Harvard-MIT Mathematics Tournament
February12,2011
2011Solutions Packet
More packets available at registration desk
14th Annual Harvard-MIT Mathematics Tournament
Saturday12February2011
Algebra&Calculus Individual Test
1.Let a,b,and c be positive real numbers.Determine the largest total number of real roots that the
following three polynomials may have among them:ax2+bx+c,bx2+cx+a,and cx2+ax+b.
Answer:4If all the polynomials had real roots,their discriminants would all be nonnegative:a2≥4bc,b2≥4ca,and c2≥4ab.Multiplying the inequalities gives(abc)2≥64(abc)2,a contradiction.
Hence one of the quadratics has no real roots.The maximum of4real roots is attainable:for example,
the values(a,b,c)=(1,5,6)give−2,−3as roots to x2+5x+6and−1
2,−1
3
as roots to6x2+5x+1.
2.Josh takes a walk on a rectangular grid of n rows and3columns,starting from the bottom left corner.
At each step,he can either move one square to the right or simultaneously move one square to the left and one square up.In how many ways can he reach the center square of the topmost row?
Answer:2n−1Note that Josh must pass through the center square of each row.There are2ways to get square of row k to the center square of row k+1.So there are2n−1ways to get to the center square of row n.
3.Let f:R→R be a differentiable function such that f(0)=0,f(1)=1,and|f (x)|≤2for all real
numbers x.If a and b are real numbers such that the t of possible values of
1joinin
0f(x)dx is the open
interval(a,b),determine b−a.
Answer:3
4
Draw lines of slope±2passing through(0,0)and(1,1).The form a parallelogram with vertices(0,0),(.75,1.5),(1,1),(.25,−.5).By the mean value theorem,no point of(x,f(x))lies outside
this parallelogram,but we can construct functions arbitrarily clo to the top or the bottom of the parallelogram while satisfying the condition of the problem.So(b−a)is the area of this parallelogram,
which is3
4
.
4.Let ABC be a triangle such that AB=7,and let the angle bictor of∠BAC interct line BC at
D.If there exist points E and F on sides AC and BC,respectively,such that lines AD and EF
are parallel and divide triangle ABC into three parts of equal area,determine the number of possible integral values for BC.
Answer:13
A B
反省英文C D
E
F 7
14
Note that such E,F exist if and only if
[ADC]
[ADB]
www listeningexpress com
=2.(1) ([]denotes area.)Since AD is the angle bictor,and the ratio of areas of triangles with equal height is the ratio of their bas,
AC AB =
DC
DB
=
[ADC]
[ADB]
.
Hence(1)is equivalent to AC=2AB=14.Then BC can be any length d such that the triangle inequalities are satisfied:
d+7>14
7+14>d
Hence7<d<21and there are13possible integral values for BC.
5.Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep a
running tally of the sum of the results of all rolls made.A player wins if,after he rolls,the number on the running tally is a multiple of7.Play continues until either player wins,or el indefinitely.If Nathaniel goesfirst,determine the probability that he ends up winning.
Answer:5
quart11
For1≤k≤6,let x k be the probability that the current player,say A,will win when the number on the tally at the beginning of his turn is k modulo7.The probability that the total
is l modulo7after his roll is1
6for each l≡k(mod7);in particular,there is a1
6
chance he winsgrayscale
immediately.The chance that A will win if he leaves l on the board after his turn is1−x l.Hence for
1≤k≤6,
x k=1
6
1≤l≤6,l=k
(1−x l)+
1
6
.
Letting s= 6
l=1
x l,this becomes x k=x k−s
6
+1or5x k
6
宇航员的英文=−s
6
+1.Hence x1=···=x6,and6x k=s
for every k.Plugging this in gives11x k
6=1,or x k=6
11
.
Since Nathaniel cannot win on hisfirst turn,he leaves Obediah with a number not divisible by7.
Hence Obediah’s chance of winning is6
热血教师电影11and Nathaniel’s chance of winning is5
11
.
6.Let a b=ab+a+b for all integers a and b.Evaluate1 (2 (3 (4 ...(99 100)...))).
Answer:101!−1We willfirst show that is both commutative and associative.
•Commutativity:a b=ab+a+b=b a
•Associativity:a (b c)=a(bc+b+c)+a+bc+b+c=abc+ab+ac+bc+a+b+c and
(a b) c=(ab+a+b)c+ab+a+b+c=abc+ab+ac+bc+a+b+c.So a (b c)=(a b) c.
So we need only calculate((...(1 2) 3) 4)... 100).We will prove by induction that
((...(1 2) 3) 4)... n)=(n+1)!−1.
•Ba ca(n=2):(1 2)=2+1+2=5=3!−1
nurryrhyme
•Inductive step:
Suppo that
(((...(1 2) 3) 4)... n)=(n+1)!−1.
Then,
((((...(1 2) 3) 4)... n) (n+1))=((n+1)!−1) (n+1)
=(n+1)!(n+1)−(n+1)+(n+1)!−1+(n+1)
=(n+2)!−1
Hence,((...(1 2) 3) 4)... n)=(n+1)!−1for all n.For n=100,this results to101!−1.
7.Let f:[0,1)→R be a function that satisfies the following condition:if
x=
∞
n=1
a n
10
=.
四级算分器
is the decimal expansion of x and there does not exist a positive integer k such that a n=9for all
柒怎么读
n≥k,then
f(x)=
∞
n=1
a n
102n
.
Determine f 1 3
.
Answer:0Note that1
3=
∞
n=1
3
10n
.
Clearly f is an increasing function.Also for any integer n≥1,we e from decimal expansions that
f(1
3±1
10n
)−f(1
3
)
=±1
102n
.
Consider h such that10−n−1≤|h|<10−n.The two properties of f outlined above show that|f(1
3
+
h)−f(1
3)|<1
102n
.And from|1
h
|≤10n+1,we get
f(13+h)−f(13)
h
<110n−1.Taking n→∞gives h→0
and f (1
3)=lim n→∞1
10n−1
=0.
8.Find all integers x such that2x2+x−6is a positive integral power of a prime positive integer.
Answer:2,5Let f(x)=2x2+x−6=(2x−3)(x+2).Suppo a positive integer a divides both 2x−3 2.Then a must also divide2(x+2)−(2x−3)=7.Hence,a can either be1or7.As
a result,2x−3=7n or−7n for some positive integer n.We consider the following cas:
•(2x−3)=1.Then x=2,which yields f(x)=4,a prime power.
•(2x−3)=−1.Then x=1,which yields f(x)=−3,not a prime power.
•(2x−3)=7.Then x=5,which yields f(x)=49,a prime power.
•(2x−3)=−7.Then x=−2,which yields f(x)=0,not a prime power.
•(2x−3)=±7n,for n≥2.Then,since x+2=(2x−3)+7
2
,we have that x+2is divisible by7
but not by49.Hence x+2=±7,yielding x=5,−9.The former has already been considered, while the latter yields f(x)=147.
So x can be either2or5.
9.Let ABCDEF be a regular hexagon of area1.Let M be the midpoint of DE.Let X be the
interction of AC and BM,let Y be the interction of BF and AM,and let Z be the interction of AC and BF.If[P]denotes the area of polygon P for any polygon P in the plane,evaluate [BXC]+[AY F]+[ABZ]−[MXZY].
Answer:0
