Download Preliminary Practice - Art of Problem Solving

USA Aime 1983: Problems & Solutions
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Problems
1. Let x,y, and z all exceed 1, and let w be a positive number such that logx w = 24, logy w = 40, and
logxyz w = 12. Find logz w.
2. Let f (x) = |x − p| + |x − 15| + |x − p − 15|, where p ≤ x ≤ 15. Determine the minimum value taken
by f (x) by x in the interval 0 < p < 15.
√
3. What is the product of the real roots of the equation x2 + 18x + 30 = 2 x2 + 18x + 45?
4. A machine shop cutting tool is √
in the shape of a notched circle, as
shown. The radius of the circle is 50 cm, the length of AB is 6cm,
and that of BC is 2cm. The angle ABC is a right angle. Find
the square of the distance (in centimeters) from B to the center of the circle.
5. Suppose that the sum of the squares of two complex numbers x and y is 7 and the sum of the cubes
is 10. What is the largest real value of x + y can have?
6. Let an equal 6n + 8n . Determine the remainder upon dividing a83 by 49.
7. Twenty five of King Arthur’s knights are seated at their customary round table. Three of them
are chosen - all choices being equally likely - and are sent of to slay a troublesome dragon. Let P be
the probability that at least two of the three had been sitting next to each other. If P is written as a
fraction in lowest terms, what is the sum of the numerator and the denominator?
8. What is the largest 2-digit prime factor of the integer 200
100 ?
9. Find the minimum value of
9x2 sin2 x+4
x sin x
for 0 < x < π.
10. The numbers 1447, 1005, and 1231 have something in common. Each is a four-digit number
beginning with 1 that has exactly two identical digits. How many such numbers are there?
11. The solid shown has a square base of side length s. √The upper edge is parallel to the base and
has length 2s. All edges have length s. Given that s = 6 2, what is the volume of the solid?
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12. The length of diameter AB is a two digit integer. Reversing the
digits gives the length of a perpendicular chord CD. The distance
from their intersection point H to the center O is a positive rational
number. Determine the length of AB.
13. For {1, 2, 3, . . . , n} and each of its non-empty subsets, an alternating
sum is defined as follows. Arrange the number in the subset in decreasing
order and then, beginning with the largest, alternately add and subtract
succesive numbers. For example, the alternating sum for {1, 2, 3, 6, 9} is 9 − 6 + 3 − 2 + 1 = 6 and for
{5} it is simply 5. Find the sum of all such alternating sums for n = 7.
14. In the adjoining figure, two circles with radii 6 and 8 are drawn with their centers 12 units apart.
At P , one of the points of intersection, a line is drawn in sich a way that the chords QP and P R have
equal length. (P is the midpoint of QR) Find the square of the length of QP .
15. The adjoining figure shows two intersecting chords in a circle, with B on minor arc AD. Suppose
that the radius of the circle is 5, that BC = 6, and that AD is bisected by BC. Suppose further that
AD is the only chord starting at A which is bisected by BC. It follows that the sine of the minor arc
AB is a rational number. If this fraction is expressed as a fraction m
n in lowest terms, what is the
product mn?
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Answers
1. 060
2. 015
3. 020
4. 026
5. 004
6. 035
7. 057
8. 061
9. 012
10. 432
11. 288
12. 065
13. 448
14. 130
15. 175
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Solutions
1. Instead of applying logarithm rules, we exponentiate everything so that x24 = w, y 40 = w, and
(xyz)12 = w. Now, we need to isolate z and w in order to find logz w. Notice that x120 = w5 ,
y 120 = w3 , and (xyz)120 = w10 . With a quick substitution, we get w5 w3 z 120 = w10 and logz w = 060 .
2. First, we eliminate the absolute value. With the conditions given, |x − p| = x − p, |x − 15| = 15 − x,
and |x − p − 15| = 15 + p − x. Adding these together, the sum is equal to 30 − x, of which the minimum
value is attained when x = 15. Therefore, The answer is 015 .
3. Instead of immediately squaring, we substitute y = x2 + 18x + 30 such that our expression becomes
√
y = 2 y + 15. Solving for y, y = 10 or y = −6. The second solution is extraneous, so we’ll will go
with the first. Substituting x2 + 18x + 30 back in for y and simplifying, x2 + 18x + 20 = 0. The
product of our roots is therefore 20
1 = 020 .
4. Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Extend
a perpendicular from O to AB and label it D. Additionally, extend a perpendicular from O to the
line BC, and label it E. Let OE = x and OD = y. We’re trying to find x2 + y 2 .
Applying the Pythagorean Theorem, OA2 = OD2 + AD2 , and OC 2 = EC 2 + EO2 .
√
√
Thus, ( 50)2 = y 2 + (6 − x)2 , and ( 50)2 = x2 + (y + 2)2 . We solve this system to get x = 1 and
y = 5, resulting in an answer of 12 + 52 = 026 .
5. First, notice that x2 + y 2 = (x + y)2 − 2xy = 7 and
x3 + y 3 = (x + y)(x2 + y 2 ) − xy(x + y) = 7(x + y) − xy(x + y) = (7 − xy)(x + y) = 10. Now that
we’re left with x + y and xy, let a = x + y and b = xy. We get a2 − 2b = 7 and a(7 − b) = 10.
Because we want the largest possible a, let’s find an expression for b in terms of a.
2
a2 − 7 = 2b =⇒ b = a 2−7 . Substituting, a3 − 21a + 20 = 0. Factored, (a − 1)(a + 5)(a − 4) = 0 The
largest possible solution is therefore 004 .
6. Notice 683 + 883 = (7 − 1)83 + (7 + 1)83 . Applying the Binomial Theorem, half of our terms cancel
out and we are left with 2(783 + 3403 · 781 + · · · + 83 · 7). Alsl of the terms in this big jumble of
numbers are divisible by 49 except the final term. Therefore, our answer is 2(83)(7)
(mod 49) = 035 .
7. Some casework should do the trick. To position three knights adjacently, we can have the
rightmost knight on ony of the 25 spots, so there are 25 ways. To position two knights adjacently, we
once again have 25 positions for the rightmost knight, in addition to the 21 positions where we can
place the remaining knight,
so there are 25 · 21 ways. Dividing by the total number of ways to
25
position the knights, 3 , our probability is 25+25∗21
= 11
46 , and the answer is 057 .
(25
)
3
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200!
8. Expanding, we get 200
100 = 100!100! . Thus, our two digit prime p must satisfy 3p < 200. The largest
such prime is 061 , which is our answer.
9y 2 +4
y
9y + y4
9. First off, substitute y = x sin x so our expression becomes
sin x > 0, we have y > 0. Now we apply AM-GM to obtain
= 9y + y4 . Since x > 0 and
q
≥ 2 9y · y4 = 12 .
10. We start with the case where there are two 1’s and the other two digits are do not repeat. Letting
x denote the two unknown digits. There are 9(8) = 72 unknown digits that can fill in the spaces.
Additionally, there are 3 ways to arrange the 1’s, so we get 3(9)(8) = 216 possibilities from this case.
Next, we count the number of numbers in which 1 doesn’t repeat. Similarly, we get 216 numbers. In
total, our answer is 432 .
11. First, draw the altitude from E to ABCD, and let
√ the point of intersection be P . Additionally,
draw the altitude from E to side AD, which equals 3 q6 by the Pythagorean Theorem. We apply the
√
√
Pythagorean Theorem once more to find that EP = (3 6)2 − (3 2)2 = 6.
Next,
we complete the figure into a triangular prism. The volume is the base times the height,
√
√
6 2·12 2·6
= 432. An accurate diagram helps us see that the original figure is only 23 as large as the
2
trianglular prism we just formed, so our answer is 432( 23 ) = 288 .
10x+y
and
12. Let AB = 10x + y and CD = 10y + x. It follows that CO = AB
2 =
2
10y+x
CD
CH = 2 = 2 . Applying the Pythagorean Theorem on CO and CH,
r
2 2 q
p
10x+y
10y+x
3
9
OH =
·
11(x
+
y)(x
−
y)
=
11(x + y)(x − y).
−
=
2
2
4
2
p
Because OH is a positive rational number, the quantity 11(x + y)(x − y) cannot contain any
square roots. Therefore, x + y must equal eleven and x − y must be a perfect square (since
x + y > x − y). The only pair (x, y) that satisfies this condition is (6, 5), so our answer is 65 .
13. Let S be a non-empty subset of {1, 2, 3, 4, 5, 6}. Notice that the alternating sum of S plus the
alternating sum of S with 7 included is 7.
For
let’s consider {} and {7} a pair. There are
the6sake of convenience,
6
6
6 = 126 different sets and 64 pairs. Therefore, our answer is 64 · 7 = 448 .
+
+
·
·
·
+
=
2
0
1
6
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14. In the diagram below, we let QP = 2x,
AX = a, and BY = b. From right triangle AZB,
we see that 122 = 4x2 + (a − b)2 . From right
triangle AXP , 642 = a2 + x2 and 362 = b2 + x2 .
Solving these three equations, we find that
QP = 4x2 = 130 .
15.
Let A be any fixed point on circle O and let AD be a chord of circle O. The locus of midpoints
N of the chord AD is a circle P , with diameter AD. Generally, the circle P can intersect the chord
BC at two points, one point, or they may not have a point of intersection. By the problem condition,
however, the circle P is tangent to BC at point N.
Let √
M be the midpoint of the chord BC such that BC = 3. From right angle triangle OM B,
3
OM = OB 2 − BM 2 = 4. Thus, tan ∠BOM = BM
OM = 4 .
Notice that the distance OM equals P N + P O cos AOM = r(1 + cos AOM ) (Where r is the radius of
8
3
circle P). Evaluating this, cos ∠AOM = OM
− 1 = 2OM
R − 1 = 5 − 1 = 5 . From cos ∠AOM , we see
√r
√
1 − cos2 ∠AOM
52 − 32
4
that tan ∠AOM =
=
=
cos ∠AOM
3
3
Next, notice that ∠AOB = ∠AOM −∠BOM . We can therefore apply the tangent subtraction formula
4
−3
tan AOM − tan BOM
7
to obtain , tan AOM =
= 3 4 43 =
. It follows that sin AOM =
1 + tan AOM · tan AOM
24
1+ 3 · 4
72
7
√
= , resulting in an answer of 7 · 25 = 175 .
2
2
25
7 + 24
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