Wednesday, April 3, 2024
(in North America and South America)
Thursday, April 4, 2024
(outside of North American and South America)
©2024 University of Waterloo
For every \(x \neq 0\), we note
that \(\dfrac{x^4 + 3x^2}{x^2} = x^2 +
3\).
Therefore, when \(x = 2\), we have
\(\dfrac{x^4 + 3x^2}{x^2} = x^2 + 3 = 2^2 + 3
= 7\).
Alternatively, when \(x = 2\), we have
\(\dfrac{x^4 + 3x^2}{x^2} = \dfrac{2^4 + 3
\cdot 2^2}{2^2} = \dfrac{28}{4} = 7\).
By the Pythagorean Theorem in \(\triangle ABC\), we have \[\begin{align*}
AC^2 & = AB^2 + BC^2 \\
(t+1)^2 & = 10^2 + (t-1)^2 \\
t^2 + 2t + 1 & = 100 + t^2 - 2t + 1 \\
4t & = 100\end{align*}\] and so \(t = 25\).
Alternatively, we could remember the Pythagorean triple \(5\)-\(12\)-\(13\) and scale this triple by a factor of
\(2\) to obtain the Pythagorean triple
\(10\)-\(24\)-\(26\), noting that the difference between
\(t+1\) and \(t-1\) is \(2\) as is the difference between \(26\) and \(24\), which gives \(t + 1 = 26\) and so \(t = 25\).
Since \(\dfrac{2}{y} + \dfrac{3}{2y} =
14\), then \(\dfrac{4}{2y} +
\dfrac{3}{2y}=14\) or \(\dfrac{7}{2y} =
14\).
Therefore, \(2y = \dfrac{7}{14} =
\dfrac{1}{2}\) and so \(y =
\dfrac{1}{4}\).
Suppose that the sequence is \(a, b, c,
13, e, 36\).
Since \(36 = 13 + e\), then \(e = 36 - 13 = 23\).
Since \(e = c + 13\), then \(23 = c + 13\) and so \(c = 10\).
Since \(13 = b + c\) and \(c = 10\), then \(b = 3\).
Since \(c = a + b\) and \(c = 10\) and \(b
= 3\), then \(a = 10 - 3 =
7\).
Therefore, the first term is \(7\).
From the given information, we obtain \(5r^2 + 5r^3 = (5r)^2\) and so \(5r^2 + 5r^3 = 25r^2\).
Since \(r \neq 0\), we can divide by
\(5r^2\) to obtain \(1 + r = 5\), which gives \(r = 4\).
Suppose that Jimmy’s marks on his first, second, third, and
fourth tests were \(w\), \(x\), \(y\), and \(z\), respectively.
Since Jimmy’s average on his first, second and third tests was \(65\), then \(\dfrac{w+x+y}{3} = 65\) or \(w + x + y = 195\).
Since Jimmy’s average on his second, third and fourth tests was \(80\), then \(\dfrac{x+y+z}{3} = 80\) or \(x + y + z = 240\).
Since his mark on the fourth test was \(2\) times his mark on the first test, then
\(z = 2w\).
Thus, \(w + x + y = 195\) and \(x + y + 2w = 240\).
Subtracting the first equation from the second equation, we obtain \(w = 45\) and so his mark on the fourth test
was \(z = 2w = 90\).
Since \(y = r(x - 3)(x - r)\)
passes through \((0,48)\), then \(48 = r(0-3)(0-r)\).
Thus, \(48 = 3r^2\) and so \(r^2 = 16\) or \(r
= \pm 4\).
With \(13\%\) sales tax on an
item whose price is \(\$B\), the total
cost is \(\$(1.13B)\).
With \(5\%\) sales tax on an item whose
price is \(\$B\), the total cost is
\(\$(1.05B)\).
From the given information \(\$(1.13B) -
\$(1.05B) = \$24\) or \(1.13B - 1.05B =
24\).
Therefore, \(0.08B = 24\), which gives
\(B = 300\).
Alternatively, we could note that the difference in total prices is the
difference in the amount of tax paid. This is the difference between
\(13\%\) of the original price and
\(5\%\) of the original price; this
difference is equal to \(8\%\) of the
original price. If \(8\%\) of the
original price is equal to \(\$24\),
then \(1\%\) of the original price is
equal to \(\$3\) and so the original
price is \(\$3 \times 100 =
\$300\).
When \(n=1\), \(f(2n) = (f(n))^2\) becomes \(f(2) = (f(1))^2\).
Since \(f(1) = 3\), then \(f(2) = 3^2 = 9\).
When \(m=1\), \(f(2m+1) = 3f(2m)\) becomes \(f(3) = 3f(2)\).
Since \(f(2) = 9\), then \(f(3) = 3 \cdot 9 = 27\).
When \(n=2\), \(f(2n) = (f(n))^2\) becomes \(f(4) = (f(2))^2\).
Since \(f(2) = 9\), then \(f(4) = 9^2 = 81\).
Therefore, \(f(2) + f(3) + f(4) = 9 + 27 + 81
= 117\).
Since \(\triangle ABD\) is
right-angled at \(B\) and has \(\angle ADB = 45\degree\), then \(\angle BAD = 45\degree\).
Similarly, \(\triangle CPD\) is
right-angled and isosceles with \(\angle PCD =
45\degree\).
Further, \(\triangle APN\) and \(\triangle CBN\) are also both right-angled
and isosceles.
Since \(CB = 6\) and \(NB = CB\), then \(NB = 6\).
Since \(AB = 12\) and \(NB = 6\), then \(AN = AB - NB = 6\).
Since \(\triangle APN\) is
right-angled and isosceles, then its sides are in the ratio \(1:1:\sqrt{2}\).
Thus, \(AP = PN = \frac{1}{\sqrt{2}} AN =
\frac{6}{\sqrt{2}} = 3\sqrt{2}\).
Alternatively, if \(AP = PN = x\), then
the Pythagorean Theorem gives \(AN^2 = AP^2 +
PN^2\) and so \(6^2 = 2x^2\)
which gives \(AP^2 = x^2 = 18\).
Thus, the area of \(\triangle APN\) is
\(\frac{1}{2} \cdot AP \cdot PN = \frac{1}{2}
\cdot 3\sqrt{2} \cdot 3\sqrt{2} = 9\).
The line with equation \(y = -3x +
6\) has \(y\)-intercept \(6\), which means that \(OB = 6\).
To find the \(x\)-intercept of this
line, we set \(y = 0\) and obtain the
equation \(-3x + 6 = 0\) which gives
\(3x = 6\) or \(x = 2\). This means that \(OA = 2\).
Since \(\triangle ABO\) is right-angled
at \(O\), its area is \(\frac{1}{2} \cdot OB \cdot OA = \frac{1}{2} \cdot
6 \cdot 2 = 6\).
Since the area of \(\triangle ACD\) is
\(\frac{1}{2}\) of the area of \(\triangle ABO\), then the area of \(\triangle ACD\) is \(3\).
Next, we note that the line with equation \(y
= mx + 1\) has \(y\)-intercept
\(1\); thus, \(OD = 1\).
This means that the area of \(\triangle
ADO\) is \(\frac{1}{2} \cdot OD \cdot
OA = \frac{1}{2} \cdot 1 \cdot 2 = 1\).
We can determine the area of \(\triangle BCD\) by subtracting the areas of \(\triangle ACD\) and \(\triangle ADO\) from that of \(\triangle ABO\), which tells us that the area of \(\triangle BCD\) is \(6 - 3 - 1 = 2\).
Now, we can consider \(BD\), which
has length \(6 - 1 = 5\), as the base
of \(\triangle BCD\); the corresponding
height of \(\triangle BCD\) is the
distance from \(C\) to the \(y\)-axis, which we call \(h\).
Thus, \(\frac{1}{2} \cdot 5 \cdot h =
2\) and so \(h =
\frac{4}{5}\).
This means that \(C\) has \(x\)-coordinate \(\frac{4}{5}\).
Since \(C\) is on the line with
equation \(y = -3x + 6\), we have \(y = -3 \cdot \frac{4}{5} + 6 =
\frac{18}{5}\).
Therefore, the coordinates of \(C\) are
\((\frac{4}{5},\frac{18}{5})\).
Suppose that \(AP = w\), \(PD = x\), \(AS = y\), and \(SB = z\).
We use the notation \(|APXS|\) to
represent the area of \(APXS\), and so
on.
Thus, \(|APXS| = wy\), \(|PDRX| = xy\), \(|SXBQ| = wz\), and \(|XRCQ| = xz\).
Then, \[|APXS| \cdot |XRCQ| = wy\cdot xz = xy
\cdot wz = |PDRX| \cdot |SXQB|\] If \(|APXS| = 2\), \(|PDRX| = 3\), and \(|SWQB| = 6\), then \(a = |XRQC| = \dfrac{2 \cdot 6}{3} =
4\).
If \(|APXS| = 2\), \(|PDRX| = 6\), and \(|SWQB| = 3\), then \(a = |XRQC| = \dfrac{2 \cdot 3}{6} =
1\).
If \(|APXS| = 6\), \(|PDRX| = 2\), and \(|SWQB| = 3\), then \(a = |XRQC| = \dfrac{6 \cdot 3}{2} =
9\).
Since we are told that there are three possible values for \(a\), then these are 1, 4 and 9.
(Can you explain why there are exactly three such values?)
The \(x\)-intercepts of the
parabola with equation \(y = x^2 - 4tx + 5t^2
- 6t\) are \[x = \dfrac{4t \pm
\sqrt{(-4t)^2 - 4(5t^2 - 6t)}}{2}\] The distance, \(d\), between these intercepts is their
difference, which is \[d = \dfrac{4t +
\sqrt{(-4t)^2 - 4(5t^2 - 6t)}}{2} - \dfrac{4t - \sqrt{(-4t)^2 - 4(5t^2 -
6t)}}{2} = \sqrt{(-4t)^2 - 4(5t^2 - 6t)}\] From this we see that
\(d\) is as large as possible exactly
when the discriminant is as large as possible. Here, the discriminant,
\(\Delta\), is \[\Delta = (-4t)^2 - 4(5t^2 - 6t) = 16t^2 - 20t^2 +
24t = -4t^2 + 24t\] Completing the square, \[\Delta = -4(t^2 - 6t) = -4(t^2 - 6t + 9 - 9) =
-4(t^2 - 6t + 9) + 36 = -4(t-3)^2 + 36\] Since \((t-3)^2 \geq 0\), then \(\Delta \leq 36\) and \(\Delta = 36\) exactly when \((t-3)^2 = 0\) or \(t = 3\).
Therefore, the discriminant is maximized when \(t = 3\), which means that the distance
between the \(x\)-intercepts is as
large as possible when \(t = 3\).
Every multiple of \(21\) is of
the form \(21k\) for some integer \(k\).
For such a multiple to be between 10 000 and 100 000, we need \(10\,000 < 21k < 100\,000\) or \(\frac{10\,000}{21} < k <
\frac{100\,000}{21}\).
Since \(\frac{10\,000}{21} \approx
476.2\) and \(\frac{100\,000}{21}
\approx 4761.9\) and \(k\) is an
integer, then \(477 \leq k \leq 4761\).
(Note that \(k\) is greater than \(476.2\) and is an integer, so must be at
least \(477\); similarly, \(k\) is at most \(4761\).)
We also want the units digit of \(21k\)
to be \(1\).
This means that the units digit of \(k\) itself is \(1\), since the units digit of the product
of \(21\) and \(k\) is equal to units digit of \(k\) because the units digit of \(21\) is \(1\).
Therefore, the possible values of \(k\)
are \(481, 491, 501, \ldots, 4751,
4761\).
There are \(429\) such values. To see
this, we can see that counting the integers in this list is the same as
counting the integers in the list \(48, 49,
50, \ldots, 475, 476\). This list is equivalent to removing the
integers from \(1\) to \(47\) from the list of integers from \(1\) to \(476\), giving \(476 - 47 = 429\) integers.
Thus, \(M = 429\).
Solution 1:
We can partition the \(N\) students at Strickland S.S. into four groups:
\(a\) students who are in the physics club and are in the math club
\(b\) students who are in the physics club and are not in the math club
\(c\) students who are not in the physics club but are in the math club
\(d\) students who are not in the physics club and are not in the math club
In Math Club | Not in Math Club | |
---|---|---|
In Physics Club | \(a\) | \(b\) |
Not in Physics Club | \(c\) | \(d\) |
From the given information, there are \(\frac{2}{5}N\) students in the physics
club. In other words, \(a+b =
\frac{2}{5}N\).
Among the students in the physics club, twice as many are not in the
math club as are in the math club. This means that \(b = \frac{2}{3} \cdot \frac{2}{5}N =
\frac{4}{15}N\) and \(a =
\frac{1}{3}\cdot \frac{2}{5}N = \frac{2}{15}N\).
From the given information, there are \(\frac{1}{4}N\) students in the math club.
In other words, \(a + c =
\frac{1}{4}N\).
Since \(a = \frac{2}{15}N\), then \(c = \frac{1}{4}N - \frac{2}{15}N =
\frac{7}{60}N\).
Since \(a + b + c + d = N\), then \(d = N - a - b - c = N - \frac{4}{15}N -
\frac{2}{15}N - \frac{7}{60}N = \frac{29}{60}N\).
Lastly, we know that \(500 < N <
600\).
Since each of \(a\), \(b\), \(c\), and \(d\) is an integer, then \(N\) must be divisible by 60.
Therefore, \(N = 540\) and so the
number of students not in either club is \(d =
\frac{29}{60} \cdot 540 = 261\).
Solution 2:
Since there are \(N\) students at
Strickland S.S., then \(\frac{2}{5}N\)
are in the physics club and \(\frac{1}{4}N\) are in the math club.
Since each of \(\frac{2}{5}N\) and
\(\frac{1}{4}N\) must be an integer,
then \(N\) must be divisible by \(5\) and must be divisible by \(4\).
Since \(5\) and \(4\) share no common divisor larger than
\(1\), then \(N\) must be divisible by \(5 \cdot 4 = 20\).
Thus, we let \(N = 20m\) for some
positive integer \(m\).
In this case, \(\frac{2}{5}N = 8m\)
students are in the physics club and \(\frac{1}{4}N = 5m\) students are in the
math club.
Now, among the \(8m\) students in the
physics club, twice as many are not in the math club as are in the math
club.
In other words, \(\frac{1}{3}\) of the
\(8m\) students in the physics club are
in the math club. This means that \(m\)
must be divisible by \(3\), since \(3\) is a prime number and \(8\) is not divisible by \(3\).
Therefore, \(m = 3k\) for some positive
integer \(k\), which means that \(N = 20m = 60k\) and \(\frac{2}{5}N = 8m = 24k\) and \(\frac{1}{4}N = 5m = 15k\).
Since \(500 < N < 600\) and \(N\) is a multiple of 60, then \(N = 540\), which means that \(k = 9\).
Thus, the number of students in the physics club is \(24k = 216\), of whom \(\frac{1}{3} \cdot 216 = 72\) are in the
math club and \(\frac{2}{3} \cdot 216=
144\) are not in the math club.
Also, the number of students in the math club is \(15k = 135\).
Finally, we know that
there are \(540\) students at the school,
\(72\) of whom are in both the physics club and the math club,
\(144\) of whom are in the physics club and not in the math club, and
\(135 - 72 = 63\) are in the math club and not in the physics club.
Therefore, the number of students in neither club is \(540 - 72 - 144 - 63 = 261\).
Suppose that the length of the track is \(2L \text{ m}\), that Arun’s constant speed
is \(a\text{ m/s}\), and that Bella’s
constant speed is \(b\text{
m/s}\).
When Arun and Bella run over the same interval of time, the ratio of the
distances that they run is equal to the ratio of their speeds.
Consider the interval of time from the start to when they first meet. In
the diagram, \(A\) is Arun’s starting
point, \(B\) is Bella’s starting point,
and \(P\) is this first meeting
point.
Since Arun has run \(100\text{ m}\)
and together they have covered half of the length of the track, then
Bella has run \((L - 100)\text{
m}\).
Thus, \(\dfrac{a}{b} =
\dfrac{100}{L-100}\).
From their first meeting point \(P\) to
their second meeting point, which we label \(Q\), Bella runs \(150\text{ m}\).
Over this time, Arun runs from \(P\)
to \(B\) to \(Q\).
Since Bella runs \(150 - 100 = 50\text{
m}\) past \(A\), then \(QB = (L - 50)\text{ m}\) (because \(AB = L\text{ m}\) and \(AQ = 50\text{ m}\)) and so Arun runs \((L - 100)\text{ m} + (L - 50)\text{ m}\)
which is equal to \((2L - 150)\text{
m}\).
Thus, over this second interval of time, \(\dfrac{a}{b} = \dfrac{2L-150}{150}\).
Equating expressions for \(\dfrac{a}{b}\) and solving, \[\begin{align*}
\dfrac{100}{L-100} & = \dfrac{2L-150}{150} \\
100 \cdot 150 & = (L-100)(2L-150) \\
15\,000 & = 2L^2 - 350L + 15\,000 \\
350L & = 2L^2\end{align*}\] Since \(L \neq 0\), then \(2L = 350\), and so the total length of the
track is \(350\text{ m}\).
Checking, if the length of the track is \(350\text{ m}\), then half of the length is
\(75\text{ m}\).
This means that from the start to \(P\), Arun runs \(100\text{ m}\) and Bella runs \(75\text{ m}\).
Also, from \(P\) to \(Q\), Bella runs \(150\text{ m}\) and Arun runs \(200\text{ m}\).
Note that \(\dfrac{100}{75} =
\dfrac{200}{150}\) so these numbers are consistent with the given
information.
Using exponent laws, the following equations are equivalent:
\[\begin{align*}
4^{1 + \cos^3 \theta} & = 2^{2 - \cos\theta} \cdot 8^{\cos^2\theta}
\\
(2^2)^{1 + \cos^3 \theta} & = 2^{2 - \cos\theta} \cdot
(2^3)^{\cos^2\theta} \\
2^{2 + 2\cos^3 \theta} & = 2^{2 - \cos\theta} \cdot
2^{3\cos^2\theta} \\
2^{2 + 2\cos^3 \theta} & = 2^{2 - \cos\theta + 3\cos^2\theta} \\
2 + 2\cos^3 \theta & = 2 - \cos\theta + 3\cos^2\theta \\
2\cos^3\theta - 3\cos^2\theta + \cos\theta & = 0 \\
\cos\theta(2\cos^2 \theta - 3 \cos\theta + 1) & = 0\\
\cos\theta(2\cos \theta - 1)(\cos\theta - 1) & =
0\end{align*}\] and so \(\cos \theta =
0\) or \(\cos \theta = 1\) or
\(\cos \theta = \frac{1}{2}\).
Since \(0\degree \leq \theta \leq
360\degree\), the solutions are \(\theta = 90\degree, 270\degree, 0\degree,
360\degree, 60\degree, 300\degree\).
Listing these in increasing order, the solutions to the original
equation are \[\theta = 0\degree, 60\degree,
90\degree, 270\degree, 300\degree, 360\degree\]
We join \(B\) to \(E\) and \(A\) to \(D\).
Since \(MC\) is tangent to the circles
with centres \(A\) and \(B\) at \(D\) and \(E\), respectively, then \(AD\) and \(BE\) are perpendicular to \(MC\).
Since the radius of the circle with centre \(B\) is \(3\), then \(AB =
3\) and \(BE =3\).
Since the radius of the circle with centre \(A\) is \(4\), then \(AD =
4\) and \(AT = 4\).
Let \(CB = x\) and \(MT = y\).
We note that \(\triangle CEB\),
\(\triangle CDA\) and \(\triangle CTM\) are all similar, since they
are right-angled at \(E\), \(D\) and \(T\), respectively, and share a common angle
at \(C\).
Since \(\triangle CEB\) and \(\triangle CDA\) are similar, then \(\dfrac{CB}{CA} = \dfrac{BE}{AD}\) and so
\(\dfrac{x}{x+3} = \dfrac{3}{4}\) which
gives \(4x = 3x + 9\) and so \(x = 9\).
By the Pythagorean Theorem, \(CE = \sqrt{CB^2
- BE^2} = \sqrt{9^2 - 3^2} = \sqrt{72} = 6\sqrt{2}\).
Since \(\triangle CEB\) and \(\triangle CTM\) are similar, then \(\dfrac{BE}{CE} = \dfrac{MT}{CT}\) and so
\(\dfrac{3}{6\sqrt{2}} =
\dfrac{y}{9+3+4}\) which gives \(y =
\dfrac{16 \cdot 3}{6 \sqrt{2}} = \dfrac{8}{\sqrt{2}} =
4\sqrt{2}\).
Finally, the area of \(\triangle
MNC\) is equal to \(\dfrac{1}{2} \cdot
MN \cdot CT\).
If we joined \(B\) to \(G\), we would see that \(\triangle CEB\) is congruent to \(\triangle CGB\) (each is right-angled, they
have a common hypotenuse, and \(BE =
BG)\). This means that \(\angle BCE =
\angle BCG\), which in turn means that \(MT = TN\).
Since \(MT = TN\), then \(MN = 2 \cdot 4\sqrt{2} = 8 \sqrt{2}\) and
so the area of \(\triangle MNC\) is
\(\dfrac{1}{2} \cdot 8\sqrt{2} \cdot
16\) or \(64\sqrt{2}\).
First, we note that \[\log_3 z =
\dfrac{\log_{10} z}{\log_{10} 3} = \dfrac{2 \log_{10} z}{2 \log_{10} 3}
= \dfrac{\log_{10}(z^2)}{\log_{10}(3^2)} =
\dfrac{\log_{10}(z^2)}{\log_{10} 9} = \log_9 (z^2)\] Similarly,
\(\log_4 y = \log_{16}(y^2)\) and \(\log_5 x = \log_{25}(x^2)\).
We also note from the original system of equations that \(x > 0\) and \(y > 0\) and \(z > 0\).
Therefore, we can re-write the original system of equations as \[\begin{align*}
\log_9 x + \log_9 y + \log_9 (z^2) & = 2 \\
\log_{16} x + \log_{16}(y^2) + \log_{16} z & = 1 \\
\log_{25}(x^2) + \log_{25} y + \log_{25} z & =
0\end{align*}\] Using logarithm rules, this is equivalent to the
system \[\begin{align*}
\log_9 (xyz^2) & = 2 \\
\log_{16} (xy^2z) & = 1 \\
\log_{25} (x^2yz) & = 0 \end{align*}\] and to the system
\[\begin{align*}
xyz^2 & = 9^2 = 81 \\
xy^2z & = 16^1 = 16 \\
x^2yz & = 25^0 = 1\end{align*}\] Multiplying these three
equations together, we obtain \(x^4 y^4 z^4 =
1296\) and so \((xyz)^4 =
6^4\).
Thus, \(xyz = 6\).
Since \(xyz^2 = 81\) and \(xyz = 6\), then \(z = \dfrac{xyz^2}{xyz} = \dfrac{81}{6} =
\dfrac{27}{2}\).
Similarly, \(y = \dfrac{xy^2z}{xyz} =
\dfrac{16}{6} = \dfrac{8}{3}\) and \(x
= \dfrac{x^2yz}{xyz} = \dfrac{1}{6}\).
Therefore, \((x,y,z) = \left(\dfrac{1}{6},
\dfrac{8}{3}, \dfrac{27}{2}\right)\).
We can check by substitution that this triple does satisfy the original
system of equations.
Suppose that a sequence of \(n\)
steps includes \(p\) steps in the
positive direction and \(m\) steps in
the negative direction.
This means that \(n = p + m\) (total
number of steps) and \(d = p - m\)
(final position).
Since \(n = 9\) and \(d = 5\), then \(p
+ m = 9\) and \(p - m = 5\)
which give \(p = 7\) and \(m = 2\).
Thus, we need to count the number of sequences of \(9\) steps that include \(7\) steps in the positive direction and
\(2\) steps in the negative
direction.
This is the same as creating a sequence of \(9\) letters, \(7\) of which are R and \(2\) of which are L.
There are \(\displaystyle \binom{9}{2} =
\dfrac{9 \cdot 8}{2} = 36\) such sequences.
(Alternatively, we could see that there are \(36\) sequences by counting directly: there
are \(8\) sequences where the first L
appears in the first position, \(7\)
where the first L appears in the second position, and so on.)
Solution 1:
Since \(n = 9\) and \(d = 3\), then \(p
+ m = 9\) and \(p - m = 3\)
which gives \(p = 6\) and \(m = 3\).
This means that we can think about sequences of \(9\) letters that include \(6\) R’s and \(3\) L’s.
In this context, a direction change happens when the sequence changes
from a block of one letter to a block of another letter (that is, when
the sequence has an occurrence of "RL" or "LR").
For the sequence to change directions an even number of times, the
number of blocks of letters in the sequence is odd.
For the number of blocks to be odd, the sequence must begin and end with
the same letter.
If the sequence of \(9\) letters begins
and ends with an R, the \(7\) letters
in between include \(4\) R’s and \(3\) L’s; there are \(\displaystyle\binom{7}{3} = \dfrac{7 \cdot 6 \cdot
5}{3 \cdot 2 \cdot 1} = 35\) such sequences.
If the sequence of \(9\) letters begins
and ends with an L, the \(7\) letters
in between include \(6\) R’s and \(1\) L; there are \(\displaystyle\binom{7}{1}= 7\) such
sequences.
In total, there are \(35 + 7 = 42\)
such sequences.
Solution 2:
As in Solution 1, we want to count the number of sequences of \(9\) letters that include \(6\) R’s and \(3\) L’s and that have an odd number of
blocks.
Since there are \(3\) L’s, there can be
at most \(3\) blocks of L’s, and so at
most \(7\) blocks.
Case 1: There are \(3\) blocks
Suppose that the block order is R-L-R.
Here, there are \(3\) L’s in the block
of L’s and \(6\) R’s distributed
between two blocks of R’s.
There are \(5\) ways to distribute the
R’s: \(1+5\), \(2+4\), \(3+3\), \(4+2\), \(5+1\).
This means that there are \(5\) such
sequences.
Suppose that the block order is L-R-L.
The block of R’s includes \(6\) R’s and
there are \(3\) L’s to distribute
between two blocks.
There are \(2\) ways to do this: \(1+2\) or \(2+1\).
This means that there are \(2\) such
sequences.
In total, there are \(5 + 2 = 7\)
sequences in this case.
Case 2: There are \(7\) blocks
Since there cannot be \(4\) blocks
of L’s, the block order must be R-L-R-L-R-L-R.
Each block of L’s includes exactly \(1\) L, and there are \(2\) additional R’s to distribute after one
R is placed in each block.
There are \(4\) ways to put these two
R’s in the same block, and \(6\) ways
to put them in separate blocks (1st and 2nd, 1st and 3rd, 1st and 4th,
2nd and 3rd, 2nd and 4th, 3rd and 4th).
Therefore, there are \(10\) different
sequences in this case.
Case 3: There are \(5\) blocks
Suppose that the block order is R-L-R-L-R.
The two blocks of L’s include \(3\) L’s
in total; there are \(2\) ways to
distribute these (\(1+2\) or \(2+1\)).
The three blocks of L’s include \(6\)
R’s in total.
Starting with one R in each block, there are \(3\) additional R’s to distribute.
There are \(3\) ways in which the \(3\) R’s can go in the same block: \(3+0+0\), \(0+3+0\), \(0+0+3\).
There are \(6\) ways in which the \(3\) R’s can go in two blocks: \(2+1+0\), \(2+0+1\), \(1+2+0\), \(0+2+1\), \(1+0+2\), \(0+1+2\).
There is \(1\) way in which the \(3\) R’s can go in three different blocks:
\(1+1+1\).
Thus, there are \(2\) ways to
distribute the L’s and \(3+6+1=10\)
ways to distribute the R’s and so \(2 \cdot 10
= 20\) such sequences.
Suppose that the block order is L-R-L-R-L.
Each block of L’s includes exactly \(1\) L, and the two blocks of R’s include
exactly \(6\) R’s.
There are \(5\) ways to distribute
these R’s, as we saw in a similar situation in Case 1.
This means that there are \(5\) such
sequences.
In total, there are \(20 + 5 = 25\)
sequences in this case.
Combining the cases, there are \(10 + 7 + 25 = 42\) such sequences.
Consider the sequences of length \(n\) that end at \(x = d\) with \(d
\geq 0\).
Suppose that such a sequence includes \(p\) steps in the positive direction and
\(m\) steps in the negative
direction.
Since \(p + m = n\) (total number of
steps) and \(p - m = d\) (ending
position), then \(2p = n + d\) (which
gives \(p = \dfrac{n+d}{2}\)) and \(2m = n - d\) (which gives \(m = \dfrac{n-d}{2}\)).
This means that all sequences of length \(n\) that end at \(x = d\) with \(d
\geq 0\) correspond to the same values of \(p\) and \(m\), and so we treat \(p\) and \(m\) as fixed in what follows.
Before proceeding to the general case, we deal with some specific small
values.
When \(n = 1\), there is exactly \(1\) sequence that ends with \(d \geq 0\). This sequence moves \(1\) step to the right. Since there is an odd number of sequences when \(n = 1\), it cannot be the case that half of the sequences have an even number of changes of direction.
When \(n = 2\), we can either have
\(p = 2\) and \(m = 0\) (giving \(d = 2\)) or \(p =
1\) and \(m = 1\) (giving \(d = 0\)).
When \(n = 2\) and \(d = 2\), there is only \(1\) sequence, and so it cannot be the case
that half of the sequences in this category have an even number of
changes of direction.
When \(n = 2\) and \(d = 0\), there are \(2\) sequences (RL or LR), each of which has
\(1\) change of direction, so it is not
the case that half of the sequences in this category have an even number
of changes of direction.
Therefore, we assume that \(n \geq
3\). Since \(n = p + m\) and
\(p \geq m\), then \(p \geq 2\) as well.
We think about sequences of \(n\)
letters that include \(p\) R’s and
\(m\) L’s.
As in (b) Solution 1, the number of direction changes is even exactly
when the number of blocks of letters is odd, which is exactly when the
sequence begins and ends with the same letter.
For such a sequence to begin and end with R, the \(n-2\) letters between include \((p-2)\) R’s and \(m\) L’s. (Note that \(n - 2 \geq 0\) and \(p - 2 \geq 0\) since \(n \geq 2\) and \(p \geq 0\).)
For such a sequence to begin and end with L, the \(n-2\) letters between include \(p\) R’s and \((m-2)\) L’s. (Note that \(n - 2 \geq 0\). It is possible, though,
that \(m - 2 < 0\), in which case we
adopt the convention that there are 0 such sequences. In this case, we
have \(p = n - m = (n-2) - (m-2) >
n-2\) and so \(\displaystyle\binom{n-2}{p} = 0\), making
the calculations that follow consistent with this convention.)
Thus, there are \(\displaystyle\binom{n-2}{p-2} +
\binom{n-2}{p}\) such sequences.
This is true for exactly half of all sequences when the following
equivalent equations are true: \[\begin{align*}
\binom{n-2}{p-2} + \binom{n-2}{p} & = \dfrac{1}{2}\binom{n}{p} \\
\dfrac{(n-2)!}{(p-2)!m!} + \dfrac{(n-2)!}{p!(m-2)!} & = \dfrac{n!}{2
\cdot p!m!} \\
\dfrac{(n-2)!}{(p-2)!m!} + \dfrac{(n-2)!}{p!(m-2)!} & = \dfrac{n
\cdot (n-1) \cdot (n-2)!}{2 \cdot p!m!} \\
\dfrac{1}{(p-2)!m!} + \dfrac{1}{p!(m-2)!} & = \dfrac{n(n-1)}{2 \cdot
p!m!} \\
\dfrac{p!}{(p-2)!} + \dfrac{m!}{(m-2)!} & = \dfrac{n(n-1)}{2}\\
2p(p-1) + 2m(m-1) & = n(n-1) \\
2p^2 + 2m^2 & = n^2 - n + 2p + 2m \\
(p^2 + 2mp + m^2) + (p^2 - 2mp + m^2) & = n^2 - (p+m) + 2p + 2m \\
(p+m)^2 + (p-m)^2 & = n^2 + (p+m) \\
n^2 + d^2 & = n^2 + n \\
d^2 & = n\end{align*}\] Therefore, exactly half of these
sequences have an even number of changes of direction exactly when \(n = d^2\).
To finish the problem, we now need to count the number of perfect
squares (that is, possible values for \(n\)) in the correct range.
Since \(2 \leq n \leq 2024\) and \(d \geq 0\) and \(n\) is a perfect square, then the facts
that \(1^2 = 1\), \(2^2 = 4\), \(44^2=1936\), and \(45^2 = 2025\) tell us that there are \(43\) perfect squares from \(2\) to \(2024\), inclusive, and so \(43\) such pairs \((d,n)\). These are the pairs \((d,d^2)\) for \(d
= 2, 3, \ldots, 43, 44\).
Here is the general set-up for this problem along with the specific instance in (a) where \(s = 1\):
  Â
Since \(SP\) and \(TQ\) divide \(\triangle ABC\) into four regions of equal
area, then \(\triangle APS\), which is
made up of two of these regions, has area equal to one-half of the area
of \(\triangle ABC\).
Since \(S\) and \(C\) coincide, then \(P\) is the midpoint of \(AB\), which means that \(P\) has coordinates \((-\frac{1}{2}, 2)\).
Alternatively, we could note that the area of \(\triangle ABC\) is \(\frac{1}{2} \cdot 2 \cdot 4 = 4\) and so
the area of \(\triangle APS\) must be
2.
If \(P\) has \(y\)-coordinate \(p\), then \(\frac{1}{2} \cdot 2 \cdot p = 2\) which
gives \(p = 2\).
Since \(AB\) has slope 4 and \(y\)-intercept 4, its equation is \(y = 4x + 4\).
Since \(P\) lies on \(AB\) and has \(y\)-coordinate 2, its \(x\)-coordinate satisfies \(2 = 4x + 4\) and so \(4x = -2\) or \(x
= -\frac{1}{2}\).
Thus, the coordinates of \(P\) are
\((-\frac{1}{2},2)\).
We note that the area of \(\triangle
ABC\) is \(\frac{1}{2} \cdot 2 \cdot 4
= 4\). When the triangle is divided into four equal areas, each
of these areas must be 1.
Suppose that \((s,t)\) is a balancing
pair.
This is true exactly when
the area of \(\triangle SXT\) is 1, and
the area of \(\triangle APS\) is 2, and
the area of \(\triangle CQT\) is 2.
(Bullets 1 and 2 tell us that the area of quadrilateral \(APXT\) is \(1\).
Bullets 1 and 3 tell us that the area of quadrilateral \(CQXS\) is \(1\).
Since three of the areas are \(1\),
then the fourth must be \(1\).)
\(\triangle SXT\) has base \(TS = s+t\). If its height is \(h\), then \(\dfrac{1}{2}(s+t)h = 1\) and so \(h = \dfrac{2}{s+t}\).
\(\triangle APS\) has base \(AS = 1+s\). If its height is \(p\), then \(\dfrac{1}{2}(s+1)p = 2\) and so \(p = \dfrac{4}{s+1}\).
\(\triangle CQT\) has base \(TC = 1+t\). If its height is \(q\), then \(\dfrac{1}{2}(t+1)q = 2\) and so \(q = \dfrac{4}{t+1}\).
The line through \(A\) and \(B\) has slope 4 and \(y\)-intercept \(4\), so its equation is \(y = 4x + 4\).
The line through \(C\) and \(B\) has slope \(-4\) and \(y\)-intercept \(4\), so its equation is \(y = -4x + 4\).
Since \(P\) lies on the line with
equation \(y = 4x + 4\) and the \(y\)-coordinate of \(P\) is \(\dfrac{4}{s+1}\), then the \(x\)-coordinate of \(P\) satisfies \(\dfrac{4}{s+1} = 4x + 4\) which gives \(\dfrac{1}{s+1} = x + 1\), from which we
obtain \(x = \dfrac{1}{s+1} - 1 = \dfrac{1 - s
- 1}{s+1} = -\dfrac{s}{s+1}\).
Therefore, \(P\) has coordinates \(\left(-\dfrac{s}{s+1},
\dfrac{4}{s+1}\right)\).
Since \(Q\) lies on the line with
equation \(y = -4x + 4\) and the \(y\)-coordinate of \(Q\) is \(\dfrac{4}{t+1}\), then the \(x\)-coordinate of \(Q\) satisfies \(\dfrac{4}{t+1} = -4x + 4\) which gives
\(\dfrac{1}{t+1} = -x + 1\) or \(x = 1- \dfrac{1}{t+1} = \dfrac{t+1-1}{t+1} =
\dfrac{t}{t+1}\).
Therefore, \(Q\) has coordinates \(\left(\dfrac{t}{t+1},
\dfrac{4}{t+1}\right)\).
Next, we find the coordinates of \(X\)
by finding the equations of the lines through \(P\) and \(S\), and through \(Q\) and \(T\).
The slope of the line through \(S\) and
\(P\) is \[\dfrac{\dfrac{4}{s+1} - 0}{-\dfrac{s}{s+1} - s} =
\dfrac{4}{-s-s(s+1)} = -\dfrac{4}{s^2+2s}\] Since this line
passes through \(S(s,0)\), its equation
is \(y = -\dfrac{4}{s^2+2s}(x -
s)\).
The slope of the line through \(T\) and
\(Q\) is \[\dfrac{\dfrac{4}{t+1} - 0}{\dfrac{t}{t+1} - (-t)}
= \dfrac{4}{t+t(t+1)} = \dfrac{4}{t^2+2t}\] Since this line
passes through \(S(-t,0)\), its
equation is \(y = \dfrac{4}{t^2+2t}(x +
t)\).
To find the \(x\)-coordinate of \(X\), we find the point of intersection of
the line through \(S\) and \(P\) and the line through \(T\) and \(Q\); thus, we solve \[\begin{align*}
-\dfrac{4}{s^2+2s}(x - s) & = \dfrac{4}{t^2+2t}(x + t) \\
\dfrac{4s}{s^2+2s} - \dfrac{4t}{t^2 + 2t} & =
\left(\dfrac{4}{s^2+2s} + \dfrac{4}{t^2+2t}\right) x \\
4s(t^2+2t) - 4t(s^2+2s) & = (4(t^2+2t) + 4(s^2+2s)) x \\
x & = \dfrac{st^2 - s^2t}{t^2+2t+s^2+2s}\end{align*}\]
Therefore, the \(y\)-coordinate of
\(X\) is obtained from \[\begin{align*}
y & = \dfrac{4}{t^2 + 2t}\left(\dfrac{st^2 - s^2t}{t^2+2t+s^2+2s} +
t\right) \\
& = \dfrac{4}{t^2+2t} \cdot \dfrac{st^2 - s^2t + t^3 + 2t^2 + ts^2 +
2st}{t^2+2t+s^2+2s} \\
& = \dfrac{4}{t^2+2t} \cdot \dfrac{t^3 + 2t^2 + st^2 +
2st}{t^2+2t+s^2+2s} \\
& = \dfrac{4}{t^2+2t} \cdot \dfrac{t(t^2 + 2t) + s(t^2 +
2t)}{t^2+2t+s^2+2s} \\
& =\dfrac{4(s+t)}{t^2+2t+s^2+2s} \\\end{align*}\] Finally,
the \(y\)-coordinate of \(X\) is equal to the height \(h\) from earlier, so \[\begin{align*}
\dfrac{2}{s+t} & = \dfrac{4(s+t)}{t^2+2t+s^2+2s} \\
\dfrac{1}{s+t} & = \dfrac{2(s+t)}{t^2+2t+s^2+2s} \\
t^2+2t+s^2+2s & = 2(s+t)^2 \\
t^2 + 2t + s^2 + 2s & = 2s^2 + 4st + 2t^2 \\
-4st + 2s + 2t & = s^2 + t^2 \end{align*}\] Therefore, the
desired relationship is true, with \(d =
-4\), \(e = f = 2\) and \(g = 0\).
We look for pairs of the form \((s,t) =
(s, ks)\) where \(s\) and \(k\) are rational numbers.
From the relationship from (b), \[\begin{align*}
s^2 + k^2s^2 & = -4ks^2 + 2s + 2ks \\
s^2(k^2+4k+1) & = (2k + 2)s \\
s(k^2 + 4k + 1) & = 2k + 2 & \text{(since $s > 0$)} \\
s & = \dfrac{2k+2}{k^2 + 4k+1}\end{align*}\] Thus, \(t = ks = \dfrac{2k^2+2k}{k^2+4k+1}\).
Since \(k > 0\), then \(s > 0\).
We need to have \(s \leq t\). This is
equivalent to \[\begin{align*}
\dfrac{2k+2}{k^2 + 4k + 1} & \leq \dfrac{2k^2 + 2k}{k^2 + 4k + 1} \\
2k + 2 & \leq 2k^2 + 2k & \text{(since $k^2 + 4k + 1 > 0$)}
\\
2 & \leq 2k^2 \\
1 & \leq k & \text{(since $k > 0$)}\end{align*}\]
Thus, when \(k \geq 1\), we have \(0 < s \leq t\).
Finally, we need \(t \leq 1\). This is
equivalent to \[\begin{align*}
\dfrac{2k^2 + 2k}{k^2 + 4k + 1} & \leq 1 \\
2k^2 + 2k & \leq k^2 + 4k + 1 & \text{(since $k^2 + 4k + 1 >
0$)} \\
k^2 - 2k - 1 & \leq 0 \end{align*}\] The roots of \(k^2 - 2k - 1 = 0\) are \(k = \dfrac{2 \pm \sqrt{(-2)^2 -4(1)(-1)}}{2} = 1
\pm \sqrt{2}\).
Thus, \(k^2 - 2k - 1 \leq 0\) when
\(1 - \sqrt{2} \leq k \leq 1 +
\sqrt{2}\).
Since \(k > 0\), then \(t \leq 1\) when \(0 < k \leq 1 + \sqrt{2}\).
Therefore, when \(1 \leq k \leq 1 +
\sqrt{2}\), we have \(0 < s \leq t
\leq 1\).
We note that there are infinitely many rational numbers in any interval
of non-zero length; in particular, there are infinitely many rational
numbers \(k\) that satisfy \(1 \leq k \leq 1 + \sqrt{2}\).
We do need to confirm that the values of \(\dfrac{2k+2}{k^2 + 4k+1}\) are different
for different values of \(k\), which
will confirm that there are infinitely many different values of \(s\) as \(k\) assumes these infinitely many values
itself.
To do this, we show that if \(\dfrac{2k+2}{k^2
+ 4k+1} = \dfrac{2j+2}{j^2 + 4j+1}\), then \(k = j\); this fact will allow us to
conclude that if \(k \neq j\), then
\(\dfrac{2k+2}{k^2 + 4k+1} \neq
\dfrac{2j+2}{j^2 + 4j+1}\).
When \(k > 0\) and \(j > 0\), then following equations are
equivalent: \[\begin{align*}
\dfrac{2k+2}{k^2 + 4k+1} & = \dfrac{2j+2}{j^2 + 4j+1}\\
(2k+2)(j^2+4j+1) & = (2j+2)(k^2+4k+1) \\
(k+1)(j^2+4j+1) & = (j+1)(k^2+4k+1) \\
j^2 k + 4jk + k + j^2 + 4j + 1 & = jk^2 + 4jk + j + k^2 + 4k + 1 \\
j^2k - jk^2 + 3j - 3k + j^2 - k^2 & = 0 \\
(j-k)(jk + j + k + 3) & = 0\end{align*}\] Since \(jk + j + k + 3 > 0\), then it must be
the case that \(j - k = 0\) and so
\(j = k\).
This means that if \(k \neq j\), then
\(\dfrac{2k+2}{k^2 + 4k+1} \neq
\dfrac{2j+2}{j^2 + 4j+1}\) and so there are infinitely many
different values of \(s\).
Therefore, the pairs \((s,t) = \left(\dfrac{2k
+ 2}{k^2 + 4k + 1}, \dfrac{2k^2 + 2k}{k^2 + 4k + 1}\right)\)
where \(k\) is a rational number with
\(1 \leq k \leq 1 + \sqrt{2}\) is an
infinite family of solutions to the relationship from (b).