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CD EXAM Solutions
Texas A&M High School Math Contest
October 23, 2010
1. The perimeter of a rectangle is 100 and its diagonal has length x. What is the area of
this rectangle (in terms of x)?
Solution. Let us l, w be the lengths of sides. l + w = 50, (l + w)2 = 502 = 2500. Also,
(l + w)2 = x2 + 2lw, and thus lw = (2500 − x2 )/2.
2. Twenty five percent of a class of forty students were girls. When some new girls joined
the class, the percentage of girls increased to forty percent. How many new girls joined
the class?
Solution. There were 10 girls in the class, after the addition of x more girls, we have
10+x
= 0.4, from where x = 10. So ten more girls joined the class.
40+x
Answer. 10.
3. At the end of a school competition, each contestant shook hands with every other
contestant. The winner, a girl, shook hands with 3 times as many girls as boys.
The runner-up, a boy, shook hands with 4 times as many girls as boys. How many
contestants were there altogether?
Solution. Let the number of girls be x and let the number of boys be y. Then 3y = x−1
and 4(y − 1) = x, so y = 5, x = 16, and x + y = 21.
Answer. 21
4. Consider the set S = {8, 5, 1, 13, 34, 3, 21, 2}. Susan makes a list as follows: for each
two-element subset of S, she writes on her list the greater of this subset’s elements.
Find the sum of the numbers on her list.
Solution. Each element of the set will appear in 7 two-element subsets, once with each
other number. 34 will be the greater number in 7 subsets. 21 will be the greater number
in 6 subsets. 13 will be the greater number in 5 subsets. 8 will be the greater number
in 4 subsets. 5 will be the greater number in 3 subsets. 3 will be the greater number in
2 subsets. 2 will be the greater number in 1 subsets. 1 will be the greater number in 0
subsets. Therefore the desired sum is 34·7+21·6+13·5+8·4+5·3+3·2+2·1+1·0 = 484.
Answer. 484
5. Find the sum of all positive integers n for which n2 − 19n + 99 is a perfect square.
Solution. Suppose there is some x such that n2 − 19n + 99 = x2 . Completing the
square, we have that (n − 19/2)2 + 99 − (19/2)2 = x2 , that is, (n − 19/2)2 + 35/4 = x2 .
Multiplying both sides by 4 and rearranging, we see that (2x)2 −(2n−19)2 = 35. Thus,
(2x − 2n + 19)(2x + 2n − 19) = 35. Now, 35 has two pairs of positive factors: {1, 35}
and {5, 7}. Respectively, these yield 9 and 3 for x, which results in n = 1, 9, 10, 18.
The sum is therefore 38.
Answer. 38.
6. ABCD is a rhombus with 6 A = 30◦ . If the area of the rhombus is x cm
perimeter of the rhombus is x cm, find x in centimeters.
2
and the
Solution. Let a be the length of a side of the rhombus. Then the height opposite to
6 A = 30◦ is DE = a/2. The area is x = a(a/2) = a2 /2 and the perimeter is x = 4a.
Now 4a = a2 /2 and a = 8, so x = 4a = 4(8) = 32.
Answer. 32 cm
7. Let α and β be solutions of the quadratic equation x2 + bx + 3 = 0. Find all values of
b such that α2 + β 2 = 3.
Solution. Now α + β = −b, αβ = 3, (α)2 + (β)2 = 3, so b2 = (α + β)2 = (α)2 + (β)2 +
2αβ = 3 + 2(3) = 9. Thus b = ±3.
Answer. ±3
8. Find the sum of all the coefficients of the polynomial obtained after expanding and
collecting the terms of the product (1 − 3x + 3x2 )743 (1 + 3x − 3x2 )744 .
Solution. The expansion will look like that (1 − 3x + 3x2 )743 (1 + 3x − 3x2 )744 =
A0 + A1 x + ... + An xn where n = 743 · 2 + 744 · 2 = 2974. Let x = 1, then the expansion
becomes 1743 · 1744 = A0 + ... + An , so the sum of the coefficients is 1.
9. How many digits are there in the number 2100 after it has been expanded (multiplied
out)?
Solution. 2100 = 102410 > 100010 = 1030 , so 2100 has no fewer than 31 digits (it is more
than 1 followed by 30 zeros). But also 102410 /100010 < 102510 /100010 = (41/40)10 .
41 41
41
41
Finally, (41/40)10 = 40
· 40 ... · 40
< 41
· 40 · 39 ... · 32
= 31
< 10. Hence, 2100 = 102410 <
40 39 38
31
10 · 100010 , so 2100 has less than 32 digits.
Alternative solution. If one knows the value of log 2 or if a calculator is allowed (not
for this contest!), then the solution is shorter: log 2100 = 100 log 2 = 30.103, so 2100
must have 31 digit.
10. Find a six-digit number which is increased by a factor of six when its last three digits
are removed and placed at the beginning of the number without changing their order.
Solution. Let N be the unknown number. Let us denote the number formed by the
first three digits of N by p, and the number formed by the last three digits of N by
q. Then N = 1000p + q and it is given that 6(1000p + q) = 1000q + p. So, on the
one hand, (1000q + p) − (1000p + q) = 5(1000p + q) = 5N , and on the other hand,
(1000q + p) − (1000p + q) = 999(q − p). Thus, 5N = 999(q − p) which means that N is
divisible by 999. Furthermore, p + q = (1000p + q) − 999p = N − 999p, and so it follows
that p + q must also be divisible by 999. But both p and q are three digit numbers;
obviously neither is 999. Thus, p + q = 999. Finally, (1000q + p) + (1000p + q) =
1001(p + q) = 6N + N = 7N . So, 7N = (1001)(999) = 999999, N = 142857.
11. Given a right triangle with the legs of lengths a and b, find the length of the bisector
of the right angle.
Solution. Let 6 ABC be the right angle, BD - its bisector. Draw DE, the altitude for
the triangle 4BDC and DF , the altitude for the triangle 4BDA. Denote x = BE.
Then CE = b − x. Next, since triangles 4BDE and 4BDF are isosceles right
triangles (45◦ , 45◦ , 90◦ ) and congruent to each other, we have ED = BF = DF = x.
x
Also, F A = a − x. Triangles 4DEC and 4AF D are similar, so b−x
= a−x
, from
x
ab
where x = a+b . Finally, using Pythagorean Theorem for the triangle 4BF D, we find
√
ab 2
.
a+b
√
ab 2
a+b
that BD =
Answer.
12. John traveled at a constant speed from town A to town B and returned immediately
to town A. At the same time, Bill traveled at a constant speed on the same road from
town B to town A and returned immediately to town B. They first met at point X, 12
km from town B. On their way back, they met again at a point Y, 8 km from town A.
Find the distance from town A to town B.
Solution. Let the distance between A and B be x, and the distance between X and
Y be y. The ratio of the distances traveled by John and Bill until each meeting must
be the same (since each one traveled at a constant speed). So (draw a line diagram)
before the first meeting John traveled 8 + y km and Bill - 12 km. Before the second
meeting John traveled 12 + 12 + y = 24 + y km and Bill - y + 8 + 8 = y + 16 km. Thus,
8+y
= 24+y
. This leads to the quadratic equation y 2 + 12y − 160 = 0, which positive
12
y+16
solution is y = 8. Thus, the total distance between A and B is x = 8 + 8 + 12 = 28
km.
Answer. 28 km.
13. A, B, C are digits. The 3-digit number ACB is divisible by 3, BAC is divisible by 4,
BCA is divisible by 5, and CBA has an odd number of factors. Find ABC.
Solution. Since BCA is divisible by 5, A = 0 or 5. But ACB is a 3-digit number, so A
= 5. Since 4 divides BAC, 10A + C = 50+C is a multiple of 4, so C = 2 or 6. Since
3 divides ACB, 3 also divides A + B + C = 5+B + C. If C = 2, then B = 2 or 5 or
8. If C = 6, then B = 1 or 4 or 7. 1 Thus there are 6 possibilities for CBA: 225, 255,
285, 615, 645, 675. Of these only 225, a perfect square, has an odd number of factors.
Hence ABC is 522.
14. Let a, b, c, d, e be a five-term geometric sequence where a, b, c, d, e are integers and
0 < a < b < c < d < e < 100. What is the sum of all possible values for c?
n
Solution. The common ratio of the sequence is a fraction which we write as m
in
4
n
4
lowest terms with n > m. Then e = a × m4 . So m divides a. Hence a = km4 ,
where k is a positive integer, and the sequence becomes km4 , km3 n, km2 n2 , kmn3 , kn4
with kn4 < 100. If n > 4, then kn4 ≥ n4 > 256 > 100. So n ≤ 3. If n = 3,
then 81k < 100, so k = 1. Also m = 1 or 2. If m = 1, then the sequence is
(1, 3, 9, 27, 81). If m = 2, then the sequence is (16, 24, 36, 54, 81). If n = 2, then
m = 1 and 16k < 100. So k = 1, 2, 3, 4, 5 or 6. Thus there are six sequences:
(1, 2, 4, 8, 16), (2, 4, 8, 16, 32), (3, 6, 12, 24, 48), (4, 8, 16, 32, 64), (5, 10, 20, 40, 80) and (6, 12, 24, 48, 96).
The sum of all values of c is 9 + 36 + 4 + 8 + 12 + 16 + 20 + 24 = 129.
15. Instead of walking along two adjacent sides of a rectangular field, John took a short
cut along the diagonal and saved a distance equal to one half of the longer side of the
rectangle. Find the ratio of the shorter side of the field to the longer side.
Solution. p
Let x be the length
y - of the shorter. Then the diagonal’s
p of the longer side,
1
2
2
2
2
length is x + y and x + y = x + y − 2 x. Thus, x2 + y 2 = 14 x2 + xy + y 2 and so
3 2
x = xy. Finally, xy = 34 .
4
Answer.
3
4
16. An urn contains 16 balls of two colors, red and blue. Four balls are drawn randomly
from the urn without replacement. The probability of getting exactly two red and two
1
. The urn contains more red than blue balls. How many blue balls were
blue balls is 20
originally in the urn?
Solution. Suppose that the number of blue balls is r. Then there are 16 − r red
balls.
The probability
of getting exactly two red and two blue balls in this case is



16
−
r
r

× 

1
2
2


= , which leads to (16 − r)(15 − r)r(r − 1) = 2 × 13 × 14. On the
16 
20

4
left, there are four natural factors and a pair of them, r − 1, r are consecutive natural
numbers. On the right, the consecutive natural factors can only be 1 and 2 or 13 and
14. Substitution to the equation shows that both r = 2 and 14 are solutions. But since
there are more red balls then blue balls, r cannot be 14.
Answer. 2
17. Two circles sit in a wedge y = ± x3 , as shown in Fig below.
The radius of the inner, smaller circle is 1. The radius of the outer circle is R. Find R.
Solution. Draw the radii BE and CD of both circles to the points of tangency. Then
draw perpendiculars EF, DG to the horizontal line (see figure below).
D
E
O
F B
G C
Triangles OEF and OBE are similar (right
√ triangles with a common acute angle). So
OF
OE
= 3 = BE , so OE = 3. Then
OB = 10. Next, triangles
OBE and OCD are also
EF
√
√
R+1+
10
10+1
DC
R
OC
similar, so OB = EB or √10 = 1 , from where R = √10−1 .
Answer.
√
√10+1
10−1
18. The function is given by the table
x
1
f(x) 4
2 3 4
1 3 5
5
2
If u0 = 4, and un+1 = f (un ) for n ≥ 0, what is the value of u1234 ?
Solution. Checking the pattern for the first few values we see that u0 = 4, u1 = f (4) =
5, u2 = f (5) = 2, u3 = 1, u4 = f (1) = 4, thus u4j+k = uk , u1234 = u2 = 2
Answer. 2
19. For what value(s) of the number a do the equations x2 − ax + 1 = 0 and x2 − x + a = 0
have a common real solution?
Solution. Let z be a common real solution. Then z 2 − az + 1 = z 2 − z + a = 0, so
(1 − a)z = a − 1, thus a = 1 or z = −1. If a = 1, then x2 − x + 1 = 0 has no real
solutions. So z = −1 and (−1)2 − a(−1) + 1 = 0, thus 1 + a + 1 = 0 and so a = −2.
Answer. -2
20. The roots of x3 + 2x2 − 11x − 12 = 0 are -4, -1, and 3. What are the roots of
(x + 5)3 + 2(x + 5)2 − 11(x + 5) − 12 = 0?
Solution. The roots are found from x + 5 = −4, x + 5 = −1, x + 5 = 3, thus the roots
are −9, −6, −2.
Answer. −9, −6, −2
21. Given two circles of radius R with a common tangent line and a common point, find
the radius of the smaller circle inscribed between the two given circles and having the
same tangent line.
Solution. Let A be the center of one of the larger circles, and let AB be the perpendicular from this center to the common tangent line (to the three circles). Also, let C
be the center of the smaller circle, and let CD be the perpendicular to AB. Then in
the right triangle ACD the legs are R, R − r, and the hypothenuse is R + r, where r
denotes the radius of the smaller circle. So (R + r)2 = R2 + (R − r)2 . Expanding and
canceling terms, we get 4Rr = R2 and thus r = R4 .
A
D
B
C