Math Contest, Fall 2009
CD EXAM SOLUTIONS
1. ABCD is a rectangle with AB = 5 cm and BC = 8 cm. Points E and F are chosen
on AD and DC respectively such that AE = 5 cm and F C = 2 cm. Find the area of
the triangle BEF .
Solution. The area is the area of the rectangle without the areas of the three corner
triangles: SBEF = 5 ∗ 8 − 5 ∗ 5/2 − 2 ∗ 8/2 − 3 ∗ 3/2 = 15 cm2 .
Answer. 15 cm2 .
2. A lab worker prepared a solution by by adding some salt to water. Then he found that
the solution was too concentrated and he diluted all 1350 g of it with 750 g of water.
But the new solution was too dilute and he added 1.2 g of salt into it, obtaining a
solution containing 0.9% salt. Find the percentage of salt in the original solution.
Solution. Let x be the amount of salt in the original solution. Then, after adding
750 g of water, the solution (2100 g of it) contains still x g of salt. After adding
1.2 g of salt into the new solution, it contains x + 1.2 g of salt in 2101.2 g. Thus,
(x+1.2)(100%)
= 0.9%, and so x = 17.7108 g. The concentration of salt in the original
2101.2
17.7108
solution is 1350 = 0.013119111... = 0.0131191̄ and thus the required percentage is
1.31191̄%.
Answer. 1.31191̄%
3. A, B, and C are three containers and each of them contains a certain amount of
detergent. If 300 ml of detergent in A is poured into B then B will have twice the
amount in A. After this, if 300 ml of detergent in B is poured into C, then C will
have twice the amount in B. After this, if 300 ml of detergent in C is poured into A,
then C and A have the same amount of detergent. Find the amount of detergent each
container has at the beginning.
Solution. Let A has x ml of detergent. Then after 300 ml of detergent in A has been
poured into B, A will have x − 300 ml and B will have 2(x − 300) ml. After this, 300
ml of detergent in B is poured into C, then B will have 2x − 900 ml and C will have
2(2x − 900) ml. Next, 300 ml of detergent in C is poured into A, then C will have
2(2x − 900) − 300 = 4x − 2100 ml and A will have x ml. Since now C and A have the
same amount of detergent, x = 4x − 2100, x = 700 ml. So A at first and at last has
700 ml, B has 2x − 900 = 500 ml, and C has 700 ml (same as A).
Answer: 700 ml, 500 ml, 700 ml
4. The bases AD and BC of a trapezoid ABCD have lengths a and b respectively (a > b).
Let M N be the line segment whose endpoints divide the sides AB and CD such that
AM : M B = DN : N C = p : q. Find the length of the segment M N in terms of a, b, p,
and q.
Solution.
Draw the line from B parallel to CD and let F be the point of intersection of this line
with AD. Let E be the intersection of M N and BF . Then, AF = a − b, F D = b and
AF
BA
= M
or
M N = M E + EN . Next, since triangles BM E and ABF are similar, BM
E
(a−b)q
(a−b)q
p+q
aq+bp
a−b
= M E . Thus M E = p+q and thus M N = p+q + b = p+q .
q
Answer.
aq+bp
p+q
5. The diagonals of a convex quadrilateral ABCD intersect at a point P . If the areas of
the triangles ABP, BCP, CDP are known to be equal to a, b, c respectively, what is
the area of the triangle ADP ?
Solution.
Since both triangles on the left (below BD) have the same height and both triangles
on the right (above BD) have the same height, the ratio of their areas is the same
as the ratio of their bases: SADP : SABP = DP : BP = SCDP : SBCP , so SADP =
(SCDP )(SABP )
= acb
SBCP
Answer.
ac
b
6. The average age of m boys and n girls is p. If the average age of the boys is q, what is
the average age of the girls in terms of p, q, m and n?
Solution. Let b1 , ...bm be the boys ages and a1 , ...an be the girls ages. Then b1 +...+bm =
qm, and b1 + ... + bm + a1 + .. + an = p(m + n). Thus a1 + ... + an = p(m + n) − qm
n
and finally, a1 +...+a
= p(m+n)−qm
n
n
Answer.
p(m+n)−qm
n
7. Solve the equation 2x + 1 + x2 − x3 + x4 − x5 + ... = 13/6 where |x| < 1.
Solution. The sum of the geometric series x2 − x3 + x4 − x5 + ... with the first term
x2
x2
x2 and the ratio −x is equal to 1+x
. Thus the equation becomes 2x + 1 + 1+x
= 13/6
1
7
which solutions are 2 , − 9 .
Answer. 1/2, -7/9
8. The smallest interior angle of a convex polygon is 120◦ . If the angles of this polygon
form an arithmetic sequence with the difference of 5◦ , how many sides does a polygon
have?
Solution. The sum of interior angles is 120+(120+5)+...+(120+5(n−1)) = 180(n−2),
so 120n + 5n(n−1)
= 180(n − 2) and thus n = 16 or n = 9. It is easy to see that n=16
2
does not work, since the angle of 120+15*5=195 degrees means that the polygon is
not convex. So n = 9.
Answer. 9
9. Find all real values of a for which the equations x2 + ax + 8 = 0 and x2 + x + a = 0
share a solution.
Solution. a = −x2 − x, so x2 − x3 − x2 + 8 = 0, or x3 = 8, x = 2. Thus 22 + 2 + a =
0, a = −6.
Answer. -6
10. If 4x + 4−x = 23, what is the value of 2x + 2−x ?
Solution. (2x + 2−x )2 = 4x + 4−x + 2 = 25, so 2x + 2−x = 5
Answer. 5
11. In a triangle ABC the point K on BC divides the side BC such that BK : KC = 1 : 3.
The point L on the side AC divides the side AC such that AL : LC = 2 : 5. If O is
the point of intersection of AK and BL, find the ratio BO : OL.
Solution. Put masses into the vertices so that the point K is the center of mass of BC
and L is the center of mass of AC. For instance, B has a mass of m1 = 3 g, C - a mass
of m2 = 1 g so that m1 (BK) = m2 (CK). Then the mass at A has to be m3 = 25 g
since m2 (CL) = m3 (AL). Then the center of mass of a triangle is at the point O (since
it must be on both AK and BK). Thus BO(3) = OL(1 + 52 ) and BO : OL = 7 : 6.
Answer. 7 : 6
12. Find the natural number n such that 32 × 35 × ... × 33n−1 = 326 .
Solution. 32+5+...+(3n−1) = 3
(3n+1)n
2
= 326 , so
(3n+1)n
2
= 26, n = 4, − 13
. So n = 4.
3
Answer. 4
13. Let ABCD be a tetrahedron (regular triangular pyramid) and let E be a point inside
the face ABC. Denote by s the sum of distances from E to the faces DAB, DBC, DCA
and by S the sum of distances from E to the edges AB, BC, CA. Find the ratio Ss .
Solution. Denote by a the length of each edge. Denote by u, v, w the distances from
E to AB, BC, AC and by x, y, z the distances from E to DAB, DBC, DCA. For the
volume of the tetrahedron we have the following: VABCD = VEABD +VEDBC +VEADC =
1
(AreaDAB · x + AreaDBC · y + AreaDCA · z) = 31 (AreaDAB )(x + y + z) = 31 AreaDAB · h
3
where h is the altitude of the tetrahedron. The altitude
√
√h can be found as a leg of a
a 3
1a 3
right triangle
with another
leg being equal to 3 2 = √ 6 and the hypotenuse being
√
√
a 3
a 6
equal to 2 , so h = 3 . Thus, s = x + y + z = h = a 3 6 . Next,
S = u + v + w is the
√
altitude of the face ABC and is equal to
Answer.
√
a 3
.
2
Finally,
√
2 2
3
s
S
=
a 6
3
√
a 3
2
=
√
2 2
3
14. Which of the following numbers is a perfect square:
a) 44 55 66 b) 44 56 65 c) 45 54 66 d) 46 54 65 e) 46 55 64
Solution. For 4x 5y 6z = 22x+z 3z 5y to be a perfect square, 2x + z, z, y must be all even.
Choice C)
Answer. C
15. Participation in the local soccer league this year is 10% higher than last year. The
number of males increased by 5% and the number of females increased by 20%. What
fraction of the soccer league is now female?
Solution. Let for this year m be the number of men, w be the number of women. Then
m
w
last year the numbers of men and women respectively were 1.05
and 1.2
, while the total
m
w
m+w
m
1
1
1
1
m+w
= 47 , m+w
=
was 1.1 . Thus 1.05 + 1.2 = 1.1 and w ( 1.05 − 1.1 ) = 1.1 − 1.2 . So m
w
w
7
11
w
4
+ 1 = 4 and thus m+w = 11
4
Answer.
4
11
16. In an isosceles triangle ABC with AB = BC = 4 cm, length of the median AD is 3
cm. Find the length of the base AC.
Solution. Consider the altitude H = BE of the triangle ABC and the altitude h =
DF of the triangle ADC. Denote √
by AE = x. We need to√ find AC = 2x. Using
2
Pythagoras Theorem, we get H = 162 − x2 and thus h = 16−x
. √Since AF = 3x
,
2
2
16−x2 2
3x 2
2
2
2
from the right triangle ADF we get AF + DF = AD or ( 2 ) + ( 2 ) = 9. So
q
√
x2 = 52 , x = 52 , 2x = 10
Answer.
√
10 cm
17. In triangle ADE, B is a point on AD and C is a point on AE such that AB = BC =
CD = DE and the measure of 6 ADE is 120◦ . Find the measure of 6 BAC.
Solution. Denote 6 BAC = 6 BCA = x, 6 DCE = 6 DEC = y. Then 6 CBD =
CDB = 2x, 6 ABC = 180 − 2x, 6 CDE = 180 − 2y. So, 6 ADE = 180 − 2y + 2x = 120
and 6 BCD = 180−x−y = 180−4x The system has a unique solution x = 15, y = 45◦ .
6
Answer. 15◦
18. Let a, b, c be three numbers (not necessarily different) chosen randomly and independently from the set 1,2,3,4,5. Find the probability that the number ab + c is even.
Solution. The sum is even in the following two cases:
I. both ab and c are even. This occurs in the following three subcases: (a- even, b-odd,
c-even); (a- odd, b-even, c-even); (a-even, b-even, c-even); which gives 12 + 12 + 8 = 32
cases.
II. both ab and c are odd. This occurs only if each of the three numbers is odd which
gives 3 ∗ 3 ∗ 3 = 27 cases.
The total number of triples chosen from the set 1, 2, 3, 4, 5 is 5 ∗ 5 ∗ 5, so the probability
is P = (32 + 27)/125 = 59/125.
Answer. 59/125
19. A group of Aggies chosen for the Aggie band had 100 females and 80 males. A group
of Aggies chosen for the Aggie orchestra had 80 females and 100 males. There are 60
females in both band and orchestra. There are 230 students who are either in the band
or in the orchestra or in both of them. What is the number of males who are in the
band but not in the orchestra?
Solution. There are 180 + 180 − 230 = 130 students in both band and orchestra, out of
which 60 are females and the rest, 130 − 60 = 70 are males. Since there are 80 males
in the band and 70 of those are also in the orchestra, the remaining 10 males are in
the band only.
Answer. 10
20. What is the shortest distance from the point A(−7, 2) to the circle given by an equation
x2 + y 2 − 10x − 14y − 151 = 0 ?
Solution. The equation x2 +y 2 −10x−14y−151 = 0 transforms into (x−5)2 +(y−7)2 =
225. Thus the center and the radius of the circle are C(5, 7), R = 15. The distance
from the point A(−7, 2) to the center of the circle is CA = 13, so A is inside the circle,
thus the unknown distance id d = 15 − 13 = 2
Answer. 2
21. How many 9’s are there in the decimal expansion of 999998999992 ?
Solution. Let x = 99999899999. Notice that if we add 100001 to x, we’d get
99999899999
+100001
100000000000
In other words, x + 105 + 1 = 1011 . From this, we can conclude x = 1011 − 105 − 1. So
2
x2 = (1011 − 105 − 1) = 1022 + 1010 + 1 + 2 · 105 − 2 · 1016 − 2 · 1011 :
Now,
1022 = 10000000000000000000000 (1 is followed by 22 zeroes),
1022 + 1010 + 1 + 2 · 105 = 10000000000010000200001, and finally,
1022 + 1010 + 1 + 2 · 105 − 2 · 1016 − 2 · 1011 = 9999979999810000200001
So the digit 9 appears nine times in the decimal expansion of 999998999992 .
Answer. 9
22. Find an equation with integer coefficients whose roots include the number
√
√
Solution. Let x = 2 + 3 3. Then we can write
√
√
3
x= 2+√
3√
√
x2 = 2 √
+ 2 2 3 3 +√ 3 9
√ √
x3 = 2 2 + 3 · 2 · 3 3 + 3 · 2 · 3 9 + 3
√
√
3
Now,
3
=
x
−
,
√
√2 √
√
√
√
3
9 = x2 − 2 − 2 2 3 3 = x2 − 2 − 2 2(x − 2) = x2 + 2 − 2 2x.
√
2+
√
3
3.
Substituting
these √
into the expression
for x√3 , we obtain
√
√
x3 = 2 2 + 6(x√− 2) + 3 · 2(x2 + 2 − 2 2x) + 3, from which we find
x3 + 6x − 3 = 2(3x2 + 2). Now we square both sides and transport all terms to the
left side:
x6 + 36x2 + 9 + 12x4 − 6x3 − 36x = 2(9x4 + 12x2 + 4)
x6 − 6x4 + 12x2 − 36x + 1 = 0.