2024 Canadian Senior
Mathematics Contest
Solutions

Wednesday, November 13, 2024
(in North America and South America)

Thursday, November 14, 2024
(outside of North American and South America)

©2024 University of Waterloo

Part A

1. Solution 1

   Evaluating, $\sqrt{{10}^2+2\cdot 10\cdot 11+{11}^2}=\sqrt{100+220+121}=\sqrt{441}=21$.

   Solution 2

   Since $x^2+2xy+y^2=(x+y{)}^2$, then $$\sqrt{{10}^2+2\cdot 10\cdot 11+{11}^2}=\sqrt{(10+11{)}^2}=\sqrt{{21}^2}=21$$

   Answer: $21$

2. Label the shorter tree as $AB$ and the taller tree as $CD$, as shown.

   

   Draw a horizontal line from $A$ to $P$ on $CD$.
   Since $AB$ and $PD$ are vertical and $AP$ and $BD$ are horizontal, $ABDP$ is a rectangle which means that $PD=5\text{ m}$ and $AP=24\text{ m}$.
   Since $CD=15\text{ m}$ and $PD=5\text{ m}$, then $CP=CD-PD=10\text{ m}$.
   By the Pythagorean Theorem, $$AC=\sqrt{A{P}^2+C{P}^2}=\sqrt{(24\text{ m}{)}^2+(10\text{ m}{)}^2}=\sqrt{676{\text{ m}}^2}=26\text{ m}$$ since $AC>0$.
   Since the bird flies $26\text{ m}$ at $4\text{ m/s}$, its time to complete the flight is ${\displaystyle \frac{26\text{ m}}{4\text{ m/s}}}=6.5\text{ s}$.

   Answer: $6.5\text{ s}$

3. Since Team Why had scored $3$ goals at the end of the game, then $y\le 3$.
   Since Team Zed had scored $2$ goals at the end of the game, then $z\le 2$.
   Therefore, $0\le y\le 3$ and $0\le z\le 2$.
   The only additional restriction is that $y$ and $z$ are both integers.
   Thus, there are $4$ possible values for $y$ and $3$ possible values for $z$, which means that there are $4\cdot 3=12$ possible pairs $(y,z)$.
   These pairs are $$(y,z)=(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2),(3,0),(3,1),(3,2)$$ These pairs correspond to the possible scores at half-time.

   Answer: $12$

4. We are told that $a$, $b$, $c$, $d$ are positive integers with $abc=d$ and $d\le 8$.
   We determine the number of possible quadruples $(a,b,c,d)$ by looking at each possible value of $d$.

   If $d=1$, then $abc=1$. Since $a$, $b$, $c$ are positive integers, then $a=b=c=1$. This means that there is $1$ quadruple in this case, namely $(1,1,1,1)$.

   If $d=2$, then $abc=2$. Since $2$ is prime, then one of $a$, $b$ and $c$ equals $2$ and the other two variables equal $1$.
   This gives $3$ quadruples: $(a,b,c,d)=(2,1,1,2),(1,2,1,2),(1,1,2,2)$.
   Similarly, when $d=3$, $d=5$ and $d=7$ (the other possible prime values of $d$), there are $3$ quadruples.

   If $d=4$, then $abc=4$. We need to determine the ways in which $4$ can be factored as the product of three positive integers.
   These ways are $1\cdot 1\cdot 4=4$ and $1\cdot 2\cdot 2=4$.
   In each case, there are 3 ways to arrange the factors on the left side, which are the possible values of $a$, $b$ and $c$: $$1\cdot 1\cdot 4=1\cdot 4\cdot 1=4\cdot 1\cdot 1=4$$ $$1\cdot 2\cdot 2=2\cdot 1\cdot 2=2\cdot 2\cdot 1=4$$ Thus, there are 6 quadruples in this case.

   If $d=6$, then $abc=6$. The possible factorizations of $6$ into three positive factors are $1\cdot 1\cdot 6=6$ and $1\cdot 2\cdot 3=6$.
   In the first case, there are again $3$ quadruples.
   In the second case, there are $6$ ways of arranging the factors: $$1\cdot 2\cdot 3=1\cdot 3\cdot 2=2\cdot 1\cdot 3=2\cdot 3\cdot 1=3\cdot 1\cdot 2=3\cdot 2\cdot 1=6$$ Thus, if $d=6$, there are $9$ quadruples.

   If $d=8$, then $abc=8$. The possible factorizations of $8$ into three positive factors are $1\cdot 1\cdot 8=8$ and $1\cdot 2\cdot 4=8$ and $2\cdot 2\cdot 2=8$.
   These cases give 3, 6 and 1 quadruples, for a total of $10$.

   Combining all of the cases, there are $1+3+3+3+3+6+9+10=38$ quadruples.

   Answer: $38$

5. We use coordinate geometry.
   Suppose that $F$ is the origin and $E$ lies on the positive $x$-axis.
   The coordinates of $F$ are $(0,0)$. Since $FE=6$, the coordinates of $E$ are $(6,0)$.
   Next, we find the coordinates of $B$.
   We note that $FA=AB=2$.
   Since the six interior angles of hexagon $ABCDEF$ are equal, then the measure of each is equal to $\frac{1}{6}\cdot 720\mathrm{°}=120\mathrm{°}$. (The sum of the measures of the interior angles of any hexagon is $720\mathrm{°}$.)
   This means that $\mathrm{△}FAB$ is isosceles with $FA=AB=2$ and $\mathrm{\angle }FAB=120\mathrm{°}$.
   Since $\mathrm{\angle }FAB=120\mathrm{°}$, then $\mathrm{\angle }AFB=\mathrm{\angle }ABF=\frac{1}{2}(180\mathrm{°}-120\mathrm{°})=30\mathrm{°}$.
   Since $\mathrm{\angle }AFE=120\mathrm{°}$ and $\mathrm{\angle }AFB=30\mathrm{°}$, then $\mathrm{\angle }BFE=\mathrm{\angle }AFE-\mathrm{\angle }AFB=90\mathrm{°}$, which means that $BF$ is vertical and so $B$ lies on the positive $y$-axis.
   If we draw a perpendicular from $A$ to a point $M$ on $BF$, then, because $\mathrm{△}FAB$ is isosceles, $M$ is the midpoint of $BF$ and $AM$ bisects $\mathrm{\angle }FAB$.

   

   This means that $\mathrm{△}BAM$ is a $30\mathrm{°}$-$60\mathrm{°}$-$90\mathrm{°}$ triangle.
   Thus, $BM=\frac{\sqrt{3}}{2}AB=\sqrt{3}$ and so $BF=2\sqrt{3}$ which means that the coordinates of $B$ are $(0,2\sqrt{3})$.
   (We could have determined the length of $BF$ using the cosine law in $\mathrm{△}BAF$.)
   Also, $AM=\frac{1}{2}AB=1$. This means that the coordinates of $A$ are $(-1,\sqrt{3})$.
   Using similar arguments, the coordinates of $C$ are $(6,2\sqrt{3})$ and the coordinates of $D$ are $(7,\sqrt{3})$.
   We can determine the area of hexagon $PQRSTU$ by starting with the area of rectangle $BCEF$ (which is $6\cdot 2\sqrt{3}=12\sqrt{3}$) and subtracting the areas of $\mathrm{△}UFT$, $\mathrm{△}BPQ$, $\mathrm{△}CRQ$, $\mathrm{△}SET$, $\mathrm{△}BCQ$, and $\mathrm{△}FET$.
   By symmetry, the areas of $\mathrm{△}UFT$, $\mathrm{△}BPQ$, $\mathrm{△}CRQ$, $\mathrm{△}SET$ are all equal.
   Also, by symmetry, the areas of $\mathrm{△}BCQ$ and $\mathrm{△}FET$ are equal.
   We determine the area of $\mathrm{△}UFT$ and the area of $\mathrm{△}FET$.
   To do this, we determine the coordinates of $U$ and of $T$.
   The line segment $FD$ passes through $F(0,0)$ and $D(7,\sqrt{3})$.
   Thus, it has slope $\frac{\sqrt{3}}{7}$ and equation $y=\frac{\sqrt{3}}{7}x$.
   The line segment $AE$ passes through $A(-1,\sqrt{3})$ and $E(6,0)$.
   Thus, it has slope $\frac{\sqrt{3}}{-7}$ and equation $y=-\frac{\sqrt{3}}{7}(x-6)$ or $y=-\frac{\sqrt{3}}{7}x+\frac{6\sqrt{3}}{7}$.
   This means that the coordinates of $U$ (which is the $y$-intercept of this line) are $(0,\frac{6\sqrt{3}}{7})$.
   To find the coordinates of $T$, we find the point of intersection of $AE$ and $FD$.
   Equating expressions for $y$, we obtain $\frac{\sqrt{3}}{7}x=-\frac{\sqrt{3}}{7}x+\frac{6\sqrt{3}}{7}$ which gives $\frac{2\sqrt{3}}{7}x=\frac{6\sqrt{3}}{7}$ or $x=3$. (It should not be surprising that the $x$-coordinate of $T$ is $x=3$.)
   When $x=3$, we obtain $y=\frac{3\sqrt{3}}{7}$, and so the coordinates of $T$ are $(3,\frac{3\sqrt{3}}{7})$.
   So $\mathrm{△}FET$ has base $FE=6$ and height equal to the distance from $T$ to $FE$ (which is $\frac{3\sqrt{3}}{7}$). Thus, the area of $\mathrm{△}FET$ is $\frac{1}{2}\cdot 6\cdot \frac{3\sqrt{3}}{7}=\frac{9\sqrt{3}}{7}$.
   Also, $\mathrm{△}UFT$ has vertical base $UF$ (which has length $\frac{6\sqrt{3}}{7}$) and horizontal height equal to the distance from $T$ to $UF$ (which is 3). Thus, the area of $\mathrm{△}UFT$ is $\frac{1}{2}\cdot \frac{6\sqrt{3}}{7}\cdot 3=\frac{9\sqrt{3}}{7}$. (Can you see a reason why this should be equal to the area of $\mathrm{△}FET$?)
   Finally, this means that the area of $PQRSTU$ is equal to $12\sqrt{3}-6\cdot \frac{9\sqrt{3}}{7}$ or $\frac{84\sqrt{3}}{7}-\frac{54\sqrt{3}}{7}$ which equals $\frac{30\sqrt{3}}{7}=\frac{\sqrt{2700}}{7}$.
   Therefore, $(n,t)=(2700,7)$.

   Answer: $(2700,7)$

6. A list of $m$ distinct positive integers has a sum that is at least $1+2+\cdots +(m-1)+m$ which is equal to $\frac{1}{2}m(m+1)$.
   We note that $\frac{1}{2}\cdot 63\cdot 64=2016$ and $\frac{1}{2}\cdot 64\cdot 65=2080$.
   In other words, the sum $1+2+\cdots +63+64$ is greater than $2024$ and any sum of more than $64$ distinct positive integers is greater still.
   This means that a Gleeson list consists of at most $63$ integers.
   Note that $1+2+\cdots +62+63=\frac{1}{2}\cdot 63\cdot 64=2016$ and so $1+2+\cdots +62+63+8=2024$ which means that $1+2+\cdots +61+62+71=2016$.
   This means that there is at least one Gleeson list of length $63$ and there cannot be a Gleeson list of length greater than $63$, so $M$, the maximum length of a Gleeson list, is $63$.
   We need to determine the number of Gleeson lists of length $M=63$.
   Suppose that $a_1,a_2,a_3,\dots ,a_{62},a_{63}$ is a Gleeson list.
   That is, $a_1,a_2,a_3,\dots ,a_{62},a_{63}$ are positive integers with $a_1<a_2<\cdots <a_{62}<a_{63}$ and $$a_1+a_2+\cdots +a_{62}+a_{63}=2024$$ For each $j$ with $1\le j\le 63$, let $b_j=a_j-j$, which means that $a_j=j+b_j$.
   That is, we write a generic Gleeson list as $$1+b_1,2+b_2,\dots ,62+b_{62},63+b_{63}$$ where the $b_j$’s are the "extra" in each term.
   For each $j$ with $1\le j\le 62$, since $a_j$ and $a_{j+1}$ are integers with $a_{j+1}>a_j$, then $a_{j+1}\ge a_j+1$.
   This means that $j+1+b_{j+1}\ge j+b_j+1$ or $b_{j+1}\ge b_j$.
   This means that $b_1,b_2,\dots ,b_{62},b_{63}$ is a list of integers with $b_1\le b_2\le \cdots \le b_{62}\le b_{63}$.
   Since $a_1=1+b_1\ge 1$, then $b_1\ge 0$ and so $0\le b_1\le b_2\le \cdots \le b_{62}\le b_{63}$.
   Also, since $a_1+a_2+\cdots +a_{62}+a_{63}=2024$ then $$(1+b_1)+(2+b_2)+\cdots +(62+b_{62})+(63+b_{63})=2024$$ Since $1+2+\cdots +62+63=2016$, then $$b_1+b_2+\cdots +b_{62}+b_{63}=8$$ In other words, to count the number of Gleeson lists of length $63$, we need to count the number of non-decreasing lists of non-negative integers $b_1,b_2,\dots ,b_{62},b_{63}$ with a sum of $8$.
   We determine these lists by categorizing them using the number of non-zero terms, in each case with the understanding that the terms before these are all $0$.

   • $1$ non-zero term: $8$

   • $2$ non-zero terms: $1$, $7$; $2$, $6$; $3$, $5$; $4$, $4$

   • $3$ non-zero terms: $1$, $1$, $6$; $1$, $2$, $5$; $1$, $3$, $4$; $2$, $2$, $4$; $2$, $3$, $3$

   • $4$ non-zero terms: $1$, $1$, $1$, $5$; $1$, $1$, $2$, $4$; $1$, $1$, $3$, $3$; $1$, $2$, $2$, $3$; $2$, $2$, $2$, $2$

   • $5$ non-zero terms: $1$, $1$, $1$, $1$, $4$; $1$, $1$, $1$, $2$, $3$; $1$, $1$, $2$, $2$, $2$

   • $6$ non-zero terms: $1$, $1$, $1$, $1$, $1$, $3$; $1$, $1$, $1$, $1$, $2$, $2$

   • $7$ non-zero terms: $1$, $1$, $1$, $1$, $1$, $1$, $2$

   • $8$ non-zero terms: $1$, $1$, $1$, $1$, $1$, $1$, $1$, $1$

   In each case, we determine the lists by starting with the largest possible entries at the right, and adjusting to the left in a systematic way.
   There are $22$ such lists, and so $22$ Gleeson lists of length $63$.

   Answer: $22$

Part B

1. 1. Since $64=2^6$, then $2^{x+1}=2^6$ which gives $x+1=6$ and so $x=5$.

   2. Comparing exponents in the two equations, we obtain $t+u=8$ and $u-3=2$.
      From the second equation, $u=5$.
      Substituting $u=5$ into $t+u=8$ gives $t=3$.
      Therefore, $t=3$ and $u=5$.

   3. Since $m$ and $r$ are integers, we can compare exponents to obtain $2m+r=7$ and $3m-r=3$.
      (We can do this because every positive integer greater than 1 can be uniquely factored as a product of prime numbers.)
      Adding these two equations, we obtain $(2m+r)+(3m-r)=7+3$ or $5m=10$, which gives $m=2$.
      Substituting into the first equation gives $2\cdot 2+r=7$ and so $r=3$.
      Therefore, $m=2$ and $r=3$.

   4. Solution 1

      First, we note that $80=16\cdot 5=2^4{5}^1$.
      Since $2^{p+q}{5}^{p-q}=2^4{5}^1$ and $p$ and $q$ are integers, then comparing exponents, we obtain $p+q=4$ and $p-q=1$.
      Adding these equations, we obtain $2p=5$ which gives $p=2.5$.
      This means that there are no integers $p$ and $q$ that satisfy this equation.

      Solution 2

      First, we note that $80=16\cdot 5=2^4{5}^1$.
      This means that 80 includes exactly one factor of $5$.
      Since $2^{p+q}{5}^{p-q}=80$, then the left side also contains only one factor of $5$.
      This means that $p-q=1$, which means that $p$ and $q$ are either odd and even, or even and odd.
      This in turn means that $p+q$ is the sum of an even integer and an odd integer, so is odd.
      But $80$ includes an even number of factors of $2$ and we know that the left side includes an odd number of factors of $2$.
      This is a contradiction, and so there are no integers $p$ and $q$ that satisfy this equation.
      

2. 1. Solution 1

      Using the quadratic formula, the solutions of the quadratic equation $x^2-2x-1=0$ are $$x={\displaystyle \frac{2\pm \sqrt{(-2{)}^2-4(1)(-1)}}{2}}={\displaystyle \frac{2\pm \sqrt{8}}{2}}={\displaystyle \frac{2\pm 2\sqrt{2}}{2}}=1\pm \sqrt{2}$$ We set $r=1+\sqrt{2}$ and $s=1-\sqrt{2}$.
      Then $2r+s=2(1+\sqrt{2})+(1-\sqrt{2})=3+\sqrt{2}$ and $r+2s=(1+\sqrt{2})+2(1-\sqrt{2})=3-\sqrt{2}$.
      Therefore, a quadratic equation that has $x=3+\sqrt{2}$ and $x=3-\sqrt{2}$ as solutions is $$(x-(3+\sqrt{2}))(x-(3-\sqrt{2}))=0$$ or $$x^2-((3+\sqrt{2})+(3-\sqrt{2}))x+(3+\sqrt{2})(3-\sqrt{2})=0$$ Since $(3+\sqrt{2})(3-\sqrt{2})=3^2-(\sqrt{2}{)}^2=9-2=7$, this simplifies to $x^2-6x+7=0$.
      Thus, $b=-6$ and $c=7$.

      Solution 2

      If the quadratic equation $x^2+Bx+C=0$ has solutions $x=r$ and $x=s$, then the quadratic expressions $x^2+Bx+C$ and $(x-r)(x-s)$ must be the same because both have a leading coefficient of $1$, which means that $$x^2+Bx+C=x^2-rx-sx+rs=x^2-(r+s)x+rs$$ and so $r+s=-B$ and $rs=C$.

      Here, the quadratic equation $x^2-2x-1=0$ has roots $x=r$ and $x=s$, which means that $r+s=2$ and $rs=-1$.
      In this case, $$\begin{array}{rl}(2r+s)+(r+2s)& =3r+3s=3(r+s)=3\cdot 2=6\\ (2r+s)(r+2s)& =2r^2+5rs+2s^2\\ & =2r^2+4rs+2s^2+rs\\ & =2(r^2+2rs+s^2)+rs\\ & =2(r+s{)}^2+rs\\ & =2\cdot 2^2+(-1)=7\end{array}$$ Therefore, a quadratic equation with roots $x=2r+s$ and $x=r+2s$ is $x^2-6x+7=0$.
      Thus, $b=-6$ and $c=7$.

   2. Solution 1

      Suppose that the polynomial $f(x)=x^2+mx+p$ has two distinct positive real roots $x=r$ and $x=s$.
      This means that $r+s=-m$ and $rs=p$, as shown in part (a) Solution 2).
      In this case, $m^2-2p=(-m{)}^2-2p=(r+s{)}^2-2rs=r^2+2rs+s^2-2rs=r^2+s^2$ and $p^2=(rs{)}^2=r^2{s}^2$.
      This means that we can rewrite $g(x)$ as $g(x)=x^2-(r^2+s^2)x+r^2{s}^2$ which factors as $g(x)=(x-r^2)(x-s^2)$. (Can you see why this is true by looking at the sum and product of roots?)
      Since $g(x)=(x-r^2)(x-s^2)$, then $g(x)$ has roots $r^2$ and $s^2$.
      We recall that $r$ and $s$ are positive and distinct.
      Since neither equals 0, then $r^2$ and $s^2$ are positive.
      Since $r\ne \pm s$, then $r^2$ and $s^2$ are distinct.
      This means that $g(x)$ has two distinct positive real roots.

      Solution 2

      We proceed by using the sum and product of roots as shown in part (a) Solution 2.
      Suppose that the roots of $f(x)=x^2+mx+p$ are $x=r$ and $x=s$.
      Since $r$ and $s$ are positive, then $r+s>0$ and $rs>0$.
      Since $r+s=-m$, then $m<0$. Since $rs=p$, then $p>0$.
      Since $f(x)$ has two distinct real roots, then its discriminant is positive.
      In particular, $m^2-4p>0$.
      Now, the discriminant of $g(x)=x^2-(m^2-2p)x+p^2$ is $$\begin{array}{rl}\mathrm{\Delta }& =(m^2-2p{)}^2-4(1)(p^2)\\ & =m^4-4m^{2}p+4p^2-4p^2\\ & =m^4-4m^{2}p\\ & =m^2(m^2-4p)\end{array}$$ Since $m^2>0$ (because $m<0$) and $m^2-4p>0$, then $\mathrm{\Delta }=m^2(m^2-4p)>0$.
      This means that $g(x)$ has two distinct real roots.
      Next, the sum of the (real) roots of $g(x)$ is $m^2-2p$ and their product is $p^2$.
      We note that $m^2-2p=(m^2-4p)+2p$ which is the sum of two positive real numbers and so is positive.
      Also, $p^2$ is positive since $p>0$.
      Since the product of the roots of $g(x)$ is positive, the roots are either both positive or both negative.
      Since the sum of the roots of $g(x)$ is positive, the roots cannot both be negative, and so both are positive.
      In summary, $g(x)$ has two distinct positive real roots, as required.

   3. Solution 1

      Suppose that a cubic equation $x^3+A{x}^2+Bx+C=0$ has roots $x=r$, $x=s$, and $x=t$.
      As we saw in (a), this means that $$\begin{array}{rl}{x}^3+A{x}^2+Bx+C& =(x-r)(x-s)(x-t)\\ & =(x^2-(r+s)x+rs)(x-t)\\ & =x^3-(r+s)x^2+rsx-t{x}^2+(rt+st)x-rst\\ & =x^3-(r+s+t)x^2+(rs+rt+st)x-rst\end{array}$$ and so $A=-(r+s+t)$ and $B=rs+rt+st$ and $C=-rst$.
      Also, if $A=-(r+s+t)$ and $B=rs+rt+st$ and $C=-rst$, then a cubic equation with roots $r$, $s$ and $t$ is $x^3+Ax+Bx+C=0$.

      Consider the polynomial $f_1(x)=x^3+A_1{x}^2+B_{1}x+C_1=x^3-6x^2+10x-5$.
      Since $f(1)=1-6+10-5=0$, then $x=1$ is a root.
      Factoring, we obtain $$f_1(x)=(x-1)(x^2-5x+5)$$ By the quadratic formula, the roots of $x^2-5x+5=0$ are $$x={\displaystyle \frac{5\pm \sqrt{(-5{)}^2-4(1)(5)}}{2}}={\displaystyle \frac{5\pm \sqrt{5}}{2}}$$ We note that ${\displaystyle \frac{5+\sqrt{5}}{2}}\approx 3.62$ and ${\displaystyle \frac{5-\sqrt{5}}{2}}\approx 1.38$.
      This means that $f_1(x)$ has three distinct positive real roots.

      Suppose that, for some positive integer $n\ge 2$, we have $$\begin{array}{rl}{A}_{n-1}& =-(r+s+t)\\ B_{n-1}& =rs+rt+st\\ C_{n-1}& =-rst\end{array}$$ for some distinct, positive real numbers $r$, $s$ and $t$.
      Then $$\begin{array}{rl}{A}_n& =2B_{n-1}-(A_{n-1}{)}^2\\ & =2(rs+rt+st)-(-(r+s+t){)}^2\\ & =2(rs+rt+st)-(r+s+t{)}^2\\ & =2rs+2rt+2st-(r^2+s^2+t^2+2rs+2rt+2st)\\ & =-(r^2+s^2+t^2)\\ B_n& =(B_{n-1}{)}^2-2A_{n-1}{C}_{n-1}\\ & =(rs+rt+st{)}^2-2(-(r+s+t))(-rst)\\ & =r^2{s}^2+r^2{t}^2+s^2{t}^2+2r^{2}st+2r{s}^{2}t+2rs{t}^2-(2r^{2}st+2r{s}^{2}t+2rs{t}^2)\\ & =r^2{s}^2+r^2{t}^2+s^2{t}^2\\ C_n& =-(C_{n-1}{)}^2\\ & =-(-rst{)}^2\\ & =-r^2{s}^2{t}^2\end{array}$$ Consider the polynomial $$f_n(x)=x^3+A_n{x}^2+B_{n}x+C_n=x^3-(r^2+s^2+t^2)x^2+(r^2{s}^2+r^2{t}^2+s^2{t}^2)x-r^2{s}^2{t}^2$$ This polynomial has roots $x=r^2$, $x=s^2$ and $x=t^2$, since the coefficients of the polynomial are the correct combinations of these roots.
      This means that the roots of $f_n(x)$ are the squares of the roots of $f_{n-1}(x)$.
      Here, $f_1(x)$ has three distinct positive real roots.
      Since the roots of $f_2(x)$ are the squares of the roots of $f_1(x)$, then they are real, positive and distinct.
      Continuing in this way, the roots of $f_3(x)$, $f_4(x)$, $\dots$, $f_{100}(x)$ will each be the squares of the roots of the previous polynomial in the list, and so will be real, positive and distinct.
      Therefore, the polynomial $f_{100}(x)$ has three distinct positive real roots.

      Solution 2

      From Solution 1, we know that the polynomial $f_1(x)=x^3-6x^2+10x-5$ has three distinct positive real roots, one of which is equal to 1.
      Suppose that $A$, $B$ and $C$ are integers and also that the polynomial $f(x)=x^3+A{x}^2+Bx+C$ has three distinct positive real roots, one of which is 1.
      Consider the polynomial $h(x)=x^3+(2B-A^2)x^2+(B^2-2AC)x-C^2$.
      Note that the coefficients of $h(x)$ are integers because $A$, $B$ and $C$ are integers.
      We show that $h(x)$ has three distinct positive real roots, one of which is equal to 1.

      Since $f(x)$ has $x=1$ as a root, then $f(1)=1+A+B+C=0$ which gives $B=-1-A-C$.
      Factoring out $x-1$ from $f(x)$, we obtain $$f(x)=x^3+A{x}^2+Bx+C=(x-1)(x^2+(A+1)x-C)$$ To find the quadratic factor here, we write $f(x)=(x-1)(x^2+px+q)$, expand to obtain $x^3+A{x}^2+Bx+C=x^3+(p-1)x^2+(q-p)x-q$, and compare coefficients to get $C=-q$ (giving $q=-C$) and $p-1=A$ (giving $p=A+1$).
      We note that comparing coefficients of $x$ gives $q-p=B$ or $-C-A-1=B$, which is consistent with the relationship above.
      Since $f(x)$ has three distinct positive real roots, one of which is equal to 1, then the quadratic $x^2+(A+1)x-C$ has two distinct positive real roots neither of which equals 1.
      This means that $1+(A+1)-C\ne 0$ and so $C\ne A+2$, a fact that we will use later.
      Also, we can deduce the following:

      • The product of its roots is positive; that is, $-C>0$ or $C<0$.

      • The sum of its roots is positive; that is, $-A-1>0$ or $A<-1$.

      • The discriminant is positive; that is $(A+1{)}^2+4C>0$.

      To summarize, so far, we know that $B=-1-A-C$ and $C<0$ and $A<-1$ and $(A+1{)}^2+4C>0$.

      We now examine $h(x)=x^3+(2B-A^2)x^2+(B^2-2AC)x-C^2$.
      We start by showing that $x=1$ is a root of $h(x)$ by showing that $h(1)=0$.
      Now $$\begin{array}{rl}h(1)& =1+(2B-A^2)+(B^2-2AC)-C^2\\ & =B^2+2B+1-(A^2+2AC+C^2)\\ & =(B+1{)}^2-(A+C{)}^2\\ & =(B+1+A+C)(B+1-A-C)\end{array}$$ Recall that $B=-1-A-C$ or $1+A+B+C=0$ and so $h(1)=0$ and so $x=1$ is a root of $h(x)$.
      Factoring $h(x)$ in a similar manner to how we factored $f(x)$, we obtain $$\begin{array}{rl}h(x)& =x^3+(2B-A^2)x^2+(B^2-2AC)x-C^2\\ & =(x-1)(x^2+(2B-A^2+1)x+C^2)\\ & =(x-1)(x^2+(-2-2A-2C-A^2+1)x+C^2)\\ & =(x-1)(x^2-(A^2+2A+2C+1)x+C^2)\end{array}$$ We now need to show that the quadratic polynomial $q(x)=x^2-(A^2+2A+2C+1)x+C^2$ has two distinct real positive roots neither of which equals 1.
      First, we show that the roots of $q(x)$ are real and distinct by examining the discriminant: $$\begin{array}{rl}\mathrm{\Delta }& =(A^2+2A+2C+1{)}^2-4C^2\\ & =(A^2+2A+2C+1+2C)(A^2+2A+2C+1-2C)\\ & =((A+1{)}^2+4C)(A+1{)}^2\end{array}$$ The first factor is positive since $(A+1{)}^2+4C$ is positive; the second factor is positive since $(A+1{)}^2$ is positive because $A\ne -1$.
      Therefore, $\mathrm{\Delta }>0$ which means that $q(x)$ has two distinct real roots.
      To show that $q(x)$ does not have 1 as a root, we suppose that $q(1)=0$ and obtain a contradiction: $$\begin{array}{rl}q(1)& =0\\ 1-A^2-2A-2C-1+C^2& =0\\ C^2-2C+1-A^2-2A-1+& =0\\ (C-1{)}^2-(A+1{)}^2& =0\\ (C-1+A+1)(C-1-A-1)& =0\\ (C+A)(C-A-2)& =0\end{array}$$ Since $C<0$ and $A<0$, then $C+A\ne 0$. Further, we saw above that $C\ne A+2$. Together, this means that $q(1)\ne 0$.
      This in turn means that $h(x)$ has three distinct real roots.
      Finally, we need to show that the roots of $q(x)$ are positive.
      Since $q(x)=x^2-(A^2+2A+2C+1)x+C^2$ then the product of the two real roots of $q(x)$ is $C^2$ and the sum of the roots is $A^2+2A+2C+1$.
      Since the product is $C^2$ and $C<0$, then the product is positive, which means that both roots are positive or both are negative.
      Since the sum of the roots is $$A^2+2A+2C+1=((A+1{)}^2+4C)+(-2C)$$ and each of these terms is positive, then the sum of the roots is positive, which means that they cannot both be negative, thus must both be positive.
      Therefore, $h(x)$ has three distinct positive real roots, one of which is $1$.

      We have shown that if $f(x)=x^3+A{x}^2+Bx+C$ has three distinct positive real roots, one of which is $1$, then $h(x)=x^3+(2B-A^2)x^2+(B^2-2AC)x-C^2$ has three distinct positive real roots, one of which is $1$.
      The process of moving from $f(x)$ to $h(x)$ is exactly the process of moving from $f_{n-1}(x)$ to $f_n(x)$ for any positive integer $n\ge 2$.
      Therefore, since $f_1(x)$ has three distinct positive real roots, one of which is $1$, then $f_2(x)$ has this property, which in turn means that $f_3(x)$ has this property, and so on.
      Continuing, this shows us that $f_{100}(x)$ has three distinct positive real roots.
      

3. 1. We note that $PB=PD=53$ and $BC=28$.
      By the Pythagorean Theorem in $\mathrm{△}BCP$, we have $$P{C}^2=P{B}^2-B{C}^2={53}^2-{28}^2=(53+28)(53-28)=81\cdot 25=9^2\cdot 5^2$$ Since $PC>0$, then $PC=9\cdot 5=45$.
      Since $ABCD$ is a rectangle, then $AB=DC=PD+PC=53+45=98$.

   2. Suppose that $BC=m$, an integer, and that $PD=x$.
      Note that $PB=PD=x$.
      Also, since $DC=AB=101$, then $PC=DC-PD=101-x$.
      Using the Pythagorean Theorem in $\mathrm{△}PBC$ gives the following equivalent equations: $$\begin{array}{rl}P{B}^2& =P{C}^2+B{C}^2\\ x^2& =(101-x{)}^2+m^2\\ x^2& =x^2-202x+{101}^2+m^2\\ 202x& ={101}^2+m^2\\ x& ={\displaystyle \frac{{101}^2+m^2}{2\cdot 101}}\end{array}$$ For $x$ to be an integer, we need the numerator to be divisible by $101$.
      Since ${101}^2$ is divisible by 101, we need $m^2$ to be divisible by 101.
      Since $101$ is prime, we need $m$ to be divisible by 101.
      However, $m=BC<AB=101$ and there are no positive multiples of 101 less than 101.
      This means that $m$ cannot be divisible by 101, which means that $x$, the length of $PD$, cannot be an integer.

   3. Suppose that $AB=n$, $BC=m$, and $PD=x$ for integers $m$, $n$ and $x$ with $m<n$ and $x<n$.
      Here, we have $PB=x$, $PC=n-x$, and $BC=m$.
      Using the Pythagorean Theorem in $\mathrm{△}PBC$ gives the following equivalent equations: $$\begin{array}{rl}P{B}^2& =P{C}^2+B{C}^2\\ x^2& =(n-x{)}^2+m^2\\ x^2& =x^2-2nx+n^2+m^2\\ 2nx& =n^2+m^2\\ x& ={\displaystyle \frac{n^2+m^2}{2n}}\end{array}$$ We need to determine all positive integers $m$ with $1\le m\le 100$ for which there are $7$ positive integers $n$ for which $x={\displaystyle \frac{n^2+m^2}{2n}}$ is an integer.
      We note that, since the denominator of this fraction ($2n$) is even, then for $x$ to be an integer, the numerator of this fraction ($n^2+m^2$) is also even.
      Since $n^2+m^2$ must be even, then $n$ and $m$ are both even or both odd.
      We look at two cases: $m$ odd and $m$ even.

      Case 1: $m$ is odd.

      Since $m$ is odd, then $n$ is odd.
      Here, $x={\displaystyle \frac{n^2+m^2}{2n}}={\displaystyle \frac{n}{2}}+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{m^2}{n}}$.
      Since ${\displaystyle \frac{n}{2}}$ is an odd integer divided by $2$, then for $x$ to be an integer, ${\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{m^2}{n}}$ must also equal an odd integer divided by $2$, which means that ${\displaystyle \frac{m^2}{n}}$ must equal an odd integer.
      Recalling that $n>m$, this means that we want to determine all odd integers $m$ with $1\le m\le 100$ for which exactly $7$ integers $n$ with $m<n\le m^2$ are divisors of $m^2$.
      Note that every divisor $n$ of $m^2$ with $m<n\le m^2$ will correspond to a divisor $q$ of $m^2$ with $1\le q<m$. This correspondence comes from the factor pair $qn=m^2$ and noting that $q={\displaystyle \frac{m^2}{n}}$ from which $m<n\le m^2$ gives ${\displaystyle \frac{m^2}{m^2}}\le q<{\displaystyle \frac{m^2}{m}}$.
      Additionally, $m^2$ has the divisor $m$ for which we have not yet accounted.
      This means that we want to determine all odd integers $m$ with $1\le m\le 100$ for which $m^2$ has exactly $2\cdot 7+1=15$ positive divisors.
      If $m$ has three distinct prime divisors $p_1$, $p_2$, $p_3$, then $m^2$ is divisible by the product $p_1^2{p}_2^2{p}_3^2$. This means that $m^2$ has at least $(2+1)(2+1)(2+1)=27$ positive divisors, which is impossible.
      Therefore, $m$ can have at most two distinct prime divisors.
      Suppose that $m=p_1^a$ for some prime $p_1$ and positive integer $a$.
      Then $m^2=p_1^{2a}$ and $m^2$ has $2a+1$ positive divisors.
      Since $m^2$ has 15 positive divisors, then $2a+1=15$ which gives $a=7$ and so $m=p_1^7$.
      However, the smallest odd perfect seventh power is $3^7>100$, and so there are no $m$ in this sub-case.
      Suppose that $m=p_1^a{p}_2^b$ for some odd primes $p_1\ne p_2$ and positive integers $a$ and $b$.
      Thus, $m^2=p_1^{2a}{p}_2^{2b}$ and $m^2$ has $(2a+1)(2b+1)$ divisors.
      Since $2a+1\ge 3$ and $2b+1\ge 3$ and $(2a+1)(2b+1)=15$, then $2a+1$ and $2b+1$ must be 5 and 3 in some order, which means that $a$ and $b$ must be 2 and 1 in some order.
      Thus, $m=p_1^2{p}_2$ for some odd primes $p_1$ and $p_2$.
      If $p_1=3$, then since $m=9p_2$ and $1\le m\le 100$, we can have $m=45$ (from $p_2=5$) or $m=63$ (from $p_2=7$) or $m=99$ (from $p_2=11$).
      If $p_1=5$, then since $m=25p_2$ and $1\le m\le 100$, we can have $m=75$ (from $p_2=3$).
      If $p_1\ge 7$, there are no odd primes $p_2$ for which $m\le 100$.
      Therefore, in the case that $m$ is odd, the possible values for $m$ are $m=45,63,75,99$.

      Case 2: $m$ is even.

      Since $m$ is even, then $n$ is even.
      Let $m=2M$ and $n=2N$ for some positive integers $M$ and $N$.
      Since $1\le 2M\le 100$, then $0.5\le M\le 50$ and so $1\le M\le 50$.
      Here, $x={\displaystyle \frac{4N^2+4M^2}{4N}}=N+{\displaystyle \frac{M^2}{N}}$.
      Since $N$ is an integer, then for $x$ to be an integer, ${\displaystyle \frac{M^2}{N}}$ must also equal an integer.
      Since $n>m$, then $N>M$. This means that we want to determine all integers $M$ with $1\le M\le 50$ for which exactly $7$ integers $N$ with $M<N\le M^2$ are divisors of $M^2$.
      As above, this means that we want to determine all integers $M$ with $1\le M\le 50$ for which $M^2$ has exactly $15$ positive divisors.
      Again, as above, $M$ must be of the form $p_1^2{p}_2$ for some primes $p_1$ and $p_2$. This time, we do not have the restriction that $M$ must be odd.
      If $p_1=2$, then since $M=4p_2$ and $1\le M\le 50$, we can have $M=12$ (from $p_2=3$) or $M=20$ (from $p_2=5$) or $M=28$ (from $p_2=7$) or $M=44$ (from $p_2=11$).
      If $p_1=3$, then since $M=9p_2$ and $1\le M\le 50$, we can have $M=18$ (from $p_2=2$) or $M=45$ (from $p_2=5$).
      If $p_1=5$, then since $M=25p_2$ and $1\le M\le 50$, we can have $M=50$ (from $p_2=2$).
      Since $m=2M$, the possible values of $m$ in this case are $m=24,40,56,88,36,90,100$.

      Combining the two cases, the values of $m$ are $m=24,36,40,45,56,63,75,88,90,99,100$.