Download S USC’ 2000 H M

S OLUTIONS TO USC’ S 2000 H IGH S CHOOL M ATH C ONTEST
23 32x2 = 72x2 = 9800 = 23 52 72 , we obtain x2 = 52 72=32 . Since x > 0, we deduce that
x = 5 7=3 = 35=3.
1. (a) Since
2. (b) Since
log2 (log2(log2 16)) = log2(log2 4) = log2 2 = 1, the answer is 1.
3. (e) Using that
answer is 200 .
6
2600 = 64100, 3500 = 243100 , 4400 = 256100 , 5300 = 125100 , and 6200 = 36100 , we see that the
= 35811231 and observe that 0 < x < 108 . Hence,
14
10 + x 2 = 1028 + (2x)1014 + x2 = 100000071622462 1014 + x2 :
1014
1028
1028
Since 0 < x2 < 1016 , we see that the expression x2 does not effect the leading 12 digits of the integer in the numerator.
Thus, 1:000000358112312 = 1:000000716224 : : : . Hence, x = 7, y = 1, and z = 6 so that x + y + z = 14.
4. (b) Let x
5. (e) Given
(x3
x2
5x 2)(x4 + x3 + kx2 5x + 2) = x7 4x5 14x4 5x3 + 19x2 4;
the coefficient of x2 on the left is 2k + 25 2 = 2k + 23 and the coefficient of x2 on the right is 19. These must
be equal so that 2k + 23 = 19. Hence, k = 2.
6. (d) The 100 horizontal lines divide the plane into 101 regions. If each vertical line is considered in turn, each divides
2
the plane into 101 additional regions. The total number of regions obtained is
.
101 = 10201
2 = 32768
2 5 = 85 as well as 25 3 = 323. To see that N cannot be smaller,
= a5 for some positive integer a and that for 1 a 7 the
since it is 3
7. (c) The answer is 15
observe that the conditions in the problem imply N
number a5 is not a cube of a number different from a.
8. (d) Recall that
2 sin cos = sin(2). Hence,
(sin 15Æ )2(cos15Æ )2 = (sin 15Æ cos 15Æ )2 =
sin 30Æ 2 = 1 2 = 1 :
2
4
16
= BC . Since DBC is a 45-90-45 degree triangle, DB = x. Since ABC is a 30-90-60 degree triangle
tan 30 = 1=p3, we deduce that
BC
x
p1 = tan30Æ = BC
=
=
:
AB AD + DB 2 + x
3
9. (a) Let x
Æ
and
Therefore,
2 + x = xp3, from which we deduce that BC = x = 2=(p3 1) = p3 + 1.
93 = 729 and 103 = 1000, we see that p3 829 = 9+ r where 0 < r < 1. Since 102 = 100 and 103 = 1000,
log10 829 = 2 + s where 0 < s < 1. Thus,
p3
829 + log10 829 = (9 + r) + (2 + s) = 11 + (r + s) = 5:5 + t;
2
2
2
p3
where 0 < t = (r + s)=2 < 1. In other words, ( 829 + log10 829)=2 is strictly between 5:5 and 6:5. The nearest
integer is therefore 6.
10. (b) Since
we see that
4
9 = 36
( 3 0)
(3 0)
2 + = 15
( 3 0) (5 0)
2 + = 15
9 18 + 9 = 135
13 18 171 = 0
4 9 = 36
( 3 0)
11. (d) The graph of x2 y 2
is an hyperbola with vertices at
; and ; , and the graph of x2 x y2
is a circle of radius centered at ; and therefore crossing the x-axis at
; and ; . It follows that these
graphs intersect at three points.
Alternatively, if a; b is an intersection point, then a2
a b2
so that a2
a
b2
.
2
2
2
Since also a
b
, we deduce (by adding these equations) that a
a
. In other words,
a
a
. Hence, a
or a p = . Using the equation a2
b2
, we see that in the
first case b p and in the second case b
= . Therefore, there are three intersection points, ; and
= ;
= .
4
(1 0)
( )
4 9 = 36
( + 3)(13 57) = 0
=0
(57 13 16 3 13)
=1
= 3 = 57 13
= 16 3 13
=1 =0
=1
2
=1
=0
=2
4 24 + 1 = 97
=0
=2
=1
= 5 9 13 17
1
, a direct computation gives a2
, a3
, a4
, a5
, and a6
. Here, the process begins
12. (d) Using a1
to repeat as is even and a6
(just as is even and a2
). One deduces that for every positive integer k , a4k
,
a4k+1 , a4k+2 , and a4k+3 . Thus, aj
precisely when j
; ; ; ; : : : . For j , there are
such j (the last one being ).
24
=2
6
=1
100
+ = 10
=0
=
is the same as the number of intersection points of the graphs of y jxj
13. (c) The number of real roots of x4 jxj
and y
x4 . One easily deduces that these graphs intersect in exactly two points. Thus, there are a total of two
real roots.
Alternately, if x and x4 jxj
, then x4 x
so that x4 x
. This has one positive real
solution by Descartes’ Rule of Signs (and clearly is not a solution). If x < and x4 jxj
, then x4 x
so
4
4
4
4
that x x
. Since x
x
x x
, the equation x x
has exactly one negative
real root by Descartes’ Rule of Signs (alternatively, one can use that the negative real roots of x4 x
are
precisely minus the positive real roots of x4 x
). Thus, again, there are a total of two real roots.
= 10
0
10 = 0
+
= 10
+ = 10
0
( ) ( ) 10 = + 10
+ 10 = 0
+
=0
+1 = 0
+
0
10 = 0
+ = 10
10 = 0
( + 1)(
=0
+1 = 0
= 10
10 = 0
)=0
=0
, we obtain x
x y
. The latter
14. (e) Factoring the left-hand side of the equation x2 xy x y
is satisfied by all points x; y satisfying at least one of x
and x y
. In other words, the graph of
x2 xy x y
is composed of the two lines determined by the equations x
and x y
(intersecting
at
; ).
+
( 1 1)
=0
( )
15. (b) The two shaded triangles are similar, and the ratio of the length of each side of the larger triangle to the length of
the corresponding side of the smaller triangle is . It follows that the height of the larger triangle to base AD divided
by the height of the smaller triangle to base BC is . Since the sum of the two heights is , one deduces that the height
of the larger triangle is = and the height of the smaller triangle is = . It follows that the sum of the two areas is
3
15 4
3
54
1 3 15 + 1 1 5 = 50 = 25 = 6:25:
2
4 2
4 8 4
5
16. (c) Let
f (x) = (1 + x)20 + x(1 + x)19 + x2 (1 + x)18 + + x18 (1 + x)2 :
(x) does not change if we multiply it by 1 = (1 + x) x. Thus,
f (x) = (1 + x) x (1 + x)20 + x(1 + x)19 + x2 (1 + x)18 + + x18 (1 + x)2
= (1 + x)21 x19(1 + x)2:
The coefficient of x18 when the expression x19 (1+ x)2 is expanded is clearly 0 (there is no x18 term), so the coefficient
of x18 in f (x) is simply the coefficient of x18 in (1 + x)21 . By the binomial theorem, this is
21 = 21 20 19 = 7 10 19 = 1330:
18
6
Alternatively,
can use a more direct approach. The coefficient of x18 when the expression x20 k (1 + x)k
one k
k(k 1)
is expanded is
=
k 2
2 where 2 k 20. The answer can be obtained by summing the numbers
P
k(k 1)=2 from k = 2 to k = 20. This sum can be done more efficiently by using known formulas for nk=1 k
The polynomial f
Pn
2
n
k=1 k . (By the way, if you do not know these formulas, you might consider taking f (x) = (1 + x) + x(1 +
1
2
n
2
n
1
n
) + x (1 + x) + + x (1 + x) + x and using the first argument above to derive them by looking at the
coefficients of xn 1 and xn 2 .)
and
xn
17. (e) Say the cowboys names are Bob, Dave, Kevin, and Michael. Suppose exactly two of them were shot. There are
4
ways of choosing two of them, each equally likely. Say the two shot were Dave and Kevin. The probability
2
that any one cowboy shot some other given cowboy is = . Since Dave and Kevin were the only two shot, Kevin must
have shot Dave and Dave must have shot Kevin and each of these occurred with probability = . Michael could have
shot Dave or Kevin, and he shot one of them with probability = . Similarly, Bob shot Dave or Kevin with probability
= . Therefore, the probability that Dave and Kevin both were shot and no one else was is
=6
13
13
23
23
1122= 4:
3 3 3 3 81
Taking into account that there are
exactly two cowboys being shot is
6 ways of choosing two of the four cowboys, we deduce that the probability of
6 4=81 = 8=27.
18. (d) Observe that
2N = 3N N
= (k1 106 + k2 105 + + k7 ) (k1 + k2 10 + + k7 106)
= (106 1)(k1 k7 ) + (105 10)(k2 k6 ) + (104 102 )(k3 k5 ):
Since 10k 10` is divisible by 9 for any non-negative integers k and `, we see that the last expression above is divisible
by 9. It follows that 2N and, hence, N must be divisible by 9. One checks that the only choice of answers divisible by
9 is 71053290. Note that this does not prove that there are kj as in the problem for N = 71053290 (such a proof is not
required). For N = 71053290, one can take k1 = 210, k2 = 28, k3 = 28, k4 = 70, k5 = 98, k6 = 0, and k7 = 70.
19. (c) The inequalities
p
n+1
pn < p 1
4n + 1
p
< n
p
n
1
imply
24
X
n=1
24
X
pn pn 1
4n + 1 n=1
p p
p p
p p
p p
p p
= 1 0 + 2 1 + 3 2 + 4 3 + + 24 23
p
p 1
<
= 24 < 5
and
24
X
n=1
p 1
4n + 1
4
>
24
X
n=1
p
p
n+1
pn
p
p
p
p
= 2 1 + 3 2 + 4
p p
= 25 1 = 5 1 = 4:
p
p
3 + 5
p
p
4 + + 25
p
Hence, the answer is .
= (x; y), we see that
d = (x + 5)2 + (y + 1)2 + (x + 1)2 + y2 + (x 1)2 + (y 2)2 + (x 1)2 + (y 3)2
= 4x2 + 8x + 4y2 8y + 42
= 4(x + 1)2 + 4(y 1)2 + 34:
20. (b) Setting P
24
( + 1)
0 for all x with equality precisely when x = 1 and since (y 1)2 0 for all y with equality
= 1, the minimum value of d is 34 (obtained when x = 1 and y = 1).
2
Since x
precisely when y
log (5 2) = 3
25
( ) = 5 +2
()
()
( 100) 0 (2 5) 0 (1) 0
(100) 0
( 100 2 5) (2 5 1) (1 100)
()
()
25
21. (b) By the definition of the logarithm, the solutions of
for x > = correspond to x for which
x x
x
x3 . Thus, we consider roots of the polynomial f x x3 x . Since f x is a cubic, f x has at most
roots. One easily checks that f
< , f = > , f < , and f
> . This implies that the graph of
y f x crosses the x-axis somewhere in each of the intervals
; = , = ; , and ;
. Hence, each of
these three intervals contains a root of f x . It follows that f x has exactly two roots > = .
5
3
2=
= ()
22. (d) The conditions in the problem imply
x3
a)(x b)(x c) = x3
Hence, a + b + c = 0. Also, since a, b, and c are roots of x3 64x
64x 14 = (x
a3 = 64a + 14;
b3 = 64b + 14;
(a + b + c)x2 + 14, we obtain
and
c3 = 64c + 14:
abc:
Thus,
a3 + b3 + c3 = 64(a + b + c) + 3 14 = 64 0 + 3 14 = 42:
2
23. (d) Multiplying the original equation by xy , we wish to determine the number of non-zero integer pairs x; y such
y xy. This equation is equivalent to x
y
. Thus, x
must be plus or minus a divisor
that x
of different from
. Such a choice, say d, for x
uniquely determines x (namely, x
d ) and uniquely
determines y (since y
=d, we have y
=d
). The number of choices for x
is (they are , , and
). Therefore, the answer is .
4
4
2 +2 =
2
2=4
5
(
2)(
2
= (4 ) + 2
2) = 4
24. (a) It suffices to compute the ratio of the area of AED to the area of
Æ and
from vertex D . The conditions in the problem imply \ACD
Æ
and \ADC
. By the Law of Sines,
= 30
= 120
so that AC
2
= +2
2 5
12
ECD. Let h denote their common height
ACD is isosceles. Hence, \CAD = 30Æ
CD
CD
pAC = sinAC
120Æ = sin 30Æ = 1=2
3=2
= CDp3. Thus,
area of
area of
AED = 12 h(AC CD) = CDp3 CD = p3 1:
1 hCD
ECD
CD
2
25. (a) Observe that when the product
1 + 21 + 212 + 213 + 1 + 13 + 312 + 313 + 2 3
1 (1 (1 3)) = 3 2
is expanded, one obtains the sum of the reciprocals of all numbers of the form k ` where k and ` are non-negative
integers. This then is precisely the sum asked for in the problem. Each of the factors in the product above is a
geometric series, the first equals =
=
and the second equals =
=
= . Therefore, the
answer is =
. (Note that the product of two converging series cannot always be multiplied together term-byterm and rearranged as above; however, if the series are converging geometric series this can be done. In fact, as long
as the series are converging series with positive terms, such term-by-term multiplication and rearrangement does not
alter the value of the product.)
1 (1 (1 2)) = 2
2 3 2=3
26. (b) Observe that
1 + p5 3 = 2 + p5
2
and
1 p5 3 = 2 p5:
2
Hence,
p
p
1
1
+
5
2 + 5 + 2 5 = 2 + 2 5 = 1:
p
p p
p
3
3
A more
p direct papproach can be given as follows. Set = 2 + 5 + 2 5.
(2 + 5)(2 5) = 1, we obtain
p
q
3
3 =
3
3
p
q
2+ 5 +3
p
= (2 + 5) + 3
p
q
q
3
3
3
2 q
2+ 5
p
p
q
3
p
2
q
2 + 5 ( 1) + 3
3
2
q
5 +3
p
3
p
q
2+ 5
3
p
5 ( 1) + (2
2
Clearly,
p
is real.
2
5 +
q
3
2
Noting that
p
5
3
5) = 4 3:
+ 3x 4. Observe that 1 is clearly a root. In particular, x 1 is a factor of x3 + 3x 4. Since
1)(x2 + x + 4) and the roots of x2 + x + 4 are imaginary (by the quadratic formula), we deduce
Thus, is a root of x3
x3 + 3x 4 = (x
that = 1.
5
7
27. (c) We consider the points along the diameter and the points along the half-circular arc separately. Two line
segments can intersect at an interior point of the semi-circle in each of the following three ways: (i) the line segments
have all endpoints on the arc, (ii) each line segment has one endpoint on the diameter and one endpoint on the arc,
and (iii) one line segment has both endpoints on the arc and the other has endpoint on the diameter and on the arc.
In each case, a unique intersection point is determined by the endpoints of the two line segments (in other words,
if one knows these endpoints, then the two line segments can be drawn in exactly
one way so as to intersect at an
interior point of the semi-circle). In case (i), the endpoints can be selected in 74
ways. Thus, (i) gives rise to
interior points. In case (ii), the endpoints can be selected in 72 52
ways, producing
additional interior points. In case (iii), the endpoints can be selected in 73 51
ways, producing
more interior points. Hence, there are a total of
interior points where line segments intersect.
4
4
35
4
175
1
1
4
4
4
35+210+175 = 420
1
= 35
= 21 10 = 210
= 35 5 = 175
4
4
210
=
28. (b) For j , let Sj denote the set of positive integers consisting of j digits in base . For example, S1
4 . Observe that as m
f ; ; g. The numbers in S1 [ S2 [ S3 [ S4 are precisely the numbers from to
varies over the elements in Sj , the value of s m varies over products of the form d1 d2 dj where each dj is a base
digit (i.e., some number from the set f ; ; ; g). We deduce that
123
4
( )
0123
X
s(m) = (0 + 1 + 2 + 3)j = 6j :
1 255 = 4
1
m2Sj
Thus,
255
X
m=1
s(m) = 6 + 62 + 63 + 64 = 6 (1 + 6 + 36 + 216) = 6 259 = 1554:
=
=
29. (d) Drop a perpendicular from P to side AB , and call the intersection point R. Set AR x and BR y . Similarly,
drop a perpendicular from P to side AD, and call the intersection point S . Set AS u, and DS v . The conditions
in the problem imply
=
y2 + u2 = P B 2 = 42 = 16;
u2 + x2 = P A2 = 82 = 64;
and
=
x2 + v2 = P D2 = 72 = 49:
Hence,
P C 2 = y2 + v2 = (y2 + u2 ) + (x2 + v2 )
Therefore, P C
= 1.
(u2 + x2) = 16 + 49 64 = 1:
C
30. (a) Let X , Y , and Z be the vertices of the shaded triangle as shown.
Observe that CEX , AF Y , and BDZ all have the same area, say
. Similarly, quadrilateral EAY X , quadrilateral F BZY , and quadrilateral DCXZ all have the same area, say . Let denote the area of
XY Z . Since CF B and CAF have the same altitude drawn from
C and since F B AF , we deduce that the area of CF B is twice the
area of CAF . Hence, which implies .
Now, we see that the areas of AF Y and CF A are exactly one third
of the areas of XY Z and ABC , respectively. Therefore, the ratio of
the area of XY Z to the area of ABC is equal to the ratio of the area
of AF Y to the area of CF A. We compute the latter. Observe first
that AF Y and CF A are similar (since \AF Y
\CF A and \F AY
=2
E
X
Z D
+2 + = 2(2 + )
=3
Y
A
B
F
=
= \F CA). By the Law of Cosines,
CF 2 = AC 2 + AF 2 2 AC AF cos \CAF
= (3AF )2 + AF 2 2(3AF )AF cos 60Æ
= 7AF 2:
Since the ratio of the squares of the lengths of two corresponding sides of similar triangles is the ratio of the areas of
the triangles, we deduce that the ratio of the area of AF Y to the area of CF A is = .
17