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Problem 1) Find the first integer of four consecutive positive integers such that the
difference of the sum of the squares of the last three integers and the sum of the squares
of first three integers is 171.
[Problem submitted by Kee Lam, LACC Professor of Mathematics. Source: Kee Lam]
Solution: Let n be the first of the four consecutive positive integers, then
(n + 1) is the second integer, (n + 2) is the third integer, and (n + 3) is the fourth integer.
[(n + 1)
2
2
2
] [
2
2
]
+ (n + 2) + (n + 3) − n 2 + (n + 1) + (n + 2) = 171
2
(n + 1)
2
2
2
2
+ (n + 2 ) + (n + 3) − n 2 − (n + 1) − (n + 2) = 171
(n + 3)2 − n 2 = 171
6n + 9 = 171
n = 27
Hence, the first integer of the four consecutive positive integers is 27.
Problem 2) Evaluate
1002 + 1004 + 1006 + ... + 2012 + 2014
.
1001 + 1002 + 1003 + ... + 2013 + 2014
[Problem submitted by Kee Lam, LACC Professor of Mathematics. Source: Kee Lam]
Solution: The numerator and denominator are each an arithmetic series, a series in which
the nth term is an = a1 + (n − 1)d with d being called the common difference. The sum
n(a1 + an )
of such a series is a1 + a2 + a3 + ⋅ ⋅ ⋅ + an =
.
2
In the numerator d = 2 and 2014 = 1002 + ( n − 1)2 . So, n = 507 . The numerator is
507(1002 + 2014)
= 507 ⋅ 1508 .
2
In the denominator d = 1 and 2014 = 1001 + (n − 1) . So, n = 1014 . The denominator is
1014(1001 + 2014)
= 507 ⋅ 3015 .
2
In its simplest form the fraction is
1508
.
3015
Problem 3) Solve 5 x − 5 x −2 = 120 5 .
[Problem submitted by Vin Lee, LACC Professor of Mathematics. Source: Saint Mary’s
College Mathematics Contest Problems for Junior and Senior High School by Brother
Alfred Brousseau, 1972]
3
Solution: Simplify the equation to get 5x − 5x −2 = 24 ⋅ 5 2
3
5 x (1 − 5−2 ) = 24 ⋅ 5 2
3
24
= 24 ⋅ 5 2
25
3
x 24
2
5 ⋅
= 24 ⋅ 5
25
5x ⋅
3
5x = 25 ⋅ 5 2
x
5 =5
7
x=
2
7
2
−1
Problem 4) For x ∉ {0,− 2 ,−1 / 2}define f (x ) =
. Define f (1) (x ) = f (x ),
x+ 2
f (x ) = f ( f (x )), f (x ) = f ( f ( f (x ))), and in general f (n+1) (x ) = f f n (x ) . Find and
simplify f (2014) (x ). (Do not bother to rational denominators, but simplify fractions.)
(2 )
(
(3)
)
[Problem submitted by Kent Merryfield, CSULB Professor. Source: Kent Merryfield]
Solution: f (2 ) (x ) =
− (− 1)
f
(3 )
(x ) = x + 2
− 2
− 2
⋅
x+ 2
(
− − 2
f (4 ) (x ) = x + 2
−1
−1
x+ 2
−x− 2
−x− 2
⋅
=
=
−1
2x + 1
+ 2 x + 2 − 1 + 2x + 2
x+ 2
+1
) −1
⋅
x+ 2
x+ 2
x+ 2
x+ 2
=
=
1 − 2x − 2
− 2+x+ 2
=
− 2x − 1
x
2−x− 2
=x
−1
x+ 2
We see that f (4 ) (x ) = x is the identity function. Hence, f (5 ) (x ) = f (1) (x ),
f (6 ) (x ) = f (2 ) (x ), and in general, f (n ) (x ) = f (m ) (x ) when n ≡ m (mod 4). Since
2014 ≡ 2 (mod 4), we have that f (2014) (x ) = f (2 ) (x ) =
−x− 2
2x + 1
or −
x+ 2
2x + 1
.
Comment: Linear fractional transformations compose like 2 × 2 matrices multiply. I was
looking for periodicity, so I was looking for a matrix whose eigenvalues were roots of
±1± i
unity. I was specifically looking for 8th roots of unity,
. So this function was
2
reverse-engineered from a matrix with characteristic polynomial λ2 − 2λ + 1, namely
⎡0 − 1 ⎤
. More sophisticated solutions invoking these matrices and their eigenvalues are
⎢1
2 ⎥⎦
⎣
possible.
Problem 5) Show that a triangle with sides whose lengths are
has an area given by a rational number. Find the area.
104 , 146 , and 106
[Problem submitted by Vin Lee, LACC Professor of Mathematics. Source: Saint Mary’s
College Mathematics Contest Problems for Junior and Senior High School by Brother
Alfred Brousseau, 1972]
Solution: We will use Heron’s formula, which expresses the area of a triangle in terms of
the lengths of its sides: A = s( s − a )(s − b)(s − c) .
a+b+c
which is called the semi2
104 + 146 + 106
s=
2
146 + 104 − 104
s−a =
2
104 + 106 − 146
s−b =
2
104 + 146 − 106
s−c =
2
74 + 106 146
s( s − a ) =
2
− 74 + 106 146
( s − b)( s − c ) =
2
106 ⋅ 146 − 74 2
s( s − a )(s − b)(s − c) =
= 2500
4
A = 50
Let a = 104 , b = 146 , c = 106 , and s =
perimeter. So,
Problem 6) Let m and s be the lengths of the longest and shortest altitudes, respectively,
of triangle ABC, and let P be a point in the interior of the triangle. Suppose
perpendiculars PX, PY, and PZ are dropped from P to the sides of the triangle. (If the
angles of the triangle are not acute, it may be necessary to extend the sides as shown.)
Prove that s ≤ PX + PY + PZ ≤ m and that both inequalities are strict unless the triangle
is equilateral.
[Problem submitted by Iris Magee, LACC Professor of Mathematics. Source: Iris
Magee]
Solution: Let a, b, and c be the lengths of the altitudes of triangle ABC from vertices A,
B, and C respectively, and let K denote the area of the triangle.
Then 2 K = a ⋅ BC = b ⋅ CA = c ⋅ AB. Now draw lines PA , PB , PC , as indicated, so that
ΔABC is split into the three triangles PAB, PBC, and PCA. Since the areas of these three
triangles sum to K and since PX , PY and PZ are their respective altitudes from P, we get
2K
2K
, CA =
, and
2K = (PX )(AB) + (PY )(BC ) + (PZ )(CA). Substituting BC =
a
b
2K
PX PY PZ
, we obtain, upon dividing by 2K, the equation 1 =
. Now,
AB =
+
+
c
c
a
b
(PX + PY + PZ ) ≤ 1 ≤ (PX + PY + PZ ) . Thus resulting in
1 ≥ a, b, c ≥ s , so this yields
m
s
the desired inequalities. Furthermore, if either of these is an equality, then a = b = c and
hence we would have BC = CA = AB
Problem 7) Suppose b is a real number, b > 0 , and b ≠ 1 . Find all pairs of real numbers
( x, y ) such that logb ( x − y ) = logb x − logb y . Justify your answer.
[Problem submitted by Vin Lee, LACC Professor of Mathematics. Source: Vin Lee]
Solution: Since the domain of log b is (0, ∞) , x > y > 0 .
x
Use the relationship logb x − logb y = logb to substitute into the equation given in the
y
problem getting
x
log b ( x − y ) = log b . So,
y
x
x− y=
y
2
xy − y = x
0 = y 2 − xy + x . Use the quadratic formula to solve this equation
for y in terms of x.
x ± x2 − 4x
2
2
Since y is a real number, x − 4 x ≥ 0
x ( x − 4) ≥ 0
Since x > 0 , x − 4 ≥ 0 . So, x ≥ 4 . Any solution must have y greater than zero, and this
y=
is obviously true for y =
x is y =
x + x2 − 4x
. However, we must determine for what values of
2
x − x2 − 4x
> 0:
2
x>0
0 > −4 x
x2 > x2 − 4x
x > x2 − 4x
x − x2 − 4x > 0 .
So, y =
x − x2 − 4x
> 0 for any x > 0 .
2
Therefore, logb ( x − y ) = logb x − logb y for every ( x, y ) such that x ≥ 4 and
y=
x ± x2 − 4x
.
2
Problem 8) Five squares are arranged as shown below. Demonstrate that the area of the
shaded square is equal to that of the shaded triangle.
[Problem submitted by Roger Wolf, LACC Professor of Mathematics. Source: Roger
Wolf]
Solution:
⎧
1 ⎡ b
Area of the shaded triangle = ⎨ab + b ⎢
(2b − 2a )⎤⎥ ⎫⎬ + ⎧⎨ab − 1 a ⎡⎢ b (2b − 2a )⎤⎥ ⎫⎬
2 ⎣ a + b
2 ⎣ a + b
⎦ ⎭ ⎩
⎦ ⎭
⎩
3
2
3
2
b − ab
a −a b
= 2ab +
+
a+b
a+b
3
3
2
a + b a b + ab 2
= 2ab +
−
a+b
a+b
2
2
= 2ab + a − ab + b − (ab )
(
)
= a2 + b2
= Area of the shaded square
1
3 x for x in terms of z and find the domain of the function
Problem 9) Solve z =
1
3x + x
3
x = f (z ) .
3x −
[Problem submitted by Vin Lee, LACC Professor of Mathematics. Source: Vin Lee]
Solution:
1
3x
z=
1
3x + x
3
x
3 − 3− x
z= x
3 + 3− x
z3x + z3− x = 3x − 3− x
z32 x + z = 32 x − 1
z32 x − 32 x = − z − 1
32 x ( z − 1) = − z − 1
− z −1
32 x =
z −1
1+ z
32 x =
1− z
⎛ 1 + z ⎞
2 x = log 3 ⎜
⎟
⎝ 1 − z ⎠
1
⎛ 1 + z ⎞
x = log 3 ⎜
⎟
2
⎝ 1 − z ⎠
3x −
Since the domain of log 3 is positive numbers
1+ z
> 0 which implies − 1 < z < 1 .
1− z
Problem 10) In the diagram the length of each side of the square is 2 and the bottom of
the square is the diameter of the semicircle. The circle is tangent to the semicircle and to
two sides of the square as shown in the diagram.
A) Compute (with proof) the radius of the second circle.
B) Then as we add the smaller circle, tangent to the semicircle, the first circle and the
side of the square as shown, show that the radius of the new smaller circle is exactly 1/3
the radius of the larger circle.
[Problem submitted by Iris Magee, LACC Professor of Mathematics. Source: Iris
Magee]
Solution to Part A:
Let R be the radius of the circle. Connect the center of the circle and the center of the
semicircle as shown below.
To find the radius of the circle, R, we can solve by using Pythagorean Theorem.
(2 − R )2 + (1 − R )2 = (1 + R )2
4 − 4R + R 2 + 1 − 2R + R 2 = 1 + 2R + R 2
R 2 − 8R + 4 = 0
R = 4−2 3
Solution to Part B: Draw a vertical line through the center of the small circle and form
two right triangles as shown.
The hypotenuse of the upper triangle joins the centers of the two circles and hypotenuse
of the lower triangle joins the center of the small circle and the center of the semicircle.
Let R and r be the radii of the large circle and small circle respectively, recall that
R = 4 − 2 3 . Also, of course, the radius of the semicircle is 1.
The length of the hypotenuse of the upper triangle is R + r, and its horizontal side has
length R – r. It follows by the Pythagorean Theorem that the square of the length of the
vertical side is (R + r )2 − (R − r )2 = 4rR. Similarly, for the lower triangle, the lengths of
its hypotenuse and horizontal side are respectively 1 + r and 1 – r, and hence the square
of the length of the vertical side is 4r. We see from the picture that the total of the lengths
of the vertical sides of the two triangles is 2 – R, and this yields the equation
4rR + 4r = 2 − R = 2 3 − 1 , where the last equality follows from the known value of
(
R. Simplification yields
Observe that
This yields
r
(
)
)
R + 1 = 3 − 1, and so
r=
(
)(
)
3 −1 / R +1.
( 3 − 1)
2
= 4 − 2 3 = R and since 3 − 1 > 0, we see that 3 −1 = R .
r = R / 3 and so r = R / 3 , as we wanted!
Again using the two right triangles shown above, the Solution to Part B may be
summarized as
2 = R + ( R + r) 2 − ( R − r) 2 + (1 + r) 2 − (1 − r) 2
2 − R = 4 Rr 2 + 4r Now substitute R = 4 − 2 3 from part A to get
2 3 − 2 = 2 r ( R + 1)
3 − 1 = r ( R + 1) *
Observe that
( 3 − 1)
2
= 4−2 3 = R
So, 3 − 1 = R
Substituting this result into * above gives R = r ( 3 − 1 + 1) .
Therefore, R = 3 r
r = R/3