2022 Euclid Contest
Solutions
(Grade 12)
Tuesday, April 5, 2022
(in North America and South America)
Wednesday, April 6, 2022
(outside of North American and South America)
Tuesday, April 5, 2022
(in North America and South America)
Wednesday, April 6, 2022
(outside of North American and South America)
©2022 University of Waterloo
Evaluating, \(\dfrac{3^2 - 2^3}{2^3 -
3^2} = \dfrac{9 - 8}{8-9} = \dfrac{1}{-1} = -1\).
Alternatively, since \(2^3 - 3^2 = -(3^2 -
2^3)\), then \(\dfrac{3^2 - 2^3}{2^3 -
3^2} = -1\).
Evaluating, \(\sqrt{\phantom{\left(\right.}\hspace{-2mm}\sqrt{81}+\sqrt{9}-\sqrt{64}} = \sqrt{9 + 3 - 8} = \sqrt{4} = 2\).
Since \(\dfrac{1}{\sqrt{x^2 + 7}} =
\dfrac{1}{4}\), then \(\sqrt{x^2 + 7} =
4\).
This means that \(x^2 + 7 = 4^2 = 16\)
and so \(x^2 = 9\).
Since \(x^2 = 9\), then \(x = \pm 3\).
We can check by substitution that both of these values are
solutions.
Factoring, \(2022 = 2 \cdot 1011 = 2
\cdot 3 \cdot 337\). (It turns out that 337 is a prime number,
though this fact is not needed here.)
Therefore, \(2022 = 2 \cdot 1011\) and
\(2022 = 3 \cdot 674\) and \(2022 = 6 \cdot 337\).
Thus, the three ordered pairs are \((a,b) =
(2, 1011), (3, 674), (6, 337)\).
Manipulating algebraically, the following equations are
equivalent: \[\begin{aligned}
\dfrac{2c+1}{2d+1} & = \dfrac{1}{17}\\
17(2c+1) & = 2d + 1 \\
34c + 17 & = 2d + 1 \\
34c + 16 & = 2d \\
d & = 17c + 8\end{aligned}\] Since \(c\) is an integer with \(c > 0\), then \(c \geq 1\), which means that \(17c + 8 \geq 25\).
Therefore, the smallest possible value of \(d\) is \(d=25\).
Note that, when \(d = 25\), we obtain
\(c=1\) and so \(\dfrac{2c+1}{2d+1} = \dfrac{3}{51} =
\dfrac{1}{17}\).
Solution 1
When \(x = -5\), the left side of the
equation equals 0.
This means that when \(x = -5\), the
right side of the equation must equal 0 as well.
Thus, \((-5)^2 + 3(-5) + t = 0\) and so
\(25 - 15 + t = 0\) or \(t = -10\).
Solution 2
Expanding the left side, we obtain \[(px+r)(x+5) = px^2 + rx + 5px + 5r\] Since
this is equal to \(x^2 + 3x + t\) for
all real numbers, then the coefficients of the two quadratic expressions
must be the same.
Comparing coefficients of \(x^2\), we
obtain \(p = 1\).
This means that \[x^2 + rx + 5x + 5r = x^2 +
3x + t\] Comparing coefficients of \(x\), we obtain \(r+5 = 3\) and so \(r = -2\).
This means that \[x^2 + 3x - 10 = x^2 + 3x +
t\] Comparing constant terms, we obtain \(t = -10\).
Suppose that the volume of the jug is \(V\) L.
Then \(\frac{1}{4}V + 24 =
\frac{5}{8}V\).
Multiplying by 8, we obtain \(2V + 24 \cdot 8
= 5V\) which gives \(3V = 192\)
and so \(V = 64\).
Therefore, the volume of the jug is 64 L.
Suppose that Stephanie starts with \(n\) soccer balls.
Since Stephanie can divide the \(n\)
balls into fifths and into elevenths, then \(n\) is a multiple of both 5 and 11.
Since 5 and 11 are both prime numbers, then \(n\) must be a multiple of \(5\cdot 11 = 55\).
Thus, \(n = 55k\) for some positive
integer \(k\).
In this case, \(\frac{2}{5}n = \frac{2}{5}
\cdot 55k = 22k\) and \(\frac{6}{11}n =
\frac{6}{11} \cdot 55k = 30k\).
When Stephanie has given these balls away, she is left with \(55k - 22k - 30k = 3k\) balls.
Since \(3k\) is a multiple of 9, then
\(k\) is a multiple of 3.
Therefore, the smallest possible number of balls is obtained when \(k = 3\), which means that Stephanie started
with \(n = 55\cdot 3 = 165\) soccer
balls.
Suppose that the number of students in the Junior section is
\(j\) and the number of students in the
Senior section is \(s\).
The number of left-handed Junior students is 60% of \(j\), or \(0.6j\).
The number of right-handed Junior students is 40% of \(j\), or \(0.4j\).
The number of left-handed Senior students is 10% of \(s\), or \(0.1s\).
The number of right-handed Senior students is 90% of \(s\), or \(0.9s\).
Since the total numbers of left-handed and right-students are equal, we
obtain the equation \(0.6j + 0.1s = 0.4j +
0.9s\) which gives \(0.2j =
0.8s\) or \(j = 4s\).
This means that there are 4 times as many Junior students as Senior
students, which means that \(\frac{4}{5}\) of the students are Junior
and \(\frac{1}{5}\) are Senior.
Therefore, 80% of the students in the math club are in the Junior
section.
Let \(P\) be the point with coordinates \((7,0)\) and let \(Q\) be the point with coordinates \((0,5)\).
Then \(APDQ\) is a rectangle with
width 7 and height 5, and so it has area \(7
\cdot 5 = 35\).
Hexagon \(ABCDEF\) is formed by
removing two triangles from rectangle \(APDQ\), namely \(\triangle BPC\) and \(\triangle EQF\).
Each of \(\triangle BPC\) and \(\triangle EQF\) is right-angled, because
each shares an angle with rectangle \(APDQ\).
Each of \(\triangle BPC\) and \(\triangle EQF\) has a base of length 3 and
a height of 2.
Thus, their combined area is \(2 \cdot
\frac{1}{2} \cdot 3 \cdot 2 = 6\).
This means that the area of hexagon \(ABCDEF\) is \(35
- 6 = 29\).
Since \(\triangle PQS\) is
right-angled at \(P\), then by the
Pythagorean Theorem, \[SQ^2 = SP^2 + PQ^2 =
(x+3)^2 + x^2\] Since \(\triangle
QRS\) is right-angled at \(Q\),
then by the Pythagorean Theorem, we obtain \[\begin{aligned}
RS^2 & = SQ^2 + QR^2 \\
(x+8)^2 & = ((x+3)^2 + x^2) + 8^2 \\
x^2 + 16x + 64 & = x^2 + 6x + 9 + x^2 + 64 \\
0 & = x^2 - 10x + 9 \\
0 & = (x-1)(x-9)\end{aligned}\] and so \(x = 1\) or \(x =
9\).
(We can check that if \(x = 1\), \(\triangle PQS\) has sides of lengths 4, 1
and \(\sqrt{17}\) and \(\triangle QRS\) has sides of lengths \(\sqrt{17}\), 8 and 9, both of which are
right-angled, and if \(x = 9\), \(\triangle PQS\) has sides of lengths 12, 9
and 15 and \(\triangle QRS\) has sides
of lengths 15, 8 and 17, both of which are right-angled.)
In terms of \(x\), the perimeter of
\(PQRS\) is \(x + 8 + (x+8) + (x+3) = 3x + 19\).
Thus, the possible perimeters of \(PQRS\) are 22 (when \(x = 1\)) and 46 (when \(x = 9\)).
If \(r\) is a term in the
sequence and \(s\) is the next term,
then \(s = 1 + \dfrac{1}{1+r}\).
This means that \(s - 1 =
\dfrac{1}{1+r}\) and so \(\dfrac{1}{s-1} = 1+r\) which gives \(r = \dfrac{1}{s-1} - 1\).
Therefore, since \(a_3 =
\dfrac{41}{29}\), then \[a_2 =
\dfrac{1}{a_3-1} - 1 = \dfrac{1}{(41/29)-1} - 1 = \dfrac{1}{12/29} - 1 =
\dfrac{29}{12} - 1 = \dfrac{17}{12}\] Further, since \(a_2 = \dfrac{17}{12}\), then \[a_1 = \dfrac{1}{a_2-1} - 1 = \dfrac{1}{(17/12)-1}
- 1 = \dfrac{1}{5/12} - 1 = \dfrac{12}{5} - 1 =
\dfrac{7}{5}\]
Initially, the water in the hollow tube forms a cylinder with
radius 10 mm and height \(h\) mm. Thus,
the volume of the water is \(\pi (10\mbox{
mm})^2(h\mbox{ mm}) = 100\pi h \mbox{ mm}^3\).
After the rod is inserted, the level of the water rises to 64 mm. Note
that this does not overflow the tube, since the tube’s height is 100
mm.
Up to the height of the water, the tube is a cylinder with radius 10 mm
and height 64 mm.
Thus, the volume of the tube up to the height of the water is \[\pi (10\mbox{ mm})^2(64\mbox{ mm}) =
6400\pi \mbox{ mm}^3\] This volume consists of the water that is
in the tube (whose volume, which has not changed, is \(100\pi h \mbox{ mm}^3\)) and the rod up to
a height of 64 mm.
Since the radius of the rod is 2.5 mm, the volume of the rod up to a
height of 64 mm is \(\pi (2.5\mbox{
mm})^2(64\mbox{ mm}) = 400\pi \mbox{ mm}^3\).
Comparing volumes, \(6400\pi\mbox{ mm}^3 =
100\pi h\mbox{ mm}^3 + 400\pi\mbox{ mm}^3\) and so \(100h = 6000\) which gives \(h = 60\).
We note that \(\dfrac{2x+1}{x} =
\dfrac{2x}{x} + \dfrac{1}{x} = 2 + \dfrac{1}{x}\).
Therefore, \(\dfrac{2x+1}{x} = 4\)
exactly when \(2 + \dfrac{1}{x} = 4\)
or \(\dfrac{1}{x} = 2\) and so \(x = \dfrac{1}{2}\).
Alternatively, we could solve \(\dfrac{2x+1}{x} = 4\) directly to obtain
\(2x + 1 = 4x\), which gives \(2x = 1\) and so \(x = \dfrac{1}{2}\).
Thus, to determine the value of \(f(4)\), we substitute \(x = \dfrac{1}{2}\) into the given equation
\(f\left(\dfrac{2x+1}{x}\right) = x +
6\) and obtain \(f(4) = \dfrac{1}{2} +
6 = \dfrac{13}{2}\).
Since the graph passes through \((3,5)\), \((5,4)\) and \((11,3)\), we can substitute these three
points and obtain the following three equations: \[\begin{aligned}
5 & = \log_a(3+b) + c \\
4 & = \log_a(5+b) + c \\
3 & = \log_a(11+b) + c \end{aligned}\] Subtracting the second
equation from the first and the third equation from the second, we
obtain: \[\begin{aligned}
1 & = \log_a(3+b) - \log_a(5+b) \\
1 & = \log_a(5+b) - \log_a(11+b)\end{aligned}\] Equating
right sides and manipulating, we obtain the following equivalent
equations: \[\begin{aligned}
\log_a(5+b) - \log_a(11+b) & = \log_a(3+b) - \log_a(5+b) \\
2\log_a(5+b) & = \log_a(3+b) + \log_a(11+b) \\
\log_a\left((5+b)^2\right) & = \log_a\left((3+b)(11+b)\right) &
\text{(using log laws)} \\
(5+b)^2 & = (3+b)(11+b) & \text{(raising both sides to the power
of $a$)} \\
25+10b + b^2 & = 33 + 14b + b^2 \\
-8 & = 4b \\
b & = -2\end{aligned}\] Since \(b =
-2\), the equation \(1 = \log_a(3+b) -
\log_a(5+b)\) becomes \(1 = \log_a 1 -
\log_a 3\).
Since \(\log_a 1 = 0\) for every
admissible value of \(a\), then \(\log_a 3 = -1\) which gives \(a = 3^{-1} = \frac{1}{3}\).
Finally, the equation \(5 = \log_a(3+b) +
c\) becomes \(5 = \log_{1/3}(1) +
c\) and so \(c = 5\).
Therefore, \(a = \frac{1}{3}\), \(b = -2\), and \(c
= 5\), which gives \(y =
\log_{1/3}(x-2) + 5\).
Checking:
When \(x = 3\), we obtain \(y = \log_{1/3}(3-2) + 5 = \log_{1/3} 1 + 5 = 0 + 5 = 5\).
When \(x = 5\), we obtain \(y = \log_{1/3}(5-2) + 5 = \log_{1/3} 3 + 5 = -1 + 5 = 4\).
When \(x = 11\), we obtain \(y = \log_{1/3}(11-2) + 5 = \log_{1/3} 9 + 5 = -2 +
5 = 3\).
The probability that the integer \(n\) is chosen is \(\log_{100}\left(1 +
\dfrac{1}{n}\right)\).
The probability that an integer between 81 and 99, inclusive, is chosen
equals the sum of the probabilities that the integers 81, 82, \(\ldots\), 98, 99 are selected, which equals
\[\log_{100}\left(1 + \dfrac{1}{81}\right) +
\log_{100}\left(1 + \dfrac{1}{82}\right) + \cdots + \log_{100}\left(1 +
\dfrac{1}{98}\right) + \log_{100}\left(1 + \dfrac{1}{99}\right)\]
Since the second probability equals 2 times the first probability, the
following equations are equivalent: \[\begin{aligned}
\hspace{-2cm}
\log_{100}\left(1 + \dfrac{1}{81}\right) + \log_{100}\left(1 +
\dfrac{1}{82}\right) + \cdots + \log_{100}\left(1 + \dfrac{1}{98}\right)
+ \log_{100}\left(1 + \dfrac{1}{99}\right)
& = 2\log_{100}\left(1 + \dfrac{1}{n}\right) \\
\log_{100}\left(\dfrac{82}{81}\right) +
\log_{100}\left(\dfrac{83}{82}\right) + \cdots +
\log_{100}\left(\dfrac{99}{98}\right) +
\log_{100}\left(\dfrac{100}{99}\right)
& = 2\log_{100}\left(1 + \dfrac{1}{n}\right)\end{aligned}\]
Using logarithm laws, these equations are further equivalent to \[\begin{aligned}
\log_{100}\left(\dfrac{82}{81} \cdot \dfrac{83}{82}\cdot \cdots \cdot
\dfrac{99}{98} \cdot \dfrac{100}{99}\right)
& = \log_{100}\left(1 + \dfrac{1}{n}\right)^2 \\
\log_{100}\left(\dfrac{100}{81}\right) & = \log_{100}\left(1 +
\dfrac{1}{n}\right)^2\end{aligned}\] Since logarithm functions
are invertible, we obtain \(\dfrac{100}{81} =
\left(1 + \dfrac{1}{n}\right)^2\).
Since \(n>0\), then \(1 + \dfrac{1}{n} = \sqrt{\dfrac{100}{81}} =
\dfrac{10}{9}\), and so \(\dfrac{1}{n}
= \dfrac{1}{9}\), which gives \(n =
9\).
Since \(\dfrac{AC}{AD} = \dfrac{3}{4}\), then we let \(AC = 3t\) and \(AD = 4t\) for some real number \(t > 0\).
Using the cosine law in \(\triangle
ACD\), the following equations are equivalent: \[\begin{aligned}
AD^2 & = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cdot \cos(\angle ACD) \\
(4t)^2 & = (3t)^2 + 1^2 - 2(3t)(1)(-\tfrac{3}{5}) \\
16t^2 & = 9t^2 + 1 + \tfrac{18}{5}t \\
80t^2 & = 45t^2 + 5 + 18t \\
35t^2 - 18t - 5 & = 0 \\
(7t-5)(5t + 1) & = 0\end{aligned}\] Since \(t > 0\), then \(t = \frac{5}{7}\).
Thus, \(AC = 3t = \frac{15}{7}\).
Using the cosine law in \(\triangle
ACB\) and noting that \[\cos(\angle
ACB) = \cos(180^\circ - \angle ACD) = -\cos(\angle ACD) =
\tfrac{3}{5}\] the following equations are equivalent: \[\begin{aligned}
AB^2 & = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(\angle ACB) \\
& = \left(\tfrac{15}{7}\right)^2 + 2^2 -
2(\tfrac{15}{7})(2)(\tfrac{3}{5}) \\[1mm]
& = \tfrac{225}{49} + 4 - \tfrac{36}{7} \\[1mm]
& = \tfrac{225}{49} + \tfrac{196}{49} - \tfrac{252}{49} \\[1mm]
& = \tfrac{169}{49}\end{aligned}\] Since \(AB>0\), then \(AB = \frac{13}{7}\).
The parabola with equation \(y = ax^2 +
2\) is symmetric about the \(y\)-axis.
Thus, its vertex occurs when \(x = 0\)
(which gives \(y = a\cdot 0^2 + 2 =
2\)) and so \(V\) has
coordinates \((0, 2)\).
To find the coordinates of \(B\) and
\(C\), we use the equations of the
parabola and line to obtain \[\begin{aligned}
ax^2 + 2 & = -x + 4a \\
ax^2 + x + (2-4a) & = 0\end{aligned}\] Using the quadratic
formula, \[x = \dfrac{-1 \pm \sqrt{1^2 - 4a(2
- 4a)}}{2a} = \dfrac{-1 \pm \sqrt{1 - 8a + 16a^2}}{2a}\] Since
\(1 - 8a + 16a^2 = (4a - 1)^2\) and
\(4a - 1 > 0\) (since \(a > \frac{1}{2}\)), then \(\sqrt{1-8a + 16a^2} = 4a - 1\) and so \[x = \dfrac{-1 \pm (4a-1)}{2a}\] which
means that \(x = \dfrac{4a-2}{2a} =
\dfrac{2a-1}{a} = 2 - \dfrac{1}{a}\) or \(x = \dfrac{-4a}{2a} = -2\).
We can use the equation of the line to obtain the \(y\)-coordinates of \(B\) and \(C\).
When \(x = -2\) (corresponding to point
\(B\)), we obtain \(y = - (-2) + 4a = 4a + 2\).
When \(x = 2 - \dfrac{1}{a}\)
(corresponding to point \(C\)), we
obtain \(y = - \left(2 - \dfrac{1}{a}\right) +
4a = 4a - 2 + \dfrac{1}{a}\).
Let \(P\) and \(Q\) be the points on the horizontal line
through \(V\) so that \(BP\) and \(CQ\) are perpendicular to \(PQ\).
Then the area of \(\triangle VBC\)
is equal to the area of trapezoid \(PBCQ\) minus the areas of right-angled
\(\triangle BPV\) and right-angled
\(\triangle CQV\).
Since \(B\) has coordinates \((-2,4a+2)\), \(P\) has coordinates \((-2,2)\), \(V\) has coordinates \((0,2)\), \(Q\) has coordinates \(\left(2 - \dfrac{1}{a}, 2\right)\), and
\(C\) has coordinates \(\left(2 - \dfrac{1}{a}, 4a - 2 +
\dfrac{1}{a}\right)\), then \[\begin{aligned}
BP & = (4a + 2) - 2 = 4a \\
CQ & = \left(4a - 2 + \dfrac{1}{a}\right) - 2 = 4a - 4 +
\dfrac{1}{a} \\
PV & = 0 - (-2) = 2 \\
QV & = 2 - \dfrac{1}{a} - 0 = 2 - \dfrac{1}{a} \\
PQ & = PV + QV = 2 + 2 - \dfrac{1}{a} = 4 -
\dfrac{1}{a}\end{aligned}\] Therefore, the area of trapezoid
\(PBCQ\) is \[\dfrac{1}{2}(BP + CQ)(PQ) = \dfrac{1}{2}\left(4a
+ 4a - 4 + \dfrac{1}{a}\right)\left(4 - \dfrac{1}{a}\right) = \left(4a -
2 + \dfrac{1}{2a}\right)\left(4 - \dfrac{1}{a}\right)\] Also, the
area of \(\triangle BPV\) is \(\dfrac{1}{2}\cdot BP \cdot PV =
\dfrac{1}{2}(4a)(2) = 4a\).
Furthermore, the area of \(\triangle CQV\) is \[\dfrac{1}{2} \cdot CQ \cdot QV = \dfrac{1}{2}\left(4a - 4 + \dfrac{1}{a}\right) \left(2 - \dfrac{1}{a}\right) = \left(2a - 2 + \dfrac{1}{2a}\right) \left(2 - \dfrac{1}{a}\right)\] From the given information, \[\left(4a - 2 + \dfrac{1}{2a}\right)\left(4 - \dfrac{1}{a}\right) - 4a - \left(2a - 2 + \dfrac{1}{2a}\right) \left(2 - \dfrac{1}{a}\right) = \dfrac{72}{5}\] Multiplying both sides by \(2a^2\), which we distribute through the factors on the left side as \(2a \cdot a\), we obtain \[(8a^2 - 4a + 1)(4a - 1) - 8a^3 - (4a^2 - 4a + 1)(2a - 1) = \dfrac{144}{5} a^2\] Multiplying both sides by 5, we obtain \[5(8a^2 - 4a + 1)(4a - 1) - 40a^3 - 5(4a^2 - 4a + 1)(2a - 1) = 144 a^2\] Expanding and simplifying, we obtain \[\begin{aligned} (160a^3 - 120a^2 + 40a - 5) - 40a^3 - (40a^3 - 60a^2 + 30a - 5)& = 144a^2 \\ 80a^3 - 204a^2 + 10a & = 0 \\ 2a(40a^2 - 102a + 5) & = 0 \\ 2a(20a - 1)(2a - 5) & = 0\end{aligned}\] and so \(a = 0\) or \(a = \dfrac{1}{20}\) or \(a = \dfrac{5}{2}\). Since \(a > \dfrac{1}{2}\), then \(a = \dfrac{5}{2}\).
We prove that there cannot be such a triangle.
We prove this by contradiction. That is, we suppose that there is such a
triangle and prove that there is then a logical contradiction.
Suppose that \(\triangle ABC\) is not
equilateral, has side lengths that form a geometric sequence, and angles
whose measures form an arithmetic sequence.
Suppose that \(\triangle ABC\) has side
lengths \(BC = a\), \(AC = ar\), and \(AB = ar^2\), for some real numbers \(a>0\) and \(r
> 1\). (These lengths form a geometric sequence, and we can
assume that this sequence is increasing, and that the sides are labelled
in this particular order.)
Since \(BC < AC < AB\), then the
opposite angles have the same relationships, namely \(\angle BAC < \angle ABC < \angle
ACB\).
Suppose that \(\angle BAC = \theta\),
\(\angle ABC = \theta + \delta\), and
\(\angle ACB = \theta + 2\delta\) for
some angles \(\theta\) and \(\delta\). (In other words, these angles
form an arithmetic sequence.
Since these three angles are the angles in a triangle, then their sum is
\(180^\circ\), and so \[\begin{aligned}
\theta + (\theta + \delta) + (\theta + 2\delta) & = 180^\circ \\
3 \theta + 3\delta & = 180^\circ \\
\theta + \delta & = 60^\circ\end{aligned}\] In other words,
\(\angle ABC = 60^\circ\).
We could proceed using the cosine law: \[\begin{aligned}
AC^2 & = BC^2 + AB^2 - 2\cdot BC \cdot AB \cdot \cos(\angle ABC) \\
(ar)^2 & = a^2 + (ar^2)^2 - 2a(ar^2)\cos(60^\circ) \\
a^2r^2 & = a^2 + a^2 r^4 - 2a^2r^2 \cdot \tfrac{1}{2} \\
a^2r^2 & = a^2 + a^2 r^4 - a^2r^2 \\
0 & = a^2r^4 - 2a^2r^2 + a^2 \\
0 & = a^2(r^4 - 2r^2 + 1) \\
0 & = a^2(r^2 - 1)^2 \end{aligned}\] This tells us that \(a = 0\) (which is impossible) or \(r^2 = 1\) (and thus \(r = \pm 1\), which is impossible).
Therefore, we have reached a logical contradiction and so such a
triangle cannot exist.
Alternatively, we could proceed using the sine law, noting that \[\begin{aligned}
\angle BAC & = \theta = (\theta + \delta) - \delta = 60^\circ -
\delta \\
\angle ACB & = \theta + 2\delta = (\theta + \delta) + \delta =
60^\circ + \delta\end{aligned}\] By the sine law, \[\dfrac{BC}{\sin(\angle BAC)} =
\dfrac{AC}{\sin(\angle ABC)} = \dfrac{AB}{\sin(\angle ACB)}\]
from which we obtain \[\dfrac{a}{\sin(60^\circ - \delta)} =
\dfrac{ar}{\sin(60^\circ)} = \dfrac{ar^2}{\sin(60^\circ +
\delta)}\] Since \(a \neq 0\),
from the first two parts, \[r = \dfrac{ar}{a}
= \dfrac{\sin60^\circ}{\sin(60^\circ - \delta)}\] Since \(ar \neq 0\), from the second two parts,
\[r = \dfrac{ar^2}{ar} = \dfrac{\sin(60^\circ
+ \delta)}{\sin60^\circ}\] Equating expressions for \(r\), we obtain successively \[\begin{aligned}
\dfrac{\sin60^\circ}{\sin(60^\circ - \delta)} & =
\dfrac{\sin(60^\circ + \delta)}{\sin60^\circ} \\
\sin^2 60^\circ & = \sin(60^\circ - \delta)\sin(60^\circ + \delta)
\\
\left(\tfrac{\sqrt{3}}{2}\right)^2 & = (\sin60^\circ \cos\delta -
\cos60^\circ \sin\delta) (\sin60^\circ \cos\delta + \cos60^\circ
\sin\delta) \\
\tfrac{3}{4} & = \left(\tfrac{\sqrt{3}}{2}\cos\delta -
\tfrac{1}{2}\sin\delta\right)\left(\tfrac{\sqrt{3}}{2}\cos\delta +
\tfrac{1}{2}\sin\delta\right) \\
\tfrac{3}{4} & = \tfrac{3}{4}\cos^2\delta - \tfrac{1}{4}\sin^2\delta
\\
\tfrac{3}{4} & = \tfrac{3}{4}\cos^2\delta + \tfrac{3}{4}\sin^2\delta
- \sin^2\delta \\
\tfrac{3}{4} & = \tfrac{3}{4}(\cos^2\delta + \sin^2\delta) -
\sin^2\delta \\
\tfrac{3}{4} & = \tfrac{3}{4} - \sin^2\delta \\
\sin^2\delta & = 0\end{aligned}\] which means that \(\delta = 0^\circ\). (Any other angle \(\delta\) with \(\sin \delta = 0\) would not produce angles
in a triangle.)
Therefore, all three angles in the triangle are \(60^\circ\), which means that the triangle
is equilateral, which it cannot be.
Therefore, we have reached a logical contradiction and so such a
triangle cannot exist.
The \((4,2)\)-sawtooth sequence consists of the terms \[1, \ \ \ 2, 3, 4, 3, 2, 1, \ \ \ 2, 3, 4, 3, 2, 1\] whose sum is 31.
Solution 1
Suppose that \(m \geq 2\).
The \((m,3)\)-sawtooth sequence
consists of an initial 1 followed by 3 teeth, each of which goes from 2
to \(m\) to 1.
Consider one of these teeth whose terms are \[2, 3, 4, \ldots, m-1, m, m-1, m-2, m-3, \ldots,
2, 1\] When we write the ascending portion directly above the
descending portion, we obtain \[\begin{array}{cccccc}
2, & 3, & 4, & \ldots, & m-1, & m, \\
m-1, & m-2, & m-3, & \ldots, & 2, & 1
\end{array}\] From this presentation, we can see \(m-1\) pairs of terms, the sum of each of
which is \(m+1\).
(Note that \(2 + (m-1) = 3 + (m-2) = 4 + (m-3)
= \cdots = (m-1) + 2 = m + 1\) and as we move from left to
right, the terms on the top increase by 1 at each step and the terms on
the bottom decrease by 1 at each step, so their sum is indeed
constant.)
Therefore, the sum of the numbers in one of the teeth is \((m-1)(m+1) = m^2 - 1\).
This means that the sum of the terms in the \((m,3)\)-sawtooth sequence is \(1 + 3(m^2 - 1)\), which equals \(3m^2 - 2\).
Solution 2
Suppose that \(m \geq 2\).
The \((m,3)\)-sawtooth sequence
consists of an initial 1 followed by 3 teeth, each of which goes from 2
to \(m\) to 1.
Consider one of these teeth whose terms are \[2, 3, 4, \ldots, m-1, m, m-1, m-2, m-3, \ldots,
2, 1\] This tooth includes one 1, two 2s, two 3s, and so on,
until we reach two \((m-1)\)s, and one
\(m\).
The sum of these numbers is \[1(1) + 2(2) +
2(3) + \cdots + 2(m-1) + m\] which can be rewritten as \[2(1 + 2 + 3 + \cdots + (m-1) + m) - 1 - m = 2
\cdot \tfrac{1}{2}m(m+1) - m - 1 = m^2 + m - m - 1 = m^2 -1\]
Therefore, the sum of the numbers in one of the teeth is \((m-1)(m+1) = m^2 - 1\).
This means that the sum of the terms in the \((m,3)\)-sawtooth sequence is \(1 + 3(m^2 - 1)\), which equals \(3m^2 - 2\).
From (b), the sum of the terms in each tooth is \(m^2 - 1\).
Thus, the sum of the terms in the \((m,n)\)-sawtooth sequence is \(1 + n(m^2 - 1)\).
For this to equal 145, we have \(n(m^2 - 1) =
144\).
This means that \(n\) and \(m^2-1\) form a divisor pair of 144.
As \(m\) ranges from 2 to 12, the
values of \(m^2 - 1\) are \[3, 8, 15, 24, 35, 48, 63, 80, 99, 120,
143\] (When \(m = 13\), we get
\(m^2 - 1 = 168\) and so when \(m \geq 13\), the value of \(m^2 - 1\) is too large to be a divisor of
144.)
Of these, \(3, 8, 24, 48\) are divisors
of 144 (corresponding to \(m=2,3,5,7\)), and give corresponding
divisors \(48, 18, 6, 3\).
Therefore, the pairs \((m,n)\) for
which the sum of the terms is 145 are \[(m,n)
= (2,48), (3,18), (5,6), (7,3)\]
In an \((m,n)\)-sawtooth
sequence, the sum of the terms is \(n(m^2 - 1)
+ 1\).
In each tooth, there are \((m-1) + (m-1) =
2m-2\) terms (from 2 to \(m\),
inclusive, and from \(m-1\) to 1,
inclusive).
This means that there are \(n(2m-2) +
1\) terms in the sequence.
Thus, the average of the terms in the sequence is \(\dfrac{n(m^2 - 1) + 1}{n(2m-2) +
1}\).
We need to prove that this is not an integer for all pairs of positive
integers \((m,n)\) with \(m \geq 2\).
Suppose that \(\dfrac{n(m^2 - 1) + 1}{n(2m-2)
+ 1} = k\) for some integer \(k\). We will show, by contradiction, that
this is not possible.
Since \(\dfrac{n(m^2 - 1) + 1}{n(2m-2) + 1}
= k\), then \[\begin{aligned}
\dfrac{m^2 n - n + 1}{2mn - 2n + 1} & = k \\
m^2 n - n + 1 & = 2mnk - 2nk + k \\
m^2 n - 2mnk + (2nk - n - k + 1) & = 0 \end{aligned}\] We
treat this as a quadratic equation in \(m\).
Since \(m\) is an integer, then this
equation has integer roots, and so its discriminant must be a perfect
square.
The discriminant of this quadratic equation is \[\begin{aligned}
\Delta & = (-2nk)^2 - 4n(2nk - n - k + 1) \\
& = 4n^2k^2 - 8n^2k + 4n^2 + 4nk - 4n \\
& = 4n^2(k^2 - 2k + 1) + 4n(k - 1) \\
& = 4n^2(k-1)^2 + 4n(k-1) \\
& = (2n(k-1))^2 + 2(2n(k-1)) + 1 - 1 \\
& = (2n(k-1) + 1)^2 - 1\end{aligned}\] We note that \((2n(k-1) + 1)^2\) is a perfect square and
\(\Delta\) is supposed to be a perfect
square.
But these perfect squares differ by 1, and the only two perfect squares
that differ by 1 are 1 and 0.
(To justify this last fact, we could look at the equation \(a^2 - b^2 = 1\) where \(a\) and \(b\) are non-negative integers, and factor
this to obtain \((a+b)(a-b) = 1\) which
would give \(a+b=a-b=1\) from which we
get \(a=1\) and \(b=0\).)
Since \((2n(k-1) + 1)^2 = 1\) and \(2n(k-1) + 1\) is non-negative, then \(2n(k-1) + 1 = 1\) and so \(2n(k-1) = 0\).
Since \(n\) is positive, then \(k-1=0\) or \(k=1\).
Therefore, the only possible way in which the average is an integer is
if the average is 1.
In this case, we get \[\begin{aligned}
m^2 n - 2mn + (2n - n - 1 + 1) & = 0 \\
m^2 n - 2mn + n & = 0 \\
n(m^2 - 2m + 1) & = 0 \\
n(m-1)^2 & = 0\end{aligned}\] Since \(n\) and \(m\) are positive integers with \(m \geq 2\), then \(n(m-1)^2 \neq 0\), which is a
contradiction.
Therefore, the average of the terms in an \((m,n)\)-sawtooth sequence cannot be an
integer.
Assume that the first topping is placed on the top half of the
pizza. (We can rotate the pizza so that this is the case.)
Assume that the second topping is placed on the half of the pizza that
is above the horizontal diameter that makes an angle of \(\theta\) clockwise with the horizontal as
shown. In other words, the topping covers the pizza from \(\theta\) to \(\theta + 180^\circ\).
We may assume that \(0^\circ \leq \theta
\leq 360^\circ\).
When \(0^\circ \leq \theta \leq
90^\circ\), the angle of the sector covered by both toppings is
at least \(90^\circ\) (and so is at
least a quarter of the circle).
When \(90^\circ < \theta \leq
180^\circ\), the angle of the sector covered by both toppings is
less than \(90^\circ\) (and so is less
than a quarter of the circle).
When \(\theta\) moves past \(180^\circ\), the left-hand portion of the
upper half circle starts to be covered with both toppings again. When
\(180^\circ \leq \theta <
270^\circ\), the angle of the sector covered by both toppings is
less than \(90^\circ\) (and so is less
than a quarter of the circle).
When \(270^\circ \leq \theta \leq
360^\circ\), the angle of the sector covered by both toppings at
least \(90^\circ\) (and so is at least
a quarter of the circle).
Therefore, if \(\theta\) is chosen
randomly between \(0^\circ\) and \(360^\circ\), the combined length of the
intervals in which at least \(\frac{1}{4}\) of the pizza is covered with
both toppings is \(180^\circ\).
Therefore, the probability is \(\dfrac{180^\circ}{360^\circ}\), or \(\dfrac{1}{2}\).
Suppose that the first topping is placed on the top half of the
pizza. (Again, we can rotate the pizza so that this is the case.)
Assume that the second topping is placed on the half of the pizza that
is above the diameter that makes an angle of \(\theta\) clockwise with the horizontal as
shown. In other words, the topping covers the pizza from \(\theta\) to \(\theta + 180^\circ\).
We may assume that \(0^\circ \leq \theta \leq
180^\circ\). If \(180^\circ \leq \theta
\leq 360^\circ\), the resulting pizza can be seen as a reflection
of the one shown.
Consider the third diameter added, shown dotted in the diagram above.
Suppose that its angle with the horizontal is \(\alpha\). (In the diagram, \(\alpha < 90^\circ\).) We assume that the
topping is added on the half pizza clockwise beginning at the angle of
\(\alpha\), and that this topping stays
in the same relative position as the diameter sweeps around the
circle.
For what angles \(\alpha\) will there
be a portion of the pizza covered with all three toppings?
If \(0^\circ \leq \alpha <
180^\circ\), there will be a portion covered with three toppings;
this portion is above the right half of the horizontal diameter.
If \(180^\circ \leq \alpha < 180^\circ +
\theta\), the third diameter will pass through the two regions
with angle \(\theta\) and the third
topping will be below this diameter, so there will not be a region
covered with three toppings.
If \(180^\circ +\theta \leq \alpha \leq
360^\circ\), the third topping starts to cover the leftmost part
of the region currently covered with two toppings, and so a region is
covered with three toppings.
Therefore, for an angle \(\theta\) with
\(0^\circ \leq \theta \leq 180^\circ\),
a region of the pizza is covered with three toppings when \(0^\circ \leq \alpha < 180^\circ\) and
when \(180^\circ + \theta \leq \alpha \leq
360^\circ\).
To determine the desired probability, we graph points \((\theta, \alpha)\). A particular choice of
diameters corresponds to a choice of angles \(\theta\) and \(\alpha\) with \(0^\circ \leq \theta \leq 180^\circ\) and
\(0^\circ \leq \alpha \leq 360^\circ\),
which corresponds to a point on the graph below.
The probability that we are looking for then equals the area of the
region of this graph where three toppings are in a portion of the pizza
divided by the total allowable area of the graph.
The shaded region of the graph corresponds to instances where a portion
of the pizza will be covered by three toppings.
This shaded region consists of the entire portion of the graph where
\(0^\circ \leq \alpha \leq 180^\circ\)
(regardless of \(\theta\)) as well as
the region above the line with equation \(\alpha = \theta + 180^\circ\) (that is, the
region with \(\theta + 180^\circ \leq \alpha
\leq 360^\circ\)).
Since the slope of the line is 1, it divides the upper half of the
region, which is a square, into two pieces of equal area.
Therefore, \(\frac{3}{4}\) of the graph
is shaded, which means that the probability that a region of the pizza
is covered by all three toppings is \(\dfrac{3}{4}\).
The main idea of this solution is that the toppings all overlap
exactly when there is one topping with the property that all other
toppings “begin” somewhere in that toppings semi-circle. In the rest of
this solution, we determine the probability using this fact and then
justify this fact.
Suppose that, for \(1 \leq j \leq N\),
topping \(j\) is put on the semi-circle
that starts at an angle of \(\theta_j\)
clockwise from the horizontal left-hand radius and continues to an angle
of \(\theta_j + 180^\circ\), where
\(0^\circ \leq \theta_j <
360^\circ\). By establishing these variables and this convention,
we are fixing both the angle of the diameter and the semi-circle defined
by this diameter on which the topping is placed.
Suppose that there is some region of the pizza with non-zero area that
is covered by all \(N\) toppings.
This region will be a sector with two bounding radii, each of which must
be half of a diameter that defines one of the toppings.
Suppose that the radius at the clockwise “end” of the sector is the end
of the semi-circle where topping \(X\)
is placed, and that the radius at the counter-clockwise “beginning” of
the sector is the start of the semi-circle where topping \(Y\) is placed.
This means that each of the other \(N-2\) toppings begins between (in the
clockwise sense) the points where topping \(X\) begins and where topping \(Y\) begins.
Consider the beginning angle for topping \(X\), \(\theta_X\).
To say that the other \(N-1\) toppings
begin at some point before topping \(X\) ends is the same as saying that each
\(\theta_j\) with \(j \neq X\) is between \(\theta_X\) and \(\theta_X + 180^\circ\).
Here, we can allow for the possibility that \(\theta_X + 180^\circ\) is greater than
\(360^\circ\) by saying that an angle
equivalent to \(\theta_j\) (which is
either \(\theta_j\) or \(\theta_j + 360^\circ\)) is between \(\theta_X\) and \(\theta_X + 180^\circ\).
For each \(j \neq X\), the angle \(\theta_j\) is randomly, uniformly and
independently chosen on the circle, so there is a probability of \(\dfrac{1}{2}\) that this angle (or its
equivalent) will be in the semi-circle between \(\theta_X\) and \(\theta_X + 180^\circ\).
Since there are \(N-1\) such angles,
the probability that all are between \(\theta_X\) and \(\theta_X + 180^\circ\) is \(\dfrac{1}{2^{N-1}}\).
Since there are \(N\) possible
selections for the first topping that can end the common sector, then
the desired probability will be \(\dfrac{N}{2^{N-1}}\) as long as we can show
that no set of angles can give two different sectors that are both
covered with all toppings.
To show this last fact, we suppose without loss of generality that \[0^\circ = \theta_1 < \theta_2 < \theta_3
< \cdots < \theta_{N-1} < \theta_N < 180^\circ\] (We
can relabel the toppings if necessary to obtain this order and rotate
the pizza so that topping 1 begins at \(0^\circ\).)
We need to show that it is not possible to have a \(Z\) for which \(\theta_Z, \theta_{Z+1}, \ldots, \theta_N,
\theta_1, \theta_2, \ldots, \theta_{Z-1}\) all lie in a
semi-circle starting with \(\theta_Z\).
Since \(\theta_Z < 180^\circ\) and
\(\theta_1\) can be thought of as \(360^\circ\), then this is not possible as
\(\theta_1\) and the angles after it
are all not within \(180^\circ\) of
\(\theta_Z\).
Therefore, it is not possible to have two such regions with the same set
of angles, and so the desired probability is \(\dfrac{N}{2^{N-1}}\).