BARCELONA TECH
MATHCONTEST
February 2015
1 Let x, y be two distinct positive reals and let z be their harmonic mean.
Show that segments of length x, y, z form a triangle if and only if
1+
min{x, y}
max{x, y}
>
√
2
2 Through the midpoint of the diagonal BD in the convex quadrilateral
ABCD we draw a straight line parallel to the diagonal AC. This line intersects the side AD at the interior point E. Find the value of
[ABC] + [AEC]
[CED]
3 Let abc be a prime number.
Prove that equation ax2 + bx + c = 0 does
not have rational roots.
4
We have three numbered boxes and 10000 red balls, 10000 blue ones
and 10000 yellow ones. Balls of the same color are undistinguishable. Determine in how many ways they can be distributed in the boxes satisfying:
• Each box has 10000 balls
• No two boxes have the same amount of balls of the same color
• For every two boxes A and B, there is a color c such that the number
of balls of color c in A is exactly 2015 or 2016 bigger than the number
of balls of color c in B.
Each problem will be graded from 0 to 7 marks.
It is not allowed to use any kind of electronic devices.
Time: 4 hours.
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Solutions
1. Let x, y be two distinct positive reals and let z be their harmonic mean.
Show that segments of length x, y, z form a triangle if and only if
1+
min{x, y}
max{x, y}
>
√
2
Solution. ⇒) Suppose x, y, z form a triangle, then holds x + y > z, z +
x > y and y + z > x. WLOG we assume that x < y and substituting
z = 2xy/(x + y) in z + x > y, yields
2xy
x+y
+ x > y ⇒ 2xy + x(x + y) > y(x + y)
⇒ 2xy + x2 + xy > y 2 + xy ⇒ x2 + 2xy + y 2 > 2 y 2
from which follows
x+y >y
√
2⇒1+
x
y
>
√
2⇒1+
min{x, y}
max{x, y}
>
√
2
on account that min{x, y} = x and max{x, y} = y.
⇐) WLOG we assume that x < y. Then holds x < z < y with z = 2xy/(x+
y) as can be easily checked. Furthermore, from the preceding immediately
follows that x+y > z and y +z > x. Since min{x, y} = x and max{x, y} =
y, then
1+
min{x, y}
max{x, y}
>
√
2⇒1+
x
y
>
√
2⇒x+y >y
√
2
Squaring both terms in the last inequality, yields
x2 +2xy+y 2 > 2 y 2 ⇒ 2xy+x2 +xy > y 2 +xy ⇒ 2xy+x(x+y) > y(x+y)
from which we have
2xy
x+y
+x>y ⇒z+x>y
and the segments of length x, y, z form a triangle.
2.
Through the midpoint of the diagonal BD in the convex quadrilateral
ABCD we draw a straight line parallel to the diagonal AC. This line intersects the side AD at the interior point E. Find the value of
[ABC] + [AEC]
[CED]
2
Solution.
First, we claim
that the quadrilateral ABCE has
the same area as quadrilateral
ABCM , where M is any point
on the line EF, where F is intersection of the parallel of the diagonal AC with CD. Indeed, the
triangles ACE and ACM have
a common base and equal altitudes since the points E and M
lie on a line parallel to the base
AC. On account of the claim,
we can replace [ABCE] by an
equivalent quadrilateral [ABCM ]
(same area), where M is any point
lying on EF . Likewise, we have
[CED] = [CM AD].
B
C
F
A
M
K
E
D
Let K be the midpoint of diagonal BD. Let us make M ≡ K. Next, we will
see that [ABCK] = [AKCD]. Indeed, we have that △ABK and △AKD
have the same area because BK = KD and both have the same altitude.
The same occurs with △BKC and △CKD. Therefore, for all M on line
EF the equality [ABCM ] = [AM CD] holds. In particular, when M ≡ E
we have [ABCE] = [CED] and [ACE] + [ABC] = [CDE] from which
[ABC] + [AEC]
[CED]
=1
follows.
3.
Let abc be a prime number. Prove that equation ax2 + bx + c = 0 does
not have rational roots.
Solution. We argue by contradiction. Assume that ax2 + bx + c = 0
has two rational roots. This occurs when its discriminant is a square, say
∆ = b2 − 4ac = d2 . We have d < b and abc = 100a + 10b + c. Now we
consider the number
4a · abc = 400a2 + 40ab + 4ac = 400a2 + 40ab + b2 − d2
= (20a + b)2 − d2 = (20a + b − d)(20a + b + d)
Since abc is prime, then it divides 20a + b − d or 20a + b + d as it is
well-known. But this is not possible because abc > 20a + b − d and
abc > 20a + b + b > 20a + b + d. This is a contradictions and the statement
follows.
4. We have three numbered boxes and 10000 red balls, 10000 blue ones and
10000 yellow ones. Balls of the same color are undistinguishable. Determine
in how many ways they can be distributed in the boxes satisfying:
3
• Each box has 10000 balls
• No two boxes have the same amount of balls of the same color
• For every two boxes A and B, there is a color c such that the number
of balls of color c in A is exactly 2015 or 2016 bigger than the number
of balls of color c in B.
Solution. Let us denote the boxes by B1 , B2 and B3 , respectively. Let
c(Bi , Bj ) be the color satisfying the third condition of the statement for
Bi and Bj . First, we observe that c(B1 , B2 ) and c(B2 , B1 ) cannot be the
same, because c(B1 ) > c(B2 ) and at the same time c(B2 ) > c(B1 ), which
is impossible. WLOG, assume that c(B1 , B2 ) is red and c(B2 , B1 ) is blue.
If the difference of the number of red and blue balls is the same (either 2015
or 2016), then in order for both boxes to have 10000 balls, there should be
the same number of yellow balls in boxes B1 and B2 in contradiction with
the second condition. So the difference is 2015 in one case and 2016 in the
other. WLOG, B1 has 2016 more red balls than B2 and B2 has 2015 more
blue balls than B1 . This means that B2 has one more yellow ball than B1 .
We have that c(B1 , B3 ) and c(B3 , B1 ) are different, and so are c(B2 , B3 )
and c(B3 , B2 ). Since there are only three colors, one must be repeated.
Let x, y and z be the number of balls of this color in B1 , B2 and B3 ,
respectively. If it was either red or blue, we would have that |x − y|, |y − z|
and |z − x| are all 2015 or 2016, which is impossible. Hence the color is
yellow, and y − x = 1.
Now, there are two options: z − x = 2016 and z − y = 2015 or x − z = 2015
and y − z = 2016. If we substitute in x + y + z = 10000, the first one
gives 3z − 4031 = 10000 and the second one gives 3z + 4031 = 10000.
The second one has no integer solutions and for the first one we have
14031
= 4677, x = z − 2016 = 2661 and y = z − 2015 = 2662.
z=
3
Color c(B1 , B3 ) is either red or blue. If it is blue, B2 would have 4031 more
blue balls than B3 , but that is not possible (if we had started by considering
boxes B2 and B3 instead of B1 and B2 , we would have obtained the same
conclusion that the differences are 2015, 2016 and 1). If color c(B1 , B3 ) is
red, there are 2015 more red balls in B1 than in B3 , and there are 4677
red balls in B1 , 2661 in B2 and 2662 in B3 . By the same reasoning, there
are 4677 blue balls in B2 , 2662 in B1 and 2661 in B3 . This arrangement
satisfies all the conditions of the statement.
Finally, we need to see how many configurations we discarded in the ”without loss of generality” parts. In the first one there were six possible choices
for the pair of colors, and in the second one we could assign the 2015 to either red or blue (two options). Therefore the total number of arrangements
is 6 × 2 = 12.
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