SW MO-ARML Practice
April 15, 2007
Solutions to Relay Questions
(from ARML 1998/1999/2000 Tournaments)
B
A
(1) In rectangle ABCD, the area of ∆ADE, ∆CBE, and ∆AEB
form an arithmetic sequence. If the ratio of the area of the
largest triangle to the smallest can be expressed in simplest
m
terms as , compute m + n.
n
D
E
C
Answer.
Let the areas of ∆ADE, ∆CBE, and ∆AEB be a, a + d, and a + 2d. Note that the area of
∆AEB is half of the area of the rectangle, therefore a + (a + d) = a + 2d. Thus a = d,
m a + 2d 3a 3
implying
=
=
= and m + n = 4.
n
a
a 1
(2) Let T ≡ 4 (mod 5) and is not divisible by 3. The digits T, T + 1, T + 2, and T + 3 are placed
at random in the indicated spaces 1 _ 9 _ 9 _ 8 _ . Each digit is used, one per space.
Compute the probability that the resulting number is divisible by 3 or 5.
Answer.
Since 1 + 9 + 9 + 8 = 27 is divisible by 3, the resulting number is divisible is divisible by
3 if T + (T + 1) + (T + 2 ) + (T + 3) = 4T + 6 is divisible or if T is divisible by 3, which is
not the case. Also, the resulting number is divisible by 5 if the last digit is 0 or 5. Since
the digits to be inserted are ≡ 4, 5, 6, and 7, the probability that the last digit is 0 or 5
1
is .
4
(3) Express
x
1 1
1
in terms of T if x + = and y + = T .
y
y T
x
Answer.
1 1
1
and y + = T
=
y T
x
y
⇒
= xT
T
x+
⇒ xy + 1 =
⇒
y
and xy + 1 = xT
T
x
1
= 2
y T
(4) A box 4-by-6-by-8 is resting on the floor. A box 2-by-3-by-5 is placed on top of the first
box forming a two-box tower. If A is the exposed surface of the tower, compute the least
possible value of the area A.
Answer.
The least exposed value occurs when both boxes are placed with the side of largest area
face down. Note that one 3-by-5 side of the smaller box is hidden and also it hides an
area of 3-by-5 on the top of the larger box. Also, one 6-by-8 side of the larger box is
hidden. ∴The total exposed surface area is 2(24 + 32) + 48 – 15 + 2(6 + 10) + 15 = 192.
(5) In the regular n-gon A1A2…An, the measure of ∠A2A1A4 = K. Express the value of n in
terms of K.
Answer.
It is intuitive (?) that segments A1A4 and A2A3 are parallel.
Therefore, the exterior angle of ∠A2 is congruent to
∠A2A1A4 and its measure is also K. Therefore,
( n − 2)180
360
360
+ K = 180 , implying that K =
and n =
.
n
n
K
A1
A2
A3
A4
(6) Let f be a linear function with positive slope passing through the origin. Express the least
value of f(5T) + f −1(5T) in terms of T.
Answer.
1
1
1

x ⇒ f(5T) + f −1(5T) = 5mT + 5 T = 5T  m +  .
m
m
m

1
is 2, implying that the least value of
It is intuitive (?) that the minimum value of m +
m
2
2
1 
1

−1
f(5T) + f (5T) is 10T. [ Note:  m +  =  m −  + 4 ≥ 4 . ]
m 
m

Let f(x) = mx. Then, f −1(x) =
(7) The sum of the digits of the year 1999 is 28. Let Y be the next following year in which the
sum of the digits is 28. Compute Y – 1999.
Answer.
For the year 2ABC, A + B + C = 26. This can only be done using two 9’s and one 8.
B
The earliest such Y is therefore 2899, and Y – 1999 = 900.
A
(8) Trapezoid ABCD is divided into four congruent trapezoids
as shown.
If AB = k and DC = 2k, compute the sum of the lengths of
all line segments in the figure in terms of k.
D
Answer.
k
. It becomes clear that the sum of the
2
k
15k
lengths of all line segments = 4k + 7⋅ =
.
2
2
Since EF = AB = k, we have DE = FC =
C
(9) In ∆ABC, sin ∠A = cos ∠B. If, for x in degrees, m∠A = x2 + 5x, m∠B = x 2 – x – 36, and
m∠A − m∠B = 18, compute m∠A.
Answer.
All measurements are in degrees. We know either m∠A = 90 – m∠B or
m∠A = 90 + m∠B. If m∠A = 90 – m∠B, then x2 + 5x = 90 – (x 2 – x – 36), or
2x2 + 4x – 126 = 0 ⇒ x2 + 2x – 63 = 0 ⇒ x = 7 or –9
⇒ A = 84, B = 6; or A = 36, B = 54
If m∠A = 90 – m∠B, then x2 + 5x = 90 + (x 2 – x – 36), or 6x – 54 = 0
⇒x=9
⇒ A = 126, B = 36
Since m∠A − m∠B = 18, we must have A = 36, B = 54. Therefore, m∠A = 36.
(10) Let k be the smallest of six consecutive positive integers. If the sum of the six integers is
divisible by three distinct primes, compute the smallest possible value for k.
Answer.
k + (k + 1) + ⋅⋅⋅ + (k + 5) = 6k + 15 = 3(2k + 5) is odd, so 2 is not one of the three primes.
Try 3(2k + 5) = 3⋅5⋅7, implying k = 15.
P
(11) BC is a diameter of circle O and AP is tangent to
circle O at P. If AP = 2T and AB = T, express the
length of AO in terms of T.
A
B
O
C
Answer.
Let r be the radius of circle O. Then, in the right triangle APO, (2T)2 + r 2 = (T + r)2, or
3
5
3T 2 = 2Tr. Therefore, 3T = 2r, and the length of AO is T + r = T + T = T .
2
2
 2T 
(12) For x randomly chosen from the interval  0,
 , express the probability that
 25 
15x2 + 3 < 14x in terms of T, assuming T > 10.
Answer.
1 3
The solution set of 15x2 – 14x + 3 = (5x – 3)(3x – 1) < 0 is  ,  . Therefore, the
3 5
4
20 10
.
=
probability is equal to 15 =
2T
6T 3T
25
(13) Let [x] denote the greatest integer less than or equal to x. Compute the largest integer n such
that  n 2000  > 1 .
Answer.
Since we want  n 2000  ≥ 2 , then 2000 ≥ 2n. The largest such n is 10 when 210 = 1024.
( )
(14) Express the largest solution to ( logT x ) = logT x 2 in terms of T if T > 1.
2
Answer.
( logT x )
2
− 2 logT x = 0 ⇒ logT x ( logT x − 2 ) = 0
⇒ logT x = 0 or logT x = 2 ⇒ x = T 0 = 1 or x = T 2. The answer is T 2.
(15) Assuming T is a positive power of 10. Let n ∈ {1, 2, 3, …, T}. In terms of T, express the
number of values of n for which the product (n2 – 2n + 2)(n2 + 2n + 2) is divisible by 5.
Answer.
Note that (n2 – 2n + 2)(n2 + 2n + 2) = (n2 + 2)2 – (2n)2 = n4 + 4. To be divisible by 5, n4
must end in 1 or 6, which implies that n does not end in 0 or 5 (i.e., all others will
T
work.) Therefore, there are numbers in {1, 2, 3, …, T} not ending in 0 or 5, and
5
4T
numbers such that (n2 – 2n + 2)(n2 + 2n + 2) is divisible by 5.
there are
5
(16) For 0 < x < 2π, compute the absolute value of the difference between the solutions to the
equation: sec x = 1 + cos x + cos2 x + cos3 x + …
Answer.
1
or equivalently sec x – 1 = 1, sec x = 2, cos x = 1/2.
1 − cosx
π
5π
4π
.
There are two solution between 0 and 2π: and , and their difference is
3
3
3
We need to solve sec x =
(17) Let K be a positive integer. If (a + bi)K(a – bi)K = 512, express the value of a2 + b2 in terms
of K.
Answer.
Since we want (a2 + b2)K = 512, a2 + b2 = 5121/K.
A
(18) In the given graph, arcs are measured in degrees.
Segments PA and PC are tangents to the circle.
m(arc ABC )
m(arc ADC )
= T + 1 , express
in terms of T.
If
m(arc ADC )
m(∠P)
P
D
B
C
Answer.
All measurements are in degrees. If the measure of the arc ADC = x, then the measure
of the arc ABC = 360 – x and the measure of ∠P = (360 – x – x)/2 = 180 – x. Thus,
360 - x
360
= T +1 ⇒ x =
. Thus,
x
T +2
360
m (arc ADC )
x
360
2
(T + 2)
=
=
=
= .
360
m (∠P )
180 − x 180 −
180(T + 2) − 360 T
(T + 2)