because of carelessness
or a superficial treatment.

265.

Note that 959^{2} = 919681, 919 + 681 = 40^{2};
960^{2} = 921600, 921 + 600 = 39^{2}; and 961^{2} = 923521,
923 + 521 = 38^{2}. Establish a general result of which these
are special instances.
Solution. Let b
³ 2 be a base of enumeration. Then
we wish to investigate solutions of the system
(b^{k}  u)^{2} = b^{k} v + w 
 (1) 
where k, u, v are positive integers and the integer
w satisfies 0
£ w
£ b
^{k}  1.
The numerical examples given correspond to (b, k, u) = (10, 3, 41), (10, 3, 40) and (10, 3, 39). Subtracting (2) from
(1) yields
(b^{2k}  1)  2(b^{k}  1)u = (b^{k}  1)v 

whence v = (b
^{k} + 1)
 2u and w = u
^{2}  b
^{k}. We require that
b
^{k} £ u
^{2} £ 2b
^{k}  1 in order to get a generalization.
So, to generate examples of the phenomenon, first select a
base b and a parameter k for the number of digits; then
select u to satisfy the foregoing inequality. Then one can
check, with v and w determined, the desired system of
equations holds.
Consider first the situation b = 10. When k = 1, we have that
u = 4 and we get the case 6^{2} = 36, 3 + 6 = 3^{2}. When
k = 2, we have that 10 £ u £ 14 and find that
86^{2} = 7396, 73 + 96 = 13^{2} and so on up to 90^{2} = 8100,81 + 0 = 9^{2}. When k = 3, we have that
32 £ u £ 44, and find that
956^{2} = 913936, 913 + 936 = 43^{2} and so on up to
968^{2} = 937024, 937 + 24 = 31^{2}.
Examples from base 3 are 5^{2} = (221)_{3}, (2 + 21)_{3} = 3^{2};
6^{2} = (1100)_{3}, (11 + 0)_{3} = 2^{2};
20^{2} = (112211)_{3}, (112 + 211)_{3} = 6^{2};
21^{2} = (121100)_{3}, (121 + 100)_{3} = 5^{2}.
Comment. In the above system, we could replace u  1
by u  d and get other instances. For example, with (b, k) = (10, 2), we can get the instances
(27^{2} = 729, 7 + 29 = 6^{2}), (29^{2} = 841, 8 + 41 = 7^{2}),
(30^{2}, 3^{2}), (39^{2}, 6^{2}), (40^{2}, 4^{2}), (50^{2}, 5^{2}),
(57^{2}, 9^{2}). (60^{2}, 6^{2}), (70^{2}, 7^{2}), (75^{2}, 9^{2}),
(78^{2}, 12^{2}), (80^{2}, 8^{2}) and (98^{2}, 10^{2}).
Another formulation is to note that the numerical equations
are special instances of the system n^{2} = b^{k} x + y;
(n  x)^{2} = x + y with 0 £ y £ b^{k}  1 and
0 £ n < b^{k}, where (n, b, x, y) = (959, 10, 919, 681),
(960, 10, 921, 600), (961, 10, 923, 521). These equations
imply that x(2n  x) = (b^{k}  1)x, whence x = 2n  (b^{k}  1).
Thus,
n^{2} = 2b^{k} n  b^{2k} + b^{k} + yÞ y = (b^{k}  n)^{2}  b^{k} , 

and we require that b
^{k} £ (b
^{k}  n)
^{2} < 2b
^{k}  1.
The analysis can be continued from here.
This question was, on the whole, badly done. In describing the
generalization, one needs to provide a road map whereby one can
make the appropriate subsitutions to obtain further examples.
Many solvers were content to write down some equations of which
the numerical examples were an instance without analyzing the
conditions under which the equations could be used to obtain
further examples. In effect, no further information was provided
to show where other examples might be found.

266.

Prove that, for any positive integer n,
(2n  n) divides the least common multiple of the
numbers 1, 2, 3, ¼, 2n1, 2n.
Solution. We first establish that
for each positive real x. (
ë·
û refers to
"the greatest integer not exceeding".) If, for some integer s,
2s
£ 2x
£ 2s + 1, then s
£ x < s +
^{1}/
_{2} and
ë2x
û = 2
ëx
û; if
2s + 1
£ 2x < 2s + 2, then s +
^{1}/
_{2} £ x < s + 1
and
ë2x
û = 2s + 1 = 2
ëx
û+ 1.
Let p be a prime divisor of (2n  n), so that
p £ 2n. Suppose that p^{k} is the highest power of p
that divides an integer not exceeding 2n. Then p^{k} £ 2n < p^{k+1}. The exponent of p in the prime factorization of
(2n  n) is equal to


+ 
ê ë


2n
p^{2}


ú û

+ 
ê ë


2n
p^{3}


ú û

+ ¼+ 
ê ë


2n
p^{k}


ú û


ö ø

 2 
æ è


ê ë


n
p


ú û

+ 
ê ë


n
p^{2}


ú û

+ ¼+ 
ê ë


n
p^{k}


ú û


ö ø


 
 
= 
k å
i=1


æ è


ê ë


2n
p^{i}


ú û

2 
ê ë


n
p^{i}


ú û


ö ø

£ k . 

Hence the exponent of p in the prime factorization of (2n  n) does not exceed the exponent of p in the prime factorization
of the least common multiple of the first 2n positive integers,
for each prime divisor of (2n  n). The result follows.

267.

A nonorthogonal reflection in an axis a takes
each point on a to itself, and each point P not on a
to a point P¢ on the other side of a in such a way that
a intersects PP¢ at its midpoint and PP¢ always makes
a fixed angle q with a. Does this transformation
preserves lines? preserve angles? Discuss the image of a
circle under such a transformation.
Solution. We suppose that
q ¹ 90
^{°}.
The transformation preserves lines. This is clear for any line
parallel to a. Let AB be a line through A that meets
a at P, and let A
¢ and B
¢ be the reflective images of
A and B. Since AA
¢ BB
¢ and a is a median from P
of triangles PAA
¢ and PBB
¢, it follows that A
¢, B
¢ and
P are colllinear. Thus, any point on the line AP gets carried
to a point on the line A
¢P. However, angles are not preserved.
A line perpendicular to a is carried to a line making an angle
not equal to a right angle with a (while a is kept fixed).
[What is the angle of intersection?]
Suppose that the axis of reflection is the y axis. Let A and
B be mutual images with A to the left and B to the right of
the axis, AB meeting the axis at P and the upper right (and
lower left) angles of intersection being q. If
A ~ (x, y) (with x £ 0), then P ~ (0, y  x cotq)
and B ~ (x, y  2x cotq). If B ~ (u, v)
(with u ³ 0), then P ~ (0, v  ucotq) and
A ~ (u, v  2u cotq). Thus the transformation is
given by
(x, y) ® (X, Y) º (x, y  2x cotq) . 

Consider the particular case of the circle with equation
x
^{2} + y
^{2} = 1. The image curve has equation
1 = (X)^{2} + (Y  2Xcotq)^{2} = (1 + 4cot^{2} q)X^{2}  4XYcotq+ Y^{2} 

and this does not represent a circle. Thus, circles are not preserved
under the transformation. In fact, a general circle with equation
of the form (x
 a)
^{2} + (y
 b)
^{2} = r
^{2} gets carried to an
second degree curve in the plane which turns out to be an ellipse.
Comment. A synthetic way of analyzing the image of a circle
is to note that two chords of a circle
that bisect each other must be diameters,
and so have the same length. Using this, one can argue that a circle
with one diameter along the axis and other perpendicular to the
axis does not go to a circle.

268.

Determine all continuous real functions f of
a real variable for which
f(x + 2f(y)) = f(x) + y + f(y) 

for all real x and y.
Solution 1. First, we show that f(u) = 0 if and only if
u = 0. Suppose that f(u) = 0. Then, for all x,
f(x) = f(x + 2f(u)) = f(x) + u + f(u) = f(x) + u 

so that u = 0. On the other hand, let v = f(0).
Taking (x, y) = (
2v, 0) in the condition yields that
v = f(0) = f(2v + 2v) = f(2v) + 0 + v , 

whence
0 = f(2v) = f(2v + 2f(2v)) = 0  2v + 0 = 2v 

and v = 0.
Setting y = x yields
for all x. Let g(x)
º x + 2f(x). Then
f(g(x)) = g(x) so that

1
2

[ g(g(x))  g(x) ] = g(x) 

whence
Note also that g(0) = 0 + 2f(0) = 0.
If g(x)
º 0, then f(x) =
x/2, and this is a valid solution.
Suppose that g(z) = a
¹ 0 for some z and a. Then, as
g(0) = 0 and g is continuous, g assumes all values between
0 and a (by the intermediate value theorem). But
g(a) = g(g(z)) = 3g(z) = 3a, so by the same argument,
g assumes all values between 0 and 3a. We can continue on
to argue that g assumes all values between 0 and 3
^{k}a for
each positive integer k. Thus g assumes all positive values
if a > 0 and assumes all negative values if a < 0.
Suppose that the former holds. Then, for all x ³ 0, we have
that g(x) = 3x and so x + 2f(x) = 3x, whence f(x) = x.
Therefore, when x is arbitrary and y ³ 0,
f(x + 2y) = f(x) + 2y. In particular, 0 = f(2y + 2y) = f(2y) + 2y so that f(2y) = 2y. Hence, for all x, we must
have that f(x) = x. A similar argument can be followed to show that
f(x) º x when a < 0. Therefore, the only two solutions are
f(x) = x and f(x) = x/2.
Solution 2. [S. Eastwood] Setting x = y, we find that
f(x + 2f(x)) = x + 2f(x) for all real x. Let
A = { x + f(x) : x Î R } . 

Then A is a nonvoid set. Suppose that a
Î A. Then
f(a) = a, so that 3a = a + 2f(a)
Î A. Hence
a
Î A
Þ 3a
Î A. Since x
® x + 2f(x)
is continuous, it satisfies the intermediate value theorem and
so A must be of one of the following types: A = { 0 };
A = [b,
¥), A = [
b,
¥), A =
R for some
nonnegative value of b.
Suppose that A = { 0 }. Then, for all real x, x + 2f(x) = 0
and so f(x) = x/2. This is a valid solution.
Suppose that A has a nonzero element a. Then
a = f(a) = f(a + 2a) = f(a + 2f(a)) = f(a) + a + f(a) = f(a) + 2a , 

whence f(
a) =
a and
3a =
a + 2f(
a)
Î A.
Hence A must contain numbers that are both positive and
negative, and so must consist of the whole set of reals.
Hence, for all real x, f(x) = x, and this also is valid.
Solution 3. [J. Zhao] Suppose that f(u) = f(v).
Then, for each x,
f(x) + u + f(u) = f(x + f(u)) = f(x + f(v)) = f(x) + v + f(v) 

so that u = v. Hence, f is oneone. Since f is continuous,
f is always strictly increasing or always strictly
decreasing on
R.
Suppose that f is increasing. Then

lim
x ® +¥

x + 2f(x) = +¥ 

and

lim
x ® ¥

x + 2f(x) = ¥ , 

so that x + f(x) assumes every real value (by the intermediate
value theorem). Suppose that z is any real number. Select
y such that y + 2f(y) = z. Then
f(z) = f(y + 2f(y)) = f(y) + y + f(y) = z , 

and so f(x)
º x. This works.
Suppose that f is decreasing. Suppose, if possible, that
p and q are such that p + 2f(p) < q + 2f(q).
Then f(p + 2f(p)) > f(q + 2f(q)). But, we get a contradiction
since f(p + 2f(p)) = p + 2f(p) and f(q + 2f(q)) = q + 2f(q).
Hence, there is a constant c such that, for all real x,
x + 2f(x) = c. Hence, f(x) = ^{1}/_{2}(x + c).
Plugging this into the functional equation, we find that c = 0,
and so we obtain the solution f(x) = x/2.
Comments. If we assume that f(x) is a polynomial, then
it can be shown that its degree must be 1. Assuming a solution
f(x) = cx + d for constants c and d leads to the equations
d = 0 = 2c^{2}  c  1 = (2c + 1)(c  1). Thus, it is not hard to
get a partial solution.
There were a number of approaches to ascertaining
that f(0) = 0. A. Critch began with the observation that
f(2f(0)) = f(0 + 2f(0)) = f(0) + 0 + f(0) = 2f(0). Let
a = 2f(0), so that f(a) = a. Furthermore,
f(2a) = f(a + a) = f(a + 2f(0)) = a + f(0) = (3a/2) 

and
f(2a) = f(0 + 2a) = f(0 + 2f(a)) = f(0) + a + f(a) = (5a/2) . 

This leads to a = 0.
R. Dan noted that f(2f(y)) = y + f(y), and then went on to
derive
f(2f(y) + 2f(y)) = f(2f(y)) + y + f(y) = f(2f(y)) + f(2f(y)) . 

Along with the property that f(0) = 0, one can then show that
f assumes both positive and negative values.