CD EXAM
Texas A&M High School Math Contest
November 10, 2012
Solution.
1. Find a such that
2
1
1
9 a+
−6 a+
+ 1 = 0.
36a
36a
1
. Then the given equation is equivalent to 9x2 − 6x + 1 = 0, or (3x − 1)2 = 0.
36a
1
So, x = 1/3, and hence a +
= 1/3. The last one implies 0 = 36a2 − 12a + 1 = (6a − 1)2 ,
36a
i.e a = 1/6.
Answer: 1/6
Let x = a +
2. The surface area of a cube is equal to the surface area of a sphere. What is the ratio of the
volume of the cube to the volume of the sphere.
The surface area of a cube with side x is 6x2 . The surface area of a sphere with radius R is
4πR2 . These areas are equal, so 6x2 = 4πR2 giving
1/2
x2
4π
2π
x
2π
=
=
⇒
=
.
2
R
6
3
R
3
The ratio of the volume of the cube to the volume of the sphere is
r
r
1/2 !3
−1 3/2
3 x 3
1 2π
3
2π
1 2π
2π
π
x3
=
=
=
=
=
.
4
3
4π R
4π
3
2 3
3
2 3
6
πR
3
r
Answer:
π
6
3. The cubes of two consecutive positive odd integers differ by 2402. What is the smaller of
these two integers?
For any positive integer k the numbers 2k − 1 and 2k + 1 are consecutive positive odd integers.
Then we have (2k+1)3 −(2k−1)3 = 2402. Using the formula a3 −b3 = (a−b)(a2 +b2 +ab) we get
(2k+1−(2k−1))((2k+1)2 +(2k−1)2 +(2k+1)(2k−1)) = 2402, or 2(8k 2 +2+4k 2 −1) = 2402.
The last equality implies k = 10. Thus 2k − 1 = 19.
Answer: 19
4. Find the
of the triangle
area formed by the lines x = 0, y = x, and the line through the
1 3
1 2
points − ,
and
,
.
2 2
3 3
1 3
The equation of the line through the points − ,
is x + y = 1. This line has y-intercepts
2 2
at y=1 and it intersects the line y = x at (1/2, 1/2). So, the triangle has vertices (0,0), (0, 1)
and (1/2, 1/2). Its area is equal to 1/4 of the area of square with side equals 1, i.e. 1/4.
Answer: 1/4
1
5. Find the largest real value a such that both roots of the polynomial x2 + x + a are real and
greater or equal than a.
Let x1 and x2 be the roots of the given polynomial. Then
x1 + x2 = −1
(1)
x1 x2 = a.
(2)
and
If the roots x1 and x2 are real, then by (1) they cannot be both positive. If both roots are
negative, then by (2) a > 0, hence, the roots x1 and x2 less than a. If a = 0 then one of
the roots equals −1 which contradicts the initial condition. Thus, a < 0. But for a < 0 the
discriminant 1 − 4a is positive and both roots are real. It remains to check when the smaller root
is great or equal than a, i.e.
√
−1 − 1 − 4a
≥ a.
2
The last inequality is equivalent to the following:
√
1 − 4a ≤ −2a − 1.
Squaring both sides we get
1 − 4a ≤ (1 + 2a)2
where
1 + 2a ≤ 0.
(Note that 1 − 4a ≥ 0 because a < 0.) The last two inequalities are equivalent to
a(a + 2) ≥ 0,
a ≤ −1/2.
So, a ≤ −2, and the largest a satisfying the conditions of the problem is a = −2.
Answer: a = −2
6. For the linear function f (x) = mx + b, f (1) = −2 and f (−2) = 11. If for all x, f (x + h) −
f (x) = 6, then find h.
We have
11 = f (−2) = −2m + b
−2 = f (1) = m + b
. Subtracting these equalities we get 13 = −3m, or m = −13/3. Then 6 = f (x + h) − f (x) =
m(x + h) + b − (mx + b) = mh, hence, h = 6/m = −18/13.
Answer: −18/13
7. What is the last digit of the number 20132011 ?
The last digits follow the same pattern as the last digit of the sequence 30 = 1, 31 = 3, 32 =
9, 33 = 27, 34 = 81. Since 2011 has the reminder 3 after division by 4, the last digit of the given
number is 7.
Answer: 7
2
8. Square ABCD has side length 2. A semicircle with diameter AB is constructed inside the
square, and the tangent to semicircle from C intersects side AD at E. What is the length
of CE?
Let F be the point at which CE is tangent to the semicircle, and let G be the midpoint of AB.
Because CF and CB are both tangents to the semicircle, CF = CB = 2. Similarly, EA = EF .
Let x = EA. The Pythagorean Theorem applied to the triangle CDE gives (2 − x)2 + 22 = (2 +
x)2 . It follows that x = 1/2 and CE = 2+x = 5/2.
Answer: 5/2
9. The number 8n is written on the blackboard. The sum of its digits is calculated, then the
sum of the digits of the result is calculated and so on, until we get a single digit. What is
this digit if n = 2013?
The answer is 8. Since the remainder of a natural number and of the sum of its digits when
divided by 9 are the same, the remainder of 82013 mod 9 coincides with the remainder of the final
result x. Note that the remainder of 8n when divided by 9 is equal to 8 if n is odd and it is equal
to 1 if n is even, because it is true for n = 1 and n = 2 and then repeats. Since n = 2013 is odd
the required number x has remainder 8 modulo 9, and we know that x is a digit. We conclude
that x = 8.
Answer: 8
10. Mark has three times as many nickels as quarters and three more dimes than nickels. If the
total face value of these coins is $2.40, how many coins does Mark have?
Define variables as follows: n = # nickels, d = # dimes, q = # quarters. Then the problem is
3
equivalent to the system of linear equations
n = 3q
d=n+3
5n + 10d + 25q = 240
It follows that n = 9, d = 12, q = 3, thus n + d + q = 24.
Answer: 24
11. What is the largest integer that is a divisor of
(n + 1)(n + 3)(n + 5)(n + 7)(n + 9)
for all positive integers n?
Among five consecutive odd numbers, at least one is divisible by 3 and exactly one is divisible by
5, so the product is always divisible by 15.
Answer: 15
12. A certain restaurant has a choice of 8 appetizers, 10 entrees and 6 deserts. If you plan
to order an appetizer, an entree and a desert, how many different three-course meals are
possible?
8 × 10 × 6 = 480.
Answer: 480
1
1
13. Let {an } be a sequence such that an = √ − √
. Find a1 + a2 + . . . + a63 .
n
n+1
We have
1
1
1
1
1
1
1
1
1
a1 + a2 + . . . + a63 = 1 − √ + √ − √ + √ − √ + · · · + √ − √ + √ − √
2
2
3
3
4
62
63
63
64
1
7
=1− √ =
8
64
Answer: 7/8
14. How many integers satisfy the following inequality
x4 ex
< 0.
x2 − 4x − 21
Since x4 ≥ 0 and ex > 0 the given inequality implies x2 − 4x − 21 < 0 and x 6= 0, or
(x − 7)(x + 3) < 0,x 6= 0. The solution set of the last inequality is interval (−3, 7) and there
are 8 nonzero integers on that interval.
Answer: 8
15. Find the area of the region on the xy-plane defined
 2
 x + y2 ≤
|x| + |y| ≥

y
≥
4
by the system of inequalities:
1
1
0
The last inequality gives that the region is in the upper half-plane; the first one gives interior
of the unit circle; the second one for y ≥ 0 is bounded from below by the lines x + y = 1 and
y − x = 1. The region is shaded on the picture below.
Its area is difference between the unit
area and half area of the square with side
√ semicircle
π
2
2
Thus, the region area is (π(1) − ( 2) )/2 = 2 − 1.
π
Answer:
−1
2
√
2.
16. Find the least multiple of 7 which when divided by 2, 3, 4, 5, or 6 gives a reminder of 1.
The lowest common multiple of 2, 3, 4, 5, and 6 is 60. We have to find a multiple of 7 that is
larger by 1 than a multiple of 60. Now note that 60n + 1 = 7 × 8n + 4n + 1. Then 60n + 1 is
divisible by 7 if 4n + 1 is divisible by 7. The lowest value of n that satisfies this condition is 5.
Therefore, the number we need is 301.
Answer: 301
17. If x and y satisfy the equations
6, 751x + 3, 249y = 26, 751
3, 249x + 6, 751y = 23, 249
then find x2 − y 2 .
Adding the equations we get 10, 000x + 10, 000y = 50, 000, or x + y = 5. Subtracting the given
equations we get 3502x−3502y = 3502, or x−y = 1. Thus, x2 −y 2 = (x−y)(x+y) = 5·1 = 5.
Answer: 5
18. Given isosceles trapezoid ABCD and circle with diameter on BC (the smaller base) which
is tangent to the base AD and passes through the middle points of the diagonals. What is
b
the angle A?
Denote by CK the altitude from C to AD.Note that CK is the radius of the given circle, thus
1
CK = BC.
2
5
(3)
The angle CM B is right (because BC is diameter). Hence ∆BCM ∼
= ∆CM D (because they
have two congruent sides and the angle between them). Thus,
BC = CD
b=
(3)&(4) imply CK = 12 CD, and then sin D
Answer: π/6
CK
CD
(4)
b =A
b = π/6.
= 12 , i.e. D
√
19. Solve the equation 4 · 9x−1 = 3 22x+1
We have
22 · 32(x−1) = 3 · 2(2x+1)/2 .
Both sides of the last equality are positive, thus we can apply log2 to both sides:
log2 22 · 32(x−1) = log2 3 · 2(2x+1)/2 .
Using the properties of logarithms we get
log2 22 + log2 32(x−1) = log2 3 + log2 2(2x+1)/2 ,
2 + 2(x − 1) log2 3 = log2 3 + (2x + 1)/2,
2 + 2x log2 3 − 2 log2 3 = log2 3 + x + 1/2
x(2 log2 3 − 1) = 3 log2 3 − 3/2
3
x(2 log2 3 − 1) = (2 log2 3 − 1),
2
x = 3/2
Answer: 3/2
20. Let f (x) = x2 + 12x + 30. Find the minimal x satisfying the equation f (f (f (f (f (x))))) = 0.
Notice that f (x) = (x + 6)2 − 6. Then f (f (x)) = (((x + 6)2 − 6) + 6)2 − 6 = (x + 6)4 − 6,
f (f (f (x))) = (x + 6)8 − 6 and so on. Finally,
f (f (f (f (f (x))))) = (x + 6)32 − 6,
√
and the solution of the equation (x + 6)32 = 6 is −6 ± 32 6.
√
Answer: −6 − 32 6
6
21. For how many values of n will an n sided regular polygon have interior angles with integral
degree measures?
Use the fact that for a regular n-gon the sum of the interior angles is (n − 2)180◦
number of degrees of each interior angle is
1
Thus, the
n−2
360
180 = 180 −
.
n
n
So, the number of degrees is an integer if and only if n divides 360 = 23 · 32 · 5. There are
(3 + 1)(2 + 1)(1 + 1) = 24 factors of 360, but no polygon is formed when n = 1 or when n = 2,
so there are only 22 polygons with this property.
Answer: 22
22. Three semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2.
The centers of small semicircles divide AB into four line segments of equal length, as shown.
What is the area of the shaded region that lies within the large semicircle but outside the
smaller semicircles?
The area of the larger semicircle is 12 π(2)2 = 2π. The region deleted from the larger semicircle
consists of five congruent sectors and two equilateral √triangles.
The area of each of the sectors
√
is 61 π(1)2 = π6 , and√the area of√each triangle is 12 · 1 · 23 = 43 , so the area of the shaded region
is 2π − 5 · π6 − 2 · 43 = 7π
− 23
6
Answer:
√
7π
3
−
6
2
1
The result follows from the fact that every n-gon can be partitioned into n − 2 triangles, each of which has an
angle sum of 180◦ .
7