HS 2012 solutions for problems 12 through 20
12. Sally walks on a level path from A to B at 4 mph, up a hill from B to C at 3
mph, back down to B at 6 mph and finally returns from B to A at rate 4 mph. If
her round trip time was 6 hours then her total round trip distance, in miles, was
a) 12
b) 6
c) 24
d) cannot be determined
e) none of these
Solution: If the level distance is L and the hill distance is H then, since the hike time is 6, we
have
6 = L/4 + H/3 + H/6 +L/4 = (L + H)/2.
Thus L + H = 12 and the total trip distance is 2(L + H) = 24.
C
13. An equilateral triangle is inscribed in a circle. If the triangle area is
3 , then the circle area is
O
a) 3  /2
these
b)
 3
c)
2 3 / 3
d) 3 3 / 4
e) none of
A
D
B
Solution: Let O be the circle center and x be the distance DB. Since angle OBD is 30 degrees
then OB = 2x / 3 also OD = x 3 . The circle area is  (OB) 2 = 4x 2 / 3 and the equilateral
triangle area is given by
4  /3.
3 x 2 . Since the triangle area is
3 then x = 1 and the circle area is
14. A coin,1/2 inch in diameter, is tossed onto a black and red checkerboard. The center of the
coin lands randomly on one of the squares of the board. If the squares have 2 inch sides, then the
probability that the coin lands entirely within one of the black squares is
a) 1/2
b) 9/16
c) 1/4
d) 9/32
e) none of these
Solution: The larger square in the figure is one of the 2x2 black squares and the
inner square is 3/2x3/2 and centered in the 2x2 square. For the coin to be entirely
inside the 2x2, the center of the coin must be in the smaller square. Thus if the
center is in a black square, the entire coin will be in that square 9/16 of the time.
Also the center will land in a black square 1/2 the time. Thus the entire coin will
be in a black square 9/32 of the time.
15.
The trapezoid ABCD shown has AB parallel to CD, AD perpendicular to AB,
AD = 12, AB = 18, and an area of 186. The perimeter of the trapezoid is
a) 56
b) 54
c) 62
e)
60
e) none of these
D
Solution: Let E be on AB such that AB and CE are perpendicular. Thus
186 = AD(AB + DC)/2 = 12(18+DC)/2, hence DC = 13 and EB = 5. Thus
BC  52  122  13 and perimeter = 12 + 13 + 13 +18 = 56.
A
16. If f ( x)  1  x3 and [ g ( x)]3  x  1 , then the sum values of all possible distinct values of
g ( f (2 )) is
a) 0
b) 2
Solution:
c) 4
d) 6
e) none of these
f (2)  9 Let z  g ( f (2))  g (9) . To find all possible solutions of z 3  8 .
Factoring gives z 3  8  ( z  2)( z 2  2 z  4) , and the three possible distinct values of z are



{2,  1  i 3 ,  1  i 3 }. The sum of these is 0. This could also be found without factoring by
noting that the sum of the zeros of a cubic is the negative of the coefficient of the quadratic term.
17. Let A  x 2  6 x  6 and B  x 2  4 x  60 . Find all possible distinct values of x that satisfy
the equation AB  1 . The sum of all these x values is
a) -4
b) 6
c) 2
d) 0
e) none of these
Solution: A = 1 or B = 0. If A = 1 then x = 7 or x = -1. If B = 0 then x = 6 or x = -10. Hence the
sum is 2.
18. In the 4 × 4 grid shown, three coins are randomly placed in different squares. The
probability that no two coins lie in the same row or column is
a) 9/64
b) 6/35
c) 7/40
e) 9/65
e) none of these
Solution: After placing the first coin, there are 9 possible squares
out of 15 for the second coin. After placing the second coin in
one of these 9 squares, there are there are 4 possible squares out
of 14 for the third coin. Thus the probability that no two coins lie in
the same row or column is (9/15)(4/14)=6/35
C
E
B
19. A rectangular piece of paper, PQRS, has PQ = 15 and QR = 20. The piece of
paper is glued at on the surface of a large cube so that P and R are at vertices of the
cube. (Note that PQR and PSR lie at on the front and top faces of the cube,
respectively.) The shortest distance from Q to S, as measured through the cube, is
a)
384
b)
337
c)
342
d)
386
S
P
T
U
R
Q
e) none of these
Solution: The rectangle PQRS is shown. The several triangles all have sides in the
ratio (3,4,5). Thus TQ = 12, PT = 9, PU = 16, and SU = 12. Imbed the cube in
an (x,y,z) coordinate system as shown. Thus
Q = (0,9,-12) and S = (-12,16,0). Hence the distance
QS  122  72  122  337 .
z
S
y
P
R
Q
x
20. Given the rectangle ABCD with AB = 8 inches and AD = 4 inches. The rectangle
is folded so that vertices A and C coincide. What is the length of the crease?
a) 3 6 / 2
b) 3 5 / 2
c) 4 6 / 3
d) 4 5 / 3
e) none of these.
A
Solution: Let AD = a and AB = b be the dimensions of the rectangle. The heavy
line HE is the fold and the dashed lines are edges after the fold. Note that D folds
to G, AE folds to EC, A folds to C and AH folds to HC. Since AF = CF then the
triangles AFH and CFE are congruent (A.A.S). Thus HB = DE = EG and call
their common value x. Since HC = AH = b  x , then from triangle HBC we have
b2  a 2
a2
.
 b  2x 
x 2  a 2  (b  x)2  2 x 
b
b
a2
The right triangle HKE has legs HK = a and KE = b  2 x  .
b
Thus crease length = EH =
crease length = 2 5 .
a2 
a4 a 2 2

a  b . For a = 4 and b = 8, we have
2
b
b
D
E
G
F
H
B
K
C