Download 1 - Mathactivities

BARCELONA TECH
MATHCONTEST
February 2016
1 Let p1 , p2 , . . . , pn+1 denote the first n + 1 primes. Suppose that {A, B}
is a partition of the set X = {p1 , p2 , . . . , pn }, where A = {q1 , q2 , . . . , qs }
and B = {r1 , r2 , . . . , rt }. Prove that if m = q1 q2 . . . qs + r1 r2 . . . rt < p2n+1 ,
then m is a prime.
2 Let n ≥ 1 be a positive integer. Consider a pile of 3n coins, one of which
is fake. Suppose that all coins are either white or black and that if the
fake coin is white, it is lighter than the others, and if the fake is black,
it is heavier than the others. Furthermore, assume that the number of
white coins and the number of black coins differ by at most one. Under
these conditions, prove that the fake coin can be identified and classified
as heavy or light by at most n weighings in a scale.
3
Let A be a point outside a given circle Γ with diameter BC. Find the
locus of the orthocenter H of triangle ABC when BC changes.
4 Let a, b, c be three positive numbers such that ab + bc + ca = 3abc. Prove
that
s
a+b
c(a2 + b2 )
+
s
b+c
a(b2 + c2 )
+
s
c+a
b(c2 + a2 )
≤ 3.
Each problem will be graded from 0 to 7 marks.
It is not allowed to use any kind of electronic devices.
Time: 4 hours.
1
Solutions
1. Let p1 , p2 , . . . , pn+1 denote the first n + 1 primes. Suppose that {A, B} is
a partition of the set X = {p1 , p2 , . . . , pn }, where A = {q1 , q2 , . . . , qs } and
B = {r1 , r2 , . . . , rt }. Prove that if m = q1 q2 . . . qs + r1 r2 . . . rt < p2n+1 , then
m is a prime.
Solution. Assume to the contrary, that m is not a prime number. Then
m = ab for some integers a and b with 1 < a < m and 1 < b < m. Let
p be the smallest prime that divides a and let q be the smallest prime that
divides b. WLOG we may assume that p ≤ q. We now consider two cases
according to whether p ∈ X or p ∈
/ X.
• If p ∈ X, then either p = qi for some i with 1 ≤ i ≤ s or p = rj for
some j with 1 ≤ j ≤ t, but not both ({A, B} is a partition). Suppose
that p = qi where 1 ≤ i ≤ s. Since p | a, then p | m. Also p | q1 q2 . . . qs .
Thus p | (m − q1 q2 . . . qs ) and so p | r1 r2 . . . rt . This implies that p = rj
for some j with 1 ≤ j ≤ t. Contradiction.
• If p ∈
/ X, then q ≥ p ≥ pn+1 and so m ≥ pq ≥ p2n+1 . Contradiction.
From the preceding we conclude that m is prime, and we are done.
2.
Let n ≥ 1 be a positive integer. Consider a pile of 3n coins, one of which
is fake. Suppose that all coins are either white or black and that if the fake
coin is white, it is lighter than the others, and if the fake is black, it is heavier
than the others. Furthermore, assume that the number of white coins and
the number of black coins differ by at most one. Under these conditions,
prove that the fake coin can be identified and classified as heavy or light by
at most n weighings in a scale.
Solution. For each n ≥ 1, let P (n) be the statement to be proven. We will
argue by induction. Indeed,
• Base step: For n = 1, consider 31 = 3 coins, and without loss of
generality, suppose that two are black and one is white. Put a black
on each pan and set the white aside. If the scale balances, then white
coin is counterfeit, and since it is white, is lighter than the other coins.
If the scales tips, say the left side down, then the black on the left is
heavy and fake. In any case, the counterfeit coin is identified from
among three coins and classified as heavy or light, so P (1) holds.
• Next we consider the case when n = 3, assuming, for the moment,
that P (2) has been shown. Consider 33 = 27 coins, with, say, 14
black, and 13 white. Partition these 27 coins into three groups, say
G1 = (4B, 5W ), G2 = (5B, 4W ), and G3 = (5B, 4W ). Put G2 on
the left, weighed against G3 on the right. If the scale balances, the
2
counterfeit coin is in group G1 and so P (2) applies to a set of 9 coins,
4 black and 5 white. If the left pan goes down, either one of the 5 black
from G2 is heavy, or one of the 4 white from G3 is light. The induction
hypothesis P (2) now applies to these 32 = 9 coins. Similarly, if the
scale tips to the right, the counterfeit is among the 4 white in G2 or
the 5 black in G3 , and again P (2) applies.
• Inductive step: Fix k ≥ 1 and assume that P (k) is true. Consider
a collection of 3k+1 coins, one of which is counterfeit. Assume that
there is one more black than white (the same argument works if there
3k+1 + 1
are black
is one more white than black), so suppose that
2
3k+1 − 1
are white. Partition the coins into three groups, G1 , G2 ,
and
2
and G3 , each with 3k coins, and each with a near balance of black and
3k − 1
3k + 1
white: In G1 , put
black and
white coins. In both G2
2
2
3k + 1
3k − 1
and G3 , put
black and
white coins. Putting G1 aside,
2
2
weigh G2 on the left against G3 on the right. If the scales balance,
the counterfeit coin is in G1 , and P (k) applies to G1 , finding the fake
coin in an additional k weighings, k + 1 in all, and so P (k + 1) is true
3k + 1
in this case. If the scales go down on the left, either one of the
2
3k − 1
black coins from G2 is heavy, or one of the
white coins from
2
G3 is light and these
3k + 1
2
+
3k − 1
2
= 3k
coins satisfy the hypothesis for P (k), and so the coin is found in k+11
weighings. The analogous argument works when the right pan lowers.
Finally, by mathematical induction, for each n ≥ 1, we conclude thatP (n)
holds.
3.
Let A be a point outside a given circle Γ with diameter BC. Find the
locus of the orthocenter H of triangle ABC when BC changes.
Solution. Let O be the center of Γ and let AA1 , BB1 , and CC1 be the altitudes of △ABC with orthocenter H. Let Γ1 be a circle with diameter AO.
Since ∠A1 is right, then Γ1 passes through the points A, O, A1 . Likewise,
points B, C and B1 lie on circle Γ.
The power of H with respect to Γ is HB · HB1 and the power of H with
respect to Γ1 is HA · HA1 . Since triangles BHA1 and AHB1 are similar,
then we have
HB
HA
=
HA1
HB1
⇔ HB · HB1 = HA · HA1
3
A
1
C1
B1
l
H
O A1
B
C
This means that H lies on the radical axis ℓ of Γ and Γ1 .
Conversely, it is easy to see that each point of ℓ is the orthocenter of a
triangle of the given type.
4. Let a, b, c be three positive numbers such that ab + bc + ca = 3abc. Prove
that
s
s
s
a+b
b+c
c+a
+
+
≤ 3.
c(a2 + b2 )
a(b2 + c2 )
b(c2 + a2 )
Solution. Dividing both terms of the constraint by abc, yields
1 1 1
+ + = 3. Observing the LHS of the inequality we consider the vectors
a b c
s
s
!
r
1
1
1
a+b
b+c
c+a
. Applying
,
,
u
~ = √ ,√ ,√
and ~
v=
c
a
a2 + b2
b 2 + c2
c2 + a 2
b
Cauchy’s inequality, (~
u·~
v )2 ≤ ||~
u||2 · ||~
v ||2 , yields
s
s
s
!2
a+b
b+c
c+a
+
+
c(a2 + b2 )
a(b2 + c2 )
b(c2 + a2 )
1
1
b+c
c+a
a+b
1
+ +
+ 2
+ 2
·
≤
a
b
c
a2 + b2
b + c2
c + a2
Multiplying by a + b both terms of the inequality 2ab ≤ a2 + b2 , we get
2ab(a + b) ≤ (a + b)(a2 + b2 ) from which follows
1 1
1
a+b
a+b
=
+
≤
a2 + b2
2ab
2 a
b
Likewise, we have
b+c
b 2 + c2
≤
1
2
1
b
+
1
c
and
4
c+a
c2 + a 2
≤
1
2
1
c
+
1
a
Adding up the last three inequalities, yields
a+b
a2
+
b2
+
b+c
b2
+
c2
c+a
+
c2
+
a2
≤
1
a
+
1
b
+
1
c
Finally, on account of the above, we get
s
and
a+b
c(a2 + b2 )
s
+
s
b+c
a(b2 + c2 )
a+b
c(a2 + b2 )
+
s
+
s
c+a
b(c2 + a2 )
b+c
a(b2 + c2 )
Equality holds when a = b = c = 1.
5
+
s
!2
≤
1
a
c+a
b(c2 + a2 )
+
1
b
+
≤ 3.
1
c
2
=9