2017 Euclid Contest
Solutions

Wednesday, April 6, 2017
(in North America and South America)

Friday, April 7, 2017
(outside of North American and South America)

©2017 University of Waterloo

1. 1. Since $5(2)+3(3)=19$, then the pair of positive integers that satisfies $5a+3b=19$ is $(a,b)=(2,3)$.

   2. We list the first several powers of 2 in increasing order:

      $n$	1	2	3	4	5	6	7	8	9	10	11
      $2^n$	2	4	8	16	32	64	128	256	512	1024	2048

      Each power of 2 can be found by multiplying the previous power by 2.
      From the table, the smallest power of 2 greater than 5 is $2^3=8$ and the largest power of 2 less than 2017 is $2^{10}=1024$. Since $2^n$ increases as $n$ increases, there can be no further powers in this range.
      Therefore, the values of $n$ for which $5<2^n<2017$ are $n=3,4,5,6,7,8,9,10$.
      There are 8 such values of $n$.

   3. Each of the 600 Euros that Jimmy bought cost $1.50.
      Thus, buying 600 Euros cost $600\times \mathrm{\$}1.50=\mathrm{\$}900$.
      When Jimmy converted 600 Euros back into dollars, the rate was $\mathrm{\$}1.00=0.75\text{ Euro}$.
      Therefore, Jimmy received $600\text{ Euros}\times {\displaystyle \frac{\mathrm{\$}1.00}{0.75\text{ Euros}}}={\displaystyle \frac{\mathrm{\$}600}{0.75}}=\mathrm{\$}800$.
      Thus, Jimmy had $\mathrm{\$}900-\mathrm{\$}800=\mathrm{\$}100$ less than he had before these two transactions.
      

2. 1. Since $x\ne 0$ and $x\ne 1$, we can multiply both sides of the given equation by $x(x-1)$ to obtain ${\displaystyle \frac{5x(x-1)}{x(x-1)}}={\displaystyle \frac{x(x-1)}{x}}+{\displaystyle \frac{x(x-1)}{x-1}}$ or $5=(x-1)+x$.
      Thus, $5=2x-1$ and so $2x=6$ or $x=3$. This means that $x=3$ is the only solution.
      (We can substitute $x=3$ into the original equation to verify that this is indeed a solution.)

   2. The sum of the entries in the second column is $20+4+(-12)=12$.
      This means that the sum of the entries in each row, in each column, and on each diagonal is 12.
      In the first row, we have $0+20+a=12$ and so $a=-8$.
      On the “top left to bottom right” diagonal, we have $0+4+b=12$ and so $b=8$.
      In the third column, we have entries $a=-8$ and $b=8$ whose sum is 0. Thus, the third entry must be 12.
      Finally, in the second row, we have $c+4+12=12$ and so $c=-4$.
      In summary, $a=-8$, $b=8$, and $c=-4$.
      We can complete the magic square to obtain: $$\begin{array}{|ccc|}\hline 0& 20& -8\\ -4& 4& 12\\ 16& -12& 8\\ \hline\end{array}$$

   3. 1. If ${100}^2-n^2=9559$, then $n^2={100}^2-9559=10000-9559=441$.
         Since $n>0$, then $n=\sqrt{441}=21$.

      2. From (i), $9559={100}^2-{21}^2$.
         Factoring the right side as a difference of squares, we see that $$9559=(100+21)(100-21)=121\cdot 79$$ Therefore, $(a,b)=(121,79)$ satisfies the conditions.
         (In addition, the pairs $(a,b)=(79,121),(869,11),(11,869)$ satisfy the conditions. The last two of these pairs cannot be obtained in the same way.)
         

3. 1. The area of quadrilateral $ABCD$ is the sum of the areas of $\mathrm{△}ABC$ and $\mathrm{△}ACD$.
      Since $\mathrm{△}ABC$ is right-angled at $B$, its area equals $\frac{1}{2}(AB)(BC)=\frac{1}{2}(3)(4)=6$.
      Since $\mathrm{△}ABC$ is right-angled at $B$, then by the Pythagorean Theorem, $$AC=\sqrt{A{B}^2+B{C}^2}=\sqrt{3^2+4^2}=\sqrt{25}=5$$ because $AC>0$. (We could have also observed that $\mathrm{△}ABC$ must be a “3-4-5" triangle.)
      Since $\mathrm{△}ACD$ is right-angled at $A$, then by the Pythagorean Theorem, $$AD=\sqrt{C{D}^2-A{C}^2}=\sqrt{{13}^2-5^2}=\sqrt{144}=12$$ because $AD>0$. (We could have also observed that $\mathrm{△}ACD$ must be a “5-12-13" triangle.)
      Thus, the area of $\mathrm{△}ACD$ equals $\frac{1}{2}(AC)(AD)=\frac{1}{2}(5)(12)=30$.
      Finally, the area of quadrilateral $ABCD$ is thus $6+30=36$.

   2. Let the width of each of the identical rectangles be $a$.
      In other words, $QP=RS=TW=WX=UV=VY=a$.
      Let the height of each of the identical rectangles be $b$.
      In other words, $QR=PS=TU=WV=XY=b$.
      The perimeter of the whole shape equals $$QP+PS+SX+XY+VY+UV+TU+TR+QR$$ Substituting for known lengths, we obtain $$a+b+SX+b+a+a+b+TR+b$$ or $3a+4b+(SX+TR)$.
      But $SX+TR=(TR+RS+SX)-RS=(TW+WX)-RS=a+a-a=a$.
      Therefore, the perimeter of the whole shape equals $4a+4b$.
      The perimeter of one rectangle is $2a+2b$, which we are told equals 21 cm.
      Finally, the perimeter of the whole shape is thus $2(2a+2b)$ which equals 42 cm.

   3. Solution 1:

      Suppose that the rectangular prism has dimensions $a$ cm by $b$ cm by $c$ cm.
      Suppose further that one of the faces that is $a$ cm by $b$ cm is the face with area $27{\text{ cm}}^2$ and that one of the faces that is $a$ cm by $c$ cm is the face with area $32{\text{ cm}}^2$. (Since every pair of non-congruent faces shares exactly one side length, there is no loss of generality in picking these particular variables for these faces.)
      Therefore, $ab=27$ and $ac=32$.
      Further, we are told that the volume of the prism is $144{\text{ cm}}^3$, and so $abc=144$.
      Thus, $bc={\displaystyle \frac{a^2{b}^2{c}^2}{a^{2}bc}}={\displaystyle \frac{(abc{)}^2}{(ab)(ac)}}={\displaystyle \frac{{144}^2}{(27)(32)}}=24$.
      (We could also note that $abc=144$ means $a^2{b}^2{c}^2={144}^2$ or $(ab)(ac)(bc)={144}^2$ and so $bc={\displaystyle \frac{{144}^2}{(27)(32)}}$.)
      In other words, the third type of face of the prism has area $24{\text{ cm}}^2$.
      Thus, since the prism has two faces of each type, the surface area of the prism is equal to $2(27{\text{ cm}}^2+32{\text{ cm}}^2+24{\text{ cm}}^2)$ or $166{\text{ cm}}^2$.

      Solution 2:

      Suppose that the rectangular prism has dimensions $a$ cm by $b$ cm by $c$ cm.
      Suppose further that one of the faces that is $a$ cm by $b$ cm is the face with area $27{\text{ cm}}^2$ and that one of the faces that is $a$ cm by $c$ cm is the face with area $32{\text{ cm}}^2$. (Since every pair of non-congruent faces shares exactly one side length, there is no loss of generality in picking these particular variables for these faces.)
      Therefore, $ab=27$ and $ac=32$.
      Further, we are told that the volume of the prism is $144{\text{ cm}}^3$, and so $abc=144$.
      Since $abc=144$ and $ab=27$, then $c=\frac{144}{27}=\frac{16}{3}$.
      Since $abc=144$ and $ac=32$, then $b=\frac{144}{32}=\frac{9}{2}$.
      This means that $bc=\frac{16}{3}\cdot \frac{9}{2}=24$.
      In ${\text{cm}}^2$, the surface area of the prism equals $2ab+2ac+2bc=2(27)+2(32)+2(24)=166$.
      Thus, the surface area of the prism is $166{\text{ cm}}^2$.
      

4. 1. Solution 1:

      We expand the right sides of the two equations, collecting like terms in each case: $$\begin{array}{rl}y& =a(x-2)(x+4)=a(x^2+2x-8)=a{x}^2+2ax-8a\\ y& =2(x-h{)}^2+k=2(x^2-2hx+h^2)+k=2x^2-4hx+(2h^2+k)\end{array}$$ Since these two equations represent the same parabola, then the corresponding coefficients must be equal. That is, $a=2$ and $2a=-4h$ and $-8a=2h^2+k$.
      Since $a=2$ and $2a=-4h$, then $4=-4h$ and so $h=-1$.
      Since $-8a=2h^2+k$ and $a=2$ and $h=-1$, then $-16=2+k$ and so $k=-18$.
      Thus, $a=2$, $h=-1$, and $k=-18$.

      Solution 2:

      From the equation $y=a(x-2)(x+4)$, we can find the axis of symmetry by calculating the midpoint of the $x$-intercepts.
      Since the $x$-intercepts are $2$ and $-4$, the axis of symmetry is at $x=\frac{1}{2}(2+(-4))=-1$.
      Since the vertex of the parabola lies on the axis of symmetry, then the $x$-coordinate of the vertex is $-1$.
      To find the $y$-coordinate of the vertex, we substitute $x=-1$ back into the equation $y=a(x-2)(x+4)$ to obtain $y=a(-1-2)(-1+4)=-9a$.
      Thus, the vertex of the parabola is $(-1,-9a)$.
      Since the second equation for the same parabola is in vertex form, $y=2(x-h{)}^2+k$, we can see that the vertex is at $(h,k)$ and $a=2$.
      Since $a=2$, the vertex has coordinates $(-1,-18)$, which means that $h=-1$ and $k=-18$.
      Thus, $a=2$, $h=-1$ and $k=-18$.

   2. Let the common difference in this arithmetic sequence be $d$.
      Since the first term in the sequence is 5, then the 5 terms are $5,5+d,5+2d,5+3d,5+4d$.
      From the given information, $5^2+(5+d{)}^2+(5+2d{)}^2=(5+3d{)}^2+(5+4d{)}^2$.
      Manipulating, we obtain the following equivalent equations: $$\begin{array}{rl}{5}^2+(5+d{)}^2+(5+2d{)}^2& =(5+3d{)}^2+(5+4d{)}^2\\ 25+(25+10d+d^2)+(25+20d+4d^2)& =(25+30d+9d^2)+(25+40d+16d^2)\\ 75+30d+5d^2& =50+70d+25d^2\\ 0& =20d^2+40d-25\\ 0& =4d^2+8d-5\\ 0& =(2d+5)(2d-1)\end{array}$$ Therefore, $d=-\frac{5}{2}$ or $d=\frac{1}{2}$.
      These give possible fifth terms of $5+4d=5+4(-\frac{5}{2})=-5$ and $5+4d=5+4(\frac{1}{2})=7$.
      (We note that, for these two values of $d$, the sequences are $5,\frac{5}{2},0,-\frac{5}{2},-5$ and $5,\frac{11}{2},6,\frac{13}{2},7$.)
      

5. 1. First, we determine the perfect squares between 1300 and 1400 and between 1400 and 1500.
      Since $\sqrt{1300}\approx 36.06$, then the first perfect square larger than 1300 is ${37}^2=1369$.
      The next perfect squares are ${38}^2=1444$ and ${39}^2=1521$.
      Since Dan was born between 1300 and 1400 in a year that was a perfect square, then Dan was born in 1369.
      Since Steve was born between 1400 and 1500 in a year that was a perfect square, then Steve was born in 1444.
      Suppose that on April 7 in some year, Dan was $m^2$ years old and Steve was $n^2$ years old for some positive integers $m$ and $n$. Thus, Dan was $m^2$ years old in the year $1369+m^2$ and Steve was $n^2$ years old in the year $1444+n^2$.
      Since these represent the same years, then $1369+m^2=1444+n^2$, or $m^2-n^2=1444-1369=75$.
      In other words, we want to find two perfect squares less than 110 (since their ages are less than 110) whose difference is 75.
      The perfect squares less than 110 are $1,4,9,16,25,36,49,64,81,100$.
      The two that differ by 75 are 100 and 25.
      Thus, $m^2=100$ and $n^2=25$.
      This means that the year in which the age of each of Dan and Steve was a perfect square was the year $1369+100=1469$.

   2. Solution 1:

      $\mathrm{△}ABC$ is right-angled exactly when one of the following statements is true:

      • $AB$ is perpendicular to $BC$, or

      • $AB$ is perpendicular to $AC$, or

      • $AC$ is perpendicular to $BC$.

      Since $A(1,2)$ and $B(11,2)$ share a $y$-coordinate, then $AB$ is horizontal.
      For $AB$ and $BC$ to be perpendicular, $BC$ must be vertical.
      Thus, $B(11,2)$ and $C(k,6)$ must have the same $x$-coordinate, and so $k=11$.
      For $AB$ and $AC$ to be perpendicular, $AC$ must be vertical.
      Thus, $A(1,2)$ and $C(k,6)$ must have the same $x$-coordinate, and so $k=1$.
      For $AC$ to be perpendicular to $BC$, their slopes must have a product of $-1$.
      The slope of $AC$ is ${\displaystyle \frac{6-2}{k-1}}$, which equals ${\displaystyle \frac{4}{k-1}}$.
      The slope of $BC$ is ${\displaystyle \frac{6-2}{k-11}}$, which equals ${\displaystyle \frac{4}{k-11}}$.
      Thus, $AC$ and $BC$ are perpendicular when ${\displaystyle \frac{4}{k-1}}\cdot {\displaystyle \frac{4}{k-11}}=-1$.
      Assuming that $k\ne 1$ and $k\ne 11$, we manipulate to obtain $16=-(k-1)(k-11)$ or $16=-k^2+12k-11$ or $k^2-12k+27=0$.
      Factoring, we obtain $(k-3)(k-9)=0$ and so $AC$ and $BC$ are perpendicular when $k=3$ or $k=9$.

      In summary, $\mathrm{△}ABC$ is right-angled when $k$ equals one of $1,3,9,11$.

      Solution 2:

      $\mathrm{△}ABC$ is right-angled exactly when its three side lengths satisfy the Pythagorean Theorem in some orientation. That is, $\mathrm{△}ABC$ is right-angled exactly when $A{B}^2+B{C}^2=A{C}^2$ or $A{B}^2+A{C}^2=B{C}^2$ or $A{C}^2+B{C}^2=A{B}^2$.
      Using $A(1,2)$ and $B(11,2)$, we obtain $A{B}^2=(11-1{)}^2+(2-2{)}^2=100$.
      Using $A(1,2)$ and $C(k,6)$, we obtain $A{C}^2=(k-1{)}^2+(6-2{)}^2=(k-1{)}^2+16$.
      Using $B(11,2)$ and $C(k,6)$, we obtain $B{C}^2=(k-11{)}^2+(6-2{)}^2=(k-11{)}^2+16$.
      Using the Pythagorean relationships above, $\mathrm{△}ABC$ is right-angled when one of the following is true:

      1. $$\begin{array}{rl}100+((k-11{)}^2+16)& =(k-1{)}^2+16\\ 100+k^2-22k+121+16& =k^2-2k+1+16\\ 220& =20k\\ k& =11\end{array}$$

      2. $$\begin{array}{rl}100+((k-1{)}^2+16)& =(k-11{)}^2+16\\ 100+k^2-2k+1+16& =k^2-22k+121+16\\ 20k& =20\\ k& =1\end{array}$$

      3. $$\begin{array}{rl}((k-1{)}^2+16)+((k-11{)}^2+16)& =100\\ k^2-2k+1+16+k^2-22k+121+16& =100\\ 2k^2-24k+54& =0\\ k^2-12k+27& =0\\ (k-3)(k-9)& =0\end{array}$$ and so $k=3$ or $k=9$.

      In summary, $\mathrm{△}ABC$ is right-angled when $k$ equals one of $1,3,9,11$.
      

6. 1. Extend $CA$ and $DB$ downwards until they meet the horizontal through $O$ at $P$ and $Q$, respectively.

      

      Since $CA$ and $DB$ are vertical, then $\mathrm{\angle }CPO=\mathrm{\angle }DQO=90\mathrm{°}$.
      Since $OA=20\text{ m}$, then $AP=OA\sin 30\mathrm{°}=(20\text{ m})\cdot \frac{1}{2}=10\text{ m}$.
      Since $OB=20\text{ m}$, then $BQ=OB\sin 45\mathrm{°}=(20\text{ m})\cdot \frac{1}{\sqrt{2}}=10\sqrt{2}\text{ m}$.
      Since $AC=6\text{ m}$, then $CP=AC+AP=16\text{ m}$.
      For $CD$ to be as short as possible and given that $C$ is fixed, then it must be the case that $CD$ is horizontal:

      If $CD$ were not horizontal, then suppose that $X$ is on $DQ$, possibly extended, so that $CX$ is horizontal.

      

      Then $\mathrm{\angle }CXD=90\mathrm{°}$ and so $\mathrm{△}CXD$ is right-angled with hypotenuse $CD$.
      In this case, $CD$ is longer than $CX$ or $XD$.
      In particular, $CD>CX$, which means that if $D$ were at $X$, then $CD$ would be shorter.
      In other words, a horizontal $CD$ makes $CD$ as short as possible.

      When $CD$ is horizontal, $CDQP$ is a rectangle, since it has two vertical and two horizontal sides. Thus, $DQ=CP=16\text{ m}$.
      Finally, this means that $BD=DQ-BQ=(16-10\sqrt{2})\text{ m}$.

   2. Since $\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}$, then we assume that $\cos \theta \ne 0$.
      Therefore, we obtain the following equivalent equations: $$\begin{array}{rl}\cos \theta & =\tan \theta \\ \cos \theta & ={\displaystyle \frac{\sin \theta }{\cos \theta }}\\ {\cos }^2\theta & =\sin \theta \\ 1-{\sin }^2\theta & =\sin \theta \\ 0& ={\sin }^2\theta +\sin \theta -1\end{array}$$ Let $u=\sin \theta$. This quadratic equation becomes $u^2+u-1=0$.
      By the quadratic formula, $u={\displaystyle \frac{-1\pm \sqrt{1^2-4(1)(-1)}}{2(1)}}={\displaystyle \frac{-1\pm \sqrt{5}}{2}}$.
      Therefore, $\sin \theta ={\displaystyle \frac{-1+\sqrt{5}}{2}}\approx 0.62$ or $\sin \theta ={\displaystyle \frac{-1-\sqrt{5}}{2}}\approx -1.62$.
      Since $-1\le \sin \theta \le 1$, then the second solution is inadmissible. Thus, $\sin \theta ={\displaystyle \frac{-1+\sqrt{5}}{2}}$.

7. 1. Solution 1:

      Suppose that the trains are travelling at $v$ km/h.
      Consider two consecutive points in time at which the car is passed by a train.
      Since these points are 10 minutes apart, and 10 minutes equals $\frac{1}{6}$ hour, and the car travels at 60 km/h, then the car travels $(60\text{ km/h})\cdot (\frac{1}{6}\text{ h})=10\text{ km}$.
      During these 10 minutes, each train travels $\frac{1}{6}v$ km, since its speed is $v$ km/h.
      At the first instance, Train A and the car are next to each other.
      At this time, Train B is “3 minutes" behind Train A.

      

      Since 3 minutes is $\frac{1}{20}$ hour, then Train B is $\frac{1}{20}v$ km behind Train A and the car.
      Therefore, the distance from the location of Train B at the first instance to the location where it passes the car is $(\frac{1}{20}v+10)$ km.
      But this distance also equals $\frac{1}{6}v$ km, since Train B travels for 10 minutes.
      Thus, $\frac{1}{6}v=\frac{1}{20}v+10$ or $\frac{10}{60}v-\frac{3}{60}v=10$ and so $\frac{7}{60}v=10$ or $v=\frac{600}{7}$.
      Therefore, the trains are travelling at $\frac{600}{7}$ km/h.

      Solution 2:

      Suppose that the trains are travelling at $v$ km/h.
      Consider the following three points in time: the instant when the car and Train A are next to each other, the instant when Train B is at the same location that the car and Train A were at in the previous instant, and the instant when the car and Train B are next to each other.

      

      From the first instant to the second, Train B “catches up” to where Train A was, so this must take a total of 3 minutes, because the trains leave the station 3 minutes apart.
      Since 3 minutes equals $\frac{3}{60}$ hour and the car travels at 60 km/h, then the car travels $(60\text{ km/h})\cdot (\frac{3}{60}\text{ h})=3\text{ km}$ between these two instants.
      From the first instant to the third, 10 minutes passes, since these are consecutive points at which the car is passed by trains. In 10 minutes, the car travels 10 km.
      Therefore, between the second and third instants, $10-3=7$ minutes pass. During these 7 minutes, Train B travels 10 km.
      Since 7 minutes equals $\frac{7}{60}$ hour, then $v\text{ km/h}=\frac{10\text{ km}}{7/60\text{ h}}=\frac{600}{7}\text{ km/h}$, and so the trains are travelling at $\frac{600}{7}$ km/h.

   2. From the first equation, we note that $a\ge 0$ and $b\ge 0$, since the argument of a square root must be non-negative.
      From the second equation, we note that $a>0$ and $b>0$, since the argument of a logarithm must be positive.
      Combining these restrictions, we see that $a>0$ and $b>0$.

      From the equation ${\log }_{10}a+{\log }_{10}b=2$, we obtain ${\log }_{10}(ab)=2$ and so $ab={10}^2=100$. From the first equation, obtain $$\begin{array}{rl}(\sqrt{a}+\sqrt{b}{)}^2& =8^2\\ a+2\sqrt{ab}+b& =64\\ a+2\sqrt{100}+b& =64\\ a+b& =64-2\sqrt{100}=44\end{array}$$ Since $a+b=44$, then $b=44-a$.
      Since $ab=100$, then $a(44-a)=100$ or $44a-a^2=100$ and so $0=a^2-44a+100$.
      By the quadratic formula, $$a={\displaystyle \frac{44\pm \sqrt{{44}^2-4(1)(100)}}{2\cdot 1}}={\displaystyle \frac{44\pm \sqrt{1536}}{2}}={\displaystyle \frac{44\pm 16\sqrt{6}}{2}}=22\pm 8\sqrt{6}$$ Since $b=44-a$, then $b=44-(22\pm 8\sqrt{6})=22\mp 8\sqrt{6}$.
      Therefore, $(a,b)=(22+8\sqrt{6},22-8\sqrt{6})$ or $(a,b)=(22-8\sqrt{6},22+8\sqrt{6})$.
      (We note that $22+8\sqrt{6}>0$ and $22-8\sqrt{6}>0$, so the initial restrictions on $a$ and $b$ are satisfied.)
      

8. 1. Let $\mathrm{\angle }PEQ=\theta$.
      Join $P$ to $B$.
      We use the fact that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord. We prove this fact below.
      More concretely, $\mathrm{\angle }DEP=\mathrm{\angle }PBE$ (using the chord $EP$ and the tangent through $E$) and $\mathrm{\angle }ABP=\mathrm{\angle }PEQ=\theta$ (using the chord $BP$ and the tangent through $B$).
      Now $\mathrm{\angle }DEP$ is exterior to $\mathrm{△}FEP$ and so $\mathrm{\angle }DEP=\mathrm{\angle }FPE+\mathrm{\angle }EFP=25\mathrm{°}+30\mathrm{°}$, and so $\mathrm{\angle }PBE=\mathrm{\angle }DEP=55\mathrm{°}$.
      Furthermore, $\mathrm{\angle }AQB$ is an exterior angle of $\mathrm{△}PQE$.
      Thus, $\mathrm{\angle }AQB=\mathrm{\angle }QPE+\mathrm{\angle }PEQ=25\mathrm{°}+\theta$.

      

      In $\mathrm{△}ABQ$, we have $\mathrm{\angle }BAQ=35\mathrm{°}$, $\mathrm{\angle }ABQ=\theta +55\mathrm{°}$, and $\mathrm{\angle }AQB=25\mathrm{°}+\theta$.
      Thus, $35\mathrm{°}+(\theta +55\mathrm{°})+(25\mathrm{°}+\theta )=180\mathrm{°}$ or $115\mathrm{°}+2\theta =180\mathrm{°}$, and so $2\theta =65\mathrm{°}$.
      Therefore $\mathrm{\angle }PEQ=\theta =\frac{1}{2}(65\mathrm{°})=32.5\mathrm{°}$.

      As an addendum, we prove that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord.
      

      Consider a circle with centre $O$ and a chord $XY$, with tangent $ZX$ meeting the circle at $X$. We prove that if $ZX$ is tangent to the circle, then $\mathrm{\angle }ZXY$ equals $\mathrm{\angle }XWY$ whenever $W$ is a point on the circle on the opposite side of $XY$ as $XZ$ (that is, the angle subtended by $XY$ on the opposite side of the circle).
      We prove this in the case that $\mathrm{\angle }ZXY$ is acute. The cases where $\mathrm{\angle }ZXY$ is a right angle or an obtuse angle are similar.
      Draw diameter $XOV$ and join $VY$.

      

      Since $\mathrm{\angle }ZXY$ is acute, points $V$ and $W$ are on the same arc of chord $XY$.
      This means that $\mathrm{\angle }XVY=\mathrm{\angle }XWY$, since they are angles subtended by the same chord.
      Since $OX$ is a radius and $XZ$ is a tangent, then $\mathrm{\angle }OXZ=90\mathrm{°}$.
      Thus, $\mathrm{\angle }OXY+\mathrm{\angle }ZXY=90\mathrm{°}$.
      Since $XV$ is a diameter, then $\mathrm{\angle }XYV=90\mathrm{°}$.
      From $\mathrm{△}XYV$, we see that $\mathrm{\angle }XVY+\mathrm{\angle }VXY=90\mathrm{°}$.
      But $\mathrm{\angle }OXY+\mathrm{\angle }ZXY=90\mathrm{°}$ and $\mathrm{\angle }XVY+\mathrm{\angle }VXY=90\mathrm{°}$ and $\mathrm{\angle }OXY=\mathrm{\angle }VXY$ tells us that $\mathrm{\angle }ZXY=\mathrm{\angle }XVY$.
      This gives us that $\mathrm{\angle }ZXY=\mathrm{\angle }XWY$, as required.

   2. Solution 1:

      Draw a line segment through $M$ in the plane of $\mathrm{△}PMN$ parallel to $PN$ and extend this line until it reaches the plane through $P$, $A$ and $D$ at $Q$ on one side and the plane through $N$, $B$ and $C$ at $R$ on the other side.
      Join $Q$ to $P$ and $A$. Join $R$ to $N$ and $B$.

      

      So the volume of solid $ABCDPMN$ equals the volume of solid $ABCDPQRN$ minus the volumes of solids $PMQA$ and $NMRB$.
      Solid $ABCDPQRN$ is a trapezoidal prism. This is because $NR$ and $BC$ are parallel (since they lie in parallel planes), which makes $NRBC$ a trapezoid. Similarly, $PQAD$ is a trapezoid. Also, $PN$, $QR$, $DC$, and $AB$ are all perpendicular to the planes of these trapezoids and equal in length, since they equal the side lengths of the squares.
      Solids $PMQA$ and $NMRB$ are triangular-based pyramids. We can think of their bases as being $\mathrm{△}PMQ$ and $\mathrm{△}NMR$. Their heights are each equal to 2, the height of the original solid. (The volume of a triangular-based pyramid equals $\frac{1}{3}$ times the area of its base times its height.)
      The volume of $ABCDPQRN$ equals the area of trapezoid $NRBC$ times the width of the prism, which is 2.
      That is, this volume equals $\frac{1}{2}(NR+BC)(NC)(NP)=\frac{1}{2}(NR+2)(2)(2)=2\cdot NR+4$.
      So we need to find the length of $NR$.
      Consider quadrilateral $PNRQ$. This quadrilateral is a rectangle since $PN$ and $QR$ are perpendicular to the two side planes of the original solid.
      Thus, $NR$ equals the height of $\mathrm{△}PMN$.
      Join $M$ to the midpoint $T$ of $PN$.
      Since $\mathrm{△}PMN$ is isosceles, then $MT$ is perpendicular to $PN$.

      

      Since $NT=\frac{1}{2}PN=1$ and $\mathrm{\angle }PMN=90\mathrm{°}$ and $\mathrm{\angle }TNM=45\mathrm{°}$, then $\mathrm{△}MTN$ is also right-angled and isosceles with $MT=TN=1$.
      Therefore, $NR=MT=1$ and so the volume of $ABCDPQRN$ is $2\cdot 1+4=6$.
      The volumes of solids $PMQA$ and $NMRB$ are equal. Each has height 2 and their bases $\mathrm{△}PMQ$ and $\mathrm{△}NMR$ are congruent, because each is right-angled (at $Q$ and at $R$) with $PQ=NR=1$ and $QM=MR=1$.
      Thus, using the formula above, the volume of each is $\frac{1}{3}(\frac{1}{2}(1)(1))2=\frac{1}{3}$.
      Finally, the volume of the original solid equals $6-2\cdot \frac{1}{3}=\frac{16}{3}$.

      Solution 2:

      We determine the volume of $ABCDPMN$ by splitting it into two solids: $ABCDPN$ and $ABNPM$ by slicing along the plane of $ABNP$.
      Solid $ABCDPN$ is a triangular prism, since $\mathrm{△}BCN$ and $\mathrm{△}ADP$ are each right-angled (at $C$ and $D$), $BC=CN=AD=DP=2$, and segments $PN$, $DC$ and $AB$ are perpendicular to each of the triangular faces and equal in length.
      Thus, the volume of $ABCDPN$ equals the area of $\mathrm{△}BCN$ times the length of $DC$, or $\frac{1}{2}(BC)(CN)(DC)=\frac{1}{2}(2)(2)(2)=4$. (This solid can also be viewed as “half” of a cube.)

      Solid $ABNPM$ is a pyramid with rectangular base $ABNP$. (Note that $PN$ and $AB$ are perpendicular to the planes of both of the side triangular faces of the original solid, that $PN=AB=2$ and $BN=AP=\sqrt{2^2+2^2}=2\sqrt{2}$, by the Pythagorean Theorem.)

      Therefore, the volume of $ABNPM$ equals $\frac{1}{3}(AB)(BN)h=\frac{4\sqrt{2}}{3}h$, where $h$ is the height of the pyramid (that is, the distance that $M$ is above plane $ABNP$).
      So we need to calculate $h$.
      

      Join $M$ to the midpoint, $T$, of $PN$ and to the midpoint, $S$, of $AB$. Join $S$ and $T$. By symmetry, $M$ lies directly above $ST$.
      Since $ABNP$ is a rectangle and $S$ and $T$ are the midpoints of opposite sides, then $ST=AP=2\sqrt{2}$.
      Since $\mathrm{△}PMN$ is right-angled and isosceles, then $MT$ is perpendicular to $PN$. Since $NT=\frac{1}{2}PN=1$ and $\mathrm{\angle }TNM=45\mathrm{°}$, then $\mathrm{△}MTN$ is also right-angled and isosceles with $MT=TN=1$.

      

      Also, $MS$ is the hypotenuse of the triangle formed by dropping a perpendicular from $M$ to $U$ in the plane of $ABCD$ (a distance of 2) and joining $U$ to $S$. Since $M$ is 1 unit horizontally from $PN$, then $US=1$.
      Thus, $MS=\sqrt{2^2+1^2}=\sqrt{5}$ by the Pythagorean Theorem.

      

      We can now consider $\mathrm{△}SMT$. $h$ is the height of this triangle, from $M$ to base $ST$.

      

      Now $h=MT\sin (\mathrm{\angle }MTS)=\sin (\mathrm{\angle }MTS)$.
      By the cosine law in $\mathrm{△}SMT$, we have $$M{S}^2=S{T}^2+M{T}^2-2(ST)(MT)\cos (\mathrm{\angle }MTS)$$ Therefore, $5=8+1-4\sqrt{2}\cos (\mathrm{\angle }MTS)$ or $4\sqrt{2}\cos (\mathrm{\angle }MTS)=4$.
      Thus, $\cos (\mathrm{\angle }MTS)=\frac{1}{\sqrt{2}}$ and so $\mathrm{\angle }MTS=45\mathrm{°}$ which gives $h=\sin (\mathrm{\angle }MTS)=\frac{1}{\sqrt{2}}$.
      (Alternatively, we note that the plane of $ABCD$ is parallel to the plane of $PMN$, and so since the angle between plane $ABCD$ and plane $PNBA$ is $45\mathrm{°}$, then the angle between plane $PNBA$ and plane $PMN$ is also $45\mathrm{°}$, and so $\mathrm{\angle }MTS=45\mathrm{°}$.)
      Finally, this means that the volume of $ABNPM$ is $\frac{4\sqrt{2}}{3}\cdot \frac{1}{\sqrt{2}}=\frac{4}{3}$, and so the volume of solid $ABCDPMN$ is $4+\frac{4}{3}=\frac{16}{3}$.
      

9. 1. There are $4!=4\cdot 3\cdot 2\cdot 1=24$ permutations of $1,2,3,4$.
      This is because there are 4 possible choices for $a_1$, and for each of these there are 3 possible choices for $a_2$, and for each of these there are 2 possible choices for $a_3$, and then 1 possible choice for $a_4$.
      Consider the permutation $a_1=1,a_2=2,a_3=3,a_4=4$. (We write this as 1,2,3,4.)
      Here, $|a_1-a_2|+|a_3-a_4|=|1-2|+|3-4|=1+1=2$.
      This value is the same as the value for each of 2,1,3,4 and 1,2,4,3 and 2,1,4,3 and 3,4,1,2 and 4,3,1,2 and 3,4,2,1 and 4,3,2,1.
      Consider the permutation 1,3,2,4.
      Here, $|a_1-a_2|+|a_3-a_4|=|1-3|+|2-4|=2+2=4$.
      This value is the same as the value for each of 3,1,2,4 and 1,3,4,2 and 3,1,4,2 and 2,4,1,3 and 4,2,1,3 and 2,4,3,1 and 4,2,3,1.
      Consider the permutation 1,4,2,3.
      Here, $|a_1-a_2|+|a_3-a_4|=|1-4|+|2-3|=3+1=4$.
      This value is the same as the value for each of 4,1,2,3 and 1,4,3,2 and 4,1,3,2 and 2,3,1,4 and 3,2,1,4 and 2,3,4,1 and 3,2,4,1.
      This accounts for all 24 permutations.
      Therefore, the average value is ${\displaystyle \frac{2\cdot 8+4\cdot 8+4\cdot 8}{24}}={\displaystyle \frac{80}{24}}={\displaystyle \frac{10}{3}}$.

   2. There are $7!=7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$ permutations of $1,2,3,4,5,6,7$, because there are 7 choices for $a_1$, then 6 choices for $a_2$, and so on.
      We determine the average value of $a_1-a_2+a_3-a_4+a_5-a_6+a_7$ over all of these permutations by determining the sum of all $7!$ values of this expression and dividing by $7!$.
      To determine the sum of all $7!$ values, we determine the sum of the values of $a_1$ in each of these expressions and call this total $s_1$, the sum of the values of $a_2$ in each of these expressions and call this total $s_2$, and so on.
      The sum of the $7!$ values of the original expression must equal $s_1-s_2+s_3-s_4+s_5-s_6+s_7$. This uses the fact that, when adding, the order in which we add the same set of numbers does not matter.
      By symmetry, the sums of the values of $a_1,a_2,a_3,a_4,a_5,a_6,a_7$ will all be equal. That is, $s_1=s_2=s_3=s_4=s_5=s_6=s_7$.
      This means that the desired average value equals $${\displaystyle \frac{s_1-s_2+s_3-s_4+s_5-s_6+s_7}{7!}}={\displaystyle \frac{(s_1+s_3+s_5+s_7)-(s_2+s_4+s_6)}{7!}}={\displaystyle \frac{4s_1-3s_1}{7!}}={\displaystyle \frac{s_1}{7!}}$$ So we need to determine the value of $s_1$.
      Now $a_1$ can equal each of $1,2,3,4,5,6,7$.
      If $a_1=1$, there are $6!$ combinations of values for $a_2,a_3,a_4,a_5,a_6,a_7$, since there are still 6 choices for $a_2$, 5 for $a_3$, and so on.
      Similarly, there are $6!$ combinations with $a_1$ equal to each of $2,3,4,5,6,7$.
      Thus, $s_1=1\cdot 6!+2\cdot 6!+3\cdot 6!+4\cdot 6!+5\cdot 6!+6\cdot 6!+7\cdot 6!=6!(1+2+3+4+5+6+7)=28(6!)$.
      Therefore, the average value of the expression is ${\displaystyle \frac{28(6!)}{7!}}={\displaystyle \frac{28(6!)}{7(6!)}}={\displaystyle \frac{28}{7}}=4$.

   3. There are $200!$ permutations of $1,2,3,\dots ,198,199,200$.
      We determine the average value of $$|a_1-a_2|+|a_3-a_4|+\cdots +|a_{197}-a_{198}|+|a_{199}-a_{200}|(\ast )$$ over all of these permutations by determining the sum of all $200!$ values of this expression and dividing by $200!$.
      As in (b), we let $s_1$ be the sum of the values of $|a_1-a_2|$ in each of these expressions, $s_2$ be the sum of the values of $|a_3-a_4|$, and so on.
      The sum of the $200!$ values of $(\ast )$ equals $s_1+s_2+\cdots +s_{99}+s_{100}$.
      By symmetry, $s_1=s_2=\cdots =s_{99}=s_{100}$.
      Therefore, the average value of $(\ast )$ equals ${\displaystyle \frac{100s_1}{200!}}$. So we need to determine the value of $s_1$.
      Suppose that $a_1=i$ and $a_2=j$ for some integers $i$ and $j$ between 1 and 200, inclusive.
      There are $198!$ permutations with $a_1=i$ and $a_2=j$ because there are still 198 choices for $a_3$, 197 choices for $a_4$, and so on.
      Similarly, there are $198!$ permutations with $a_1=j$ and $a_2=i$.
      Since $|i-j|=|j-i|$, then there are $2(198!)$ permutations with $|a_1-a_2|=|i-j|$ that come from $a_1$ and $a_2$ equalling $i$ and $j$ in some order.
      Therefore, we may assume that $i>j$ and note that $s_1$ equals $2(198!)$ times the sum of $i-j$ over all possible pairs $i>j$.
      (Note that there are ${\displaystyle (\genfrac{}{}{0ex}{}{200}{2})={\displaystyle \frac{200(199)}{2}}}$ choices for the pair of integers $(i,j)$ with $i>j$. For each of these choices, there are $2(198!)$ choices for the remaining entries in the permutation, which gives ${\displaystyle \frac{200(199)}{2}}\cdot 2(198!)=200(199)(198!)=200!$ permutations, as expected.)
      So to determine $s_1$, we need to determine the sum of the values of $i-j$.
      We calculate this sum, which we call $D$, by letting $j=1,2,3,\dots ,198,199$ and for each of these, we let $i$ be the possible integers with $j<i\le 200$: $$\begin{array}{rl}D& =(2-1)+(3-1)+(4-1)+\cdots +(197-1)+(198-1)+(199-1)+(200-1)\\ & +(3-2)+(4-2)+(5-2)+\cdots +(198-2)+(199-2)+(200-2)\\ & +(4-3)+(5-3)+(6-3)+\cdots +(199-3)+(200-3)\\ & ⋮\\ & +(199-198)+(200-198)\\ & +(200-199)\\ & =199(1)+198(2)+197(3)+\cdots +2(198)+1(199)\text{(grouping by columns)}\\ & =199(200-199)+198(200-198)+197(200-197)+\cdots +2(200-2)+1(200-1)\\ & =200(199+198+197+\cdots +3+2+1)-({199}^2+{198}^2+{197}^2+\cdots +3^2+2^2+1^2)\\ & =200\cdot \frac{1}{2}(199)(200)-\frac{1}{6}(199)(199+1)(2(199)+1)\\ & =100(199)(200)-\frac{1}{6}(199)(200)(399)\\ & =199(200)(100-\frac{133}{2})\\ & =199(200)\frac{67}{2}\end{array}$$ Therefore, $s_1=2(198!)D=2(198!)\cdot \frac{199(200)(67)}{2}=67(198!)(199)(200)=67(200!)$.
      Finally, this means that the average value of $(\ast )$ is ${\displaystyle \frac{100s_1}{200!}}={\displaystyle \frac{100(67)(200!)}{200!}}=6700$.
      We note that we have used the facts that, if $n$ is a positive integer, then

      • $1+2+\cdots +(n-1)+n=\frac{1}{2}n(n+1)$

      • $1^2+2^2+\cdots +(n-1{)}^2+n^2=\frac{1}{6}n(n+1)(2n+1)$

      Using sigma notation, we could have calculated $D$ as follows: $$\begin{array}{rl}D& =\sum _{i=2}^{200}\sum _{j=1}^{i-1}(i-j)\\ & =\left(\sum _{i=2}^{200}\sum _{j=1}^{i-1}i\right)-\left(\sum _{i=2}^{200}\sum _{j=1}^{i-1}j\right)\\ & =(\sum _{i=2}^{200}i(i-1))-(\sum _{i=2}^{200}\frac{1}{2}(i-1)i)\\ & =(\sum _{i=2}^{200}i(i-1))-\frac{1}{2}(\sum _{i=2}^{200}(i-1)i)\\ & =\frac{1}{2}(\sum _{i=2}^{200}(i-1)i)\\ & =\frac{1}{2}(\sum _{i=1}^{200}(i-1)i)\text{(since }(i-1)i=0\text{ when }i=1\text{)}\\ & =\frac{1}{2}(\sum _{i=1}^{200}(i^2-i))\\ & =\frac{1}{2}(\sum _{i=1}^{200}{i}^2-\sum _{i=1}^{200}i)\\ & =\frac{1}{2}(\frac{1}{6}(200)(200+1)(2(200)+1)-\frac{1}{2}(200)(200+1))\\ & =\frac{1}{2}(200)(201)(\frac{1}{6}(401)-\frac{1}{2})\\ & =100(201)\cdot \frac{398}{6}\\ & =100(201)\cdot \frac{199}{3}\\ & =100(67)(199)\end{array}$$ which equals $199(200)\frac{67}{2}$, as expected. (Can you determine a general formula when 200 is replaced with $2n$?)

10. 1. We start with the subset $\{1,2,3\}$.
       The sums of pairs of elements are $1+2=3$ and $1+3=4$ and $2+3=5$, which are all different.
       Thus, $\{1,2,3\}$ is exciting.
       We proceed to include additional elements in $\{1,2,3\}$.
       We cannot include 4 to create an exciting set, since if we did, we would have $1+4=2+3$, and so $\{1,2,3,4\}$ is boring.
       Consider the subset $\{1,2,3,5\}$.
       The sums of pairs of elements are $$1+2=31+3=41+5=62+3=52+5=73+5=8$$ which are all different.
       Thus, $\{1,2,3,5\}$ is exciting.
       We cannot include $6$ or $7$ since $2+5=1+6$ and $3+5=1+7$.
       Consider the subset $\{1,2,3,5,8\}$.
       In addition to the six sums above, we have the additional sums $1+8=9$ and $2+8=10$ and $3+8=11$ and $5+8=13$, so the 10 sums are all different.
       Therefore, $\{1,2,3,5,8\}$ is an exciting subset of $\{1,2,3,4,5,6,7,8\}$ that contains exactly 5 elements.
       (The subset $\{1,4,6,7,8\}$ is the only other exciting subset of $\{1,2,3,4,5,6,7,8\}$ that contains exactly 5 elements.)

    2. Suppose that $S$ is an exciting set that contains exactly $m$ elements.
       There are ${\displaystyle (\genfrac{}{}{0ex}{}{m}{2})={\displaystyle \frac{m(m-1)}{2}}}$ pairs of elements of $S$.
       Since $S$ is exciting, the sums of these pairs of elements are all distinct positive integers.
       This means that the largest of these sums is greater than or equal to ${\displaystyle \frac{m(m-1)}{2}}$.
       When two numbers add to ${\displaystyle \frac{m(m-1)}{2}}$ or greater, then at least one of them must be at least as large as ${\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{m(m-1)}{2}}={\displaystyle \frac{m^2-m}{4}}$.
       Therefore, there is an element of $S$ that is greater than or equal to ${\displaystyle \frac{m^2-m}{4}}$.

    3. Let $n$ be a positive integer with $n\ge 10$.
       For each integer $k$ with $1\le k\le n$, define $x_k=2n\cdot \text{rem}(k^2,n)+k$, where $\text{rem}(k^2,n)$ is the remainder when $k^2$ is divided by $n$.
       Define $T=\{x_1,x_2,\dots ,x_{n-1},x_n\}$.
       We show that $T$ is exciting exactly when $n$ is prime.

       Suppose that $a,b,c,d$ are distinct integers between 1 and $n$ with $x_a+x_b=x_c+x_d$.
       This equation is equivalent to $$(2n\cdot \text{rem}(a^2,n)+a)+(2n\cdot \text{rem}(b^2,n)+b)=(2n\cdot \text{rem}(c^2,n)+c)+(2n\cdot \text{rem}(d^2,n)+d)$$ and $$2n\cdot (\text{rem}(a^2,n)+\text{rem}(b^2,n)-\text{rem}(c^2,n)-\text{rem}(d^2,n))=c+d-a-b$$ Since $a,b,c,d$ are distinct integers between 1 and $n$, inclusive, then we have $1+2\le a+b\le (n-1)+n$, or $3\le a+b\le 2n-1$. Similarly, $3\le c+d\le 2n-1$.
       This means that $3-(2n-1)\le c+d-a-b\le (2n-1)-3$ or $-2n+4\le c+d-a-b\le 2n-4$.
       But the left side of the equation $$2n\cdot (\text{rem}(a^2,n)+\text{rem}(b^2,n)-\text{rem}(c^2,n)-\text{rem}(d^2,n))=c+d-a-b$$ is an integer that is a multiple of $2n$, so the right side ($c+d-a-b$) must be as well.
       Since $-2n+4\le c+d-a-b\le 2n-4$ and the only multiple of $2n$ between $-2n+4$ and $2n-4$ is $0\cdot 2n=0$, then $c+d-a-b=0$ or $c+d=a+b$.
       Thus, $2n\cdot (\text{rem}(a^2,n)+\text{rem}(b^2,n)-\text{rem}(c^2,n)-\text{rem}(d^2,n))=0$.
       Since $n\ne 0$, then $\text{rem}(a^2,n)+\text{rem}(b^2,n)-\text{rem}(c^2,n)-\text{rem}(d^2,n)=0$.
       Therefore, $x_a+x_b=x_c+x_d$ exactly when $$a+b=c+d\text{and}\text{rem}(a^2,n)+\text{rem}(b^2,n)=\text{rem}(c^2,n)+\text{rem}(d^2,n)$$ Suppose that $n$ is composite. We show that $T$ is boring.
       We consider three cases: $n=p^2$ for some prime $p$, $n$ is even, and all other $n$.
       Suppose that $n=p^2$ for some prime $p$. Since $n\ge 10$, then $p\ge 5$.
       Set $a=p$, $b=4p$, $c=2p$, and $d=3p$.
       Then $a+b=5p=c+d$.
       Also, since $p\ge 5$, then $0<p<2p<3p<4p<p^2$.
       Furthermore, since each of $a,b,c,d$ is divisible by $p$, then each of $a^2,b^2,c^2,d^2$ is divisible by $p^2=n$, so $\text{rem}(a^2,n)=\text{rem}(b^2,n)=\text{rem}(c^2,n)=\text{rem}(d^2,n)=0$.
       This means that $a+b=c+d$ and $\text{rem}(a^2,n)+\text{rem}(b^2,n)=\text{rem}(c^2,n)+\text{rem}(d^2,n)$, and so $x_a+x_b=x_c+x_d$, which means that $T$ is boring.

       Next, suppose that $n$ is even, say $n=2t$ for some positive integer $t\ge 5$.
       Set $a=1$, $b=t+2$, $c=2$, and $d=t+1$.
       Since $t\ge 5$, then $1\le a<b<c<d<2t$, so $a,b,c,d$ are distinct positive integers in the correct range.
       Also, $a+b=t+3=c+d$.
       To show that $x_a+x_b=x_c+x_d$, it remains to show that $$\text{rem}(1^2,2t)+\text{rem}((t+2{)}^2,2t)=\text{rem}(2^2,2t)+\text{rem}((t+1{)}^2,2t)$$ Now $\text{rem}(1^2,2t)=\text{rem}(1,2t)=1$ and $\text{rem}(2^2,2t)=\text{rem}(4,2t)=4$ since $2t>4$.
       Also, since $(t+2{)}^2=t^2+4t+4$ and so $(t+2{)}^2$ and $t^2+4$ differ by a multiple of $n=2t$, then $\text{rem}((t+2{)}^2,2t)=\text{rem}(t^2+4,2t)$.
       Similarly, since $(t+1{)}^2=t^2+2t+1$, then $\text{rem}((t+1{)}^2,2t)=\text{rem}(t^2+1,2t)$.
       Therefore, we need to show that $\text{rem}(t^2+4,2t)-\text{rem}(t^2+1,2t)=4-1=3$.
       Since $t\ge 5$, then $t^2+t>t^2+4$.
       This means that $t^2<t^2+1<t^2+2<t^2+3<t^2+4<t^2+t$; in other words, each of $t^2+1$, $t^2+2$, $t^2+3$, $t^2+4$ is strictly between two consecutive multiples of $t$, and so none of these four integers can be a multiple of $t$. This means that none of these is a multiple of $n=2t$.
       Therefore, $t^2+4$ and $t^2+1$ are bounded between the same two multiples of $n$, and so the difference between their remainders when dividing by $n$ equals the difference between the integers, which is 3.
       Thus, $x_a+x_b=x_c+x_d$, which means that $T$ is boring.

       Finally, we consider the case where $n$ is odd and composite and can be written as $n=MN$ for some odd integers $M>N>1$.
       Set $a=\frac{1}{2}(M+N)$, $b=n-a$, $c=\frac{1}{2}(M-N)$, and $d=n-c$.
       Since $M$ and $N$ are both odd, then $M+N$ and $M-N$ are even, and so $a,b,c,d$ are integers.
       Since $M>N>0$, then $a>c>0$.
       Since $N\ge 3$, then $n=MN\ge 3M>2M$ and so $M<\frac{1}{2}n$.
       Since $M>N$, then $a=\frac{1}{2}(M+N)<\frac{1}{2}(M+M)=M<\frac{1}{2}n$.
       Therefore, $0<c<a<\frac{1}{2}n$.
       Since $b=n-a$ and $d=n-c$, then $\frac{1}{2}n<b<d<n$ and so $0<c<a<\frac{1}{2}n<b<d<n$.
       This means that $a,b,c,d$ are distinct integers in the correct range.
       Also, note that $a+b=n=c+d$.
       To show that $x_a+x_b=x_c+x_d$, it remains to show that $$\text{rem}(a^2,n)+\text{rem}(b^2,n)=\text{rem}(c^2,n)+\text{rem}(d^2,n)$$ We show that $\text{rem}(a^2,n)=\text{rem}(b^2,n)=\text{rem}(c^2,n)=\text{rem}(d^2,n)$, which will provide the desired conclusion.
       Since $b=n-a$, then $b^2=n^2-2na+a^2$. Since $b^2$ and $a^2$ differ by a multiple of $n$, their remainders after division by $n$ will be equal. Similarly, $\text{rem}(c^2,n)=\text{rem}(d^2,n)$.
       Thus, it remains to show that $\text{rem}(a^2,n)=\text{rem}(c^2,n)$.
       But $$a^2-c^2=(a+c)(a-c)=(\frac{1}{2}(M+N)+\frac{1}{2}(M-N))(\frac{1}{2}(M+N)-\frac{1}{2}(M-N))=MN=n$$ Since $a^2$ and $c^2$ differ by a multiple of $n$, then $\text{rem}(a^2,n)=\text{rem}(c^2,n)$.
       Thus, $x_a+x_b=x_c+x_d$, which means that $T$ is boring.

       Suppose that $n$ is prime. We show that $T$ is exciting.
       Since $n\ge 10$, then $n$ is odd.
       Suppose that $x_a+x_b=x_c+x_d$. We will show that this is not possible.
       Recall that $x_a+x_b=x_c+x_d$ is equivalent to the conditions $a+b=c+d$ and $\text{rem}(a^2,n)+\text{rem}(b^2,n)=\text{rem}(c^2,n)+\text{rem}(d^2,n)$.
       We work with this second equation.
       When $a^2$ is divided by $n$, we obtain a quotient that we call $q_a$ and the remainder $\text{rem}(a^2,n)$.
       Note that $a^2=q_{a}n+\text{rem}(a^2,n)$ and $0\le \text{rem}(a^2,n)<n$.
       We define $q_b,q_c,q_d$ similarly and obtain $$(a^2-q_{a}n)+(b^2-q_{b}n)=(c^2-q_{c}n)+(d^2-q_{d}n)$$ or $$a^2+b^2-c^2-d^2=n(q_a+q_b-q_c-q_d)$$ Since $a+b=c+d$, then $a^2+2ab+b^2=c^2+2cd+d^2$ or $a^2+b^2-c^2-d^2=2cd-2ab$.
       Therefore, $x_a+x_b=x_c+x_d$ exactly when $a+b=c+d$ and $2cd-2ab=n(q_a+q_b-q_c-q_d)$.
       Since $d=a+b-c$, then this last equation becomes $$\begin{array}{rl}2c(a+b-c)-2ab& =n(q_a+q_b-q_c-q_d)\\ -2(c^2-ac-bc+ab)& =n(q_a+q_b-q_c-q_d)\\ -2(c(c-a)-b(c-a))& =n(q_a+q_b-q_c-q_d)\\ -2(c-a)(c-b)& =n(q_a+q_b-q_c-q_d)\end{array}$$ Since $x_a+x_b=x_c+x_d$, then $a+b=c+d$ and $-2(c-a)(c-b)=n(q_a+q_b-q_c-q_d)$.
       Therefore, $2(c-a)(c-b)$ is a multiple of $n$, which is an odd prime.
       This means that either $c-a$ or $c-b$ is a multiple of $n$.
       But $a,b,c,d$ are between 1 and $n$ inclusive and are distinct, so $1-n\le c-a\le n-1$ and $1-n\le c-b\le n-1$.
       The only multiple of $n$ in this range is $0$, so $c-a=0$ or $c-b=0$, which contradicts the fact that $a,b,c,d$ are distinct.
       Therefore, if $n$ is prime, there do not exist four distinct elements of $T$ that make $T$ boring, so $T$ is exciting.

       In summary, $T$ is exciting exactly when $n\ge 10$ is prime.