Diego

Junior problems

J67. Prove that among seven arbitrary perfect squares there are two whose difference
is divisible by 20.
Proposed by Ivan Borsenco, University of Texas at Dallas, USA

First solution by Salem Malikic, Sarajevo, Bosnia and Herzegovina
It is easy to check that perfect squares can give one of the following residues:
1, 2, 4, 8, 16

(mod 20).

By the Pigeonhole principlewe conclude that among seven perfect squares we
must have at least two that have the same residue modulo 20. Hence their
difference is divisible by 20 and our proof is complete.
Second solution by Vicente Vicario Garca, Huelva, Spain
Note that for all integer x we have x2 ≡ 1, 2, 4, 8, 16 (mod 20) and we have six
distinct possible residues. If we have seven arbitrary perfect squares x2 , x2 ,x2 ,
1
2
3
x2 , x2 , x2 , x2 , by the pigeonhole principle, there are two squares x2 and x2 with
4
5
6
7
i
j
the same residue and they satisfy the requirement.
Third solution by Vishal Lama, Southern Utah University, USA
Observe that by the Pigeonhole Principle, there are at least four perfect squares
which all have the same parity. Now, note that for any integer n, we have
n2 ≡ −1,0, 1 (mod 5). Again by the Pigeonhole Principle, out of these four
perfect squares, we have at least two perfect squares, say a2 and b2 , such that
a2 ≡ b2 (mod 5). This implies that 5 | a2 − b2 . Also, 2 | a − b and 2 | a + b since
both a and b have the same parity. Hence, 4 | a2 − b2 , but gcd(5, 4) = 1, thus
we have 20 | a2 − b2 , and we are done.
Also solved by Andrea Munaro, Italy;Arkady Alt, San Jose, California, USA;
Brian Bradie, VA, USA; Daniel Campos Salas, Costa Rica; Daniel Lasaosa,
Universidad Publica de Navarra, Spain; Ganesh Ajjanagadde Acharya Vidya
Kula, Mysore, India; Jose Hernandez Santiago, UTM, Oaxaca, Mexico; Oleh
Faynshteyn, Leipzig, Germany; G.R.A.20 Math Problems Group, Roma, Italy.

Mathematical Reflections 1 (2008)

2

J68. Let ABC be a trianglewith circumradius R. Prove that if the length of one
of the medians is equal to R, then the triangle is not acute. Characterize all
triangles for which the lengths of two medians are equal to R.
Proposed by Daniel Lasaosa, Universidad Publica de Navarra, Spain

First solution by Vicente Vicario Garca, Huelva, Spain
Let O be the circumcenter and M be the midpoint of the side BC . Withoutloss of generality we have that a ≥ b ≥ c, we have
mA =

1
2

2b2 + 2c2 − a2 , mB =

1
2

2a2 + 2c2 − b2 , mC =

1
2

2a2 + 2b2 − c2 ,

and we deduce that mA ≤ mB ≤ mC . On the other hand, if the triangle is
acute angled, then its circumcenter lies int the interior of the triangle. Note
that mA > R, because ∠AOM is obtuse, and the equality does not occur. Thus
triangle ABC isnot acute angled.
For the second part it is not difficult to see that if two medians in a triangle are
equal, then the triangle is isosceles, because
mB = mC ⇔

1
2

2a2 + 2c2 − b2 =

1
2

2a2 + 2b2 − c2 ⇔ b = c.

Let the ABC be isosceles triangle with b = c. By the Law of Sines and the Law
of Cosines we have
R=

a
,
2 sin A

cos A =

b2 + c2 − a2
2b2 − a2
=
2bc
2b2

andif mB = mC = R, we have
m2 = R2 , ⇒
B

1
(2a2 + 2b2 − b2 ) = R2 , ⇒ 2a2 + b2 = 4R2
4

(1)

and finally
R2 =

a2
a2
=
=
2
4(1 − cos2 A)
4 sin A

a2
4 1−

2b2 −c2
2b2

2

, ⇒ 4R 2 =

4b 4
.
4b2 − a2

Finally, using (1) we get
2a2 + b2 =
yielding b = c =

4b 4
, ⇒ 7a2 b2 − 2a4 = 0 ⇒ a2 (7b2 − 2a2 ) = 0,
4b2 − a2
2
7

a, and we are done.

MathematicalReflections 1 (2008)

3

Third solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain
Without loss of generality, let us assume that the length of the median from A
equals R. The square of the length of this median is given by
b2 + c2 a2
a2
−
=
+ bc cos A = R2 sin2 A + 4R2 sin B sin C cos A.
2
4
4
Equating this result to R2 and grouping terms in one side of the equality...