-
Since \((a,0)\) is on the line with
equation \(y=x+8\), then
\(0=a+8\) or
\(a=-8\).
Answer: \(-8\)
-
Simplifying,
\[\begin{aligned} x & =
\left(1-\tfrac{1}{12}\right)\left(1-\tfrac{1}{11}\right)\left(1-\tfrac{1}{10}\right)\left(1-\tfrac{1}{9}\right)\left(1-\tfrac{1}{8}\right)\left(1-\tfrac{1}{7}\right)\left(1-\tfrac{1}{6}\right)\left(1-\tfrac{1}{5}\right)\left(1-\tfrac{1}{4}\right)\left(1-\tfrac{1}{3}\right)\left(1-\tfrac{1}{2}\right)\\
& =
\left(\tfrac{11}{12}\right)\left(\tfrac{10}{11}\right)\left(\tfrac{9}{10}\right)\left(\tfrac{8}{9}\right)\left(\tfrac{7}{8}\right)\left(\tfrac{6}{7}\right)\left(\tfrac{5}{6}\right)\left(\tfrac{4}{5}\right)\left(\tfrac{3}{4}\right)\left(\tfrac{2}{3}\right)\left(\tfrac{1}{2}\right)\\
& = \tfrac{11\cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5
\cdot 4 \cdot 3 \cdot 2 \cdot 1}{12 \cdot 11\cdot 10 \cdot 9 \cdot 8
\cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} \\ & =
\tfrac{1}{12} \qquad \mbox{(dividing out common
factors)}\end{aligned}\]
Answer: \(\frac{1}{12}\)
-
The following pairs of rectangles are not touching:
\(AC, AE, CD\).
There are 3 such pairs.
Answer: 3
-
Let the side length of the square be
\(s\).
Since the diagonal of length 10 is the hypotenuse of a right-angled
triangle with two sides of the square as legs, then
\(s^2 + s^2 = 10^2\) or
\(2s^2 = 100\), which gives
\(s^2 = 50\).
Since the area of the square equals
\(s^2\), then the area is
\(50\).
Answer: 50
-
To make \(n\) as large as possible,
we make each of the digits \(a\),
\(b\),
\(c\) as large as possible, starting
with \(a\).
Since \(a\) is divisible by
\(2\), its largest possible value is
\(a=8\), so we try
\(a=8\).
Consider the two digit integer
\(8b\).
This integer is a multiple of
\(3\) exactly when
\(b=1,4,7\).
We note that \(84\) is divisible by
\(6\), but
\(81\) and
\(87\) are not.
To make \(n\) as large possible, we
try \(b=7\), which makes
\(n=87c\).
For \(n=87c\) to be divisible by 5,
it must be the case that \(c=0\) or
\(c=5\).
But \(875 = 7 \cdot 125\) so
\(875\) is divisible by 7.
Therefore, for \(n\) to be divisible
by 5 and not by 7, we choose
\(c=0\).
Thus, the largest integer \(n\) that
satisfies the given conditions is
\(n=870\).
Answer: 870
-
First, we note that
\(a^2 \geq 0\) for every real number
\(a\).
Therefore,
\((4x^2-y^2)^2 \geq 0\) and
\((7x+3y-39)^2 \geq 0\).
Since
\((4x^2-y^2)^2 + (7x+3y-39)^2 = 0\),
then it must be the case that
\((4x^2-y^2)^2 = 0\) and
\((7x+3y-39)^2 = 0\).
This means that \(4x^2-y^2 = 0\) and
\(7x+3y-39=0\).
The equation \(4x^2 - y^2 = 0\) is
equivalent to \(4x^2 = y^2\) and to
\(y = \pm 2x\).
If \(y = 2x\), then the equation
\(7x+3y-39=0\) becomes
\(7x + 6x = 39\) or
\(x = 3\).
This gives \(y = 2x = 6\).
If \(y = -2x\), then the equation
\(7x+3y-39=0\) becomes
\(7x - 6x = 39\) or
\(x = 39\).
This gives \(y = -2x = -78\).
Therefore, the solutions to the original equation are
\((x,y)=(3,6),(39,-78)\).
Answer: \((x,y)=(3,6),(39,-78)\)
-
In an arithmetic sequence with common difference
\(d\), the difference between any two
terms must be divisible by \(d\).
This is because to get from any term in the sequence to any term later
in the sequence, we add the common difference
\(d\) some number of times.
In the given sequence, this means that
\(468 - 3 = 465\) is a multiple of
\(d\) and
\(2018 - 468 = 1550\) is a multiple
of \(d\).
Thus, we want to determine the possible positive common divisors of
\(465\) and
\(1550\).
We note that
\(465 = 5 \cdot 93 = 3 \cdot 5 \cdot 31\)
and that
\(1550 = 50 \cdot 31 = 2 \cdot 5^2 \cdot 31\).
Therefore, the positive common divisors of
\(465\) and
\(1550\) are
\(1, 5, 31, 155\). (These come from
finding the common prime divisors.)
Since \(d>1\), the possible values
of \(d\) are
\(5, 31, 155\). The sum of these is
\(5+31+155=191\).
Answer: 191
-
Let points \(P\) and
\(Q\) be on
\(AB\) so that
\(GP\) and
\(EQ\) are perpendicular to
\(AB\).
Let point \(R\) be on
\(DC\) so that
\(FR\) is perpendicular to
\(DC\).
Note that each of \(\triangle BCE\),
\(\triangle EQB\),
\(\triangle EQF\),
\(\triangle FRE\),
\(\triangle FRG\), and
\(\triangle FPG\) is right-angled and
has height equal to \(BC\), which has
length \(10\) m.
Since \(\angle BEC = \angle FEG\),
then the angles of
\(\triangle BCE\) and
\(\triangle FRE\) are equal. Since
their heights are also equal, these triangles are congruent.
Since \(FREQ\) and
\(BCEQ\) are rectangles (each has
four right angles) and each is split by its diagonal into two
congruent triangles, then
\(\triangle EQF\) and
\(\triangle EQB\) are also congruent
to \(\triangle BCE\).
Similarly, \(\triangle FPG\) and
\(\triangle FRG\) are congruent to
these triangles as well.
Let \(CE=x\) m. Then
\(ER=RG=EC=x\) m.
Since \(\triangle HDG\) is
right-angled at \(D\) and
\(\angle HGD = \angle FGE\), then
\(\triangle HGD\) is similar to
\(\triangle FRG\).
Since \(HD:FR=6:10\), then
\(DG:RG=6:10\) and so
\(DG = \frac{6}{10}RG = \frac{3}{5}x\)
m.
Since \(AB=18\) m and
\(ABCD\) is a rectangle, then
\(DC=18\) m.
But
\(DC = DG + RG + ER + EC = \frac{3}{5}x + 3x =
\frac{18}{5}x\)
m.
Thus, \(\frac{18}{5}x = 18\) and so
\(x = 5\).
Since \(x = 5\), then by the
Pythagorean Theorem in
\(\triangle BCE\),
\[BE = \sqrt{BC^2 + EC^2} = \sqrt{(10\mbox{ m})^2 + (5\mbox{ m})^2}
= \sqrt{125\mbox{ m}^2} = 5\sqrt{5}\mbox{ m}\]
Now
\(FG = EF = BE = 5\sqrt{5}\mbox{ m}\)
and
\(GH = \frac{6}{10}FG = \frac{3}{5}(5\sqrt{5}\mbox{ m}) =
3\sqrt{5}\mbox{ m}\).
Therefore, the length of the path
\(BEFGH\) is
\(3(5\sqrt{5}\mbox{ m}) + (3\sqrt{5}\mbox{ m})\)
or \(18\sqrt{5}\mbox{ m}\).
Answer: \(18\sqrt{5}\mbox{ m}\)
-
The box contains \(R+B\) balls when
the first ball is drawn and
\(R+B-1\) balls when the second ball
is drawn.
Therefore, there are
\((R+B)(R+B-1)\) ways in which two
balls can be drawn.
If two red balls are drawn, there are
\(R\) balls that can be drawn first
and \(R-1\) balls that can be drawn
second, and so there are
\(R(R-1)\) ways of doing this.
Since the probability of drawing two red balls is
\(\dfrac{2}{7}\), then
\(\dfrac{R(R-1)}{(R+B)(R+B-1)} = \dfrac{2}{7}\).
If only one of the balls is red, then the balls drawn are either red
then blue or blue then red.
There are \(RB\) ways in the first
case and \(BR\) ways in the second
case, since there are \(R\) red balls
and \(B\) blue balls in the box.
Since the probability of drawing exactly one red ball is
\(\dfrac{1}{2}\), then
\(\dfrac{2RB}{(R+B)(R+B-1)} = \dfrac{1}{2}\).
Dividing the first equation by the second, we obtain successively
\[\begin{aligned} \dfrac{R(R-1)}{(R+B)(R+B-1)} \cdot
\dfrac{(R+B)(R+B-1)} {2RB} & = \dfrac{2}{7} \cdot \dfrac{2}{1}
\\ \dfrac{R-1}{2B} & = \dfrac{4}{7} \\ R & = \dfrac{8}{7}B +
1\end{aligned}\]
Substituting into the second equation, we obtain successively
\[\begin{aligned}
\dfrac{2\left(\tfrac{8}{7}B+1\right)B}{\left(\tfrac{8}{7}B+1+B\right)\left(\tfrac{8}{7}B+1+B-1\right)}
& = \dfrac{1}{2} \\
\dfrac{2\left(\tfrac{8}{7}B+1\right)B}{\left(\tfrac{15}{7}B+1\right)\left(\tfrac{15}{7}B\right)}
& = \dfrac{1}{2} \\ \dfrac{2(8B+7)}{15\left(\tfrac{15}{7}B + 1
\right)} & = \dfrac{1}{2} \qquad\mbox{(since $B \neq 0$)}\\ 32B
+ 28 & = \tfrac{225}{7}B + 15 \\ 13 & = \tfrac{1}{7}B \\ B
& = 91\end{aligned}\]
Since \(R = \tfrac{8}{7}B+1\), then
\(R = 105\) and so
\((R,B)=(105,91)\).
(We can verify that the given probabilities are correct with these
starting numbers of red and blue balls.)
Answer: \((R,B)=(105,91)\)
-
Consider the front face of the tank, which is a circle of radius 10
m.
Suppose that when the water has depth 5 m, its surface is along
horizontal line \(AB\).
Suppose that when the water has depth
\((10+5\sqrt{2})\) m, its surface is
along horizontal line \(CD\).
Let the area of the circle between the chords
\(AB\) and
\(CD\) be
\(x\mbox{ m}^2\).
Since the tank is a cylinder which is lying on a flat surface, the
volume of water added can be viewed as an irregular prism with base of
area \(x\mbox{ m}^2\) and length 30
m.
Thus, the volume of water equals
\(30x\mbox{ m}^3\). Therefore, we
need to calculate the value of
\(x\).
Let \(O\) be the centre of the
circle, \(N\) be the point where the
circle touches the ground, \(P\) the
midpoint of \(AB\), and
\(Q\) the midpoint of
\(CD\).
Join \(O\) to
\(A\),
\(B\),
\(C\), and
\(D\). Also, join
\(O\) to
\(N\) and to
\(Q\).
Since \(Q\) is the midpoint of chord
\(CD\) and
\(O\) is the centre of the circle,
then \(OQ\) is perpendicular to
\(CD\).
Since \(P\) is the midpoint of
\(AB\), then
\(ON\) passes through
\(P\) and is perpendicular to
\(AB\).
Since \(AB\) and
\(CD\) are parallel and
\(OP\) and
\(OQ\) are perpendicular to these
chords, then \(QOPN\) is a straight
line segment.
Since the radius of the circle is 10 m, then
\(OC=OD=OA=ON=OB=10\mbox{ m}\).
Since \(AB\) is 5 m above the ground,
then \(NP = 5\mbox{ m}\).
Since \(ON = 10\mbox{ m}\), then
\(OP = ON - NP = 5\mbox{ m}\).
Since \(CD\) is
\((10+5\sqrt{2})\) m above the ground
and \(ON=10\mbox{ m}\), then
\(QO = 5\sqrt{2}\mbox{ m}\).
Since \(\triangle AOP\) is
right-angled at \(P\) and
\(OP:OA=1:2\), then
\(\triangle AOP\) is a
\(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle with
\(\angle AOP = 60^\circ\). Also,
\(AP = \sqrt{3}OP = 5\sqrt{3}\mbox{ m}\).
Since \(\triangle CQO\) is
right-angled at \(Q\) and
\(OC:OQ=10:5\sqrt{2} = 2:\sqrt{2} = \sqrt{2}:1\), then \(\triangle CQO\) is a
\(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangle with
\(\angle COQ = 45^\circ\). Also,
\(CQ = OQ = 5\sqrt{2}\mbox{ m}\).
We are now ready to calculate the value of
\(x\).
The area between \(AB\) and
\(CD\) is equal to the area of the
circle minus the combined areas of the region under
\(AB\) and the region above
\(CD\).
The area of the region under
\(AB\) equals the area of sector
\(AOB\) minus the area of
\(\triangle AOB\).
The area of the region above
\(CD\) equals the area of sector
\(COD\) minus the area of
\(\triangle COD\).
Since \(\angle AOP = 60^\circ\), then
\(\angle AOB = 2\angle AOP = 120^\circ\).
Since \(\angle COQ = 45^\circ\), then
\(\angle COD = 2\angle COQ = 90^\circ\).
Since the complete central angle of the circle is
\(360^\circ\), then sector
\(AOB\) is
\(\frac{120^\circ}{360^\circ} = \frac{1}{3}\)
of the whole circle and sector
\(COD\) is
\(\frac{90^\circ}{360^\circ} = \frac{1}{4}\)
of the whole circle.
Since the area of the entire circle is
\(\pi(10\mbox{ m})^2 = 100\pi\mbox{ m}^2\), then the area of sector \(AOB\) is
\(\frac{100}{3}\pi\mbox{ m}^2\) and
the area of sector \(COD\) is
\(25\pi\mbox{ m}^2\).
Since \(P\) and
\(Q\) are the midpoints of
\(AB\) and
\(CD\), respectively, then
\(AB = 2AP = 10\sqrt{3}\mbox{ m}\)
and
\(CD = 2CQ = 10\sqrt{2}\mbox{ m}\).
Thus, the area of
\(\triangle AOB\) is
\(\frac{1}{2}\cdot AB\cdot OP = \frac{1}{2}(10\sqrt{3}\mbox{
m})(5\mbox{ m}) = 25\sqrt{3}\mbox{ m}^2\).
Also, the area of
\(\triangle COD\) is
\(\frac{1}{2}\cdot CD \cdot OQ = \frac{1}{2}(10\sqrt{2}\mbox{
m})(5\sqrt{2}\mbox{ m}) = 50\mbox{ m}^2\).
This means that the area of the region below
\(AB\) is
\(\left(\frac{100}{3}\pi - 25\sqrt{3}\right)\mbox{ m}^2\)
and the area of the region above
\(CD\) is
\((25\pi - 50)\mbox{ m}^2\).
Finally, this means that the volume of water added, in
\(\mbox{m}^3\), is
\[\begin{aligned} 30x & = 30\left(100\pi -
\left(\tfrac{100}{3}\pi - 25\sqrt{3}\right) - (25\pi - 50)\right) \\
& = 3000\pi - 1000\pi + 750\sqrt{3} - 750\pi + 1500 \\ & =
1250\pi + 1500 + 750\sqrt{3}\end{aligned}\]
Therefore,
\(a\pi + b + c\sqrt{p} = 1250\pi + 1500 + 750\sqrt{3}\)
and so
\((a,b,c,p) = (1250,1500,750,3)\).
Answer: \((1250,1500,750,3)\)
-
Evaluating,
\[\sqrt{1+2+3+4+5+6+7+8} = \sqrt{36} = 6\]
Answer: 6
-
Since the bucket is
\(\frac{2}{3}\) full and contains 9 L
of maple syrup, then if it were
\(\frac{1}{3}\) full, it would
contain
\(\frac{1}{2}(9\mbox{ L}) = 4.5\mbox{ L}\).
Therefore, the capacity of the full bucket is
\(3 \cdot (4.5\mbox{ L}) = 13.5\mbox{ L}\).
Answer: 13.5 L
-
Since the four integers are consecutive odd integers, then they differ
by 2.
Let the four integers be
\(x-6,x-4,x-2,x\).
Since the sum of these integers is 200, then
\((x-6)+(x-4)+(x-2)+x = 200\).
Simplifying and solving, we obtain
\(4x-12 = 200\) and
\(4x = 212\) and
\(x=53\).
Therefore, the largest of the four integers is 53.
Answer: 53
-
Since \(80 = 20 \cdot 4\), then to
make
\(80 = 20 \cdot 4\) thingamabobs, it
takes
\(20 \cdot 11 = 220\) widgets.
Since \(220 = 44 \cdot 5\), then to
make \(220 = 44 \cdot 5\) widgets, it
takes
\(44 \cdot 18 = 792\) doodads.
Therefore, to make 80 thingamabobs, it takes 792 doodads.
Answer: 792
-
Since
\(BP_1=P_1P_2 = P_2P_3=P_3P_4 = P_4P_5 = P_5P_6 = P_6P_7 = P_7
P_8\), then each of
\(\triangle BP_1P_2\),
\(\triangle P_1P_2P_3\),
\(\triangle P_2P_3P_4\),
\(\triangle P_3P_4P_5\),
\(\triangle P_4P_5P_6\),
\(\triangle P_5P_6P_7\), and
\(\triangle P_6P_7P_8\) is
isosceles.
Since \(\angle ABC = 5^\circ\), then
\(\angle BP_2P_1 = \angle ABC = 5^\circ\).
Next, \(\angle P_2P_1P_3\) is an
exterior angle for
\(\triangle BP_1P_2\).
Thus,
\(\angle P_2P_1P_3 = \angle P_1BP_2 + \angle
P_1P_2B=10^\circ\).
(To see this in another way,
\(\angle BP_1P_2 = 180^\circ - \angle P_1BP_2 - \angle
P_1P_2B\)
and
\[\angle P_2P_1P_3 = 180^\circ - \angle BP_1P_2 = 180^\circ -
(180^\circ - \angle P_1BP_2 - \angle P_1P_2B) = \angle P_1BP_2 +
\angle P_1P_2B\]
The first of these equations comes from the sum of the angles in the
triangle and the second from supplementary angles.)
Continuing in this way,
\[\begin{aligned} \angle P_2P_3P_1 & = \angle P_2P_1P_3 =
10^\circ \\ \angle P_3P_2P_4 & = \angle P_2P_3P_1 + \angle
P_2BP_3 = 15^\circ \\ \angle P_3P_4P_2 & = \angle P_3P_2P_4 =
15^\circ \\ \angle P_4P_3P_5 & = \angle P_3P_4P_2 + \angle
P_3BP_4 = 20^\circ\\ \angle P_4P_5P_3 & = \angle P_4P_3P_5 =
20^\circ \\ \angle P_5P_4P_6 & = \angle P_4P_5P_3 + \angle
P_4BP_5 = 25^\circ\\ \angle P_5P_6P_4 & = \angle P_5P_4P_6 =
25^\circ \\ \angle P_6P_5P_7 & = \angle P_5P_6P_4 + \angle
P_5BP_6 = 30^\circ\\ \angle P_6P_7P_5 & = \angle P_6P_5P_7 =
30^\circ \\ \angle P_7P_6P_8 & = \angle P_6P_7P_5 + \angle
P_6BP_7 = 35^\circ\\ \angle P_7P_8P_6 & = \angle P_7P_6P_8 =
35^\circ\\ \angle AP_7P_8 & = \angle P_7P_8P_6 + \angle P_7BP_8
= 40^\circ\end{aligned}\]
Answer: \(40^\circ\)
-
Suppose that the base of the pyramid has
\(n\) sides.
The base will also have
\(n\) vertices. Since the pyramid has
one extra vertex (the apex), then it has
\(n+1\) vertices in total.
The pyramid has \(n+1\) faces: the
base plus \(n\) triangular faces
formed by each edge of the base and the apex.
The pyramid has \(2n\) edges: the
\(n\) sides that form the base plus
one edge joining each of the
\(n\) vertices of the base to the
apex.
From the given information,
\(2n+(n+1) = 1915\).
Thus, \(3n = 1914\) and so
\(n = 638\).
Since the pyramid has \(n+1\) faces,
then it has \(639\) faces.
Answer: 639
-
Since \(2^{11}=2048\) and
\(2^5=32\), the eight values are
\[\begin{array}{cccc} 2^{11} + 2^5 + 2 = 2082 & 2^{11} + 2^5 -
2 = 2078 & 2^{11} - 2^5 + 2 = 2018 & 2^{11} - 2^5 - 2 = 2014
\\ -2^{11} + 2^5 + 2 = -2014 & -2^{11} + 2^5 - 2 = -2018 &
-2^{11} - 2^5 + 2 = -2078 & -2^{11} - 2^5 - 2 = -2082
\end{array}\]
The third largest value is
\(2^{11} - 2^5 + 2 = 2018\).
Answer: \(2^{11} - 2^5 + 2 = 2018\)
-
For every real number \(a\),
\((-a)^3 = -a^3\) and so
\((-a)^3+a^3 = 0\).
Therefore,
\[(-n)^3+(-n+1)^3+\cdots+(n-2)^3+(n-1)^3+n^3+(n+1)^3\]
which equals
\[((-n)^3 + n^3) + ((-n+1)^3 + (n-1)^3) + \cdots + ((-1)^3 +
1^3)+0^3 + (n+1)^3\]
is equal to \((n+1)^3\).
Since \(14^3 = 2744\) and
\(15^3 = 3375\) and
\(n\) is an integer, then
\((n+1)^3 < 3129\) exactly when
\(n+1 \leq 14\).
There are \(13\) positive integers
\(n\) that satisfy this condition.
Answer: 13
-
Using the given definition, the following equations are equivalent:
\[\begin{aligned} (2\nabla x) - 8 & = x \nabla 6 \\ (2x - 4x) -
8 & = 6x - 6x^2 \\ 6x^2 - 8x - 8 & = 0 \\ 3x^2 - 4x - 4
& = 0\end{aligned}\]
The sum of the values of \(x\) that
satisfy the original equation equals the sum of the roots of this
quadratic equation.
This sum equals \(-\tfrac{-4}{3}\) or
\(\tfrac{4}{3}\).
(We could calculate the roots and add these, or use the fact that the
sum of the roots of the quadratic equation
\(ax^2 + bx + c = 0\) is
\(-\dfrac{b}{a}\).)
Answer: \(\tfrac{4}{3}\)
-
Suppose Birgit’s four numbers are
\(a,b,c,d\).
This means that the totals \(a+b+c\),
\(a+b+d\),
\(a+c+d\), and
\(b+c+d\) are equal to
\(415\),
\(442\),
\(396\), and
\(325\), in some order.
If we add these totals together, we obtain
\[\begin{aligned} (a+b+c)+(a+b+d)+(a+c+d)+(b+c+d) & = 415 + 442
+ 396 + 325 \\ 3a+3b+3c+3d & = 1578 \\ a+b+c+d & =
526\end{aligned}\]
since the order of addition does not matter.
Therefore, the sum of Luciano’s numbers is
\(526\).
Answer: 526
-
Let \(v_1\) km/h be Krikor’s constant
speed on Monday.
Let \(v_2\) km/h be Krikor’s constant
speed on Tuesday.
On Monday, Krikor drives for 30 minutes, which is
\(\frac{1}{2}\) hour.
Therefore, on Monday, Krikor drives
\(\frac{1}{2}v_1\) km.
On Tuesday, Krikor drives for 25 minutes, which is
\(\frac{5}{12}\) hour.
Therefore, on Tuesday, Krikor drives
\(\frac{5}{12}v_2\) km.
Since Krikor drives the same distance on both days, then
\(\frac{1}{2}v_1 = \frac{5}{12}v_2\)
and so
\(v_2 = \frac{12}{5} \cdot \frac{1}{2} v_1= \frac{6}{5}v_1\).
Since
\(v_2 = \frac{6}{5}v_1 = \frac{120}{100}v_1\), then \(v_2\) is 20% larger than
\(v_1\).
That is, Krikor drives 20% faster on Tuesday than on Monday.
Answer: 20%
-
Using logarithm laws,
\[\begin{aligned} & \hspace{-1cm} \pi \log_{2018} \sqrt{2} +
\sqrt{2}\log_{2018} \pi + \pi
\log_{2018}\left(\dfrac{1}{\sqrt{2}}\right) + \sqrt{2}
\log_{2018}\left(\dfrac{1}{\pi}\right) \\ & =
\log_{2018}\left(\sqrt{2}^{\pi} \right) +
\log_{2018}\left(\pi^{\sqrt{2}} \right) +
\log_{2018}\left(\dfrac{1}{\sqrt{2}^{\pi}} \right) +
\log_{2018}\left(\dfrac{1}{\pi^{\sqrt{2}}} \right) \\ & =
\log_{2018} \left( \dfrac{\sqrt{2}^{\pi} \cdot
\pi^{\sqrt{2}}}{\sqrt{2}^{\pi} \cdot \pi^{\sqrt{2}}}\right) \\ &
= \log_{2018} (1) \\ & = 0\end{aligned}\]
Answer: 0
-
We make a table that lists, for each possible value of
\(k\), the digits, the possible
three-digit integers made by these digits, and
\(k+3\):
| \(0\) |
\(0, 1, 2\) |
\(102, 120, 201, 210\) |
\(3\) |
| \(1\) |
\(1, 2, 3\) |
\(123, 132, 213, 231, 312, 321\) |
\(4\) |
| \(2\) |
\(2, 3, 4\) |
\(234, 243, 324, 342, 423, 432\) |
\(5\) |
| \(3\) |
\(3, 4, 5\) |
\(345, 354, 435, 453, 534, 543\) |
\(6\) |
| \(4\) |
\(4, 5, 6\) |
\(456, 465, 546, 564, 645, 654\) |
\(7\) |
| \(5\) |
\(5, 6, 7\) |
\(567, 576, 657, 675, 756, 765\) |
\(8\) |
| \(6\) |
\(6, 7, 8\) |
\(678, 687, 768, 786, 867, 876\) |
\(9\) |
| \(7\) |
\(7, 8, 9\) |
\(789, 798, 879, 897, 978, 987\) |
\(10\) |
When \(k=0\), the sum of the digits
of each three-digit integer is 3, so each is divisible by 3.
When \(k=1\), only two of the
three-digit integers are even: 132 and 312. Each is divisible by 4.
When \(k=2\), none of the three-digit
integers end in 0 or 5 so none is divisible by 5.
When \(k=3\), only two of the
three-digit integers are even: 354 and 534. Each is divisible by 6.
When \(k=4\), the integer 546 is
divisible by 7. The rest are not. (One way to check this is by
dividing each by 7.)
When \(k=5\), only two of the
three-digit integers are even: 576 and 756. Only 576 is divisible by
8.
When \(k=6\), the sum of the digits
of each of the three-digit integers is 21, which is not divisible by
9, so none of the integers is divisible by 9.
When \(k=7\), none of the three-digit
integers end in 0 so none is divisible by 10.
In total, there are
\(4+2+2+1+1=10\) three-digit integers
that satisfy the required conditions.
Answer: 10
-
Suppose that \(d\) is the common
difference in this arithmetic sequence.
Since \(t_{2018} = 100\) and
\(t_{2021}\) is 3 terms further along
in the sequence, then
\(t_{2021} = 100 + 3d\).
Similarly,
\(t_{2036} = 100 + 18d\) since it is
18 terms further along.
Since \(t_{2018} = 100\) and
\(t_{2015}\) is 3 terms back in the
sequence, then
\(t_{2015} = 100 - 3d\).
Similarly,
\(t_{2000} = 100 - 18d\) since it is
18 terms back.
Therefore,
\[\begin{aligned} t_{2000} + 5t_{2015} + 5t_{2021} + t_{2036} &
= (100-18d) + 5(100-3d) + 5(100+3d) + (100+18d)\\ & = 1200 -18d
-15d + 15d + 18d\\ & = 1200\end{aligned}\]
Answer: 1200
-
The area of the square wall with side length
\(n\) metres is
\(n^2\) square metres.
The combined area of \(n\) circles
each with radius 1 metre is
\(n \cdot \pi \cdot 1^2\) square
metres or \(n\pi\) square metres.
Given that Mathilde hits the wall at a random point, the probability
that she hits a target is the ratio of the combined areas of the
targets to the area of the wall, or
\(\dfrac{n\pi\mbox{ m}^2}{n^2\mbox{ m}^2}\), which equals \(\dfrac{\pi}{n}\).
For
\(\dfrac{\pi}{n} > \dfrac{1}{2}\),
it must be the case that
\(n < 2\pi \approx 6.28\).
The largest value of \(n\) for which
this is true is \(n=6\).
Answer: \(n=6\)
-
First, we count the number of factors of 7 included in
\(200!\).
Every multiple of 7 includes least 1 factor of 7.
The product \(200!\) includes 28
multiples of 7 (since
\(28\times7=196\)).
Counting one factor of 7 from each of the multiples of 7 (these are
\(7,14,21,\dots,182,189,196\)), we
see that \(200!\) includes at least
28 factors of 7.
However, each multiple of
\(7^2=49\) includes a second factor
of 7 (since
\(49=7^2, 98=7^2\times2\), etc.)
which was not counted in the previous 28 factors.
The product \(200!\) includes 4
multiples of 49, since
\(4\times49=196\), and thus there are
at least 4 additional factors of 7 in
\(200!\).
Since \(7^3>200\), then
\(200!\) does not include any
multiples of \(7^3\) and so we have
counted all possible factors of 7.
Thus, \(200!\) includes exactly
\(28+4=32\) factors of 7, and so
\(200!=7^{32}\times t\) for some
positive integer \(t\) that is not
divisible by 7.
Counting in a similar way, the product
\(90!\) includes 12 multiples of 7
and 1 multiple of 49, and thus includes
\(13\) factors of 7.
Therefore, \(90!=7^{13}\times r\) for
some positive integer \(r\) that is
not divisible by 7.
Also, \(30!\) includes
\(4\) factors of 7, and thus
\(30!=7^{4}\times s\) for some
positive integer \(s\) that is not
divisible by 7.
Therefore,
\(\dfrac{200!}{90!30!}=\dfrac{7^{32}\times t}{(7^{13}\times
r)(7^{4}\times s)}=\dfrac{7^{32}\times t}{(7^{17}\times
rs)}=\dfrac{7^{15}\times t}{rs}\).
Since we are given that
\(\dfrac{200!}{90!30!}\) is equal to
a positive integer, then
\(\dfrac{7^{15}\times t}{rs}\) is a
positive integer.
Since \(r\) and
\(s\) contain no factors of 7 and
\(7^{15}\times t\) is divisible by
\(rs\), then it must be the case that
\(t\) is divisible by
\(rs\).
In other words, we can re-write
\(\dfrac{200!}{90!30!} = \dfrac{7^{15}\times t}{rs}\)
as
\(\dfrac{200!}{90!30!} = 7^{15}\times \dfrac{t}{rs}\)
where \(\dfrac{t}{rs}\) is an
integer.
Since each of \(r\),
\(s\) and
\(t\) does not include any factors of
7, then the integer
\(\dfrac{t}{rs}\) is not divisible by
7.
Therefore, the largest power of 7 which divides
\(\dfrac{200!}{90!30!}\) is
\(7^{15}\), and so
\(n=15\).
Answer: 15
-
Let \(BD = h\).
Since \(\angle BCA = 45^\circ\) and
\(\triangle BDC\) is right-angled at
\(D\), then
\(\angle CBD = 180^\circ - 90^\circ - 45^\circ = 45^\circ\).
This means that \(\triangle BDC\) is
isosceles with \(CD = BD = h\).
Since \(\angle BAC = 60^\circ\) and
\(\triangle BAD\) is right-angled at
\(D\), then
\(\triangle BAD\) is a
\(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle.
Therefore,
\(BD:DA = \sqrt{3}:1\).
Since \(BD=h\), then
\(DA = \dfrac{h}{\sqrt{3}}\).
Thus,
\(CA = CD + DA = h + \dfrac{h}{\sqrt{3}} = h \left(1 +
\dfrac{1}{\sqrt{3}}\right) = \dfrac{h(\sqrt{3}+1)}{\sqrt{3}}\).
We are told that the area of
\(\triangle ABC\) is
\(72 + 72 \sqrt{3}\).
Since \(BD\) is perpendicular to
\(CA\), then the area of
\(\triangle ABC\) equals
\(\dfrac{1}{2}\cdot CA \cdot BD\).
Thus,
\[\begin{aligned} \dfrac{1}{2}\cdot CA \cdot BD & = 72 + 72
\sqrt{3} \\ \dfrac{1}{2} \cdot \dfrac{h(\sqrt{3}+1)}{\sqrt{3}} \cdot
h & = 72 (1 + \sqrt{3}) \\ \dfrac{h^2}{2\sqrt{3}}& = 72 \\
h^2 & = 144 \sqrt{3}\end{aligned}\]
and so
\(BD = h = 12\sqrt[4]{3}\) since
\(h>0\).
Answer: \(12\sqrt[4]{3}\)
-
Since each word is to be 7 letters long and there are two choices for
each letter, there are
\(2^7 = 128\) such words.
We count the number of words that do contain three or more A’s in a
row and subtract this total from 128.
There is 1 word with exactly 7 A’s in a row: AAAAAAA.
There are 2 words with exactly 6 A’s in a row: AAAAAAB and BAAAAAA.
Consider the words with exactly 5 A’s in a row.
If such a word begins with exactly 5 A’s, then the 6th letter is B and
so the word has the form AAAAAB\(x\)
where \(x\) is either A or B. There
are 2 such words.
If such a word has the string of exactly 5 A’s beginning in the second
position, then it must be BAAAAAB since there cannot be an A either
immediately before or immediately after the 5 A’s.
If such a word ends with exactly 5 A’s, then the 2nd letter is B and
so the word has the form \(x\)BAAAAA
where \(x\) is either A or B. There
are 2 such words.
There are \(5\) words wth exactly 5
A’s in a row.
Consider the words with exactly 4 A’s in a row.
Using similar reasoning, such a word can be of one of the following
forms: AAAAB\(xx\), BAAAAB\(x\), \(x\)BAAAAB,
\(xx\)BAAAA.
Since there are two choices for each
\(x\), then there are
\(4+2+2+4=12\) words.
Consider the words with exactly 3 A’s in a row.
Using similar reasoning, such a word can be of one of the following
forms: AAAB\(xxx\), BAAAB\(xx\), \(x\)BAAAB\(x\), \(xx\)BAAAB,
\(xxx\)BAAA.
Since there are two choices for each
\(x\), then there appear to be
\(8+4+4+4+8=28\) such words. However,
the word AAABAAA is counted twice. (This the only word counted twice
in any of these cases.) Therefore, there are
\(27\) such words.
In total, there are
\(1+2+5+12+27 = 47\) words that
contain three or more A’s in a row, and so there are
\(128 - 47 = 81\) words that do not
contain three or more A’s in a row.
Answer: 81
-
Let \(t = x^{1/5}\). Thus,
\(x^{2/5} = t^2\) and
\(x^{3/5} = t^3\).
Therefore, the following equations are equivalent:
\[\begin{aligned} x^{3/5} - 4 & = 32 - x^{2/5} \\ t^3 + t^2 -
36 & = 0 \\ (t-3)(t^2 + 4t + 12) & = 0\end{aligned}\]
Thus, \(t = 3\) or
\(t^2 + 4t + 12 = 0\).
The discriminant of this quadratic equation is
\(4^2 - 4(1)(12) < 0\), which
means that there are no real values of
\(t\) that are solutions. This in
turn means that there are no corresponding real values of
\(x\).
This gives \(x^{1/5} = t = 3\) and so
\(x = 3^5 = 243\) is the only
solution.
Answer: \(3^5 = 243\)
-
Let \(AC=x\).
Thus \(BC = AC - 1 = x-1\).
Since \(AC = AB - 1\), then
\(AB = AC + 1 = x+1\).
The perimeter of \(\triangle ABC\) is
\(BC + AC + AB = (x-1)+x+(x+1)=3x\).
By the cosine law in
\(\triangle ABC\),
\[\begin{aligned} BC^2 & = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle
BAC) \\ (x-1)^2 & = (x+1)^2 + x^2 -
2(x+1)x\left(\tfrac{3}{5}\right) \\ x^2 -2x + 1 & = x^2 + 2x + 1
+ x^2 - \tfrac{6}{5}(x^2 + x) \\ 0 & = x^2 +4x -
\tfrac{6}{5}(x^2 +x) \\ 0 & = 5x^2 + 20x - 6(x^2 + x) \\ x^2
& = 14x\end{aligned}\]
Since \(x>0\), then
\(x = 14\).
Therefore, the perimeter of
\(\triangle ABC\), which equals
\(3x\), is
\(42\).
Answer: 42
-
Since
\(f(2x-3) - 2f(3x-10) + f(x-3) = 28-6x\)
for all real numbers \(x\), then when
\(x=2\), we obtain
\(f(2(2)-3) - 2f(3(2)-10) + f(2-3) = 28 - 6(2)\)
and so
\(f(1) -2f(-4) + f(-1) = 16\).
Since \(f\) is an odd function, then
\(f(-1) = -f(1)\) or
\(f(1)+f(-1)=0\).
Combining with
\(f(1) -2f(-4) + f(-1) = 16\), we
obtain \(-2f(-4) = 16\) and so
\(-f(-4)=8\).
Since \(f\) is an odd function, then
\(f(4) = -f(-4) = 8\).
Answer: 8
-
Let the radius of the small sphere be
\(r\) and the radius of the large
sphere be \(2r\).
Draw a vertical cross-section through the centre of the top face of
the cone and its bottom vertex.
By symmetry, this will pass through the centres of the spheres.
In the cross-section, the cone becomes a triangle and the spheres
become circles.
We label the vertices of the triangle
\(A\),
\(B\),
\(C\).
We label the centres of the large circle and small circle
\(L\) and
\(S\), respectively.
We label the point where the circles touch
\(U\).
We label the midpoint of
\(AB\) (which represents the centre
of the top face of the cone) as
\(M\).
Join \(L\) and
\(S\) to the points of tangency of
the circles to \(AC\). We call these
points \(P\) and
\(Q\).
Since \(LP\) and
\(SQ\) are radii, then they are
perpendicular to the tangent line
\(AC\) at
\(P\) and
\(Q\), respectively.
Draw a perpendicular from \(S\) to
\(T\) on
\(LP\).
The volume of the cone equals
\(\frac{1}{3}\pi \cdot AM^2 \cdot MC\). We determine the lengths of
\(AM\) and
\(MC\) in terms of
\(r\).
Since the radii of the small circle is
\(r\), then
\(QS = US = r\).
Since \(TPQS\) has three right angles
(at \(T\),
\(P\) and
\(Q\)), then it has four right
angles, and so is a rectangle.
Therefore, \(PT = QS = r\).
Since the radius of the large circle is
\(2r\), then
\(PL = UL = ML = 2r\).
Therefore,
\(TL = PL - PT = 2r - r = r\).
Since \(MC\) passes through
\(L\) and
\(S\), it also passes through
\(U\), the point of tangency of the
two circles.
Therefore,
\(LS = UL + US = 2r + r = 3r\).
By the Pythagorean Theorem in
\(\triangle LTS\),
\[TS = \sqrt{LS^2 - TL^2} = \sqrt{(3r)^2 - r^2} = \sqrt{8r^2} =
2\sqrt{2}r\]
since \(TS>0\) and
\(r>0\).
Consider \(\triangle LTS\) and
\(\triangle SQC\).
Each is right-angled,
\(\angle TLS = \angle QSC\) (because
\(LP\) and
\(QS\) are parallel), and
\(TL = QS\).
Therefore, \(\triangle LTS\) is
congruent to \(\triangle SQC\).
Thus, \(SC = LS = 3r\) and
\(QC = TS = 2\sqrt{2}r\).
This tells us that
\(MC = ML + LS + SC = 2r + 3r + 3r = 8r\).
Also, \(\triangle AMC\) is similar to
\(\triangle SQC\), since each is
right-angled and they have a common angle at
\(C\).
Therefore,
\(\dfrac{AM}{MC} = \dfrac{QS}{QC}\)
and so
\(AM = \dfrac{8r \cdot r}{2\sqrt{2}r} = 2\sqrt{2}r\).
This means that the volume of the original cone is
\(\tfrac{1}{3}\pi \cdot AM^2 \cdot MC = \tfrac{1}{3}\pi
(2\sqrt{2}r)^2 (8r) = \tfrac{64}{3}\pi r^3\).
The volume of the large sphere is
\(\frac{4}{3}\pi(2r)^3 = \frac{32}{3}\pi r^3\).
The volume of the small sphere is
\(\frac{4}{3}\pi r^3\).
The volume of the cone not occupied by the spheres is
\(\tfrac{64}{3}\pi r^3 - \tfrac{32}{3}\pi r^3 - \tfrac{4}{3}\pi r^3
= \tfrac{28}{3}\pi r^3\).
The fraction of the volume of the cone that this represents is
\(\dfrac{\tfrac{28}{3}\pi r^3}{\tfrac{64}{3}\pi r^3} =
\frac{28}{64} = \frac{7}{16}\).
Answer: \(\frac{7}{16}\)
-
Let \(f(x) = -x^2 + 2ax + a\).
Since \((x-a)^2 = x^2 - 2ax + a^2\),
then
\[-x^2 + 2ax + a = -(x^2 - 2ax + a^2) + a^2 + a = -(x-a)^2 + a^2 +
a\]
Therefore, \(M(t)\) is the maximum
value of \(-(x-a)^2 + a^2 + a\) over
all real numbers \(x\) with
\(x \leq t\).
Now the graph of
\(y = f(x) = -(x-a)^2 + a^2 + a\) is
a parabola opening downwards with vertex at
\((a,a^2 + a)\).
Since the parabola opens downwards, then the parabola reaches its
maximum at the vertex
\((a,a^2+a)\).
Therefore, \(f(x)\) is increasing
when \(x < a\) and decreasing when
\(x > a\).
This means that, when \(t < a\),
the maximum of the values of
\(f(x)\) with
\(x\leq t\) is
\(f(t)\) (because
\(f(x)\) increases until
\(f(t)\)) and when
\(t \geq a\), the maximum of the
values of \(f(x)\) with
\(x \leq t\) is
\(f(a)\) (because the maximum value
of \(f(x)\) is to the left of
\(t\)).
In other words,
\[M(t) = \left\{ \begin{array}{lll} f(t) & & t < a \\
f(a) & & t \geq a \end{array}\right.\]
Since \(a-1<a\), then
\(M(a-1) = f(a-1)\).
Since \(a+2>a\), then
\(M(a+2) = f(a)\).
Therefore,
\[\begin{aligned} M(a-1) + M(a+2) & = f(a-1) + f(a)\\ & =
(-((a-1)-a)^2 + a^2 + a) + (-(a-a)^2 + a^2 + a)\\ & = -1 + a^2 +
a - 0 + a^2 + a\\ & = 2a^2 + 2a - 1\end{aligned}\]
Answer: \(2a^2+2a-1\)
-
We find the points of intersection of
\(y = 2 \cos 3x + 1\) and
\(y = - \cos 2x\) by equating values
of \(y\) and obtaining the following
equivalent equations:
\[\begin{aligned} 2 \cos 3x + 1 & = - \cos 2x \\ 2 \cos(2x + x)
+ \cos 2x + 1 & = 0 \\ 2(\cos 2x \cos x - \sin 2x \sin x) + \cos
2x + 1 & = 0 \\ 2((2\cos^2 x - 1)\cos x - (2\sin x \cos x)\sin
x) + (2\cos^2 x - 1) + 1 & = 0 \\ 2(2\cos^3 x - \cos x - 2\sin^2
x \cos x) + 2\cos^2 x & = 0 \\ 4\cos^3 x - 2\cos x - 4(1-\cos^2
x) \cos x + 2\cos^2 x & = 0 \\ 4\cos^3 x - 2\cos x - 4\cos x +
4\cos^3 x + 2\cos^2 x & = 0 \\ 8\cos^3 x + 2\cos^2 x - 6\cos x
& = 0 \\ 4\cos^3 x + \cos^2 x - 3\cos x & = 0 \\ \cos
x(4\cos^2 x + \cos x - 3) & = 0 \\ \cos x(\cos x + 1)(4\cos x -
3) & = 0\end{aligned}\]
Therefore, \(\cos x = 0\) or
\(\cos x = -1\) or
\(\cos x = \frac{3}{4}\).
Two of these points of intersection,
\(P\) and
\(Q\), have
\(x\)-coordinates between
\(\frac{17\pi}{4} = 4\pi + \frac{\pi}{4}\)
and
\(\frac{21\pi}{4} = 5\pi + \frac{\pi}{4}\).
Since \(\cos 4\pi = 1\) and
\(\cos (4\pi + \frac{\pi}{4}) = \cos \frac{\pi}{4} =
\frac{1}{\sqrt{2}} = \frac{2\sqrt{2}}{4}< \frac{3}{4}\), then there is an angle
\(\theta\) between
\(4\pi\) and
\(\frac{17\pi}{4}\) with
\(\cos \theta = \frac{3}{4}\) and so
there is not such an angle between
\(\frac{17\pi}{4}\) and
\(\frac{21\pi}{4}\).
Therefore, we look for angles
\(x\) between
\(\frac{17\pi}{4}\) and
\(\frac{21\pi}{4}\) with
\(\cos x = 0\) and
\(\cos x = -1\).
Note that \(\cos 5\pi = -1\) and that
\(\cos \frac{18\pi}{4} = \cos \frac{9\pi}{2} = \cos \frac{\pi}{2} =
0\).
Thus, the \(x\)-coordinates of
\(P\) and
\(Q\) are
\(5\pi\) and
\(\frac{9\pi}{2}\).
Suppose that \(P\) has
\(x\)-coordinate
\(5\pi\).
Since \(P\) lies on
\(y = -\cos 2x\), then its
\(y\)-coordinate is
\(y = -\cos 10\pi = -1\).
Suppose that \(Q\) has
\(x\)-coordinate
\(\frac{9\pi}{2}\).
Since \(Q\) lies on
\(y = -\cos 2x\), then its
\(y\)-coordinate is
\(y = -\cos 9\pi = 1\).
Therefore, the coordinates of
\(P\) are
\((5\pi, -1)\) and the coordinates of
\(Q\) are
\((\frac{9\pi}{2}, 1)\).
The slope of the line through
\(P\) and
\(Q\) is
\(\dfrac{1-(-1)}{\frac{9\pi}{2} - 5\pi} = \dfrac{2}{-\frac{\pi}{2}}
= -\frac{4}{\pi}\).
The line passes through the point with coordinates
\((5\pi,-1)\).
Thus, its equation is
\(y - (-1) = -\frac{4}{\pi}(x - 5\pi)\)
or
\(y + 1 = -\frac{4}{\pi} x +20\) or
\(y = -\frac{4}{\pi}x + 19\).
This line intersects the \(y\)-axis
at \(A(0,19)\). Thus,
\(AO = 19\).
This line intersects the \(x\)-axis
at point \(B\) with
\(y\)-coordinate
\(0\) and hence the
\(x\)-coordinate of
\(B\) is
\(x = 19 \cdot \frac{\pi}{4} = \frac{19\pi}{4}\). Thus,
\(BO = \frac{19\pi}{4}\).
Since \(\triangle AOB\) is
right-angled at the origin, then its area equals
\(\frac{1}{2}(AO)(BO)=\frac{1}{2}(19)(\frac{19\pi}{4})=\frac{361\pi}{8}\).
Answer: \(\frac{361\pi}{8}\)
-
Let \(f(2n)\) be the number of
different ways to draw
\(n\) non-intersecting line segments
connecting pairs of points so that each of the
\(2n\) points is connected to exactly
one other point.
Then \(f(2) = 1\) (since there are
only 2 points) and \(f(6) = 5\) (from
the given example).
Also, \(f(4) = 2\). (Can you see
why?)
We show that
\[f(2n) = f(2n-2) + f(2) f(2n-4) + f(4) f(2n-6) + f(6) f(2n-8) +
\cdots + f(2n-4) f(2) + f(2n-2)\]
Here is a justification for this equation:
Pick one of the \(2n\) points and
call it \(P\).
\(P\) could be connected to the 1st
point counter-clockwise from \(P\).
This leaves \(2n-2\) points on the
circle. By definition, these can be connected in
\(f(2n-2)\) ways.
\(P\) cannot be connected to the
2nd point counter-clockwise, because this would leave an odd number
of points on one side of this line segment. An odd number of points
cannot be connected in pairs as required.
Similarly, \(P\) cannot be
connected to any of the 4th, 6th, 8th,
\((2n-2)\)th points.
\(P\) can be connected to the 3rd
point counter-clockwise, leaving
\(2\) points on one side and
\(2n-4\) points on the other side.
There are \(f(2)\) ways to connect
the 2 points and \(f(2n-4)\) ways
to connect the \(2n-4\) points.
Therefore, in this case there are
\(f(2)f(2n-4)\) ways to connect the
points. We cannot connect a point on one side of the line to a point
on the other side of the line because the line segments would cross,
which is not allowed.
\(P\) can be connected to the 5th
point counter-clockwise, leaving
\(4\) points on one side and
\(2n-6\) points on the other side.
There are \(f(4)\) ways to connect
the 4 points and \(f(2n-6)\) ways
to connect the \(2n-6\) points.
Therefore, in this case there are
\(f(4)f(2n-6)\) ways to connect the
points.
Continuing in this way, \(P\) can
be connected to every other point until we reach the last (\((2n-1)\)th) point which will leave
\(2n-2\) points on one side and
none on the other. There are
\(f(2n-2)\) possibilities in this
case.
Adding up all of the cases, we see that there are
\[f(2n) = f(2n-2) + f(2) f(2n-4) + f(4) f(2n-6) + f(6) f(2n-8) +
\cdots + f(2n-4) f(2) + f(2n-2)\]
ways of connecting the points.
The figures below show the case of
\(2n=10\).
We can use this formula to successively calculate
\(f(8), f(10), f(12), f(14), f(16)\):
\[\begin{aligned} f(8) & = f(6) + f(2)f(4) + f(4)f(2) + f(6)\\
& = 5 + 1(2) + 2(1) + 5\\& = 14 \\[2mm] % f(10) & = f(8)
+ f(2)f(6) + f(4)f(4) + f(6)f(2) + f(8)\\ & = 14 + 1(5) + 2(2) +
5(1) + 14\\& = 42 \\[2mm] % f(12) & = f(10) + f(2)f(8) +
f(4)f(6) + f(6)f(4) + f(8)f(2) + f(10)\\ & = 42 + 1(14) + 2(5) +
5(2) + 14(1) + 42\\& = 132 \\[2mm] % f(14) & = f(12) +
f(2)f(10) + f(4)f(8) + f(6)f(6) + f(8)f(4) + f(10)f(2) + f(12)\\
& = 132 + 1(42) + 2(14) + 5(5) + 14(2) + 42(1) + 132\\& =
429 \\[2mm] % f(16) & = f(14) + f(2)f(12) + f(4)f(10) + f(6)f(8)
+ f(8)f(6) + f(10)f(4) + f(12)f(2) + f(14) \\ & = 429 + 1(132) +
2(42) + 5(14) + 14(5) + 42(2) + 132(1) + 429 \\ & =
1430\end{aligned}\]
Therefore, there are 1430 ways to join the 16 points.
(The sequence
\(1,2,5,12,42,\ldots\) is a famous
sequence called the Catalan numbers.)
Answer: 1430
(Note: Where possible, the solutions to parts (b) and (c) of each Relay
are written as if the value of \(t\) is
not initially known, and then \(t\) is
substituted at the end.)