Download IrMO 2016 paper 2 solutions

TWENTY NINTH IRISH MATHEMATICAL OLYMPIAD
Saturday, 23 April 2016
Second Paper
Solutions and Marking Schemes
6. Proposed by Jim Leahy.
Triangle ABC has sides a = |BC| > b = |AC|. The points K and H on the
segment BC satisfy |CH| = (a + b)/3 and |CK| = (a − b)/3. If G is the
centroid of triangle ABC, prove that 6 KGH = 90◦ .
Solution
Because |CH| > |CK| and both, H and K, are between C and B, we have
|KH| = |CH| − |CK| = a+b
− a−b
= 23 b. If M is the midpoint of KH, then
3
3
|KM| = |HM| = 13 b. Next, we show that |GM| = 13 b as well. Here are two
ways to show this.
Version 1. Let E be the midpoint of AC.
A
b
E
G
b
b
b
b
b
B
H
b
M
b
K
C
+ 3b = a3 = 31 |BC| and |EG| = 13 |BE|,
Because |CM| = |CK| + |KM| = a−b
3
the lines GM and EC are parallel. This implies that
|BM|
|BC| − |CM|
|CM|
2
|GM|
=
=
=1−
= .
|EC|
|BC|
|BC|
|BC|
3
Hence, using that |EC| = b/2, we obtain |GM| = 23 |EC| = 31 b, as desired.
Version 2. Let D be the midpoint of BC.
A
b
G
B
b
b
b
b
H
D
1
b
M
b
K
b
C
Because
|CM| = |CK| + |KM| =
a−b b
a
a
+ = < = |CD|,
3
3
3
2
M is between C and D and we have
|DM| = |CD| − |CM| =
a a
a
1
− = = |DC|.
2 3
6
3
Because |DG| = 31 |DA|, the lines GM and AC are parallel, hence
|DA|
|AC|
=
=3
|GM|
|DG|
and so
b
|GM| = .
3
After establishing that |GM| = 3b , we obtain |GM| = |KM| = |HM|, thus G
is on the circle with diameter KH which implies that 6 KGH = 90◦ .
Suggested Marking Scheme:
• 2 marks for considering the midpoint of HK and finding that |KH| = 23 b
or |KM| = |HM| = 31 b,
• 3 marks for showing that GM and AC are parallel,
• 2 marks for deducing from this parallelism that |GM| =
similar triangles or the Intercept Theorem).
• The remaining 3 marks for completing the solution.
b
3
(e.g. using
• Up to 2 marks should be deducted if proofs are missing for non-obvious
relative positionings that are used (e.g. that M is between C and D as
needed in Version 2 above).
2
7. Proposed by Rachel Quinlan.
A rectangular array of positive integers has four rows. The sum of the entries
in each column is 20. Within each row, all entries are distinct. What is the
maximum possible number of columns?
Solution
Let the number of columns be k. Then the sum of all the numbers in the
array is 20k. Since each row consists of k distinct positive integers, the least
possible value of the sum of the entries in a single row is
1+2+···+k =
k(k + 1)
.
2
Thus
k(k + 1)
.
2
After dividing both sides by 2k we get 10 ≥ k + 1, and so k ≤ 9. This shows
that the number of columns cannot exceed 9.
20k ≥ 4 ×
On the other hand, the bound
shows.
1 2
2 1
8 9
9 8
of 9 can be attained as the following example
3
4
6
7
4
3
7
6
5
5
5
5
6
7
3
4
7
6
4
3
8
9
1
2
9
8
2
1
Suggested Marking Scheme:
• 6 marks for a correct proof that the answer is at most 9.
• 4 marks for a correct example with 9 columns.
Partial marks can be awarded as follows:
• 1 mark for any correct example with at least 5 but less than 9 columns
(cannot be added to the 4 marks for an example with 9 columns);
• 1 mark for the claim that the answer is 9, even in the absence of any
other relevant ingredients of a solution;
• 1 mark for noting that the sum of all entries is 20k;
• 3 marks for considering the minimum possible value of the sum of k
distinct positive integers.
3
8. Proposed by Finbarr Holland.
Suppose a, b, c are real numbers such that abc 6= 0. Determine x, y, z in terms
of a, b, c such that
bz + cy = a,
Prove also that
cx + az = b,
ay + bx = c.
1 − x2
1 − y2
1 − z2
=
=
.
a2
b2
c2
Solution
Multiply the first equation by a, the second by b and the third by c. Add the
difference between the first and second of the resulting equations to the third
one, thereby obtaining the following one:
2cay = c2 + a2 − b2 .
Since ca 6= 0, it follows that
y=
c2 + a2 − b2
.
2ca
Likewise, by cyclicity,
x=
a2 + b2 − c2
b2 + c2 − a2
, z=
.
2bc
2ab
Next, focusing on z,
1 − z 2 = (1 − z)(1 + z)
2ab − (a2 + b2 − c2 )
2ab + (a2 + b2 − c2 )
=
2ab
2ab
2
2
2
2
2ab + a + b2 − c2
2ab − a − b + c
=
2ab
2ab
2
2
2
2
(a + b) − c
c − (a − b)
=
2ab
2ab
c − (a − b) c + (a − b) (a + b) − c a + b + c
=
4a2 b2
(b + c − a)(c + a − b)(a + b − c)(a + b + c)
,
=
4a2 b2
whence
4s(s − a)(s − b)(s − c)
1 − z2
=
,
2
c
a2 b2 c2
where 2s = a + b + c. Since the expression on the right side of this equation
is symmetric in a, b, and c, we may conclude as well that
4
1 − x2
1 − y2
(b + c − a)(c + a − b)(a + b − c)(a + b + c)
=
=
,
2
2
a
b
4a2 b2 c2
whence the result.
Suggested Marking Scheme:
• Five marks for each part.
• Award 3 marks for determining correctly the value of one of x, y, z;
• 2 marks more for identifying with explanation the values of the other
two.
• Award 4 marks for correctly deriving a symmetric expression for one of
the ratios;
• 1 more mark for explaining why this is equal to the remaining ratios.
5
9. Proposed by Gordon Lessells.
Show that the number




251
√
1
√
√
− 10 3 63
3
3
252 − 5 2
1
+
√
3
√
251
√
+ 10 3 63
3
252 + 5 2
3



is an integer and find its value.
Solution
√
√
√
Let a3 = 252 and b3 = 250, then ab = 3 252 3 250 = 10 3 63. Using that
3
3
3
3
and 1 = a −b
, the given expression inside the bracket becomes
251 = a +b
2
2
a3 +b3
2
a3 −b3
2
a−b
+
− ab
a3 −b3
2
a3 +b3
2
a+b
.
+ ab
The second term is obtained from the first by replacing b with −b. Therefore,
we consider only one of the two terms. The first term is equal to
a3 + b3
a3 + b3
a3 + b3
= 2
=
= a + b.
a + ab + b2 − 2ab
a2 − ab + b2
− 2ab
a−b
a3 −b3
Hence, the second term is equal to a − b and the given number is equal to
3
(a + b) + (a − b) = (2a)3 = 8 · 252 = 2016.
Suggested Marking Scheme:
• 10 marks for a full solution.
• Proof of integrality with no evaluation: 8 marks.
Partial marks can be awarded as follows:
• 2 marks for substitutions such as a3 = 252 and b3 = 250.
• 2 marks for replacement of 251 and 1 by terms in the new variables.
• 2 marks for using the factorisation of a3 + b3 or a3 − b3 .
√
√
• 1 mark for observing that 5 3 2 = 3 250.
• Up to 3 marks if one of the two terms is simplified but not the other.
These three marks cannot be added to other partial marks.
6
10. Proposed by Jim Leahy.
Let AE be a diameter of the circumcircle of triangle ABC. Join E to the
orthocentre, H, of △ABC and extend EH to meet the circle again at D.
Prove that the nine point circle of △ABC passes through the midpoint of
HD.
Note. The nine point circle of a triangle is a circle that passes through the
midpoints of the sides, the feet of the altitudes and the midpoints of the line
segments that join the orthocentre to the vertices.
Solution
Let F be the mid pint of DH, K be the mid point of AH, L on BC the foot
of the altitude from A, and let M be the intersection point of BC and DE.
C
b
b
M
b
b
b
A
H
K
b
b
b
E
L
b
b
F
B
b
D
Because AE is a diameter of the circumcircle, 6 ACE = 6 ABE = 90◦ , so
CE is perpendicular to AC and BE is perpendicular to AB. Because BH
is on the altitude which is perpendicular to AC, we get BHkCE. Similarly,
CH is perpendicular to AB and so CHkBE. This shows that CHBE is a
parallelogram and M is the intersection point of its diagonals. Hence M is the
mid point of BC and thus is on the nine point circle.
Because AE is a diameter of the circle, 6 ADE = 90◦ . As F, K are the
mid points of DH and AH, respectively, we obtain 6 KF M = 90◦ . Because
6 KLM = 90◦ as well, the quadrilateral KF LM is cyclic.
As K, L and M are on the nine point circle, F is on that circle as well.
Suggested Marking Scheme:
• 5 marks for showing that DE meets BC at the mid point of BC.
• 4 marks for showing that KF LM is cyclic.
• 1 mark for completing the proof.
7