Solutions for the December Problems

192.

Let ABC be a triangle, D be the midpoint of AB and
E a point on the side AC for which AE = 2EC. Prove that
BE bisects the segment CD.
In the following solutions, F is the intersection point of
BE and CD.
Solution 1. Let
G be the midpoint of AE. Then AG = GE = EC and DG  BE.
In triangle ADC, DG  FE and GE = EC, from which it follows
that DF = FC, as required.
Solution 2. Let u = [ADF] = [BDF] (where [ ¼] denotes
area), v = [AFE], w = [CFE] and z = [BFC]. Then
2u + v = 2(w + z) and v = 2w, whence 2u = 2z and u = z. It
follows from this (two triangles with the same height and equal
collinear bases) that F is the midpoint of CD.
Solution 3. By Menelaus' Theorem, applied to triangle ACD and
transversal BFE,

CE EA

· 
AB BD

· 
DF FC

= 1 , 

so that
^{1}/
_{2} ·(
2) ·(DF/FC) =
1 and
DF = FC, as desired.
Solution 4. [T. Yue] Let K be the midpoint of AC; then
BC = 2DK and BC  DK. Suppose that BE produced and DK
produced meet at H. Since triangles EBC and EHK are similar
and EC = 2EK, it follows that BC = 2KH and so DH = BC.
Thus, DHCB is a parallelogram whose diagonals BH and CD must
bisect each other. The result follows.
Solution 5. Place the triangle in the cartesian plane so that
B ~ (0, 0), C ~ (3, 0) and A ~ (6a, 6b). Then
D ~ (3a, 3b), E ~ (2(a+1), 2b) and the lines BE and
and CD have the respective equations y = bx/(a+1) and
y = b(x3)/(a1). These lines intersect at the point
((3/2)(a+1), (3/2)b), and the result follows.
Solution 6. [L. Chen] [BDE] = [ADE] = ^{1}/_{2}[ABE] = [BEC]. Let M and N be the respective feet of the
perpendiculars from D and C to BE. Then [BDE] = [BEC]Þ DM = CN. Since DMF and CNF are similar right
triangles with DM = CN, they are congruent and so DF = CF.
Solution 7. [F. Chung; Y. Jean] As in the previous solution,
[BDE] = [BEC]. Therefore,
DF:FC = [DEF]:[CEF] = [DBF]:[CBF] = ([DEF]+[DBF]):([CEF]+[CBF]) 

Solution 8. [Y. Wei] Let U be a point on BC such that
DU  AC. Suppose that DU and BE intersect in V. Then
2EC = AE = 2DV, so that DV = EC. Also ÐVDF = ÐECF abd ÐDFV = ÐECF, so that triangles
DVF and CEF are congruent. Hence DF = FC.
Solution 9. Let AF produced meet BC at L. By Ceva's
Theorem,
whence BL = 2LC and, so, LE
 AB. Since the triangles
ABC and ELC are similar with factor 3, AB = 3EL.
Let EL intersect CD at M. Then the triangles
AFB and LFE are similar, so that FD = 3FM. But,
FD + FM + MC = DC = 3MC Þ 2FM = MC ÞFC = FM + MC = 3FM = FD , 

as desired.
Solution 10. [H. Lee] Let u = [( ®)  DB],
v = [( ®)  EC], a = [( ®)  BF],
la = [( ®)  FE], b = [( ®)  CF] and mb = [( ®)  FD].
Then
and
Hence
u = a  mb and v = b  la . 

Therefore, from triangle ABE,

 
= (l+ 1)a + 2b + 2la 2a  2mb 
 


Since {
a,
b } is a linearly independent set,
l = 1/3 and
m = 1, yielding the desired result.
Solution 11. [M. Zaharia]
Place masses 1, 1, 2, respectively, at the vertices
A, B, C. We locate the centre of gravity of these masses in
two ways. Since the masses at A and B have their centre of
gravity at D, we can get an equivalent system by replacing
the masses at A and B by a mass 2 at the point D. The centre
of gravity of the original setup is equal to the centre of gravity
of masses of 2 placed at each of D and C, namely at the midpoint
of CD.
On the other hand, the centre of gravity of the masses at A and
C is at E. So the centre of gravity of the original setup is equal to the centre of gravity of a mass 3 located at E and
a mass 1 located at B, namely on the segment BE (at the point
F for which BF = 3FE). Since both BE and CD contain the
centre of gravity of the original setup, the result follows.
Solution 12. Place the triangle in the complex plane with
C at 0, B at 12z and A at 12. Then D is located at
6(z+1) and E at 4. Let P be the midpoint 3(z+1) of
CD. Then, BP and PE are collinear since
12z  3(z+1) = 3(3z  1) = 3[3(z+1)  4] , 

i.e., the vector [(
®)  BP] is a
real multiple of [(
®)  PE]. The result follows.

193.

Determine the volume of an isosceles tetrahedron for which
the pairs of opposite edges have lengths a, b, c. Check your
answer independently for a regular tetrahedron.
Solution 1. The edges of the tetrahedron can be realized
as the diagonals of the six faces of a rectangular parallelepiped
with edges of length u, v, w in such a way that
a
^{2} = v
^{2} + w
^{2}, b
^{2} = u
^{2} + w
^{2} and c
^{2} = u
^{2} + v
^{2}.
The tetrahedran can be obtained from the parallelepiped by
trimming away four triangular pyramids each with three mutually
perpendicular faces (surrounding a corner of the parallelepiped)
and three pairwise orthogonal edges of lengths u, v, w.
Hence the volume of the tetrahedron is equal to
uvw  4((1/6)uvw) = (1/3)uvw . 

>From the foregoing equations, 2u^{2} = b^{2} + c^{2}  a^{2},
2v^{2} = c^{2} + a^{2}  b^{2} and 2w^{2} = a^{2} + b^{2}  c^{2}. (By
laying out the tetrahedron flat, we see that the triangle of
sides a, b, c is acute and the right sides of these equations are
indeed positive.) It follows that the volume of the tetrahedron is

Ö2 12


 __________________________________ Ö(b^{2} + c^{2}  a^{2})(c^{2} + a^{2}  b^{2})(a^{2} + b^{2}  c^{2})

. 

In the case of a regular tetrahedron of side 1, the height is equal
to Ö[(2/3)] and the area of a side is equal to Ö3/4,
and the formula checks out.
Solution 2. [D. Yu] Let the base of the tetrahedron be
triangle ABC, eith a = BC  = AD ,
b = AC  = BD , c = AB  = CD ; let P be the foot of the perpendicular
from D to the plane of ABC and let h = DP .
Then AP  = Ö[(a^{2}  h^{2})], BP  = Ö[(b^{2}  h^{2})], CP  = Ö[(c^{2}  h^{2})].
Suppose that a = ÐBCP and b = ÐACP.
Then using the Law of Cosines on triangles BCP, ACP and
ABC, we obtain that
cosa = 
a^{2} + c^{2}  b^{2}



cosb = 
b^{2} + c^{2}  a^{2}



and
cos(a+ b) = 
a^{2} + b^{2}  c^{2} 2ab

, 

whence

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

= 


(a^{2} + c^{2}  b^{2})(b^{2} + c^{2}  a^{2})  
 ________________________ Ö4a^{2}(c^{2}  h^{2})  (a^{2} + c^{2}  b^{2})^{2}


 ________________________ Ö4b^{2}(c^{2}  h^{2})  (b^{2} + c^{2}  a^{2})^{2}


4ab(c^{2}  h^{2})

. 

Shifting terms and squaring leads to
[2(a^{2} + b^{2}  c^{2})(c^{2}  h^{2})  (a^{2} + c^{2}  b^{2})(b^{2} + c^{2}  a^{2})]^{2} = [4a^{2}(c^{2}  h^{2})  (a^{2} + c^{2}  b^{2})^{2}][4b^{2}(c^{2}  h^{2})  (b^{2} + c^{2}  a^{2})^{2}] . 

With u = b
^{2} + c
^{2}  a
^{2}, v = c
^{2} + a
^{2}  b
^{2}, w = a
^{2} + b
^{2}  c
^{2}, z = c
^{2}  h
^{2}, this can be rendered

= [2wz  uv]^{2}  [4a^{2}z  v^{2}][4b^{2}z  u^{2}] 
 
= z[4(w^{2}  4a^{2}b^{2})z  4(uvw  a^{2}u^{2}  b^{2}v^{2})] 


so that
c^{2}  h^{2} = z = 
a^{2} u^{2} + b^{2} v^{2}  uvw 4a^{2}b^{2}  w^{2}



and
h^{2} = 
4a^{2}b^{2}c^{2} + uvw  a^{2}u^{2}  b^{2}v^{2}  c^{2}w^{2} 4a^{2}b^{2}  w^{2}

. 

Now

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


where S is the area of triangle ABC.
Now consider the numerator of h^{2}.
Its value when w = a^{2} + b^{2}  c^{2}
is set equal to 0 is 4a^{2}b^{2}c^{2}  a^{2}u^{2}  b^{2}v^{2} = 4a^{2}b^{2}c^{2}  a^{2}(2b^{2})  b^{2}(2a^{2}) = 0, so that w divides the
numerator. So also do u and v. Hence the numerator of degree
6 in a, b, c must be a multiple of uvw, also of degree 6
in a, b, c. Hence the numerator is a multiple of uvw.
Comparing the coefficients of a^{6} (say) gives that the
numerator must be 2uvw. Hence
h^{2} = 
2uvw 16S^{2}

= 
uvw 8S^{2}

. 

The volume V of the tetrahedron satisfies
V^{2} = 
æ ç
è


Sh 3


ö ÷
ø

2

= 
S^{2}h^{2} 9

= 
uvw 72

, 

whence
The checking for the tetrahedron proceeds as before.

194.

Let ABC be a triangle with incentre I. Let M
be the midpoint of BC, U be the intersection of AI produced with
BC, D be the foot of the perpendicular from I to BC and
P be the foot of the perpendicular from A to BC. Prove that
PD DM  = DU PM  . 

Solution 1. Suppose that the lengths of the sides of the
triangle are a, b and c, using the conventional notation.
Then the distance from B of the following points on the side
BC are given by (B, 0), (C, a), (M, a/2), (U, ca/(b+c)),
(D, (a+c
b)/2) and (P, ccosB) = (P, (a
^{2} + c
^{2}  b
^{2})/(2a)).
One can then verify the desired relation by calculation.
Solution 2. [L. Chen] Let the side lengths of the triangle
be a, b, c, as conventional, and, wolog, suppose that c < b.
Let u = BP  and v = PC . Then, equating
two expressions for the area of the triangle, with r = ID  as the
inradius, we find that AP  = 2rs/a. From similar
triangle, we have that

PU  DU 

= 
AP  ID 

= 
2s a

= 1 + 
b+c a

, 

whence
Now
PM
 = (a/2)
 u = (v
u)/2 and
DM
 = (b
c)/2. Hence
By Pythagoras' Theorem, c
^{2}  u
^{2} = b
^{2}  v
^{2}, whence

v  u b  c

= 
b + c v + u

= 
b + c a



and the result follows.

195.

Let ABCD be a convex quadrilateral and let the midpoints
of AC and BD be P and Q respectively, Prove that
AB ^{2} + BC ^{2} + CD ^{2} +DA ^{2} = AC ^{2} + BD ^{2} +4 PQ ^{2} . 

Solution 1. Let X denote the vector from an origin to
a point X. Then, vectorially, it can be verified that

+ (B  C)·(B  C)+ (C  D)·(C  D) + (D  A)·(D  A) 
 
(A  C)·(A  C)  (B  D)·(B  D) 
 
= 2A ·B  2B ·C  2C ·D  2D ·A +2A ·C + 2B ·D + A^{2} + B^{2} + C^{2} + D^{2} 
 
= 4 
æ ç
è


A + C 2

 
B + D 2


ö ÷
ø

· 
æ ç
è


A + C 2

 
B + D 2


ö ÷
ø

, 


which yields the desired result.
Solution 2. [T. Yin] We use the result that for any
parallelogram KLMN, 2KL ^{2}+ 2LM ^{2} = KM ^{2} + LN ^{2}. This
is straightforward to verify using the Law of Cosines, for
example. Let W, X, Y, Z be the respective midpoints of
the sides AB, BC, CD, DA. Using the fact that all of
WXYZ, PXQZ and PWQY are parallelograms, we have that
AB ^{2} + BC ^{2}+ CD ^{2} + DA ^{2} 

= 4[ PX ^{2} + PW ^{2} +PZ ^{2} + PY ^{2} ] 
 
= 2[ PQ ^{2} + XZ ^{2}+ PQ ^{2} + WY ^{2} ] 
 
= 4PQ ^{2} + 2 [ XZ ^{2} + WY ^{2} ] 
 
= 4PQ ^{2} + 4 [ WZ ^{2} + WX ^{2} ] 
 
= 4 PQ ^{2} + BD ^{2} + AC ^{2} . 



196.

Determine five values of p for which the polynomial
x^{2} + 2002x  1002p has integer roots.
Answer. Here are some values of (p; u, v) with u and
v the corresponding roots:
(0; 0,
2002),
(4; 2, 2004), (784; 336, 2338),
(1780; 668, 2670), (3004; 2002, 3004),
(3012; 1004, 3006),
(4460; 1338, 3340),
(8012; 2004, 4006), (8024; 2006, 4008),
(556; 334, 1668),
(1000; 1000, 1002).
Solution 1. If x satisfies the equation x^{2} + 2002x  1002p = 0, then we must have
p = x(x + 2002)/(1002). If we choose integers x for which
x(x + 2002) is a multiple of 1002, then this value of p will
be an integer that yields a quadratic with two integer roots,
namely x and 2002x. One way to do this is to select
either x º 0 or x º 2 (mod 1002). Observing that
1002 = 2 ×3 ×167, we can also try to make
x º 0 (mod 167) and x º 2 (mod 6). For example,
x = 668 works. We can also try x º 2 (mod 167) and
x º 0 (mod 6); in this case, x = 336 works.
Solution 2. The discriminant of the quadratic is 4 times
1001^{2} + 1002p. Suppose that p is selected to make this
equal to a square q^{2}. Then we have that
1002p = q^{2}  1001^{2} = (q  1001)(q + 1001) . 

We select q so that either q
 1001 or q + 1001 is divisible
by 1002. For example q = 2003, 1, 3005, 4007 all work. We can also
make one factor divisible by 667 and the other by 6.

197.

Determine all integers x and y that satisfy
the equation x^{3} + 9xy + 127 = y^{3}.
Solution 1. Let x = y + z. Then the equation becomes
(3z + 9)y
^{2} + (3z
^{2} + 9z)y + (z
^{3} + 127) = 0, a quadratic in
y whose discriminant is equal to

 
= (3z + 9) [ z^{2} (3z + 9)  4(z^{3} + 127) ] 
 
=  (3z + 9)(z^{3}  9z^{2} + 508) . 


Note that z
^{3}  9z
^{2} + 508 = z
^{2}(z
 9) + 508 is nonnegative if
and only if z
³ 5 (z being an integer) and that
3z + 9 is nonnegative if and only if z
³ 3. Hence the
discriminant is nonnegative if and only if z =
3,
4,
5.
>From the quadratic equation, we have that z
^{3} + 127
º 0
(mod 3). The only possibility is z =
4 and this
leads to the equation 0 =
3y
^{2} + 12y + 63 =
3(y
 7)(y + 3) and the solutions (x, y) = (3, 7),(
7,
3).
Solution 2. The equation can be rewritten
(x  y)[(x  y)^{2} + 3xy] + 9xy = 127 

or
where u = x
 y and v = xy.
Hence
3v =  
u^{3} + 127 u + 3

=  
é ê
ë

(u^{2}  3u + 9) + 
100 u + 3


ù ú
û

. 

Therefore, u
^{3} + 127
º 0 (mod 3), so that u
º 2
(mod 3), and u + 3 divides 100. The candidates are
u = 103, 28, 13, 7, 4, 1, 2, 17, 47 . 

Checking these out leads to the posible solutions.

198.

Let p be a prime number and let f(x) be a polynomial
of degree d with integer coefficients such that f(0) = 0 and
f(1) = 1 and that, for every positive integer n,
f(n) º 0 or f(n) º 1, modulo p. Prove that
d ³ p  1. Give an example of such a polynomial.
Solution. Since the polynomial is nonconstant, d
³ 1,
so that the result holds for p = 2. Henceforth, assume that
p is an odd prime. Let 0
£ k
£ p
 2. Consider the
polynomial
p_{k} (x) = 
x(x1)(x2)¼(x  k + 1)(x  k  1)¼(x  p + 2) k!(pk2)!(1)^{pk}

. 

We have that p
_{k}(k) = 1 and p
_{k}(x) = 0 when x = 0, 1, 2,
¼,k
 1, k + 1,
¼, p
 2. Let
g(x) = 
p2 å
k=0

f(k)p_{k} (x) . 

Then the degree of g(x) does not exceed p
 2 and
g(x) = f(x) for x = 0, 1, 2,
¼, p
2; in fact, g(x) is
the unique polynomial of degree less than p
1 that agrees with
f at these p
1 points (why?).
Now
g(p  1) = 
p2 å
k=0

(1)^{pk} 
(p1)! k!(pk1)!

f(k) = 
p2 å
k=0

(1)^{pk} 
æ ç
è

p1
k

ö ÷
ø

f(k) . 

Since ((p
1)  k) = (p  k)
 ((p
1)  (k
1))
and (p  k)
º 0 (mod p) for 1
£ k
£ p
1, and
induction argument yields that ((p
1)  k)
º (
1)
^{k}
for 1
£ k
£ p
1, so that
g(p1) º (1)^{p} 
p2 å
k=0

f(k) 

(mod p). Since f(0) = 0 and f(1) = 1, it follows that
å_{k=0}^{p2} f(k) is congruent to some number between 1 and
p
2 inclusive, so that g(p
1) \not
º 0 and
g(p
1) \not
º 1 (mod p). Hence f(p
1)
¹ g(p
1), so that
f and g are distinct polynomials. Thus, the degree of g
exceeds p
2 as desired.
By Fermat's Little Theorem, the polynomial x^{p1} satisfies
the condition.
Solution 2. [M. GuayPaquet] Let
h(x) = f(x) + f(2x) + ¼+ f((p1)x) . 

Then h(1) \not
º 0 (mod p) and h(0) = 0. The degree of
h is equal to d, the degree of f.
Let x \not º 0 (mod p). Then (x, 2x, 3x, ¼, (p1)x)
is a permutation of (1, 2, 3, ¼, p1), so that
h(x) º h(1) (mod p).
Suppose that g(x) = h(x)  h(1). The degree of g is equal to d,
g(0) º h(1) \not º 0 (mod p) and
g(x) º 0 whenever x \not º 0 (mod p). Therefore,
g(x) differs from a polynomial of the form k(x1)(x2)¼(x[`(p1)]) by a polynomial whose coefficients are multiples
of p. Since k \not º 0 (mod p) (check out the value at
0), the coefficient of x^{k1} must be nonzero, and so
d ³ p1, as desired.