Solutions.

297.

The point P lies on the side BC of triangle
ABC so that PC = 2BP, ÐABC = 45^{°} and ÐAPC = 60^{°}. Determine ÐACB.
Solution 1. Let D be the image of C under a reflection
with axis AP. Then
ÐAPC =
ÐAPD =
ÐDPB = 60
^{°}, PD = PC = 2BP, so that
ÐDBP = 90
^{°}. Hence AB bisects the angle DBP,
and AP bisects the angle DPC, whence
A is equidistant from BD, PC and PD.
Thus, AD bisects ÐEDP, where E lies on BD produced.
Thus


 
 
= 
1
2

(180^{°}  ÐBDP) = 
1
2

(180^{°}  30^{°}) = 75^{°} . 

Solution 2. [Y. Zhao] Let Q be the midpoint of PC and
R the intersection of AP and the right bisector of PQ, so
that PR = QR and BR = CR. Then ÐRPQ = ÐRQP = 60^{°} and triangle PQR is equilateral. Hence
PB = PQ = PR = RQ = QC and ÐPBR = ÐPRB = ÐQRC = ÐQCR = 30^{°}.
Also, ÐRBA = 15^{°} = ÐPAB = ÐRAB, so
AR = BR = CR. Thus, ÐRAC = ÐRCA. Now
ÐARC = 180^{°}  ÐPRQ  ÐQRC = 90^{°},
so that ÐRCA = 45^{°} and ÐACB = 75^{°}.
Solution 3. [R. Shapiro] Let H be the foot of the
perpendicular from C to AP. Then CPH is a 306090
triangle, so that BP = ^{1}/_{2}PC = PH and
ÐPBH = ÐPHB = 30^{°} = ÐPCH. Hence,
BH = HC. As
ÐHAB = ÐPAB = 180^{°}  120^{°}  45^{°} = 15^{°} = ÐABP  ÐHBP = ÐABH , 

AH = BH = HC. Therefore,
ÐHAC =
ÐHCA = 45
^{°}.
Thus,
ÐACB =
ÐHCA +
ÐPCH = 75
^{°}.
Solution 4. From the equation expressing tan30^{°}
in terms of tan15^{°}, we find that
tan15^{°} = 2  Ö3 and sin15^{°} = [(Ö3  1)/(2Ö2)]. Let ÐACP = q,
so that
ÐPAC = 180^{°}  60^{°}  q = 120^{°}  q . 

Suppose, wolog, we set
BP
 = 1, so that
PC
 = 2. Then by the Law of Sines in triangle
ABP,
AP  = 
sin45^{°}
sin15^{°}

= Ö3 + 1. 

By the Law of Sines in triangle APC,

sinq
Ö3 + 1

= 
sin(120^{°}  q)
2

= 
Ö3cosq
4

+ 
sinq
4



whence (3
 Ö3)sin
q = (3 +
Ö3)cos
q.
Hence
tanq = 2 + Ö3 = (2  Ö3)^{1} = (tan15^{°})^{1} , 

so that
q = 75
^{°}.

298.

Let O be a point in the interior of a quadrilateral
of area S, and suppose that
2S = OA ^{2} + OB ^{2} + OC ^{2}+ OD ^{2} . 

Prove that ABCD is a square with centre O.
Solution.


+ OB ^{2} + OC ^{2}+ OD ^{2} 
 
 
= 
1
2

(OA ^{2} + OB ^{2}) + 
1
2

(OB ^{2} + OC ^{2}) + 
1
2

(OC ^{2} + OD ^{2}) + 
1
2

(OD ^{2} + OA ^{2}) 
 
 
³ OA OB + OB OC + OC OD + OD OA  
 
 
³ 2[AOB] + 2[BOC] + 2[COD] + 2[DOA] = 2S 

with equality if and only if OA = OB = OC = OD and all the
angles AOB, BOC, COD and DOA are right. The result follows.

299.

Let s(r) denote the sum of all the divisors
of r, including r and 1. Prove that there are infinitely
many natural numbers n for which
whenever 1 £ k < n.
Solution 1. Let u
_{m} =
s(m)/m for each positive integer
m. Since d
« 2d is a oneone correspondence
between the divisors of m and some even divisor of 2m,
s(2m)
³ 2
s(m) + 1, so that
u_{2m} = 
s(2m)
2m

³ 
2s(m) + 1
2m

> 
s(m)
m

= u_{m} 

for each positive integer m.
Let r be a given positive integer, and select s £ 2^{r} such that
u_{s} ³ u_{k} for 1 £ k £ 2^{r} (i.e., u_{s} is the
largest value of u_{k} for k up to and including 2^{r}). Then,
as u_{2s} > u_{s}, it must happen that 2^{r} £ 2s £ 2^{r+1} and
u_{2s} ³ u_{k} for 1 £ k £ 2^{r}.
Suppose that n is the smallest positive integer t for which
2^{r} £ t and u_{k} £ u_{t} for 1 £ k £ 2^{r}. Then
2^{r} £ n £ 2s £ 2^{r+1}. Suppose that 1 £ k £ n.
If 1 £ k £ 2^{r}, then u_{k} £ u_{n} from the definition of
n. If 2^{r} < k < n, then there must be some number k¢ not
exceeding 2^{r} for which u_{k} < u_{k¢} £ u_{n}. Thus, n has
the desired property and 2^{r} £ n £ 2^{r+1}. Since such
n can be found for each positive exponent r, the result follows.
Comment. The sequence selected in this way starts off:
{ 1, 2, 4, 6, 12, ¼}.
Solution 2. [P. Shi] Define u_{m} as in Solution 1.
Suppose, if possible, that there are
only finitely many numbers n satisfying the condition of the
problem. Let N be the largest of these, and let u_{s} be the
largest value of u_{m} for 1 £ m £ N. We prove by induction
that u_{n} £ u_{s} for every positive integer n. This holds for
n £ N. Suppose that n > N. Then, there exists an integer
r < n for which u_{r} > u_{n}. By the induction hypothesis,
u_{r} £ u_{s}, so that u_{n} < u_{s}. But this contradicts the
fact (as established in Solution 1) that u_{2s} > u_{s}.

300.

Suppose that ABC is a right triangle with
ÐB < ÐC < ÐA = 90^{°}, and let
K be its circumcircle. Suppose that the tangent
to K at A meets BC produced at D and that E is
the reflection of A in the axis BC. Let X be the foot
of the perpendicular from A to BE and Y the midpoint of AX.
Suppose that BY meets K again in Z. Prove that
BD is tangent to the circumcircle of triangle ADZ.
Solution 1. Let AZ and BD intersect at M, and AE and
BC intersect at P. Since PY joints the midpoints of two
sides of triangle AEX, PY
 EX. Since
ÐAPY =
ÐAEB =
ÐAZB =
ÐAZY, the quadrilateral AZPY is
concyclic. Since
ÐAYP =
ÐAXE = 90
^{°},
AP is a diameter of the circumcircle of AZPY and BD is a tangent
to this circle. Hence MP
^{2} = MZ ·MA.
Since
ÐPAD = ÐEAD = ÐEBA = ÐXBA, 

triangles
PAD and XBA are similar. Since
ÐMAD = ÐZAD = ÐZBA = ÐYBA, 

it follows that
ÐPAM = ÐPAD  ÐMAD = ÐXBA  ÐYBA = ÐXBY 

so that triangles PAM and XBY are similar. Thus

PM
AP

= 
XY
XB

= 
XA
2XB

= 
PD
2PA

Þ PD = 2PMÞ MD = PM . 

Hence MD
^{2} = MP
^{2} = MZ ·MA and the desired result follows.
Solution 2. [Y. Zhao] As in Solution 1, we see
that there is a circle
through the vertices of AZPY and that BD is tangent to this
circle. Let O be the centre of the circle K. The triangles
OPA and OAD are similar, whereupon OP ·OD = OA^{2}.
The inversion in the circle K interchanges P and
D, carries the line BD to itself and takes the circumcircle
of triangle AZP to the circumcircle of triangle AZD.
As the inversion preserves tangency of circles and lines, the
desired result follows.

301.

Let d = 1, 2, 3. Suppose that M_{d} consists of the
positive integers that cannot be expressed as the sum of
two or more consecutive terms of an arithmetic progression
consisting of positive integers with common difference d.
Prove that, if c Î M_{3}, then there exist integers a Î M_{1} and b Î M_{2} for which c = ab.
Solution. M
_{1} consists of all the powers of 2, and
M
_{2} consists of 1 and all the primes. We prove these assertions.
Since k + (k+1) = 2k+1, every odd integer exceeding 1 is the
sum of two consecutive terms. Indeed, for each positive integers
m and r,
(mr) + (mr+1) + ¼+ (m1) + m + (m+1) + ¼+ (m+r1)+ (m+r) = (2r + 1)m , 

and,
m + (m + 1) + ¼+ (m + 2r  1) = r[2(m+r)  1] , 

so that it can be deduced that every positive integer with at least one
odd positive divisor exceeding 1 is the sum of consecutives, and no power
of 2 can be so expressed.
(If m < r in the first sum, the negative terms in the sum are cancelled by
positive ones.) Thus, M
_{1} consists solely of all the powers of 2.
Since 2n = (n + 1) + (n  1), M_{2} excludes all even numbers
exceeding 2. Let k ³ 2 and m ³ 1. Then
m + (m + 2) + ¼+ (m + 2(k1)) = km + k(k1) = k(m + k  1) 

so that M
_{2} excludes all multiples of k from k
^{2} on.
Since all such numbers are composite, M
_{2} must include all primes.
Since each composite number is at least as large as the square
of its smallest nontrivial divisor, each composite number must
be excluded from M
_{2}.
We now examine M_{3}. The result will be established if we show
that M_{3} does not contain any number of the form
2^{r} u v where r is a nonnegative integer and u, v are
odd integers with u ³ v > 1. Suppose first that
r ³ 1 and let a = 2^{r}u  ^{3}/_{2}(v1). Then
a ³ 2u  
3
2

(v1) ³ 
v
2

+ 1 > 1 

and
a + (a + 3) + ¼+ [a + 3(v1)] = v[a + (3/2)(v1)] = 2^{r} uv . 

Since m + (m+3) = 2m + 3, we see that M_{3} excludes all
odd numbers exceeding 3, and hence all odd composite numbers.
Hence, every number in M_{3} must be the product of a power of
2 and an odd prime or 1.
Comment. The solution provides more than necessary.
It suffices to show only that M_{1} contains all powers of
2, M_{2} contains all primes and M_{3} excludes all numbers with
a composite odd divisor.

302.

In the following, ABCD is an arbitrary convex
quadrilateral. The notation [ ¼] refers to the area.


(a) Prove that ABCD is a trapezoid if and only if
[ABC] ·[ACD] = [ABD] ·[BCD] . 



(b) Suppose that F is an interior point of the
quadrilateral ABCD such that ABCF is a parallelogram.
Prove that
[ABC] ·[ACD] + [AFD] ·[FCD] = [ABD] ·[BCD] . 

Solution 1. (a) Suppose that AB is not parallel to CD.
Wolog, let these lines meet at E with A between E and
B, and D between E and C. Let P, Q, R, S be the respective
feet of the perpendiculars from A to CD, B to CD,
C to AB, D to AB produced. Then
[ABC]·[ACD] = [ABD][BCD] ÛAB CR CD AP  = AB DS CD BQ Û CR : DS = BQ : AP . 

By similar triangles, we find that CE : DE = CR : DS = BQ : AP = BE : AE. The dilation with centre E and factor
AE
/
BE
 takes B to A, C to D and
so the segment BC to the parallel segment AD. Thus ABCD is
a trapezoid.
(b) Let the quadrilateral be in the horizontal plane of
threedimensional space and let F be at the origin of vectors.
Suppose that u = [( ®)  FA],
v = [( ®)  FC], and pu  qv = [( ®)  FD], where p and q are nonnegative scalars.
We have that [( ®)  FB] = u + v.
Then


 
 
= u ×v + u ×(pu + qv) + v ×(pu + qv)  
 
 

2[ABD] = (pu + qv + u) ×v  = (1 + p) u ×v  ; 

2[BCD] = (pu + qv + v) ×u  = (1 + q) u ×v  . 

The result follows.
Solution 2. [Y. Zhao] Observe that, since
(A + C) + (B + D) = 360^{°},


= 
1
2

[ cos(A  C)  cos(A + C)  cos(B  D)+ cos(B + D)] 
 
 
= 
1
2

[ cos(A  C)  cos(B  D) ] = 
1
2

[ cos(B + A  B  C)  cos(B + A + B + C) ] 
 
 

(a) Hence


 4[ABC][ACD] = (AB ·DA sinA)(BC ·CD sinC) (AB ·BC sinB)(CD ·DA sinD) 
 
 
= (AB ·BC ·CD ·DA) (sinA sinC  sinB sinD) 
 
 
= (AB ·BC ·CD ·DA) sin(B + A) sin(B + C) . 

The left side vanishes if and only if A + B = C + D = 180
^{°}
or B + C = A + D = 180
^{°},
i.e.,
AD
 BC or AB
 CD.
(b) From (a), we have that


 4[ABC][ACD] = (AB ·BC ·CD ·DA) sin(A + B)sin(B + C) 
 
 
= (AB ·BC ·CD ·DA) sin(A + B  180^{°})sin(B + C  180^{°}) 
 
 
= (FC ·AF ·CD ·DA) (sin(ÐBAD  ÐBAF)sin(ÐBCD  ÐBCF)) 
 
 
= [(DA ·AF)sinÐDAF][ (DC ·CF) sinÐDCF] 
 
 

as desired.

303.

Solve the equation
tan^{2} 2x = 2 tan2x tan3x + 1 . 

Solution 1. Let u = tanx and v = tan2x.
Then
v^{2}  2v 
æ è


u + v
1  uv


ö ø

 1 = 0 

Û v^{2}  uv^{3}  2uv  2v^{2}  1 + uv = 0 

Û0 = uv + 1 + v^{2} + uv^{3} = (uv + 1)(1 + v^{2}) 

Now v = 2u(1
 u
^{2})
^{1}, so that 2u = v
 u
^{2}v = u + v and
u = v. But then u
^{2} =
1 which is impossible. Hence the
equation has no solution.
Solution 2.


= tan^{2} 2x  2 tan3x tan2x  1 
 
 
= tan^{2} 2x  2 tan3x tan2x + tan^{2} 3x  sec^{2} 3x 
 
 
= (tan2x  tan3x)^{2}  sec^{2} 3x 
 
 
= (tan2x  tan3x  sec3x)(tan2x  tan3x + sec3x) . 

Hence, either tan2x = tan3x + sec3x or tan2x = tan3x
 sec3x. Suppose that the former holds. Multiplying the equation
by cos2x cos3x yields sin2x cos3x = sin3x cos2x + cos2x. Hence


= cos2x + (sin3x cos2x  sin2x cos3x) 
 
 
= 1  2sin^{2} x + sinx = (1  sinx)(1 + 2sinx) , 

whence
modulo 2
p. But tan3x is not defined at any of these angles,
so the equation fails. Similarly, in the second case, we obtain
0 = (2 sinx
 1)(sinx + 1) so that
modulo 2
p, and the equation again fails. Thus, there are no
solutions.
Solution 3. Let t = tanx, so that tan2x = 2t(1  t^{2})^{1}
and tan3x = (3t  t^{3})(1  3t^{2})^{1}. Substituting for t
in the equation and clearing fractions leads to
4t^{2}(1  3t^{2}) = 4t(3t  t^{3})(1  t^{2}) + (1  t^{2})^{2}(1  3t^{2}) 

Û4t^{2}  12t^{4} = (12t^{2}  16t^{4} + 4t^{6}) + (1  5t^{2} + 7t^{4}  3t^{6}) 

Û0 = t^{6} + 3t^{4} + 3t^{2} + 1 = (t^{2} + 1)^{3} . 

There are no real solutions to the equation.
Solution 4. The equation is undefined if 2x or 3x is an
odd multiple of p/2. We exclude this case. Then the equation
is equivalent to

sin^{2} 2x  cos^{2} 2x
cos^{2} 2x

= 
2 sin2x sin3x
cos2x cos3x



or


= 
2 sin2x sin3x
cos2x cos3x

+ 
cos4x
cos^{2} 2x


 
 
= 
sin4x sin3x + cos4x cos3x
cos^{2} 2x cos3x


 
 
= 
cosx
cos^{2} 2x cos3x

. 

Since cosx vanishes only if x is an odd multiple of
p, we
see that the equation has no solution.
Solution 5. [Y. Zhao] Observe that, when tan(A  B) ¹ 0,
1 + tanA tanB = 
tanA  tanB
tan(A  B)

. 

In particular,
1 + tanx tan2x = 
tan2x  tanx
tanx

and 1 + tan2x tan3x = 
tan3x  tan2x
tanx

. 

There is no solution when x
º 0 (mod
p), so we exclude this
possibility. Thus


= (1 + tan2x tan3x) + (tan2x tan3x  tan^{2} 2x) 
 
 
= (tan3x  tan2x)(cotx + tan2x) = cotx (tan3x  tan2x)(1 + tanx tan2x) 
 
 
= cot^{2} x (tan3x  tan2x)(tan2x  tanx) 
 
 
= cot^{2} x 
æ è


sinx
cos2x cos3x


ö ø


æ è


sinx
cosx cos2x


ö ø

. 

This has no solution.
Solution 6. For a solution, neither 2x nor 3x can be a
multiple of p/2, so we exclude these cases. Since
tan4x = 
2 tan2x
1  tan^{2} 2x

, 

we find that
cot4x = 
1  tan^{2} 2x
2 tan2x

=  tan3x , 

whence 1 + tan3x tan4x = 0. Now
tan4x  tan3x = (1 + tan3x tan4x)tanx = 0 , 

so that 4x
º 3x (mod
p). But we have excluded this.
Hence there is no solution to the equation.