2022 Hypatia Contest
Solutions
(Grade 11)
Tuesday, April 12, 2022
(in North America and South America)
Wednesday, April 13, 2022
(outside of North American and South America)
Tuesday, April 12, 2022
(in North America and South America)
Wednesday, April 13, 2022
(outside of North American and South America)
©2022 University of Waterloo
Regular hexagon \(ABCDEF\) has
side length \(2x\), and so \(AB=2x\).
Since \(\triangle OAB\) is equilateral,
then \(OA=OB=AB=2x\).
The radius of the circle is equal to \(OA\) and thus is \(2x\).
Since \(M\) is the midpoint of
\(AB\) and \(OA=OB\), then \(OM\) is perpendicular to \(AB\).
Since \(M\) is the midpoint of \(AB\), then \(AM=\frac12AB=x\).
Using the Pythagorean Theorem in right-angled \(\triangle OAM\), we get \(OA^2=OM^2+AM^2\) or \((2x)^2=OM^2+x^2\), and so \(OM^2=3x^2\) or \(OM=\sqrt{3}x\) (since \(OM>0\)).
Alternatively, notice that \(\triangle
OAM\) is a \(30^{\circ}\)-\(60^{\circ}\)-\(90^{\circ}\) triangle, and so \(AM:OA:OM=1:2:\sqrt{3}=x:2x:\sqrt{3}x\).
The diagonals \(AD\), \(BE\) and \(CF\) divide \(ABCDEF\) into six congruent equilateral
triangles.
Thus the area of \(ABCDEF\) is six
times the area of \(\triangle OAB\)
(having base \(AB\) and height \(OM\)), or \[6\times\tfrac12\times AB\times OM=3\times
2x\times \sqrt{3}x=6\sqrt{3}x^2\]
The area of the shaded region is determined by subtracting the
area of \(ABCDEF\) from the area of the
circle with centre \(O\) and radius
\(2x\).
Thus, the area of the shaded region is \[\pi(2x)^2-6\sqrt{3}x^2=4\pi
x^2-6\sqrt{3}x^2=(4\pi-6\sqrt{3})x^2\] The area of this shaded
region is 123, and so \((4\pi-6\sqrt{3})x^2=123\) or \(x^2=\dfrac{123}{4\pi-6\sqrt{3}}\).
Since \(x>0\), we get \(x=\sqrt{\dfrac{123}{4\pi-6\sqrt{3}}}\) and
so \(x=7.5\) when rounded to the
nearest tenth.
With 1 kg of muffin batter, exactly 24 mini muffins and 2 large muffins can be made. Thus with 2 kg of muffin batter, exactly \(2\times24=48\) mini muffins and \(2\times2=4\) large muffins can be made, and so \(n=4\).
Solution 1
With 2 kg of muffin batter, exactly 36 mini muffins and 6 large muffins can be made. With 2 kg of muffin batter, exactly 48 mini muffins and 4 large muffins can be made. Adding these, we get that with \(2 \textrm{ kg}+2 \textrm{ kg}= 4 \textrm{ kg}\) of muffin batter, exactly \(36+48=84\) mini muffins and \(6+4=10\) large muffins can be made, and so \(x=4\).
Solution 2
Let \(m\) be the number of kilograms
of muffin batter needed to make 1 mini muffin. Let \(\ell\) be the number of kilograms of muffin
batter needed to make 1 large muffin. Since 1 kg of muffin batter makes
exactly 24 mini muffins and 2 large muffins, then \(1=24m+2\ell\).
Since 2 kg of muffin batter makes exactly 36 mini muffins and 6 large
muffins, then \(2=36m+6\ell\).
Subtracting the second equation from 3 times the first equation, we get
\(3\times1-2=3\times(24m+2\ell)-(36m+6\ell)\)
or \(1=36m\), and so \(m=\dfrac{1}{36}\).
Substituting \(m=\dfrac{1}{36}\) into the second equation, we get \(2=36\left(\dfrac{1}{36}\right)+6\ell\) or \(1=6\ell\), and so \(\ell=\dfrac16\).
Since \(\dfrac{1}{36}\) kg of muffin
batter is needed to make 1 mini muffin, then \(84\cdot\dfrac{1}{36}=\dfrac73\) kg of
muffin batter is needed to make 84 mini muffins.
Since \(\dfrac{1}{6}\) kg of muffin
batter is needed to make 1 large muffin, then \(10\cdot\dfrac{1}{6}=\dfrac53\) kg of muffin
batter is needed to make 10 large muffins.
Thus, exactly 84 mini muffins and 10 large muffins can be made with
\(\dfrac73+\dfrac53=\dfrac{12}{3}=4\)
kg of muffin batter, and so \(x=4\).
In part (b) Solution 2, it was determined that \(\frac{1}{36}\) kg of muffin batter is
needed to make 1 mini muffin, and \(\frac{1}{6}\) kg of muffin batter is needed
to make 1 large muffin.
Therefore, the number of kilograms of muffin batter needed to make 1
large muffin is 6 times the number of kilograms of batter needed to make
1 mini muffin \(\left(6\times\frac{1}{36}=\frac16\right)\).
In other words, the batter needed to make 1 large muffin can make 6 mini
muffins.
Then, using the amount of batter that makes 7 large muffins, \(6\times7=42\) mini muffins can be
made.
If the first number in a sequence is 3 and the sequence is generated by the function \(x^2-3x+1\), then the second number in the sequence is \[3^2-3(3)+1=1,\] and the third number in the sequence is \[1^2-3(1)+1=-1,\] and the fourth number in the sequence is \[(-1)^2-3(-1)+1=5.\] The first four numbers in the sequence are \(3, 1, -1, 5\).
Let the first and second numbers in the sequence generated by the
function \(x^2-4x+7\) be \(f\) and \(s\), respectively.
Then, the first three numbers in the sequence are \(f,s,7\).
Since the third number in the sequence is 7, then \(s^2-4s+7=7\).
Solving this equation, we get \(s^2-4s=0\) or \(s(s-4)=0\), which has solutions \(s=0\) and \(s=4\), and so the first three numbers in
the sequence could be \(f,0,7\) or
\(f,4,7\).
If the second number in the sequence is 0, then \(f^2-4f+7=0\).
The discriminant of this equation is \((-4)^2-4(1)(7)=-12\) (less than zero) and
so there are no real solutions.
Thus, there is no first number in this sequence for which the second
number is 0.
If the second number in the sequence is 4, then \(f^2-4f+7=4\), or \(f^2-4f+3=0\) and so \((f-1)(f-3)=0\), which has solutions \(f=1\) and \(f=3\).
Therefore, if 7 is the third number in a sequence generated by the
function \(x^2-4x+7\), then the first
three numbers in the sequence could be \(1,4,7\) or \(3,4,7\), and so the possible first numbers
in the sequence are 1 and 3.
The first two numbers in the sequence are \(c,c\), and so \(c^2-7c-48=c\).
Solving this equation, we get \(c^2-8c-48=0\) or \((c+4)(c-12)=0\), which has solutions \(c=-4\) and \(c=12\).
The first number in the sequence is \(a\) and the second number is \(b\), and so \[a^2-12a+39=b.\] The second number in the
sequence is \(b\) and the third number
is \(a\), and so \[b^2-12b+39=a.\] Subtracting the second
equation from the first and simplifying, we get \[\begin{align*}
(a^2-12a+39)-(b^2-12b+39)&=b-a\\
a^2-b^2-12a+12b&=b-a\\
a^2-b^2-11a+11b&=0\\
(a-b)(a+b)-11(a-b)&=0\\
(a-b)(a+b-11)&=0\end{align*}\] Since \(a\neq b\), then \(a-b\neq 0\) and so \(a+b-11=0\) or \(b=11-a\).
Substituting into the first equation, we get \(a^2-12a+39=11-a\) or \(a^2-11a+28=0\). Factoring gives \((a-4)(a-7)=0\) and so the possible values
of \(a\) are 4 and 7.
(Note that the two possible sequences are \(4,7,4, \dots\) and \(7,4,7,\dots\).)
Written as a product of its prime factors, \(240=2^43^15^1\) and so \[f(240)=(1+4)(1+1)(1+1)=(5)(2)(2)=20\]
Suppose \(f(N)=6\) and \(N\) is refactorable. Then 6 divides \(N\).
\(N\) is divisible by 6 exactly when
its prime factors include at least one 2 and at least one 3 (since \(6=2\times3\)).
Assume that \(N\) contains an
additional prime factor \(p\), distinct
from 2 and 3.
In this case, the divisors of \(N\) are
\(1,2,3,6,p,2p,3p\), and \(6p\), which is too many divisors.
Thus, it must be that \(N\) is a
positive integer of the form \(N=2^u3^v\), where \(u\) and \(v\) are positive integers, and so \(f(N)=(1+u)(1+v)=6\).
Since \(u\) and \(v\) are positive integers, then \(1+u\geq2\) and \(1+v\geq2\) and so there are exactly two
possibilities: \(u=1\) and \(v=2\) or \(u=2\) and \(v=1\).
When \(u=1\) and \(v=2\), \(N=2^13^2=18\) and when \(u=2\) and \(v=1\), \(N=2^23^1=12\).
The refactorable numbers \(N\) with
\(f(N)=6\) are 12 and 18.
Since \(N\) is refactorable,
then \(f(N)=256=2^8\) divides \(N\).
Thus for some integer \(w\geq8\), \(N\) is of the form \(N=2^wp_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}\),
where \(2<p_1<p_2<
\cdots<p_k\) are prime numbers, and \(a_1,a_2,\cdots,a_k\) are positive
integers.
In this case, \(f(N)=(1+w)(1+a_1)(1+a_2)\cdots(1+a_k)=2^8\),
and so each of the factors \(1+w, 1+a_1,
1+a_2, \cdots ,1+a_k\) is a power of 2 (since their product is
\(2^8\)).
This means that each of the exponents \(w,a_1,a_2,\cdots ,a_k\) is one less than a
power of two.
Since \(w\geq8\), then \(1+w\geq 9\) and so \(1+w\geq 2^4\) (the smallest power of 2
greater than 8).
Since \(f(N)=(1+w)(1+a_1)(1+a_2)\cdots(1+a_k)=2^8\),
then \(2^4\leq 1+w\leq 2^8\) and \(1\leq (1+a_1)(1+a_2)\cdots(1+a_k)\leq
2^4\), and so \(w\) must be
equal to \(15, 31, 63, 127\), or 255,
and each of \(a_1,a_2,\cdots,a_k\) must
be equal to \(1,3,7\), or 15.
For example, if \(1+w=2^8\), then \(w=256-1=255\) and \(N=2^{255}\).
If \(1+w=2^4\), then \(w=15\) and \(N\) is of the form \(N=2^{15}p_1^{a_1}p_2^{a_2}\cdots
p_k^{a_k}\).
Thus, the smallest \(N\) having the
greatest number of distinct prime factors is \(N=2^{15}3^{1}5^{1}7^{1}11^{1}\), and we may
confirm that \(f(N)=(1+15)(1+1)(1+1)(1+1)(1+1)=2^4\cdot2\cdot2\cdot2\cdot2=2^8\),
as required.
Comparing these first two values of \(N\), we recognize that \(2^{15}3^{1}5^{1}7^{1}11^{1}\) is
significantly less than \(2^{255}\).
Further, we recognize that
all remaining possible values of \(N\) have exactly 2, 3 or 4 distinct prime factors,
where exponents are equal, smaller prime factors give smaller values of \(N\),
the greatest exponents must occur on the smallest prime factors, and
we recall that each exponent is one less than a power of 2.
In the table below, we use the above information to determine the
smallest possible values of \(N\)
having 1, 2, 3, 4, and 5 distinct prime factors.
Further, we compare the size of each of these values of \(N\) within each of the 5 groups.
| Number of distinct prime factors of \(N\) | Values of \(N\) |
|---|---|
| 1 | \(2^{255}\) |
| 2 | \(2^{15}3^{15}<2^{31}3^7<2^{63}3^3<2^{127}3^1\) |
| 3 | \(2^{15}3^{3}5^3<2^{15}3^75^1<2^{31}3^35^1<2^{63}3^15^1\) |
| 4 | \(2^{15}3^{3}5^17^1<2^{31}3^15^17^1\) |
| 5 | \(2^{15}3^{1}5^17^111^1\) |
Finally, we compare the smallest values of \(N\) from each of the 5 rows in the table
above.
Since \(2^{15}3^{3}5^17^1<2^{15}3^{1}5^17^111^1<2^{15}3^{3}5^3<2^{15}3^{15}<2^{255}\),
then the smallest refactorable number \(N\) with \(f(N)=256\), is \(N=2^{15}3^{3}5^17^1=30\,965\,760\).
Let \(m\) be a positive integer
of the form \(m=p_1^{a_1}p_2^{a_2}\cdots
p_k^{a_k}\), where \(p_1<p_2<
\cdots<p_k\) are prime numbers, and \(a_1,a_2,\cdots,a_k\) are positive
integers.
We begin by stating a value of \(N\)
and then proceed to show that it works.
For each \(m=p_1^{a_1}p_2^{a_2}\cdots
p_k^{a_k}\), choose \(N=p_1^{\left(p_1^{a_1}-1\right)}p_2^{\left(p_2^{a_2}-1\right)}\cdots
p_k^{\left(p_k^{a_k}-1\right)}\).
On the contest paper, the useful fact states that \(2^n\geq n+1\) for all positive integers
\(n\).
Since \(p\geq2\) for all prime numbers
\(p\), then \(p^n\geq 2^n\geq n+1\).
Since \(p^n\geq n+1\), then \(p^n-1\geq n\), and so \(p_i^{a_i}-1\geq a_i\) for all integers
\(i\) where \(1\leq i \leq k\).
Thus, \(p_1^{a_1}\) divides \(p_1^{\left(p_1^{a_1}-1\right)}\) since each
is a power with base \(p_1\) and \(p_1^{a_1}-1\geq a_1\).
Similarly, \(p_2^{a_2}\) divides \(p_2^{\left(p_2^{a_2}-1\right)}\), and so
on.
Therefore, \(m=p_1^{a_1}p_2^{a_2}\cdots
p_k^{a_k}\) divides \(N=p_1^{\left(p_1^{a_1}-1\right)}p_2^{\left(p_2^{a_2}-1\right)}\cdots
p_k^{\left(p_k^{a_k}-1\right)}\).
Further, \(f(N)=\left(1+p_1^{a_1}-1\right)\left(1+p_2^{a_2}-1\right)\cdots\left(1+p_k^{a_k}-1\right)=p_1^{a_1}p_2^{a_2}\cdots
p_k^{a_k}=m\).
Thus for every integer \(m>1\),
there exists a refactorable number \(N\) such that \(f(N)=m\).