Download High School Math Contest University of South Carolina February 5, 2011 Solutions

High School Math Contest
University of South Carolina
February 5, 2011
Solutions
1. (b) The left-hand side is an even number unless m or n is zero. If m = 0, we have 2n ≥ 1 and
2m − 2n ≤ 0. If n = 0, then 2m = 64, i.e. m = 6. Thus (6, 0) is the only such pair.
2
2. (d) If the relation ab = 1 then either a = ±1, or b = 0 (and a 6= 0). The equation 2x 3−5 = 1
2
yields solutions x = ±2. If 2x 3−5 = −1, then x = ±1, but in this case x2 − 2x is odd, so
2 x2 −2x
2x −5
= −1. From the last option we obtain x2 − 2x = 0, i.e. x = 0 or x = 2. Thus
3
we have three solutions: x = 0, ±2.
3. (d) We can write x51 + 51 = Q(x) · (x + 1) + R, where Q(x) is the quotient and R is the
remainder. When x = −1, this gives R = (−1)51 + 51 = 50.
4. (d) Let x be the total number of people. Then the total number of handshakes is 5x
, where we
2
5x
have to divide by 2 since every handshake is counted twice. Thus 2 = 60, which leads to
x = 24.
5. (b) Every power of six ends in 6. Also note that 324 = 816 . Since the last digit of 81 is 1, any
power of it will also end in 1. (Alternatively, for the powers of 3 we have 31 = 3, 32 = 9,
33 = 27, 34 = 81, 35 = 243, ... – the units digits repeat in cycles of length 4. Since 24 is
divisible by 4, the final digit of 324 is the same as the final digit of 34 = 81, i.e. 1.) Thus the
units digit of 625 − 324 is 6 − 1 = 5.
√
6. (c) Obviously, x > y. If r is the radius of the circle, then x = 2πr and y = 4 2r. Hence
x
π
= 2√
< 3.2
≈ 1.14 < 23 .
y
2.8
2
7. (c) The product has 2011 factors. In order for it to be negative, we need to have an odd
number of negative factors (notice that all the exponents are odd) and, hence, an even number
of positive factors. But the number of positive factors is simply equal to x. Since there are
1005 even positive integers less than 2011, the answer is 1005 (for integers x ≥ 2011 the
product is obviously positive).
1
8. (c) Draw the perpendicular DO from D onto the side BC. Then BO = OC = AD. Let
X be the intersection point of P R and DO, and Y be the intersection of QS and DO. Then
DR
the triangles 4DXR and 4DOC are similar, so that XR
= DC
= 13 . In the same fashion,
OC
YS
DS
2
4DY S is similar to 4DOC, so OC = DC = 3 . Since AQ = P B = 23 AB, the bases of the
triangles 4AQR and 4P BS have equal lengths and we only need to compare the heights.
We have
4
OC
BO + 31 OC
PR
P X + XR
4
a1
=
=
=
= 35
= .
2
a2
QS
QY + Y S
5
BO + 3 OC
OC
3
1
= 2 log1 3
log2 9
2
log2 3
log5 2
= log
=
log3 2
25
9. (c) We have x = log9 2 =
1
y.
2
We also have log5 3
= 12 log3 2, so log3 2 = 2x. Also, log5 2 = 21 log5 4 =
=
y
.
4x
Hence
1
1
+
log3 6 log5 6
1
1
=
+
log3 2 + log3 3 log5 2 + log5 3
1
1
4x
1
=
+ 1
+
y =
2x + 1
2x + 1 y(2x + 1)
y + 4x
2
4x + y
=
.
y(2x + 1)
log6 15 = log6 3 + log6 5 =
p
√
√
10. (d) Write
29
−
12
5
=
a
+
b
√
√ n and square both sides of this relation. One obtains
2
2
29 − 12 5 = (a + b n) + 2ab n. If we take n = 5, we need to find a and b which satisfy
a2 + 5b2 = 29 and 2ab = −12. It is easy to see that the solutions are a = −3 and b√= 2.
(In fact, a = 3 and b = −2 also satisfy the equations, but in this case the number a + b n is
negative.)
2
11. (a) Let F (n) be the number of ways to seat n students in a row of n chairs as described in the
problem. The student who has the ticket to seat number n has two options. He/she can either
sit in his/her own chair, in which case the remaining n − 1 students can arrange themselves
in F (n − 1) ways. Alternatively, if this student takes the (n − 1)st seat, the student with the
ticket to that chair is forced to take the nth seat, and the other n − 2 students can be seated
in F (n − 2) ways. Thus, F (n) = F (n − 1) + F (n − 2), i.e. this is a sequence of Fibonacci
numbers. It is easy to see that F (1) = 1 and F (2) = 2. Therefore, F (10) = 89, the 10th
Fibonacci number.
1
12. (a) For 0 < x < 90◦ , we have ln tan(90◦ − x) = ln tan x = − ln tan x . Thus
ln tan 1◦ + ln tan 89◦ = ln tan 2◦ + ln tan 88◦ = ln tan 45◦ = 0, so the sum
above is zero.
13. (b) Notice that F A = AE. Indeed, 4ABC ∼
= 4CDA, while F A and AE are equal to
the corresponding heights of these triangles. Denote h = F A = AE. Triangles 4ADF
and 4BAE are similar, since ∠DF A = ∠AEB = 90◦ and ∠F AD = 180◦ − ∠DAB
−
√
DF
AE
a
h
◦
∠BAE = 90 − ∠BAE = ∠EBA. It follows that F A = EB , i.e. h = b , so h = ab and
√
F E = 2h = 2 ab.
14. (d) Since 1 −
k2 − 1
(k − 1)(k + 1)
1
=
=
, we have
2
2
k
k
k2
1
1
1
1− 2
1 − 2 ··· 1 −
2
3
20112
1·3 2·4 3·5
2009 · 2011 2010 · 2012
2012
= 2 · 2 · 2 · ··· ·
·
=
.
2
2
2
3
4
2010
2011
2 · 2011
3
15. (a) Denote by vA , vB , and vT the speeds of Allan, Bill, and the train, respectively. The relative
speeds of the train with respect to Allan an Bill are then vT − vA and vT − vB . It follows from
−vA
9
.
the condition that the train passes Allan in 10 seconds and Bill in 9 seconds that vvTT −v
= 10
B
The relative speed of Allan with respect to Bill is vA − vB . We have
vA − vB
vT − vB
vT − vA
9
1
=
−
=1−
= .
vT − vB
vT − vB vT − vB
10
10
Thus, Allan’s speed with respect to Bill is ten times smaller than the train’s, so, in order to
reach Bill, it would take Allan ten times the time it would take the train, i.e. 200 minutes.
16. (b) We have 1020 = 103·6+2 . Since 103 = 1001 − 1, we find that 103·6 = (1001 − 1)6 =
1001 · k + 1, where k is some integer. Indeed, when we multiply out (1001 − 1)6 , all the
terms will be divisible by 1001 except for the last one which is (−1)6 = 1. Then 1020 =
(1001 · k + 1) · 100 = 1001 · 100k + 100, in other words, it has remainder 100 when divided
by 1001.
3
17. (b) We see that f (−1/2) = 1. Next, f (−1) = (3/2)− 2 < 1 and f (−3/2) = 2−2 = 1/4
√ < 1.
− 21
Therefore, options (c) √
and (d) are excluded. We are left with f (0) = (1/2)
= 2 and
6
− 61
f (1/3) = (1/6) = 6 which are both greater than 1. To compare these two, notice that
1
1
1
6 < 8, hence, 6 6 < 8 6 = 2 2 .
18. (c) It is easy to see (e.g., by performing long multiplication) that the smallest such number is
91. Indeed, 90 · 1111 = 99990 has 5 digits, while 91 · 1111 = 101101 is a six-digit number.
Thus, the answer is 9 + 1 = 10.
19. (c) Since ∠ACB + ∠AP B = 90◦ + 90◦ = 180◦ , the quadrilateral AP BC can be inscribed
in a circle. But then ∠P CB = ∠P AB = 45◦ since they are inscribed angles with the same
endpoints.
4
20. (d) For the number to be divisible by 9 the sum of its digits has to be divisible by 9. Thus, any
nine-digit number with all digits equal is divisible by 9 (this gives 9 possibilities). A threedigit or a six-digit number would have to have the common digit be either 3, 6, or 9 (2 · 3 = 6
possibilities). For all other numbers of digits (1, 2, 4, 5, 7, 8) the only available choice of the
common digit is 9 – this yields 6 more numbers. Thus, all together we have 9 + 6 + 6 = 21
such numbers.
21. (e) It is easy to see that the triangles with specified areas are similar to each other and to
4XY Z. Then, since the ratio of their areas is 9 : 25 : 16, the ratio of the lengths of their
bases parallel to the side XZ is 3 : 5 : 4. Denote these bases by 3x, 5x, and 4x, respectively.
The three unmarked subregions are parallelograms. Therefore, one can see that XZ = 3x +
5x + 4x = 12x. Since 4XY Z is similar to any of the smaller triangles, say the one of area
2
25, its area is 12x
· 25 = 122 = 144.
5x
22. (a) Label the points with numbers from 1 to 6. Let Aj be the event that all six points can be
covered by some arc of length 1 and point number j is the first one in this arc if one counts
clockwise, j = 1, . . . , 6. The event Aj happens exactly if all the remaining 5 points fall onto
the semicircle of length 1 which “starts” at the j th point. The probability of each point being
1
placed onto this semicircle is 21 . The probability of Aj is therefore 215 = 32
. It is also easy to
see that the events Aj are mutually exclusive. Hence, the probability in question is equal to
P
6
3
6
j=1 P(Aj ) = 32 = 16 .
23. (c) Label the vertices of the triangle A, B, and C and denote the radius of the circles by
r. Let O1 and O2 be the centers of the circles tangent to the side AB, and let O1 P and
O2 Q be the perpendiculars
drawn from these centers to the side AB. Then AP = QB =
√
◦
r · cot 30 = 3r, and
√P Q = O1 O2 = 2r since O1 O2 QP is a rectangle. Thus a = AB =
AP + P Q + QB = 2 3r + 2r, so
√
a
3−1
r= √
=
a.
4
2( 3 + 1)
5
24. (e) This expression is a quadratic polynomial of x. At the same time, it is equal to 1 at
three different points: x = a, b, and c. This can happen only if this quadratic polynomial is
identically equal to 1.
25. (b) Denote the area of 4XY Z by x. Observe that the areas of triangles 4AY R, 4BZP , and
4CXQ are the same, denote them by y. The areas of the quadrilaterals AY ZP , BZXQ, and
CXY R are also equal, denote them by z. Then we have x + 3y + 3z = 1. Since triangles
4ACP and 4P CB have the same height drawn from point C, we see that the area of the
former is twice the area of the latter, i.e. 2z + x + y = 2 · (2y + z), which implies that x = 3y.
By the Law of Cosines
CP 2 = CB 2 + BP 2 − 2 · CB · CP · cos 60◦
= (3BP )2 + BP 2 − 3 · BP 2 = 7BP 2 .
It is easy to see that 4ABR is similar to√4ZBP , since they have one common angle and
∠BP Z = ∠BRA. Since AR = CP = 7 · BP , the area of 4ABR is 7 times the area of
4ZBP , i.e. 2y + z = 7y, or z = 5y. From the equation x + 3y + 3z = 1 we now obtain
1
3y + 3y + 15y = 1, i.e. y = 21
. Hence x = 3y = 71 .
26. (d) Notice that for n ≥ 5 the number n! is divisible by both 2 and 5. Hence it is divisible by ten
and ends with a 0. Therefore, we only need to find the last digit of the sum 1! + 2! + 3! + 4! =
1 + 2 + 6 + 24 = 33, which is 3.
27. (d) We have 4! · 5! · 6! = (2 · 3 · 22 ) · (2 · 3 · 22 · 5)(·2 · 3 · 22 · 5 · (2 · 3)) = 210 · 34 · 52 . The
prime factorization of a perfect square dividing 4! · 5! · 6! has to consist of even powers of the
primes 2, 3, and 5, and the exponents should not exceed the corresponding exponents in the
factorization of 4! · 5! · 6!. Thus, there are 6 choices for the exponent in the power of 2 (0, 2,
4, 6, 8, 10), 3 choices for the power of 3 (0, 2, 4), and two choices for the power of 5 (0, 2).
The total number of possibilities is then 6 · 3 · 2 = 36.
6
28. (d) Let a, b, c, and d be the lengths of the perpendiculars drawn from the point O to the sides
AB, BC, CD, and DA, respectively. Then, by the Pythagorean theorem, we have OA2 =
a2 + d2 , OB 2 = a2 + b2 , OC 2 = b2 + c2 , and OD2 = c2 + d2 . Adding the third equation to the
2
first, and the fourth to the second, we√find that OA2 +OC
= OB 2 +OD2 = a2 +b2 +c2 +d2 .
√
Thus 102 + 52 = 92 + x2 , and x = 100 + 25 − 81 = 44.
29. (b) All together there are 27 ways to color the vertices garnet or black – we shall refer to them
as “patterns”. Let us count how many patterns correspond to the same coloring.
Assume first that the coloring is non-trivial (not all vertices black or all garnet). In this case, a
rotation of a pattern necessarily produces a different pattern – this follows from the fact 7 is a
prime number.
Indeed, if a rotation of a pattern by k vertices (k < 7) yields the same pattern, then you can
continue this process, each time obtaining the same pattern. Since k and 7 are relatively prime,
there exists n such that nk has remainder 1 when divided by 7. This means that we have now
rotated the initial pattern by one vertex and still obtained the same pattern. Thus, all rotations
of the pattern have to be the same. But this is only possible if the coloring is monochromatic
(all vertices are of the same color).
So we see that each non-trivial coloring corresponds to 7 non-trivial patterns. There are 27 − 2
7
such patterns, and, therefore, 2 7−2 = 18 colorings. In addition we have two trivial colorings
(all black or all garnet). Thus, the total number of colorings is 18 + 2 = 20.
30. (c) The total area of the four small circles is 4 × π4 = π – equal to the area of the large
circle. Thus, the total area of the overlaps is equal to the area of the part of the large circle
which is not covered by the small circles. Indeed, by the inclusion-exclusion principle π =
(4 · π/4 − 4a2 ) + 4a1 , so a1 = a2 .
7