2025 Fryer Contest
Solutions
(Grade 9)
Thursday, April 3, 2025
(in North America and South America)
Friday, April 4, 2025
(outside of North American and South America)
Thursday, April 3, 2025
(in North America and South America)
Friday, April 4, 2025
(outside of North American and South America)
©2025 University of Waterloo
During the first week, Azizi sold a total of \(15+23+18+15+7=78\) chocolate bars.
During the second week, Azizi sold a total of \(7+15+x+23+x=45+2x\) chocolate bars, and so \(45+2x=73\). Solving \(45+2x=73\), we get \(2x=28\), and so \(x=14\).
During the third week, Azizi sold \(y\) chocolate bars on Monday, \(y+6\) on Tuesday, \(y+12\) on Wednesday, \(y+18\) on Thursday, and \(y+24\) on Friday.
In total, he sold \(y+(y+6)+(y+12)+(y+18)+(y+24)=5y+60\)
chocolate bars, and so \(5y+60=100\).
Solving \(5y+60=100\), we get \(5y=40\), and so \(y=8\).
(Can you explain why the total number sold, \(5y+60\), is equal to \(5\) times the number that Azizi sold on
Wednesday?)
We organize the given results in a table similar to the example shown. A winning player is awarded \(W=3\) points for each game won.
| Game Results | Points Awarded | ||
|---|---|---|---|
| Ava | Beau | Cato | |
| Ava loses to Beau | \(1\) | \(1\) | —— |
| Beau and Cato tie | —— | \(0\) | \(3\) |
| Ava and Cato tie | \(1\) | —— | \(1\) |
When the tournament has finished, Ava has been awarded \(2\) points, Beau has been awarded \(1\) point, and Cato has been awarded \(4\) points, and so \(S=2+1+4=7\). Alternately, \(2\) games end in a tie, and in each game ending in a tie, \(2\) points are awarded (\(1\) point to each player). Thus, the \(2\) tie games contribute \(2\times2=4\) points to \(S\). The remaining game ended with a winner, and so \(3\) points are awarded (\(3\) to the winning player, \(0\) to the losing player). Thus, this game contributes \(3\) points to \(S\), and so \(S=4+3=7\).
Each game that ends in a tie contributes \(2\) points to \(S\) (1 point to each player).
If all \(3\) games end in a tie, then
\(S=3\times2=6\), as required.
Next, we confirm that all \(3\)
games ending in a tie is the only possibility.
Each game that ends with a winner contributes \(4\) points to \(S\) (\(4\) to the winning player, \(0\) to the
losing player).
If all \(3\) games end with a player winning, then \(S=3\times4=12\).
If \(2\) games end with a player
winning, and \(1\) game ends in a tie,
then \(S=2\times4+2=10\).
If \(1\) game ends with a player
winning, and \(2\) games end in a tie,
then \(S=4+2\times2=8\).
Thus, if \(W=4\) and \(S=6\), the only possibility is that all
\(3\) games end in a tie.
The tournament ends with exactly one of the three games ending in
a tie, and so exactly two of the games end with a player winning.
The game ending in a tie contributes \(2\) to \(S\) (\(1\) point to each player).
A game that ends with a player winning contributes \(W\) to \(S\), and so two games that end with a
player winning contribute \(2W\) to
\(S\).
In this case, \(S=2W+2\) or \(2W+2=24\) which gives \(2W=22\), and so \(W=11\).
As was demonstrated in (b), there are four outcomes that must be
considered when determining the possible values of \(S\). The tournament can end with exactly
\(0\), \(1\), \(2\), or \(3\) ties.
If the tournament ends with \(0\) ties,
then each of the \(3\) games has a
winning player, and thus \(S=3W\). In
this case, \(3W=21\) and so \(W=7\).
If the tournament ends with \(1\) tie,
then \(2\) games have a winning player,
and thus\(S=2W+2\). In this case, \(2W+2=21\) or \(2W=19\) which is not possible since \(W\) is an integer.
If the tournament ends with \(2\) ties,
then \(1\) game has a winning player,
and thus\(S=W+2\times2\). In this case,
\(W+4=21\) or \(W=17\).
Finally, if the tournament ends with \(3\) ties, then \(0\) games have a winning player, and thus
\(S=3\times2=6\), and so \(S\) cannot equal \(21\).
If the tournament finishes with \(S=21\), the possible integer values of
\(W\) are \(7\) and \(17\).
A positive integer \(n>1\) is
a perfect square exactly when the exponent on each prime factor in the
prime factorization of \(n\) is
even.
We note that \(2^3\times3^2\) has an
odd exponent on the prime factor 2, and an even exponent on the prime
factor \(3\).
Thus, \(2^3\times3^2\times j\) is a
perfect square exactly when \(j\) is
equal to the product of an odd number of factors of \(2\) and an even number of each additional
prime factor greater than \(2\)
(including possibly having no additional prime factors).
Since \(j\leq20\), then the prime
factorization of \(j\) must contain
either one factor of \(2\) or three
factors of \(2\), and \(j\) cannot contain five or more factors of
\(2\) since \(2^5>20\).
When \(j=2\), the given product, \(2^3\times3^2\times2=2^4\times3^2=(2^2\times3)\times(2^2\times3)\),
is a perfect square.
When \(j=2^3=8\), the given product,
\(2^3\times3^2\times2^3=2^6\times3^2=(2^3\times3)\times(2^3\times3)\),
is also a perfect square.
Is it possible for the prime factorization of \(j\) to contain one \(2\) and an even number of another prime
factor greater than \(2\)?
The smallest prime number greater than \(2\) is \(3\), and when \(j=2\times3^2=18<20\), the given product
\(2^3\times3^2\times2\times3^2=2^4\times3^4=(2^2\times3^2)\times(2^2\times3^2)\),
is a perfect square.
The next smallest possible value of \(j\) for which its prime factorization
contains exactly one 2 is \(2\times5^2\), which is greater than \(20\). (Note that \(j=2\times3^4\) is also greater than \(20\).)
The next smallest possible value of \(j\) for which its prime factorization
contains exactly three factors of \(2\)
is \(2^3\times3^2\), which is also
greater than \(20\).
Therefore, the positive integers \(j\)
which satisfy the given conditions are \(2\), \(8\)
and \(18\).
Since \(3600=60^2\) and \(60=2^2\times3\times5\), then \(3600=(2^2\times3\times5)^2=2^4\times3^2\times5^2\).
Thus, each divisor of \(3600\) contains
at most the prime factors \(2\), \(3\) and \(5\), and cannot contain any other prime
factors.
Further, the prime factorization of each divisor of \(3600\) contains at most four factors of
\(2\), two factors of \(3\), and two factors of \(5\).
Suppose that \(k=2^x\times3^y\times5^z\). Then \(20\times
k=2^2\times5\times2^x\times3^y\times5^z=2^{x+2}\times3^y\times5^{z+1}\)
is a divisor of \(3600=2^4\times3^2\times5^2\) exactly when
\(0\leq x\leq 2\), \(0\leq y\leq 2\), and \(0\leq z\leq 1\), for integers \(x,y,z\).
As noted in part (a), \(20\times
k=2^{x+2}\times3^y\times5^{z+1}\) is a perfect square exactly
when each of the exponents, \(x+2\),
\(y\), and \(z+1\) is even.
Since \(0\leq x\leq 2\), then \(x+2\) is even when \(x=0\) or \(x=2\).
Since \(0\leq y\leq 2\), then \(y\) is even when \(y=0\) or \(y=2\).
Since \(0\leq z\leq 1\), then \(z+1\) is even when \(z=1\).
There are \(2\) choices for \(x\), \(2\)
choices for \(y\), and 1 choice for
\(z\), and so there are \(2\times2\times1=4\) possible values of
\(k\).
When \((x,y,z)=(0,0,1)\), we get \(k=2^0\times3^0\times5^1=5\).
When \((x,y,z)=(0,2,1)\), we get \(k=2^0\times3^2\times5^1=45\).
When \((x,y,z)=(2,0,1)\), we get \(k=2^2\times3^0\times5^1=20\).
When \((x,y,z)=(2,2,1)\), we get \(k=2^2\times3^2\times5^1=180\).
The positive integers \(k\) satisfying
the given conditions are \(5\), \(45\), \(20\), and \(180\).
Since \(2025=45^2\) and \(45=3^2\times5\), then \(2025=(3^2\times5)^2=3^4\times5^2\).
Since \(a^2\) and \(b^2\) are perfect squares, and \(a^2\times b^2\times c=2025\), then \(a^2\) and \(b^2\) are each equal to perfect square
divisors of \(2025\).
The perfect square divisors of \(2025=3^4\times5^2\) are: 1, \(3^2\), \(3^4\), \(5^2\), \(3^2\times5^2\), and \(3^4\times5^2\).
We count the number of ordered triples of positive integers \((a,b,c)\) by considering the following
\(2\) cases: \((1)\) At least one of \(a^2\) or \(b^2\) is equal to \(1\); \((2)\) Both \(a^2\) and \(b^2\) are not equal to \(1\).
Case 1: At least one of \(a^2\) or \(b^2\) is equal to \(1\).
Suppose that \(a^2=1\). Then \(b^2\) can be equal to each of the perfect
square divisors of \(2025\) previously
listed.
That is, \(b^2\) can be equal to each
of the \(6\) values: \(1\), \(3^2\), \(3^4\), \(5^2\), \(3^2\times5^2\), and \(3^4\times5^2\).
For each of these values of \(b^2\),
\(c=\dfrac{2025}{a^2\times b^2}\). For
example, when \(a^2=1\) and \(b^2=1\), then \(c=\dfrac{3^4\times5^2}{1\times
1}=3^4\times5^2\), and so \((a,b,c)=(1,1,3^4\times5^2)\) is a possible
ordered triple.
When \(a^2=1\) and \(b^2=3^2\), then \(c=\dfrac{3^4\times5^2}{1\times
3^2}=3^2\times5^2\), and so \((a,b,c)=(1,3,3^2\times5^2)\) is a possible
ordered triple.
Continuing in this way with \(a^2=1\),
we get the following 6 ordered triples \((a,b,c)\): \[(1,1,3^4\times5^2), (1,3,3^2\times5^2),
(1,3^2,5^2), (1,5,3^4), (1,3\times5,3^2), (1,3^2\times5, 1)\] For
each of the ordered triples above, with the exception of \((1,1,3^4\times5^2)\), we get a new ordered
triple by switching \(a\) and \(b\).
Each of these \(5\) new ordered triples
can be determined by letting \(b^2=1\)
and following the process above, and thus each satisfies the given
conditions.
There are a total of \(5\times2+1=11\)
ordered triples in this case.
Case 2: Both \(a^2\) and \(b^2\) are not equal to \(1\).
If one of \(a^2\) or \(b^2\) is equal to \(3^4\times5^2\), then the other must be
equal to \(1\) (since \(2025=3^4\times5^2\)).
In Case 2, both \(a^2\) and \(b^2\) are not equal to \(1\), and so each is also not equal to \(3^4\times5^2\).
Removing these from our list of perfect square divisors, the possible
values of \(a^2\) and \(b^2\) that remain are: \(3^2\), \(3^4\), \(5^2\), and \(3^2\times5^2\).
If \(a^2=3^2\), then \(b^2\) can be equal to \(3^2\) or \(5^2\) or \(3^2\times5^2\).
If \(a^2=3^4\), then \(b^2\) can be equal to \(5^2\).
If \(a^2=5^2\), then \(b^2\) can be equal to \(3^2\) or \(3^4\).
And finally, if \(a^2=3^2\times5^2\),
then \(b^2\) can be equal to \(3^2\).
In each case, the value of \(c\) can be
determined as it was in Case 1.
Doing so gives the following \(7\)
ordered triples \((a,b,c)\): \[(3,3,5^2), (3,5,3^2), (3,3\times5,1), (3^2,5,1),
(5,3,3^2), (5,3^2, 1),(3\times5,3,1)\] The number of ordered
triples of positive integers \((a,b,c)\) so that \(a^2\times b^2\times c=2025\)is \(11+7=18\).
We begin by placing \(E(4,4)\) on \(CD\), as shown.
To determine the area of each of the three triangles, consider the
base of each to be \(AB\).
Points \(C\), \(D\) and \(E\) each have the same \(y\)-coordinate, \(4\), and so each lies on the horizontal
line \(y=4\).
The height of each of the three triangles is the vertical distance
between the line \(y=4\) and the \(x\)-axis (the line through \(A\) and \(B\)), which is \(4\).
Since \(AB=7\), then the area of
each of the three triangles is \(\frac12\times
7\times4=14\).
The sum of the areas of \(\triangle
ABC\), \(\triangle ABD\), and
\(\triangle ABE\) is \(3\times14=42\).
\(\triangle CDG\) has non-zero
area, so \(G\) cannot lie on \(CD\).
On \(AB\), there are \(8\) possible
locations for \(G\).
These are the points \((k,0)\) for the
integers \(0\leq k\leq 7\).
On \(AD\), \((1,2)\) is the only additional point for
which the coordinates are both integers. Similarly, \((8,2)\) is the only additional possibility
for \(G\) on \(BC\). The points \((1,2)\) and \((8,2)\) are labelled \(M\) and \(N\) respectively, as shown.
In total, there are \(10\) possibilities for the point \(G\).
Each such triangle has three vertices chosen from the \(18\) points with integer coordinates on the
perimeter of \(ABCD\), provided that
the three vertices are not all on the same line.
These \(18\) points with integer
coordinates are the \(8\) points on
\(AB\) (including \(A\) and \(B\)), the \(8\) points on \(CD\) (including \(C\) and \(D\)), and the \(2\) points \(M(1,2)\) and \(N(8,2)\).
Each such triangle is described by exactly one of the following
cases:
the number of triangle vertices at \(M\) and \(N\) is \(0\)
the number of triangle vertices at \(M\) and \(N\) is \(1\)
the number of triangle vertices at \(M\) and \(N\) is \(2\)
Case 1: the number of triangle vertices at \(M\) and \(N\) is \(0\).
In this case, \(2\) vertices are on
\(AB\) and \(1\) vertex is on \(CD\), or vice versa.
Consider the triangles with \(2\)
vertices on \(AB\) and 1 vertex on
\(CD\).
Consider the base of each such triangle to lie along \(AB\).
There is \(1\) such triangle with
vertices at \(A(0,0)\) and \(B(7,0)\), and thus has base length 7.
There are \(2\) such triangles with
base length \(6\). One of these
triangles has vertices at \(A(0,0)\)
and \((6,0)\), and the other has
vertices at \((1,0)\) and \(B(7,0)\).
Continuing in this way, there are \(3\)
triangles with base length \(5\), \(4\) triangles with base length \(4\), \(5\)
triangles with base length \(3\), \(6\) triangles with base length \(2\), and \(7\) triangles with base length \(1\).
Each of these triangles has height \(4\) since all points on \(CD\) are a vertical distance of \(4\) from any base that lies along \(AB\) (as in part (a)).
Suppose \(Q\) is one such point on
\(CD\) having integer
coordinates.
The sum of the areas of all triangles having \(2\) vertices on \(AB\) and \(1\) vertex at \(Q\) is \[\frac12\times4\times(1(7)+2(6)+3(5)+4(4)+5(3)+6(2)+7(1))=\frac12\times4\times84=168\]
Also, for each of these bases, there are \(8\) possibilities for the third vertex that
lies on \(CD\). These are the points
\((k,4)\) for integers \(2\leq k\leq 9\).
Thus, the sum of the areas of all triangles having \(2\) vertices on \(AB\) and \(1\) vertex on \(CD\) is \(168\times8=1344\).
In a similar way, the sum of the areas of all triangles having \(2\) vertices on \(CD\) and \(1\) vertex on \(AB\) is also \(1344\), and so the sum of the areas of all
triangles in Case 1 is \(1344\times2=2688\).
Case 2: the number of triangle vertices at \(M\) and \(N\) is \(1\).
This case can be divided into the following two subcases:
\(2\) vertices are on \(AB\) (or \(2\) vertices are on \(CD\)), and \(1\) vertex is \(M\) or \(N\)
\(1\) vertex is on \(AB\), \(1\) vertex is on \(CD\), and \(1\) vertex is \(M\) or \(N\)
Subcase 2(i): \(2\) vertices on \(AB\) (or \(2\) vertices on \(CD\)), and \(1\) vertex is \(M\) or \(N\).
Consider the triangles with \(2\)
vertices on \(AB\) and \(1\) vertex at either \(M\) or \(N\).
Consider the base of each such triangle to lie along \(AB\).
The numbers and lengths of these bases are the same as in Case 1.
Each of these triangles has height \(2\) since \(M\) and \(N\) are each a vertical distance of \(2\) from any base that lies along \(AB\).
The sum of the areas of all triangles having \(2\) vertices on \(AB\) and \(1\) vertex at \(M\) is \[\frac12\times2\times(1(7)+2(6)+3(5)+4(4)+5(3)+6(2)+7(1))=\frac12\times2\times84=84\]
Also, for each of these bases, the third vertex could also be \(N\).
Thus, the sum of the areas of all triangles having \(2\) vertices on \(AB\) and \(1\) vertex at either \(M\) or \(N\) is \(84\times2=168\).
In a similar way, the sum of the areas of all triangles having \(2\) vertices on \(CD\) and \(1\) vertex at either \(M\) or \(N\) is \(168\), and so the sum of the areas of all
triangles in Subcase 2(i) is \(168\times2=336\).
Subcase 2(ii): \(1\) vertex is on \(AB\), \(1\) vertex is on \(CD\), and \(1\) vertex is \(M\) or \(N\).
Consider two fixed points with integer coordinates, point \(P\) on \(AB\), and point \(Q\) on \(CD\).
In the diagram, \(\triangle PMQ\) and
\(\triangle PNQ\) are two such
triangles described by Subcase 2(ii).
The sum of the areas of \(\triangle PMQ\) and \(\triangle PNQ\) is equal to the sum of the areas of \(\triangle PMN\) and \(\triangle QMN\).
Consider the base of both \(\triangle
PMN\) and \(\triangle QMN\) to
be \(MN\), which has length \(7\). Then the height of each triangle is
\(2\).
Thus, the sum of the areas of \(\triangle
PMN\) and \(\triangle QMN\) is
\(2\times\frac12\times2\times7=14\).
There are \(8\) possible locations for
\(P\) and \(8\) possible locations for \(Q\), and thus \(8\times8=64\) different pairs of triangles
\(PMN\) and \(QMN\) whose areas have a sum of \(14\). (You should confirm for yourself that
this is true even when \(P\) is \((0,0)\) or \((7,0)\) and/or when \(Q\) is \((2,4)\) or \((9,4)\).) The sum of the areas of all
triangles in Subcase 2(ii) is \(14\times64=896\).
Case 3: the number of triangle vertices at \(M\) and \(N\) is \(2\).
In this case, \(1\) vertex is on
\(AB\) (or on \(CD\)), \(1\) vertex is \(M\), and \(1\) vertex is \(N\).
Consider the triangles with vertices \(M\), \(N\), and \(1\) vertex on \(AB\).
Consider the base of each such triangle to be \(MN=7\). Then each triangle has height \(2\).
Since there are \(8\) possible vertices
on \(AB\), then the sum of all such
triangles is \(\frac12\times2\times7\times8=56\). In a
similar way, the sum of the areas of all triangles having vertices \(M\), \(N\), and \(1\) vertex on \(CD\) is also \(56\), and so the sum of the areas of all
triangles in Case 3 is \(56\times2=112\).
The sum of the areas of all triangles whose vertices have integer coordinates and lie on the perimeter of \(ABCD\) is \(2688+336+896+112=4032\).