Download 2008 Mississippi Mu Alpha Theta Inter-School Test

2008 Mississippi Mu Alpha Theta Inter-School Test (Key)
1. The equations x 2 + 2ax + b = 0 and x 2 + bx + 16a = 0 have real roots. If a and b are positive
real numbers, find the smallest possible value of a + b.
Since the roots of x 2 + 2ax + b = 0 are both real we know that the discriminant 4 a 2 – 4b is
nonnegative, that is, a 2 ≥ b. Similarly, b 2 ≥ 64a from the second equation. Therefore, a 4 ≥ b 2 ≥
64a. Since a is positive, this last inequality implies that a 3 ≥ 64 or a ≥ 4. Thus b 2 ≥ 64(4) or b ≥
16 since b is positive. Using a = 4 and b = 16 produces real identical solutions for both
equations. As a consequence, the minimum value of a + b is 20.
2. Find all integer solutions of the equation 2 2 x  5 2 y = 39.
The equation 2 2 x  5 2 y = 39 is equivalent to ( 2 x  5 y )( 2 x  5 y ) = 3(13) or ( 2 x  5 y )( 2 x  5 y ) =
39(1). Since x and y are both integers we know that both 2 x and 5 y are integers. Solving 2 x  5 y
= 3 and 2 x  5 y = 13 simultaneously results in x = 3 and y = 1. There is no integer solution for
the other possible system of equations ( 2 x  5 y = 13 and 2 x  5 y = 3) nor for ( 2 x  5 y )( 2 x  5 y )
= 39(1). Therefore, x = 3 and y = 1 is the only integer solution of the equation 2 2 x  5 2 y = 39.
3. Let q1 ( x) and r1 be the quotient and remainder, respectively, when the polynomial f(x) = x 16
is divided by x +
1
2
. Let q2 ( x) and r2 be the quotient and the remainder, respectively, when
q1 ( x) is divided by x +
Let a = –
1
2
1
2
. Find r2 .
. Since x 16 = (x – a) q1 ( x) + r1 , the remainder theorem implies that r1 = f(a) =
a 16 . Since q1 ( x) = (x – a) q2 ( x) + r2 , the remainder theorem implies again that r2 = q1 (a) .
Solving x 16 = (x – a) q1 ( x) + a 16 for q1 ( x) we obtain
q1 ( x) =
x 16  a 16
= ( x 8  a 8 )( x 4 + a 4 )( x 2 + a 2 ) (x + a)
xa
Therefore,
15
r2 = q1 (a) = 16 a
15
1
 1
=16    = –
2048
 2
4. Let ABC be a right triangle such that the square of the hypotenuse equals to 5/2 of the
product of the legs. Find the tangent of the greatest acute angle.
Let c be the hypotenuse of the triangle and a and b its legs. The condition and the Pythagorean
Theorem imply that 2 a 2 – 5ba + 2 b 2 = 0, which is a quadratic equation in a (or b). Using the
quadratic formula to solve for a in terms of b we obtain, a =
5b  3b
. Thus a = 2b or a = b/2.
4
Using a = 2b results in tan A = 2 and tan B = 1/2. Using a = b/2 results in tan A = 1/2 and tan B
= 2. Therefore, the tangent of the greatest acute angle is 2.
5. Let ABF, BCD, and ACE be outward equilateral triangles constructed on the sides of an
arbitrary triangle ΔABC. Prove that segments AD, BE, and CF are congruent.
E
C
D
A
B
F
Since AE = AC,  EAB   CAF (both equal to 60 +  CAB)), and AB = AF, we have that
ΔABE  ΔAFC by the side angle side congruence criterion. Therefore, BE = CF since
corresponding sides of congruent triangles are congruent. Similarly, CF = AD since ΔBCF
2
 ΔBDA by the same congruence criterion (BC = BD,  CBF   DBA, and BF = BA). We
conclude that segments AD, BE, and CF are congruent.
6. Find the point of intersection of the angle bisectors of the interior angles of a triangle whose
vertices are (-2, -3), (32, -17), and (
1 29
,
).
2 2
The equations of the sides of the triangle (see figure below) are
l: 7x + 17y + 65 = 0
m: x + y – 15 = 0
n: 7x – y + 11 = 0
The point of intersection of the angles bisectors of a triangle is the center (h, k) of the inscribed
circle in the triangle. Let d 1 , d 2 , and d 3 be the distances from point (h, k) to lines l, m, and n,
respectively. Thus,
d1 =
7h  17k  65
d2 =
 338
h  k  15
2
d3 =
7h  k  11
 50
Point (h, k) is equidistant from the sides of the triangle and all distances are negative (point (h,
k)) and the origin are on the same side of each line). Then, d 1 = d 2 = d 3 .
Since d 1 = d 2 ,
Since d 2 = d 3 ,
7h  17k  65
 13 2
h  k  15
2
=
=
h  k  15
2
7h  k  11
5 2
Simplifying,
3
2h + 3k – 13 = 0
3h + k – 16 = 0
Solving the last two equations simultaneously produces h = 5, k = 1.
20
C
10
n
-20
O (h, k)
m
20
A
-10
l
B
-20
7. Let ABCD be a non-convex quadrilateral with angle C as the interior angle greater than 180º.
Let E be any point in the plane. Prove that AE + BE + CE + DE ≥ AC + BC + DC.
D
C
E
A
B
We have that BE + DE ≥ BC + CD (a proof of this lemma will follow) and AE+ CE ≥ AC. These
two inequalities imply that AE + BE + CE + DE ≥ BC + CD + AC. A proof that BE + DE ≥ BC
+ CD follows: let F be the point of intersection of lines BC and DE (See figure below). Applying
4
the triangle inequality twice we get BE + DE ≥ BF + FD ≥ BC + CD. The argument can be
modified for different positions of E.
D
C
F
E
A
B
8. The speed of light depends on the medium through which it travels. Fermat principle states
that the light travels from one point to another following a path for which the time of travel is
a minimum. A ray of light passes though a medium with velocity v1 from a point A to some
point B in a second medium with velocity v2 . If 1 and  2 are, respectively, the angle of
incidence and the angle of refraction, prove that the path followed by the ray of light satisfies
sin  1
v
= 1 .
sin  2
v2
A
1
C
2
B
Let t denote the time of travel of the light from point A to point B and c the distance from O to D
(See figure below). So
t=
AC CB
+
v1
v2
5
a2  x2
+
v1
=
b 2  (c  x ) 2
v2
This equation represents a function of x with domain [0, c]
dt
x
 (c  x )
=
+
dx
v1 a 2  x 2
v 2 b 2  (c  x ) 2
=
sin  2
sin 1
–
v2
v1
Restricting x to the interval [0, c], we know that t has a negative derivative at x = 0 and a positive
derivative at x = c. The Intermediate Value Theorem for Derivatives implies that there is a point
x 0 in the interval [0, c] such that
dt
dt
= 0. This point is unique because
is an increasing
dx
dx
function of x. At this point
sin  2
sin 1
=
v2
v1
Therefore
sin  1
v
= 1
sin  2
v2
y
A
1
a
O
C
x
D
2
b
B
6
9. Let a, b, and c be nonnegative real numbers. Prove that ab + bc + ca ≥
3abc(a  b  c ) .
Since a, b, and c be nonnegative real numbers, this inequality is equivalent to
a 2 b 2 + b 2 c 2 + c 2 a 2 + 2abc(a + b + c) ≥ 3abc (a + b + c)
This, in turn, is equivalent to
a 2 b 2 + b 2 c 2 + c 2 a 2 – a 2 bc – a b 2 c – ab c 2 ≥ 0
This last inequality can be rewritten as
(ab  bc) 2 + (bc  ca) 2 + (ca  ab) 2 ≥ 0.
10. Let S n = 1 – 2 + 3 – 4 + … + (1) n 1 n, n = 1, 2, … . Find S159 + S170 .
S159 = 1 – 2 + 3 – 4 + … + 159 = 1 + 1(158)/2 = 1 + 79 = 80
S170 = 1 – 2 + 3 – 4 + … – 170 = – 1(170)/2 = – 85
Therefore,
S159 + S170 = – 5
11. Let N be a natural number whose representation in base 9 ( abc9 ) has three digits. If N =
abc9 = cba7 , find the representation of N in base 10.
abc9 = 81a + 9b + c = 49c + 7b + a = cba7
Thus,
7
80a + 2b – 48c = 0
Solving for b
b = 8(3c – 5a)
In other words, b is a multiple of 8. Since b ≤ 6, we conclude that 3c – 5a = 0 or 3c = 5a.
Therefore, b = 0. Since 3c is a multiple of 5 and 0 < c < 7 (c appears is the first digit of N in base
7) we obtain c = 5. Thus a = 3. Hence, N = 81(3) +9(0) + 5 = 248.
12. The probability of getting heads when tossing a biased coin is
1
. Find the probability of
4
obtaining an even number of heads when the coin is tossed 100 times.
Let p the probably of getting an even number of heads and let q be the probability of getting an
odd number of heads. The probability of getting k heads and 100 – k heads is
100   1   3 

    
k   4  4 
k
100 k
Therefore
100   1   3 
    
p = 
0   4   4 
0
100   1   3 
    
q= 
1

  4 4
1
100
99
100   1   3 
     + … +
+ 
2   4  4 
2
98
100   1   3 
     + … +
+ 
3

 4 4
3
97
100   1 

  
100   4 
100
100   1   3 

    
99

 4  4
99
0
3
  and
4
1
As a consequence,
p–q=
100  1   3 
   
(1) k 

k 0
 k  4   4 
100
k
100 k
8
Applying the binomial theorem we obtain
100  1   3 
   
(1) 

k 0
 k  4   4 
k
100
k
100 k
 1 3
=   
 4 4
1
= 
2
1
Since p + q = 1 and p – q =  
2
p=
100
100
100
we obtain
1
1 
1  100 
2 2 
59049
13. Let [x] denote the greatest integer less or equal to x. Find
 log
N 1
3
N .
 1 for 3  N  3 2

2
3
 2 for 3  N  3
 3 for 33  N  3 4
log 3 N  
...
 8 for 38  N  39

9 for 39  N  310

10
10 for N  3
The requested sum is
59049
 log
N 1
3
N  = 1( 3 2 – 3) + 2( 33 – 3 2 ) + 3( 3 4 – 33 ) + … + 9 ( 310 – 39 ) + 10
= – 3 – 3 2 – 33 – … – 39 + (9) 310 + 10
9
= (9) 310 – (3 + 3 2 + 33 + … + 39 ) + 10
= (9) 310 – (1 + 3 + 3 2 + 33 + … + 39 ) + 11
 310  1 

= (9) 310 – 
 2 
= 501917
14. If p and q are the roots of the equation ax 2 + bx + c = 0, find
1
1
+ 2.
2
p
q
1
1
p2  q2
+ 2 =
p2
q
p2q2
p 2  q 2  2 pq  2 pq
=
p2q2
( p  q) 2  2 pq
=
p2q2
Since p + q = 
b
c
and pq =
the last equality becomes
a
a
2
c
 b
   2
a
a
= 
2
c
a2
 b 2  2ca  2

a
a 2 

=
c2
10
=
b 2  2ac
c2
15. If the lengths of the sides of a triangle ABC are 15, 36, and 39, find the length of the radius
of the inscribed circle.
By the converse of the Pythagorean Theorem, ΔABC is a right triangle. Let r be the radius of the
inscribed circle. Applying the power of a point theorem we obtain AF = 15 – r and BF = 36 – r.
Thus
15 – r + 36 – r = 39
Therefore, r = 6.
B
36-r
36-r
r O
D
r
C
r
r
r
F
15-r
E 15-r
A
11
2008 Mississippi Mu Alpha Theta Inter-School Test (Key)
Tie breakers
1. Prove that 
1 ( x  y )(1  xy) 1
≤
≤
(Hint: Use a trigonometric substitution such as
2 (1  x 2 )(1  y 2 ) 2
x = tan a).
Let x = tan a and y = tan b. Thus
x + y = tan a + tan b =
=
sin a
sin b
+
cos a
cos b
sin( a  b)
cos a cos b
1 – xy = 1 – tan a tan b = 1 –
=
1 + x2 =
1
cos 2 a
1 + y2 =
1
cos 2 b
The inequality 
sin a sin b
cos a cos b
cos( a  b)
cos a cos b
1 ( x  y )(1  xy) 1
≤
≤
is equivalent to
2 (1  x 2 )(1  y 2 ) 2
12
–1≤
2 sin( a  b) cos( a  b) cos 2 a cos 2 b
≤1
(cos a cos b)(cos a cos b)
Using the identity sin 2A = 2 sin A cos B we obtain the equivalent inequality
– 1 ≤ sin 2(a + b) ≤ 1
 n  1
2. Prove whether the sequence whose nth term is a n = 
 converges.
 n 1
n
Since the limit leads to the indeterminate form 1 we take logarithms to change this form to 0/0
so we can apply l’Hôpital rule
lim ln a
n
n
=
 n 1
lim n ln  n  1 
n
=
lim
n
 n 1
ln 

 n 1
1/ n
Applying l’ Hôpital rule we obtain
=
lim
 2 /( n 2  1)
 1/ n 2
lim
2n 2
n2 1
n
=
n
=2
Since a n = e ln an and e x is continuous we have that
lim
n
an =
lim e
ln a n
= e2
n
13
3. Let ABC any triangle. Prove that (cos A)(cos B)(cos C) ≤
The inequality (cos A)(cos B)(cos C) ≤
1
.
8
1
is equivalent to 1 – 8(cos A)(cos B)(cos C) ≥ 0. Since
8
(cos B)(cos C) = [cos (B + C) + cos (B – C)]/2 we obtain
1 – 4cos A cos (B + C) – 4cos A cos (B – C)] ≥ 0
Observe that cos (B + C) = – cos (180 – B – C) = – cos A. Thus the previous inequality is
equivalent to
1 – 4cos A cos (B – C) + 4 cos 2 A ≥ 0
Using the trigonometric identity sin 2 (B – C) + cos 2 (B – C) = 1 we obtain
1 – 4cos A cos (B – C) + 4 cos 2 A
= sin 2 (B – C) + cos 2 (B – C) – 4cos A cos (B – C) + 4 cos 2 A
= sin 2 (B – C) + [cos (B – C) – 2cos A] 2 ≥ 0
In conclusion, 1 – 4cos A cos (B – C) + 4 cos 2 A ≥ 0
14