Problem 1
There are two bags. The first bag contains two white balls and one black ball,
while the second bag contains one white ball and two black balls. We randomly
draw a ball from the first bag and put it into the second bag. Now if we draw a
ball from the second bag, what is the probability that the ball drawn is white?
[Problem submitted by Roger Wolf, Math Department Chairman, LACC]
Solution for Problem 1
Let W1 , W2 stand for ‘the first draw is White’, ‘the second draw is White’ respectively.
B1 stands for ‘the first draw is Black’.
P( W2 ) = P( W1 , W2 ) + P( B1 , W2 )
= P( W2 W1 )P( W1 ) + P( W2 B1 )P( B1 )
1 1
1 2
g + g
4 3
2 3
5
=
12
=
Problem 2
The figure below shows square ABCD, an electric circuit in which each edge of
the square contains a switch that is closed or open with probability p and 1-p,
respectively. If a current is fed into vertex A, what is the probability that it will be
transmitted through the edges of the square to vertex C?
[Problem submitted by Luis Zambrano, LACC Associate Professor of Mathematics,]
Solution for Problem 2
The answer is p 2 (2 − p 2 ) . Let 1, 2, 3, 4 represent the event of a closed switch in each
respective edge of the square. As the switches are independent, a closed circuit is obtained
whenever two successive switches from A to C are closed. This can happen one of two
possible ways: hence, we want to compute P(1,2 OR 3,4). Now by the addition rule of
probability, we have
P(1,2 OR 3,4)= P(1,2) +P(3,4) – P(1,2,3,4), which yields the above result.
Problem 3
We draw cards, one at a time, at random and successively from an ordinary deck
of 52 cards with replacement. What is the probability that an ace appears before
a face card?
[Problem submitted by Luis Zambrano, LACC Associate Professor of Mathematics.]
Solution A for Problem 3
Let E be the desired event. Then
P(E) = P( A)P( E A) + P(F )P(E F ) + P( N )P(E N ) . where A, F, and N are the events of drawing
an ace, face card, and neither, respectively. Then,
P( E ) =
4
12
36
⋅ 1 + ⋅ 0 + P( E N ) .
52
52
52
But P(E/N) =P(E) since the probability of E before or after a selection of a card that is neither
an ace or a face card is the same.
P( E ) =
4
12
36
1
⋅ 1 + ⋅ 0 + P( E ). Solving for P(E) yields P( E ) = .
52
52
52
4
Solution B for Problem 3
The probability in the problem will not change if we use only the 4 aces and the 12 face cards.
Then the favorable event is that the first draw is an ace. Since there are 4 aces in 16 cards, the
1
answer is .
4
Problem 4
Find the minimum value of the function y = max{ x − 2 , x − 3 }
[Problem submitted by Anatoliy Nikolachuk, LACC Professor of Mathematics.]
Solution for Problem 4
At the intersection of y =
(
x−2
2
)
2
= x − 3 → x − 2 = x2 − 6 x + 9
7− 5
7+ 5
, x2 =
2
2
7− 5
5 −1
=
= y ( x1 ) = 3 − x1 = 32
2
x 2 − 7 x + 11 = 0 → x1 =
ymin
x − 2 and y = x − 3 ,
x−2 = x −3 .
Problem 5
217 in what base is equal to 574 in base 9?
[Problem submitted by Juergen Pahl, LACC Professor of Mathematics.]
Solution for Problem 5
574 (base 9) = 5 ⋅ 81 + 7 ⋅ 9 + 4 = 472 (base 10)
217 (base x) = 2 ⋅ x 2 + 1 ⋅ x + 7 (base 10)
2 x 2 + x + 7 = 472
2 x 2 + x − 465 = 0
(2 x + 31)( x − 15) = 0
x = 15 is the base.
Problem 6
⎛ 1 ⎞
Simplify the following expression: ⎜ ⎟
⎝ 2 ⎠
log 4
1
x2
[Problem submitted by Anatoliy Nikolachuk, LACC Professor of Mathematics.]
Solution for Problem 6
⎛ 1 ⎞
⎜ 2 ⎟
⎝ ⎠
log 4
1
x2
2
( )
= 2−1
( log 4 x =
− log 4 x 2
log x
2
log 2
2
=
=2
log 4 x
2log x
2log 2
2
=
=2
log 2 x
log x
log 2
= x
= log 2 x )
Problem 7
In ΔABC , AB and AC have equal lengths, P is a point inside the triangle and
m∠APB > m∠APC . Prove that m∠BAP > m∠PAC .
[Problem submitted by Steve Lee, LACC Professor of Mathematics]
Solution for Problem 7
Intuitively it is obvious that the length of PC is greater than that of PB, but the proof should not
be based on intuition. Therefore we need to prove it.
Construct ∠ CAQ such that m ∠ CAQ=m ∠ BAP, and the length of AQ equal to that of AP.
∴ Δ BAP ≅ Δ CAQ. Therefore m ∠ AQC= m ∠ APB> m ∠ APC.
That is m ∠ AQC> m ∠ APC…….(1) AP=AQ implies m ∠ AQP= m ∠ APQ…….(2)
Subtract (2) from (1) we get m ∠ AQC- m ∠ AQP > m ∠ APC-m ∠ APQ. That is in ΔPQC ,
m ∠ PQC> m ∠ QPC, which implies PC>QC=PB
Then compare Δ ABP and Δ ACP. AB=AC, AP=AP and PC>PB. ∴ m∠BAP > m∠PAC .
Problem 8
A road is 300 miles long. A car, A, starts at noon from one end and drives toward
the other end at 50 mph. At the same time another car, B, driving uniformly at
100 mph, starts from the other end together with a fly traveling 150 mph. When
the fly meets car A, it immediately turns and flies toward car B. At what time does
the fly meet car B?
[Problem submitted by Kevin Windsor, LACC Associate Professor of Mathematics,]
Solution for Problem 8
A
d3
d1
d2
d1 = length of the 300 miles the fly flew that B did not cover while driving.
d2 = length of the 300 miles B drove until the fly contacted B.
d3 = length of the 300 miles A drove until the fly contacted A.
t1 is the time the fly flew until it contacted B.
It is the same as the time, t2, that B drove until the fly contacted B.
2d + d2
d
2d1 + d2 d2
t1 = 1
and t 2 = 2
⇒
=
⇒ 4d1 = d2
150
100
150
100
t3 is the time that the fly flew until it contacted A.
It is the same as the time, t4, that A drove until the fly contacted A.
Also, d1 + d2 + d3 = 300 ⇒ d1 + d2 = 300 − d3 .
d
300 − d3 d3
d + d2
d + d2 d3
t3 = 1
and t 4 = 3 ⇒ 1
=
⇒
=
⇒ d3 = 75
150
50
150
50
150
50
Solving for d1 and d2,
d1 + d2 + d3 = 300 ⇒ d1 + 4d1 + 75 = 300 ⇒ d1 = 45 ⇒ d2 = 180
It took the fly t1 hours to contact B,
2d + d2
2 ⋅ 45 + 180
t1 = 1
⇒
⇒ 1.8 hours or 1 hour and 48 mintues
150
150
When does the fly meet B? At 1:48PM.
Fly
B
Problem 9
Prove the following: Given the equation ay 3 + by = cx + d , where the coefficients a,
b, c and d are all integers. If ( x0 , y0 ) is an integral solution (both x0 and y0 are
integers) of the equation, then there must exist infinitely many integral solutions.
[Problem submitted by Steve Lee, LACC Professor of Mathematics]
Solution for Problem 9
Since ( x0 , y0 ) is a solution, we get ay03 + by0 = cx0 + d . This will be used in the following
derivation.
3
a ( y0 + nc ) + b ( y0 + nc )
2
3
= ( ay03 + by0 ) + a(3 y02 nc + 3 y0 ( nc ) + ( nc ) ) + bnc
=( cx0 + d )+ c[a (3 y02 n + 3 y0 n 2c + n3c 2 ) + bn]
= c[ x0 + a(3 y02 n + 3 y0 n 2c + n3c 2 ) + bn] + d
Therefore we have shown ( xn , yn ) is an integral solution, where
xn = [ x0 + a (3 y02 n + 3 y0 n 2c + n3c 2 ) + bn] and yn = y0 + nc .
Problem 10
(
)
Solve the equation: x 4 + 1 = 2x x2 + 1
[Problem submitted by Iris Magee, LACC Professor of Mathematics]
Solution for Problem 10
0 = (x4 +1) – 2x( x2 +1 )
0 = x4 – 2x3 – 2x + 1
0 = ( x2 – x + 1)2 – 3x2
So [(x2 – x + 1) - 3x ][(x2 – x + 1)+
3x ]=0
Thus x2 – ( 1+ 3 )x + 1 = 0 or x2 – ( 1- 3 )x + 1 = 0
Using the quadratic formula, we get four solutions:
( 1+ 3) ± 2 3
,
2
( 1- 3) ±
(
2
)
2 3 i