Download The University of Melbourne–Department of Mathematics and

The University of Melbourne–Department of Mathematics and
Statistics
School Mathematics Competition, 2015
INTERMEDIATE DIVISION: SOLUTIONS
Time allowed: Three hours
These questions are designed to test your ability to analyse a problem and to express yourself
clearly and accurately. The following suggestions are made for your guidance:
(1) Considerable weight will be attached by the examiners to the method of presentation of
a solution. Candidates should state as clearly as they can the reasoning by which they
arrived at their results. In addition, more credit will be given for an elegant than for a
clumsy solution.
(2) The seven questions are not of equal length or difficulty. Generally, the later
questions are more difficult than the earlier questions.
(3) It may be necessary to spend considerable time on a problem before any real progress is
made.
(4) You may need to do considerable rough work but you should then write out your final
solution neatly, stating your arguments carefully.
(5) Credit will be given for partial solutions; however a good answer to one question will
normally gain you more credit than sketchy attempts at several questions.
Textbooks, electronic calculators and computers are NOT allowed. Otherwise normal examination conditions apply.
(1) Mia buys three sizes of pens from the shop. Large pens cost $1.05 each, medium pens
cost $1.00 each and small pens 95 cents each. Mia spends $20.15 and buys more than
one of each size of pen. What is the total number of pens that she buys?
Solution. Mia buys approximately 20 pens, so the answer will be close to 20. Let L, M
and S be the number of large, medium and small pens that Mia buys.
105L + 100M + 95S = 2015
⇔
21L + 20M + 19S = 403
Suppose Mia buys 20 pens, i.e. L + M + S = 20. Then
3 = 403 − 400 = 21L + 20M + 19S − 20(L + M + S) = L − S
Then (L, M, S) = (n + 3, 17 − 2n, n) for n = 2, ..., 7 (e.g. (L, M, S) = (5, 13, 2)) gives
solutions with a total of 20 pens.
To show that Mia must buy 20 pens, assume that she buys ≤ 19 pens. Then
21L + 20M + 19S ≤ 21(L + M + S) ≤ 21 × 19 = 399 < 403
so Mia has not spent enough. Similarly, assume that she buys ≥ 21 pens. Then
21L + 20M + 19S ≥ 21 × 2 + 20 × 2 + 19(L − 2 + M − 2 + S) ≥ 42 + 40 + 19 × 17 = 405 > 403
where we have used the fact that L and M are at least 2, and we see Mia has spent too
much.
(2) Nina noticed that the sum of the squares of the digits of her (two-digit) age is equal to
her age plus the (positive) difference between the two digits. How old is Nina?
Solution. Suppose Nina is 10a + b years old where 0 < a < 10 and 0 ≤ b < 10. If a ≥ b
then we would have
a2 + b2 = 10a + b + a − b = 11a
⇔
b2 = a(11 − a).
For a = 1, 2, 3, 4, 5, a(11 − a) = 10, 18, 24, 30 is not a square. (And a = 6, 7, 8, 9 gives
the same as a = 5, 4, 3, 2.)
If a < b then we would have
a2 + b2 = 10a + b + b − a = 9a + 2b
⇔
(b − 1)2 = a(9 − a) + 1.
For a = 1, 2, 3, 4, a(9 − a) + 1 = 9, 15, 19, 21 (and a = 5, 6, 7, 8, 9 gives the same as
a = 4, 3, 2, 1, 0.) The only square is 9 so a = 1 and (b − 1)2 = 9 i.e. b = 4. Hence Nina
is 14 and
12 + 42 = 14 + 4 − 1.
(3) What is the maximum possible area of a triangle with one side of length 12 and the
second side length three times the third side length?
h a
x
3a
12
Solution. The area is A = 6h where h is the height drawn in the picture. (The
assumption that the triangle is obtuse where sides of length a and 12 meet is correct,
but one can easily deal with the other case.)
(i) h2 + x2 = a2 ,
(ii) h2 + (12 + x)2 = 9a2 ,
1
(i) − (ii) ⇒ x = a2 − 6
3
substitute x into (i) to get:
1
1
1
h = − a4 + 5a2 − 36 = − (a2 − 9)(a2 − 36) = −
9
9
9
2
9 + 36
a −
2
2
2
+
81
81
≤
4
4
hence the maximum height is h = 9/2 and the maximum area is A = 6h = 27.
(4) Find the smallest positive multiple of 45 that has digits consisting of only 0s and 1s.
Solution. The sum of the digits of any multiple of 9 is also divisible by 9. Since any
multiple of 45 is divisible by 9 the number of 1s must be one of 9, 18, etc. Any multiple
of 45 is divisible by 5 so it must end in a 5 or 0, hence 0. Thus
1111111110
is divisible by 9 and 5, hence divisible by 45. It is the smallest such number because
any other number consisting of only 0s and 1s with fewer than 9 1s is not divisible by
9, or with ten digits and 9 1s would end in 1 and not be divisible by 5.
(5) Theo picked up his grandfather’s watch and was very curious to see the circular face
with hour, minute and second hands moving to indicate
the time. He noticed that at twelve o’clock the three
11 12 1
10
2
hands of the clock point in the same direction. At what
9
3
time up to the nearest minute, not within five minutes
8
4
of twelve o’clock, are the three hands closest? In other
7 6 5
words, when is the sum of the two smaller angles between
hands smallest?
Solution. The hour and minute hands meet 10 times between 12 o’clock and 12 o’clock.
It is enough to consider only half these meetings, since by symmetry the same angles
occur when time goes backwards. The problem is equivalent to determining the time
when the hour and minute hands meet such that the second hand is closest to the hour
and minute hands, since one can move forward or backward by less than a minute so
that the second hand is close to the other two. The hour and minute hands meet between 1 and 2 when the minute hand points to m minutes and the hour hand, which
1
m
moves at 12
th the speed, has moved 12
minutes from 1. Hence
m
5
⇒ m=5
12
11
5
and the second hand is a little less than 11 around the clock from the other two hands.
Similarly, between 2 and 3
m
10
m − 10 =
⇒ m = 10 ,
12
11
between 3 and 4
4
m
⇒ m = 16 ,
m − 15 =
12
11
between 4 and 5
m
9
m − 20 =
⇒ m = 21 .
12
11
4
Between 3 and 4 the second hand is closest to the other two hands since 11
of a minute
occurs at approximately 22 seconds which is about 5 seconds clockwise from the other
two hands. (All others occur at least 15 seconds away.) Moving the second hand back
until it meets the hour hand will give the minimum angle and it occurs within a minute
of 16 minutes past 3 or 3:16. By symmetry it also occurs at about 8:44.
m−5=
(6) The numbers 1 to 5 in the left diagram satisfy conditions (i) and (ii) below. Is it possible
to put the numbers 1 to 9 in the middle diagram and 1 to 14 in the right diagram, each
satisfying (i) and (ii)?
5
1
4
3
4
4
2
3
5
2
6
9
1
7
9
8
(i) each number is used at least once;
(ii) as you travel clockwise around each small triangle the numbers decrease only once.
Solution. Each small triangle contains exactly one edge where it decreases as you travel
around it clockwise. If we fill in a hexagon, obtained from the middle example above
by removing the three vertices of the largest triangle, each internal edge of the hexagon
must be decreasing for one of the neighbouring triangles unless its two vertices are given
the same number. This can happen exactly once and in the example above we used 9
twice. If we had not used a number twice in the hexagon, then as we travel clockwise
around the boundary, the boundary vertices would be strictly increasing which contradicts arriving back where we started. Hence, we need to use a repeated number in the
interior of any hexagon. But there are three hexagons in the third diagram, and they
do not share an internal edge. Hence we cannot fill this hexagon with the numbers 1
to 14 satisfying (i) and (ii). Alternatively, there are 18 interior edges in the third diagram, hence 17 decreasing edges (if we use the repeated number). But there are only 16
small triangles so at one of them must have at least 2 decreasing edges which violates (ii).
(7) The three points A, B and C in the diagram are vertices of an equilateral triangle.
Given any point P on the circle containing A, B and C, consider the three distances
AP , BP and CP . Prove that the sum of the two shorter distances gives the longer
distance.
•P
A•
Q
•
•B
•
C
Solution. The line AB intersects P C at the point Q. We have similar triangles
4AP C ∼ 4QAC
because ∠AP C =
π
3
= ∠QAC and ∠P CA = ∠ACQ. Hence
AP
QA
=
PC
AC
Similarly BP/P C = QB/BC from the similar triangles 4BP C ∼ 4QBC. Hence
AP + BP
QA QB
QA + QB
AB
=
+
=
=
=1
PC
AC BC
AC
AC
since AB = BC = AC. So we have shown AP + BP = P C.