Download 1 - Art of Problem Solving

1. 1. Determine all solutions to the system of equations
x+y+z=2
x2 − y2 − z2 = 2
x − 3y2 + z = 0
Let a be x+z
From equation 1) a+y=2
From equation 2) a-3y²=0
Subtracting the LS and RS of the second from those of the first
3y²+y-2=0
(3y-2)(y+1)=0
y=2/3 or y= -1
Case 1: y=2/3
Substituting y=2/3 into equation 1) 2/3+a=2
a=4/3
Factoring equation 2) -y²+a(x-z)=2
-4/9 + 4/3(x-z)=2
x -z=11/6
From our definition of a
x+z=4/3
Adding the previous two equations
2x=19/6
x=19/12
Substituting that into equation 1)
19/ +2/ +z=2
12
3
1
z= - /4
This gives us one solution to (x,y,z) of (19/12, 2/3, -1/4)
Case 2: y= -1
Substituting y= -1 into equation 1) -1+a=2
a=3
Factoring equation 2)-y²+a(x-z)=2
-1+3(x-z)=2
x-z=1
From our definition of a
x+z=3
2x=4
x=2
Substituting that into equation 1)
2-1+z=2
z=1
This gives us our second solution of (2, -1, 1)
Therefore, the solutions (x,y,z) to the three equations are (19/12, 2/3, -1/4), and (2, 1, 1).
2. Triangle ABC is right-angled at C with AC = b and BC = a. If d is the
length of the altitude from
C to AB, prove that 1/a2 + 1/b2 = 1/d2.
Let T be the area of the triangle.
T=bh/2
In one case, the base is b, and the height is a, so T=ab/2.
In another case, the base can be c, and the height is thus d, so T=cd/2.
Since ab/2=T=cd/2
ab=cd
Squaring both sides, we find that (ab)²=(cd)².
Divide both sides by (abd)² [none of a, b, or d are 0, so it is permissible]
This results in 1/d2=c²/a²b²
Due to the pythagorean theorem, c²=a²+b².
1/ 2=(a²+b²)/a²b²
d
By applying the distributive property, we arrive at
1/ 2=1/a²+1/b²
d
which is the desired identity.
3. Prove that there does not exist a polynomial f(x) with integer coefficients
for which f(2008) = 0
and f(2010) = 1867.
Let a polynomial f(x) exist such that f(2008)=0 and f(2010)=1867.
This polynomial is of the form anxm + an-1 xm-1 + an-2 xm-2 ...a2x2 + a1x1 + a0.
If that polynomial exists, a similar polynomial f(x+2008) would exist, which would
be congruent to f(x), but shifted 2008 units to the left.
Let this new polynomial be g(x).
g(0)=f(2008)=0
g(0)=an0m + an-1 0m-1 + an-2 0m-2 ...a202 + a101 + a0.
0=0+a0.
a0=0.
Thus, g(x)=anxm + an-1 xm-1 + an-2 xm-2 ...a2x2 + a1x1
g(2)=f(2010)=1867.
1867=an2m + an-1 2m-1 + an-2 2m-2 ...a222 + a121.
Factoring out a two on the right side of the equation
1867=2(an2m-1 + an-1 2m-2 + an-2 2m-3 ...a22 + a1)
Every term on the inside of the bracket must be an integer, thus the sum of them
must be an integer.
The sum would have to be equal to 1867/2 for the equation to hold. According to
the above statement that cannot be true, thus g(x) does not exist, and thus f(x)
does not exist.
4. Three fair six-sided dice are thrown. Determine the probability that the
sum of the numbers on the
three top faces is 6.
Let P(A) be the desired probability.
P(A)=(The number of ways the sum of the numbers on the top three faces can be
six)/(The number of ways three fair six-sided dice can be thrown)
P(A)=10/63
P(A)=5/108
5. Determine all positive integers n for which n(n + 9) is a perfect square.
Since n cannot be 0, n(n+9) is equivalent to n²[(n+9)/n].
For that to be a perfect square integer, either both n² and (n+9)/n must be perfect
squares, or (n+9) and n must both be perfect squares.
For the former case, the latter term factors to 1+9/n, so n would have to be an
integer factor of 9.
Applying all three integer factors of 9 to the original equation
1 produces 10 (not a perfect square)
9 produces 162 (not a perfect square)
3 produces 36 (a perfect square, and thus one of the solutions.
For the latter case, for n+9 and n to be perfect squares, n must either be 0 or 16
(since perfect squares differ by increasing odd numbers, so 0+1+3+5=0+9
produces 9, and 16+9=25).
Since n is positive, 0 is disallowed, but n=16 works.
Our final solution set is
n={3, 16}
6. Triangle ABC is right-angled at C. AQ is drawn parallel to BC with Q and
B on opposite sides of AC
so that when BQ is drawn, intersecting AC at P, we have PQ = 2AB. Prove
that <ABC = 3<PBC.
A basic fact regarding right triangles is that in all right triangles, the median on
the hypotenuse is half of the hypotenuse.
Let <PBC=z.
Let M be the midpoint of PQ.
Let x be the length of AB.
Observe the diagram.
Note that since AQ is parallel to BC, due to the Alternate Angle Theorem,
<PAQ=<ABC=90 degrees.
A line segment connecting M and A has been drawn. This being the median on
the hypotenuse, it is equal to half the hypotenuse.
Therefore, MA=x.
Due to the Alternate Angle Theorem, <AQM=<PBC=z
Because ΔAQM is isosceles, <MAQ=<AQM=z
Due to the external angle theorem, <AMP=2z
Because ΔABM is isosceles, <ABP=<AMP=2z
<ABC=<PBC+<ABP.
<ABC=<PBC+2z
<ABC=<PBC+2<PBC
<ABC=3<PBC, the desired equation.
7. A rectangular sheet of paper is folded so that two diagonally opposite
corners come together. If the
crease formed is the same length as the longer side of the sheet, what is
the ratio of the longer side
of the sheet to the shorter side?
Note that the crease is denoted by the line EF.
From ΔEFG, cosx=W/L
From ΔDBC, tanx=W/L
cosx=tanx
a/h=o/a
This is a case of the kepler triangle, and as such the two non-hypotenuse sides
are in the ratio of root(phi):1, or [(1+51/2)/2]1/2:1.
{ To go into this slightly further, the three sides of the triangle must be in the ratio
1 : x : x² (x/1=x, x²/x=x), and since it's a right triangle, 1²+x²=x4
x4-x2-1=0
x²=[1+(5)1/2]/2
Since x must be positive, x=[(1+51/2)/2]1/2
x:1 is the ratio mentioned above. }
8. Determine an infinite family of quadruples (a, b, c, d) of positive integers,
each of which is a solution to a4 + b5 + c6 = d7.
I conjecture that (266+105n, 253+84n, 244+70n, 238+60n) for any non-negative integer n
provides an infinite family of solutions as n varies. Since allowing n to be
fractional or negative might make a, b, c, or d a non-integer real, they must be
disallowed. To demonstrate that this is true, let us substitute these values into the
equation:
L.S. =(266+105n))4 + (253+84n)5 + (244+70n)6
=2264+420n + 2265+420n + 2264+420n
=2420n[2264 + 2264 + 2265]
=2420n[2(2264) + 2265]
=2420n[2265 + 2265]
=2420n[2(2265)]
=2420n(2266)
R.S. = (238+60n)7
= 2266+420n
=2420n(2266)
RS=LS, thus the equation holds true for any non-negative integer value n.
9. Suppose that m and k are positive integers. Determine the number of
sequences x1, x2, x3, . . . , xm−1, xm
with
xi an integer for i = 1, 2, 3, . . . ,m,
1 ≤ xi ≤ k for i = 1, 2, 3, . . . ,m,
x1 ≠ xm, and
no two consecutive terms equal.
Let the function f(m) be the number of sequences that satisfy all the above
conditions for any given m.
Let the function g(m) be the number of sequences that satisfy all the above
conditions, ignoring the third (i.e. the first and last terms are permitted to be the
same)
Let us find an equation for g(m) first. Assuming the first term has k possibilities,
each successive term may have k-1 possibilities since the value for the previous
box is disallowed.
Therefore, g(m)=k(k-1)m-1
Now, there are two possibilities for a sequence that satisfies all four conditions either the first and second-last term are equal, or they are not.
In the case of the former, there are k-1 possibilities for the last term. If the last
term is removed from these sequences, we are left with all sequences of m-1
terms that are included in g but not in f. The number of these sequences is thus
g(m-1) - f(m-1).
In the case of the latter, there are k-2 possibilities for the last term.
f(m) is the sum of these two cases.
f(m)=(k-1)*[g(m-1) - f(m-1)] + (k-2)*f(m-1)
Substituting in for the known value of g(m),
f(m)=(k-1)*[k(k-1)m-2 - f(m-1)] + (k-2)*f(m-1)
f(m)=k(k-1)m-1 - (k-1) * f(m-1) + (k-2) * f(m-1)
f(m)=k(k-1)m-1 + (k-2-k+1) * f(m-1)
f(m)=k(k-1)m-1 - f(m-1)
We observe this as a geometric series for the geometric sequence of the form ar n
where a=+k(k-1), r= -(k-1), and n=(m-1), going from 2->n [f(1)=0]. The reason
that the a value can be positive or negative is that the largest term must be
positive. As such, when m is odd, k is positive. When m is even, k is negative.
Thus, the closed form solution for f(m) is
If m is odd:
k[(1-k)m - (1-k)]/(-k)
= - (1-k)m -k+1
If m is even:
-k[(1-k)m - (1-k)]/(-k)
= (1-k)m + k - 1
This leaves us with
f(m)= / (k-1)m - k +1
\ (k-1)m + k - 1
m mod 2=1
m mod 2=0
This further simplifies to our answer,
f(m)=(k-1)m +(k-1)*[(-1)m]
10. Ten boxes are arranged in a circle. Each box initially contains a positive
number of golf balls. A move consists of taking all of the golf balls from
one of the boxes and placing them into the boxes that follow it in a
counterclockwise direction, putting one ball into each box. Prove that if the
next move always starts with the box where the last ball of the previous
move was placed, then after some number of moves, we get back to the
initial distribution of golf balls in the boxes.
Let a configuration be a distribution of golf balls, as well as a specific starting
box.
As well, let it be noted that if one follows a standard ball-placing procedue with no
random factors, starting from a given position, only one possible outcome can
exist.
1 Any given move will lead to a unique configuration.
Let there be a new type of move, called an anti-move.
An anti-move functions by picking up one ball from the starting box, and then
picking up one ball from each of the boxes following it clockwise, going around as
many times as necessary until an empty space is reached, then all the picked-up
balls are placed into that box. This is the inverse of a move, so effectively, this
undoes a normal move.
2 Each anti-move also leads to a unique configuration. (For the same reason as
1)
This means that there is a unique configuration (due to 1) A that leads to a
configuration B with a move, and thus B leads to A (and only A, because of 2)
with an anti-move. As such, no other configuration C could lead to B with a move,
because B would then have to lead to C with an anti-move, which is impossible.
Because only A can lead to B, to get to B, one would have to get to A. If B->C,
then to get to C, one would have to get to B, thus one would have to get to A,
and so on and so forth.
Thus, from a starting position, one performs a move.
Each time a move is performed, one of two outcomes is achieved:
A) The original position is achieved, a satisfactory ending.
B) The original position is not achieved and a new configuration is achieved (an
old configuration cannot be achieved, because it would have had to go through
the original position). Since this is unsatisfactory, perform a move, and the result
will follow either A) or B).
This loop must eventually end because there are a finite number of
configurations, so eventually there will be no more 'old' configurations to achieve,
and thus the original position must be the next move. Therefore, for any given
configuration of golfballs, after some number of moves, the original position will
eventually be reached.