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[ "The largest possible $C$ is $C=\\frac{1}{2}$.\n\nFor $0<C \\leqslant \\frac{1}{2}$, Turbo can simply choose an arbitrary point $P$ (different from its starting point) to avoid. When Turbo is at an arbitrary point $A$ different from $P$, the two arcs $A P$ have total length 1; therefore, the larger of the two the arcs (or either arc in case $A$ is diametrically opposite to $P$ ) must have length $\\geqslant \\frac{1}{2}$. By always choosing this larger arc (or either arc in case $A$ is diametrically opposite to $P$ ), Turbo will manage to avoid the point $P$ forever.\n\nFor $C>\\frac{1}{2}$, we write $C=\\frac{1}{2}+a$ with $a>0$, and we choose the sequence\n\n$$\n\\frac{1}{2}, \\quad \\frac{1+a}{2}, \\quad \\frac{1}{2}, \\quad \\frac{1+a}{2}, \\quad \\frac{1}{2}, \\ldots\n$$\n\nIn other words, $c_{i}=\\frac{1}{2}$ if $i$ is odd and $c_{i}=\\frac{1+a}{2}<C$ when $i$ is even. We claim Turbo must eventually visit all points on the circle. This is clear when it crawls in the same direction two times in a row; after all, we have $c_{i}+c_{i+1}>1$ for all $i$. Therefore, we are left with the case that Turbo alternates crawling clockwise and crawling counterclockwise. If it, without loss of generality, starts by going clockwise, then it will always crawl a distance $\\frac{1}{2}$ clockwise followed by a distance $\\frac{1+a}{2}$ counterclockwise. The net effect is that it crawls a distance $\\frac{a}{2}$ counterclockwise. Because $\\frac{a}{2}$ is positive, there exists a positive integer $N$ such that $\\frac{a}{2} \\cdot N>1$. After $2 N$ crawls, Turbo will have crawled a distance $\\frac{a}{2}$ counterclockwise $N$ times, therefore having covered a total distance of $\\frac{a}{2} \\cdot N>1$, meaning that it must have crawled over all points on the circle.\n\nNote: Every sequence of the form $c_{i}=x$ if $i$ is odd, and $c_{i}=y$ if $i$ is even, where $0<x, y<C$, such that $x+y \\geqslant 1$, and $x \\neq y$ satisfies the conditions with the same argument. There might be even more possible examples.", "To show that $C\\le \\frac12$\n\nWe consider the following related problem:\n\nWe assume instead that the snail Chet is moving left and right on the real line. Find the size $M$ of the smallest (closed) interval, that we cannot force Chet out of, using a sequence of real numbers $d_{i}$ with $0<d_{i}<1$ for all $i$.\n\nThen $C=1 / M$. Indeed if for every sequence $c_{1}, c_{2}, \\ldots$, with $c_{i}<C$ there exists a point that Turbo can avoid, then the circle can be cut open at the avoided point and mapped to an interval of size $M$ such that Chet can stay inside this interval for any sequence of the from $c_{1} / C, c_{2} / C, \\ldots$, see Figure 5 . However, all sequences $d_{1}, d_{2}, \\ldots$ with $d_{i}<1$ can be written in this form. Similarly if for every sequence $d_{1}, d_{2}, \\ldots$, there exists an interval of length smaller or equal $M$ that we cannot force Chet out of, this projects to a subset of the circle, that we cannot force Turbo out of using any sequence of the form $d_{1} / M, d_{2} / M, \\ldots$. These are again exactly all the sequences with elements in $[0, C)$.\n\n<img_3381>\n\nFigure 5: Chet and Turbo equivalence\n\nClaim: $M \\geqslant 2$.\n\nProof. Suppose not, so $M<2$. Say $M=2-2 \\varepsilon$ for some $\\varepsilon>0$ and let $[-1+\\varepsilon, 1-\\varepsilon]$ be a minimal interval, that Chet cannot be forced out of. Then we can force Chet arbitrarily close to $\\pm(1-\\varepsilon)$. In partiular, we can force Chet out of $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$ by minimality of $M$. This means that there exists a sequence $d_{1}, d_{2}, \\ldots$ for which Chet has to leave $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$, which means he ends up either in the interval $\\left[-1+\\varepsilon,-1+\\frac{4}{3} \\varepsilon\\right)$ or in the interval $\\left(1-\\frac{4}{3} \\varepsilon, 1-\\varepsilon\\right]$.\n\nNow consider the sequence,\n\n$$\nd_{1}, 1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon, d_{2}, 1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon, d_{3}, \\ldots\n$$\n\nobtained by adding the sequence $1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon$ in between every two steps. We claim that this sequence forces Chet to leave the larger interval $[-1+\\varepsilon, 1-\\varepsilon]$. Indeed no two consecutive elements in the sequence $1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon$ can have the same sign, because the sum of any two consecutive terms is larger than $2-2 \\varepsilon$ and Chet would leave the interval $[-1+\\varepsilon, 1-\\varepsilon]$. It follows that the $\\left(1-\\frac{7}{6} \\varepsilon\\right)$ 's and the $\\left(1-\\frac{2}{3} \\varepsilon\\right)$ 's cancel out, so the position after $d_{k}$ is the same as before $d_{k+1}$. Hence, the positions after each $d_{k}$ remain the same as in the original sequence. Thus, Chet is also forced to the boundary in the new sequence.\n\nIf Chet is outside the interval $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$, then Chet has to move $1-\\frac{7}{6} \\varepsilon$ towards 0 , and ends in $\\left[-\\frac{1}{6} \\varepsilon, \\frac{1}{6} \\varepsilon\\right]$. Chet then has to move by $1-\\frac{2}{3} \\varepsilon$, which means that he has to leave the interval $[-1+\\varepsilon, 1-\\varepsilon]$. Indeed the absolute value of the final position is at least $1-\\frac{5}{6} \\varepsilon$. This contradicts the assumption, that we cannot force Chet out of $[-1+\\varepsilon, 1-\\varepsilon]$. Hence $M \\geqslant 2$ as needed." ]

[ "$\\frac{1}{2}$" ]

[ "Let the length of $A E=E C$ be $x$.\n\nThen $A F=x-25$.\n\nIn, $\\triangle B C F, \\frac{x+25}{B F}=\\tan \\left(59^{\\circ}\\right)$.\n\nIn $\\triangle A B F, \\frac{x-25}{B F}=\\tan \\left(41^{\\circ}\\right)$.\n\nSolving for $B F$ in these two equations and equating,\n\n$$\nB F=\\frac{x+25}{\\tan 59^{\\circ}}=\\frac{x-25}{\\tan 41^{\\circ}}\n$$\n\nso $\\quad\\left(\\tan 41^{\\circ}\\right)(x+25)=\\left(\\tan 59^{\\circ}\\right)(x-25)$\n\n$$\n\\begin{aligned}\n25\\left(\\tan 59^{\\circ}+\\tan 41^{\\circ}\\right) & =x\\left(\\tan 59^{\\circ}-\\tan 41^{\\circ}\\right) \\\\\nx & =\\frac{25\\left(\\tan 59^{\\circ}+\\tan 41^{\\circ}\\right)}{\\tan 59^{\\circ}-\\tan 41^{\\circ}} \\\\\nx & \\doteq 79.67 .\n\\end{aligned}\n$$\n\n<img_3420>\n\nTherefore the length of $A E$ is 79.67 ." ]

[ "79.67" ]

[ "Join $B Y$. Since $B C$ is a diameter, then $\\angle B Y C=90^{\\circ}$. Since $A B=B C, \\triangle A B C$ is isosceles and $B Y$ is an altitude in $\\triangle A B C$, then $A Y=Y C=15$.\n\nLet $\\angle B A C=\\theta$.\n\nSince $\\triangle A B C$ is isosceles, $\\angle B C A=\\theta$.\n\nSince $B C Y X$ is cyclic, $\\angle B X Y=180-\\theta$ and so $\\angle A X Y=\\theta$.\n\n<img_3475>\n\nThus $\\triangle A X Y$ is isosceles and so $X Y=A Y=15$.\n\nTherefore $X Y=15$.", "Join $B Y . \\angle B Y C=90^{\\circ}$ since it is inscribed in a semicircle.\n\n$\\triangle B A C$ is isosceles, so altitude $B Y$ bisects the base.\n\nTherefore $B Y=\\sqrt{25^{2}-15^{2}}=20$.\n\nJoin $C X . \\angle C X B=90^{\\circ}$ since it is also inscribed in a semicircle.\n\n<img_3739>\n\nThe area of $\\triangle A B C$ is\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(A C)(B Y) & =\\frac{1}{2}(A B)(C X) \\\\\n\\frac{1}{2}(30)(20) & =\\frac{1}{2}(25)(C X) \\\\\nC X & =\\frac{600}{25}=24 .\n\\end{aligned}\n$$\n\nFrom $\\triangle A B Y$ we conclude that $\\cos \\angle A B Y=\\frac{B Y}{A B}=\\frac{20}{25}=\\frac{4}{5}$.\n\nIn $\\Delta B X Y$, applying the Law of Cosines we get $(X Y)^{2}=(B X)^{2}+(B Y)^{2}-2(B X)(B Y) \\cos \\angle X B Y$.\n\nNow (by Pythagoras $\\triangle B X C$ ),\n\n\n\n$$\n\\begin{aligned}\nB X^{2} & =B C^{2}-C X^{2} \\\\\n& =25^{2}-24^{2} \\\\\n& =49 \\\\\nB X & =7 .\n\\end{aligned}\n$$\n\nTherefore $X Y^{2}=7^{2}+20^{2}-2(7)(20) \\frac{4}{5}$\n\n$$\n=49+400-224\n$$\n\n$$\n=225 \\text {. }\n$$\n\nTherefore $X Y=15$." ]

[ "15" ]

[ "Placing the information on the coordinate axes, the diagram is indicated to the right.\n\nWe note that $P$ has coordinates $(a, b)$.\n\nBy symmetry (or congruency) we can label lengths $a$ and $b$ as shown. Thus $Q$ has coordinates $(2-a, 2-b)$.\n\nSince $P D=P Q, a^{2}+b^{2}=(2-2 a)^{2}+(2-2 b)^{2}$\n\n$$\n\\begin{aligned}\n& 3 a^{2}+3 b^{2}-8 a-8 b+8=0 \\\\\n& \\left(a-\\frac{4}{3}\\right)^{2}+\\left(b-\\frac{4}{3}\\right)^{2}=\\frac{8}{9}\n\\end{aligned}\n$$\n\n<img_3483>\n\n$P$ is on a circle with centre $O\\left(\\frac{4}{3}, \\frac{4}{3}\\right)$ with $r=\\frac{2}{3} \\sqrt{2}$.\n\nThe minimum angle for $\\theta$ occurs when $D P$ is tangent to the circle.\n\nSo we have the diagram noted to the right.\n\nSince $O D$ makes an angle of $45^{\\circ}$ with the $x$-axis then $\\angle P D O=45-\\theta$ and $O D=\\frac{4}{3} \\sqrt{2}$.\n\nTherefore $\\sin (45-\\theta)=\\frac{\\frac{2}{3} \\sqrt{2}}{\\frac{4}{3} \\sqrt{2}}=\\frac{1}{2}$ which means $45^{\\circ}-\\theta=30^{\\circ}$ or $\\theta=15^{\\circ}$.\n\nThus the minimum value for $\\theta$ is $15^{\\circ}$.\n\n<img_3977>", "Let $A B=B C=C D=D A=1$.\n\nJoin $D$ to $B$. Let $\\angle A D P=\\theta$. Therefore, $\\angle P D B=45-\\theta$.\n\nLet $P D=a$ and $P B=b$ and $P Q=\\frac{a}{2}$.\n\n\n\nWe now establish a relationship between $a$ and $b$.\n\nIn $\\triangle P D B, b^{2}=a^{2}+2-2(a)(\\sqrt{2}) \\cos (45-\\theta)$\n\n$$\n\\text { or, } \\quad \\cos (45-\\theta)=\\frac{a^{2}-b^{2}+2}{2 \\sqrt{2} a}\n$$\n\n<img_3307>\n\nIn $\\triangle P D R,\\left(\\frac{a}{2}\\right)^{2}=a^{2}+\\left(\\frac{\\sqrt{2}}{2}\\right)^{2}-2 a \\frac{\\sqrt{2}}{2} \\cos (45-\\theta)$\n\nor, $\\cos (45-\\theta)=\\frac{\\frac{3}{4} a^{2}+\\frac{1}{2}}{a \\sqrt{2}}$\n\nComparing (1) and (2) gives, $\\frac{a^{2}-b^{2}+2}{2 \\sqrt{2} a}=\\frac{\\frac{3}{4} a^{2}+\\frac{1}{2}}{a \\sqrt{2}}$.\n\nSimplifying this, $a^{2}+2 b^{2}=2$\n\n$$\n\\text { or, } \\quad b^{2}=\\frac{2-a^{2}}{2}\n$$\n\nNow $\\cos (45-\\theta)=\\frac{a^{2}+2-\\left(\\frac{2-a^{2}}{2}\\right)}{2 a \\sqrt{2}}=\\frac{1}{4 \\sqrt{2}}\\left(3 a+\\frac{2}{a}\\right)$.\n\nNow considering $3 a+\\frac{2}{a}$, we know $\\left(\\sqrt{3 a}-\\sqrt{\\frac{2}{a}}\\right)^{2} \\geq 0$\n\n$$\n\\text { or, } \\quad 3 a+\\frac{2}{a} \\geq 2 \\sqrt{6}\n$$\n\nThus, $\\cos (45-\\theta) \\geq \\frac{1}{4 \\sqrt{2}}(2 \\sqrt{6})=\\frac{\\sqrt{3}}{2}$\n\n$$\n\\cos (45-\\theta) \\geq \\frac{\\sqrt{3}}{2}\n$$\n\n$\\cos (45-\\theta)$ has a minimum value for $45^{\\circ}-\\theta=30^{\\circ}$ or $\\theta=15^{\\circ}$.", "Join $B D$. Let $B D$ meet $P Q$ at $M$. Let $\\angle A D P=\\theta$.\n\nBy interior alternate angles, $\\angle P=\\angle Q$ and $\\angle P D M=\\angle Q B M$.\n\nThus $\\triangle P D M \\cong \\triangle Q B M$ by A.S.A., so $P M=Q M$ and $D M=B M$.\n\nSo $M$ is the midpoint of $B D$ and the centre of the square.\n\n\n\nWithout loss of generality, let $P M=1$. Then $P D=2$.\n\nSince $\\theta+\\alpha=45^{\\circ}$ (see diagram), $\\theta$ will be minimized when $\\alpha$ is maximized.\n\n<img_3749>\n\nConsider $\\triangle P M D$.\n\nUsing the sine law, $\\frac{\\sin \\alpha}{1}=\\frac{\\sin (\\angle P M D)}{2}$.\n\nTo maximize $\\alpha$, we maximize $\\sin \\alpha$.\n\nBut $\\sin \\alpha=\\frac{\\sin (\\angle P M D)}{2}$, so it is maximized when $\\sin (\\angle P M D)=1$.\n\nIn this case, $\\sin \\alpha=\\frac{1}{2}$, so $\\alpha=30^{\\circ}$.\n\nTherefore, $\\theta=45^{\\circ}-30^{\\circ}=15^{\\circ}$, and so the minimum value of $\\theta$ is $15^{\\circ}$.", "We place the diagram on a coordinate grid, with $D(0,0)$, $C(1,0), B(0,1), A(1,1)$.\n\nLet $P D=P Q=Q B=a$, and $\\angle A D P=\\theta$.\n\nDrop a perpendicular from $P$ to $A D$, meeting $A D$ at $X$.\n\nThen $P X=a \\sin \\theta, D X=a \\cos \\theta$.\n\nTherefore the coordinates of $P$ are $(a \\sin \\theta, a \\cos \\theta)$.\n\nSince $P D \\| B Q$, then $\\angle Q B C=\\theta$.\n\nSo by a similar argument (or by using the fact that $P Q$ are symmetric through the centre of the square), the coordinates of $Q$ are $(1-a \\sin \\theta, 1+a \\cos \\theta)$.\n\n<img_3706>\n\nNow $(P Q)^{2}=a^{2}$, so $(1-2 a \\sin \\theta)^{2}+(1-2 a \\cos \\theta)^{2}=a^{2}$\n\n$$\n2+4 a^{2} \\sin ^{2} \\theta+4 a^{2} \\cos ^{2} \\theta-4 a(\\sin \\theta+\\cos \\theta)=a^{2}\n$$\n\n\n\n$$\n\\begin{aligned}\n2+4 a^{2}-a^{2} & =4 a(\\sin \\theta+\\cos \\theta) \\\\\n\\frac{2+3 a^{2}}{4 a} & =\\sin \\theta+\\cos \\theta \\\\\n\\frac{2+3 a^{2}}{4 \\sqrt{2} a} & =\\frac{1}{\\sqrt{2}} \\sin \\theta+\\frac{1}{\\sqrt{2}} \\cos \\theta=\\cos \\left(45^{\\circ}\\right) \\sin \\theta+\\sin \\left(45^{\\circ}\\right) \\cos \\theta \\\\\n\\frac{2+3 a^{2}}{4 \\sqrt{2} a} & =\\sin \\left(\\theta+45^{\\circ}\\right)\n\\end{aligned}\n$$\n\n$$\n\\text { Now } \\begin{aligned}\n\\left(a-\\sqrt{\\frac{2}{3}}\\right)^{2} & \\geq 0 \\\\\na^{2}-2 a \\sqrt{\\frac{2}{3}}+\\frac{2}{3} & \\geq 0 \\\\\n3 a^{2}-2 a \\sqrt{6}+2 & \\geq 0 \\\\\n3 a^{2}+2 & \\geq 2 a \\sqrt{6} \\\\\n\\frac{3 a^{2}+2}{4 \\sqrt{2} a} & \\geq \\frac{\\sqrt{3}}{2}\n\\end{aligned}\n$$\n\nand equality occurs when $a=\\sqrt{\\frac{2}{3}}$.\n\nSo $\\sin \\left(\\theta+45^{\\circ}\\right) \\geq \\frac{\\sqrt{3}}{2}$ and thus since $0^{\\circ} \\leq \\theta \\leq 90^{\\circ}$, then $\\theta+45^{\\circ} \\geq 60^{\\circ}$ or $\\theta \\geq 15^{\\circ}$.\n\nTherefore the minimum possible value of $\\angle A D P$ is $15^{\\circ}$." ]

[ "$15$" ]

[ "By the Pythagorean Theorem in $\\triangle E A D$, we have $E A^{2}+A D^{2}=E D^{2}$ or $12^{2}+A D^{2}=13^{2}$, and so $A D=\\sqrt{169-144}=5$, since $A D>0$.\n\nBy the Pythagorean Theorem in $\\triangle A C D$, we have $A C^{2}+C D^{2}=A D^{2}$ or $A C^{2}+4^{2}=5^{2}$, and so $A C=\\sqrt{25-16}=3$, since $A C>0$.\n\n(We could also have determined the lengths of $A D$ and $A C$ by recognizing 3-4-5 and 5-12-13 right-angled triangles.)\n\nBy the Pythagorean Theorem in $\\triangle A B C$, we have $A B^{2}+B C^{2}=A C^{2}$ or $A B^{2}+2^{2}=3^{2}$, and so $A B=\\sqrt{9-4}=\\sqrt{5}$, since $A B>0$." ]

[ "$\\sqrt{5}$" ]

[ "Join $B$ to $D$.\n\n<img_3655>\n\nConsider $\\triangle C B D$.\n\nSince $C B=C D$, then $\\angle C B D=\\angle C D B=\\frac{1}{2}\\left(180^{\\circ}-\\angle B C D\\right)=\\frac{1}{2}\\left(180^{\\circ}-60^{\\circ}\\right)=60^{\\circ}$.\n\nTherefore, $\\triangle B C D$ is equilateral, and so $B D=B C=C D=6$.\n\nConsider $\\triangle D B A$.\n\nNote that $\\angle D B A=90^{\\circ}-\\angle C B D=90^{\\circ}-60^{\\circ}=30^{\\circ}$.\n\nSince $B D=B A=6$, then $\\angle B D A=\\angle B A D=\\frac{1}{2}\\left(180^{\\circ}-\\angle D B A\\right)=\\frac{1}{2}\\left(180^{\\circ}-30^{\\circ}\\right)=75^{\\circ}$.\n\nWe calculate the length of $A D$.\n\nMethod 1\n\nBy the Sine Law in $\\triangle D B A$, we have $\\frac{A D}{\\sin (\\angle D B A)}=\\frac{B A}{\\sin (\\angle B D A)}$.\n\nTherefore, $A D=\\frac{6 \\sin \\left(30^{\\circ}\\right)}{\\sin \\left(75^{\\circ}\\right)}=\\frac{6 \\times \\frac{1}{2}}{\\sin \\left(75^{\\circ}\\right)}=\\frac{3}{\\sin \\left(75^{\\circ}\\right)}$.\n\nMethod 2\n\nIf we drop a perpendicular from $B$ to $P$ on $A D$, then $P$ is the midpoint of $A D$ since $\\triangle B D A$ is isosceles. Thus, $A D=2 A P$.\n\nAlso, $B P$ bisects $\\angle D B A$, so $\\angle A B P=15^{\\circ}$.\n\nNow, $A P=B A \\sin (\\angle A B P)=6 \\sin \\left(15^{\\circ}\\right)$.\n\nTherefore, $A D=2 A P=12 \\sin \\left(15^{\\circ}\\right)$.\n\nMethod 3\n\nBy the Cosine Law in $\\triangle D B A$,\n\n$$\n\\begin{aligned}\nA D^{2} & =A B^{2}+B D^{2}-2(A B)(B D) \\cos (\\angle A B D) \\\\\n& =6^{2}+6^{2}-2(6)(6) \\cos \\left(30^{\\circ}\\right) \\\\\n& =72-72\\left(\\frac{\\sqrt{3}}{2}\\right) \\\\\n& =72-36 \\sqrt{3}\n\\end{aligned}\n$$\n\nTherefore, $A D=\\sqrt{36(2-\\sqrt{3})}=6 \\sqrt{2-\\sqrt{3}}$ since $A D>0$.", "Drop perpendiculars from $D$ to $Q$ on $B C$ and from $D$ to $R$ on $B A$.\n\n<img_3835>\n\nThen $C Q=C D \\cos (\\angle D C Q)=6 \\cos \\left(60^{\\circ}\\right)=6 \\times \\frac{1}{2}=3$.\n\nAlso, $D Q=C D \\sin (\\angle D C Q)=6 \\sin \\left(60^{\\circ}\\right)=6 \\times \\frac{\\sqrt{3}}{2}=3 \\sqrt{3}$.\n\nSince $B C=6$, then $B Q=B C-C Q=6-3=3$.\n\nNow quadrilateral $B Q D R$ has three right angles, so it must have a fourth right angle and so must be a rectangle.\n\nThus, $R D=B Q=3$ and $R B=D Q=3 \\sqrt{3}$.\n\nSince $A B=6$, then $A R=A B-R B=6-3 \\sqrt{3}$.\n\nSince $\\triangle A R D$ is right-angled at $R$, then using the Pythagorean Theorem and the fact that $A D>0$, we obtain\n\n$$\nA D=\\sqrt{R D^{2}+A R^{2}}=\\sqrt{3^{2}+(6-3 \\sqrt{3})^{2}}=\\sqrt{9+36-36 \\sqrt{3}+27}=\\sqrt{72-36 \\sqrt{3}}\n$$\n\nwhich we can rewrite as $A D=\\sqrt{36(2-\\sqrt{3})}=6 \\sqrt{2-\\sqrt{3}}$." ]

[ "$6\\sqrt{2-\\sqrt{3}}$" ]

[ "We \"complete the rectangle\" by drawing a horizontal line through $C$ which meets the $y$-axis at $P$ and the vertical line through $B$ at $Q$.\n\n<img_3215>\n\n\n\nSince $C$ has $y$-coordinate 5 , then $P$ has $y$-coordinate 5 ; thus the coordinates of $P$ are $(0,5)$.\n\nSince $B$ has $x$-coordinate 4 , then $Q$ has $x$-coordinate 4 .\n\nSince $C$ has $y$-coordinate 5 , then $Q$ has $y$-coordinate 5 .\n\nTherefore, the coordinates of $Q$ are $(4,5)$, and so rectangle $O P Q B$ is 4 by 5 and so has area $4 \\times 5=20$.\n\nNow rectangle $O P Q B$ is made up of four smaller triangles, and so the sum of the areas of these triangles must be 20 .\n\nLet us examine each of these triangles:\n\n- $\\triangle A B C$ has area 8 (given information)\n- $\\triangle A O B$ is right-angled at $O$, has height $A O=3$ and base $O B=4$, and so has area $\\frac{1}{2} \\times 4 \\times 3=6$.\n- $\\triangle A P C$ is right-angled at $P$, has height $A P=5-3=2$ and base $P C=k-0=k$, and so has area $\\frac{1}{2} \\times k \\times 2=k$.\n- $\\triangle C Q B$ is right-angled at $Q$, has height $Q B=5-0=5$ and base $C Q=4-k$, and so has area $\\frac{1}{2} \\times(4-k) \\times 5=10-\\frac{5}{2} k$.\n\nSince the sum of the areas of these triangles is 20 , then $8+6+k+10-\\frac{5}{2} k=20$ or $4=\\frac{3}{2} k$ and so $k=\\frac{8}{3}$." ]

[ "$\\frac{8}{3}$" ]

[ "Since $\\angle H L P=60^{\\circ}$ and $\\angle B L P=30^{\\circ}$, then $\\angle H L B=\\angle H L P-\\angle B L P=30^{\\circ}$.\n\nAlso, since $\\angle H L P=60^{\\circ}$ and $\\angle H P L=90^{\\circ}$, then $\\angle L H P=180^{\\circ}-90^{\\circ}-60^{\\circ}=30^{\\circ}$.\n\n<img_3808>\n\nTherefore, $\\triangle H B L$ is isosceles and $B L=H B=400 \\mathrm{~m}$.\n\nIn $\\triangle B L P, B L=400 \\mathrm{~m}$ and $\\angle B L P=30^{\\circ}$, so $L P=B L \\cos \\left(30^{\\circ}\\right)=400\\left(\\frac{\\sqrt{3}}{2}\\right)=200 \\sqrt{3}$ m.\n\nTherefore, the distance between $L$ and $P$ is $200 \\sqrt{3} \\mathrm{~m}$.", "Since $\\angle H L P=60^{\\circ}$ and $\\angle B L P=30^{\\circ}$, then $\\angle H L B=\\angle H L P-\\angle B L P=30^{\\circ}$.\n\nAlso, since $\\angle H L P=60^{\\circ}$ and $\\angle H P L=90^{\\circ}$, then $\\angle L H P=180^{\\circ}-90^{\\circ}-60^{\\circ}=30^{\\circ}$. Also, $\\angle L B P=60^{\\circ}$.\n\nLet $L P=x$.\n\n<img_3794>\n\nSince $\\triangle B L P$ is $30^{\\circ}-60^{\\circ}-90^{\\circ}$, then $B P: L P=1: \\sqrt{3}$, so $B P=\\frac{1}{\\sqrt{3}} L P=\\frac{1}{\\sqrt{3}} x$.\n\n\n\nSince $\\triangle H L P$ is $30^{\\circ}-60^{\\circ}-90^{\\circ}$, then $H P: L P=\\sqrt{3}: 1$, so $H P=\\sqrt{3} L P=\\sqrt{3} x$.\n\nBut $H P=H B+B P$ so\n\n$$\n\\begin{aligned}\n\\sqrt{3} x & =400+\\frac{1}{\\sqrt{3}} x \\\\\n3 x & =400 \\sqrt{3}+x \\\\\n2 x & =400 \\sqrt{3} \\\\\nx & =200 \\sqrt{3}\n\\end{aligned}\n$$\n\nTherefore, the distance from $L$ to $P$ is $200 \\sqrt{3} \\mathrm{~m}$." ]

[ "$200 \\sqrt{3}$" ]

[ "In order to determine $C D$, we must determine one of the angles (or at least some information about one of the angles) in $\\triangle B C D$.\n\nTo do this, we look at $\\angle A$ use the fact that $\\angle A+\\angle C=180^{\\circ}$.\n\n\n\n<img_3524>\n\nUsing the cosine law in $\\triangle A B D$, we obtain\n\n$$\n\\begin{aligned}\n7^{2} & =5^{2}+6^{2}-2(5)(6) \\cos (\\angle A) \\\\\n49 & =61-60 \\cos (\\angle A) \\\\\n\\cos (\\angle A) & =\\frac{1}{5}\n\\end{aligned}\n$$\n\nSince $\\cos (\\angle A)=\\frac{1}{5}$ and $\\angle A+\\angle C=180^{\\circ}$, then $\\cos (\\angle C)=-\\cos \\left(180^{\\circ}-\\angle A\\right)=-\\frac{1}{5}$.\n\n(We could have calculated the actual size of $\\angle A$ using $\\cos (\\angle A)=\\frac{1}{5}$ and then used this to calculate the size of $\\angle C$, but we would introduce the possibility of rounding error by doing this.)\n\nThen, using the cosine law in $\\triangle B C D$, we obtain\n\n$$\n\\begin{aligned}\n7^{2} & =4^{2}+C D^{2}-2(4)(C D) \\cos (\\angle C) \\\\\n49 & =16+C D^{2}-8(C D)\\left(-\\frac{1}{5}\\right) \\\\\n0 & =5 C D^{2}+8 C D-165 \\\\\n0 & =(5 C D+33)(C D-5)\n\\end{aligned}\n$$\n\nSo $C D=-\\frac{33}{5}$ or $C D=5$. (We could have also determined these roots using the quadratic formula.)\n\nSince $C D$ is a length, it must be positive, so $C D=5$.\n\n(We could have also proceeded by using the sine law in $\\triangle B C D$ to determine $\\angle B D C$ and then found the size of $\\angle D B C$, which would have allowed us to calculate $C D$ using the sine law. However, this would again introduce the potential of rounding error.)" ]

[ "5" ]

[ "From the diagram, the $x$-intercepts of the parabola are $x=-k$ and $x=3 k$.\n\n\n\n<img_3883>\n\nSince we are given that $y=-\\frac{1}{4}(x-r)(x-s)$, then the $x$-intercepts are $r$ and $s$, so $r$ and $s$ equal $-k$ and $3 k$ in some order.\n\nTherefore, we can rewrite the parabola as $y=-\\frac{1}{4}(x-(-k))(x-3 k)$.\n\nSince the point $(0,3 k)$ lies on the parabola, then $3 k=-\\frac{1}{4}(0+k)(0-3 k)$ or $12 k=3 k^{2}$ or $k^{2}-4 k=0$ or $k(k-4)=0$.\n\nThus, $k=0$ or $k=4$.\n\nSince the two roots are distinct, then we cannot have $k=0$ (otherwise both $x$-intercepts would be 0 ).\n\nThus, $k=4$.\n\nThis tells us that the equation of the parabola is $y=-\\frac{1}{4}(x+4)(x-12)$ or $y=-\\frac{1}{4} x^{2}+$ $2 x+12$.\n\nWe still have to determine the coordinates of the vertex, $V$.\n\nSince the $x$-intercepts of the parabola are -4 and 12 , then the $x$-coordinate of the vertex is the average of these intercepts, or 4.\n\n(We could have also used the fact that the $x$-coordinate is $-\\frac{b}{2 a}=-\\frac{2}{2\\left(-\\frac{1}{4}\\right)}$.)\n\nTherefore, the $y$-coordinate of the vertex is $y=-\\frac{1}{4}\\left(4^{2}\\right)+2(4)+12=16$.\n\nThus, the coordinates of the vertex are $(4,16)$." ]

[ "$4,(4,16)$" ]

[ "We proceed directly.\n\nOn the first pass from left to right, Josephine closes all of the even numbered lockers, leaving the odd ones open.\n\nThe second pass proceeds from right to left. Before the pass, the lockers which are open are $1,3, \\ldots, 47,49$.\n\nOn the second pass, she shuts lockers 47, 43, 39, .., 3 .\n\nThe third pass proceeds from left to right. Before the pass, the lockers which are open are $1,5, \\ldots, 45,49$.\n\nOn the third pass, she shuts lockers $5,13, \\ldots, 45$.\n\nThis leaves lockers 1, 9, 17, 25, 33, 41, 49 open.\n\nOn the fourth pass, from right to left, lockers 41, 25 and 9 are shut, leaving 1, 17, 33, 49.\n\nOn the fifth pass, from left to right, lockers 17 and 49 are shut, leaving 1 and 33 open.\n\nOn the sixth pass, from right to left, locker 1 is shut, leaving 33 open.\n\nThus, $f(50)=33$." ]

[ "33" ]

[ "The length of $P Q$ is equal to $\\sqrt{(0-5)^{2}+(12-0)^{2}}=\\sqrt{(-5)^{2}+12^{2}}=13$.\n\nIn a similar way, we can see that $Q R=R S=S P=13$.\n\nTherefore, the perimeter of $P Q R S$ is $4 \\cdot 13=52$.\n\n(We can also see that if $O$ is the origin, then $\\triangle P O Q, \\triangle P O S, \\triangle R O Q$, and $\\triangle R O S$ are congruent because $O Q=O S$ and $O P=O R$, which means that $P Q=Q R=R S=S P$.)" ]

[ "52" ]

[ "Suppose that $B$ has coordinates $(r, s)$ and $C$ has coordinates $(t, u)$.\n\nSince $M(3,9)$ is the midpoint of $A(0,8)$ and $B(r, s)$, then 3 is the average of 0 and $r$ (which gives $r=6)$ and 9 is the average of 8 and $s$ (which gives $s=10$ ).\n\nSince $N(7,6)$ is the midpoint of $B(6,10)$ and $C(t, u)$, then 7 is the average of 6 and $t$ (which gives $t=8$ ) and 6 is the average of 10 and $u$ (which gives $u=2$ ).\n\nThe slope of the line segment joining $A(0,8)$ and $C(8,2)$ is $\\frac{8-2}{0-8}$ which equals $-\\frac{3}{4}$.", "Since $M$ is the midpoint of $A B$ and $N$ is the midpoint of $B C$, then $M N$ is parallel to $A C$. Therefore, the slope of $A C$ equals the slope of the line segment joining $M(3,9)$ to $N(7,6)$, which is $\\frac{9-6}{3-7}$ or $-\\frac{3}{4}$." ]

[ "$-\\frac{3}{4}$" ]

[ "We begin by determining the length of $A B$ in terms of $x$.\n\nSince $A B D E$ is a rectangle, $B D=A E=2 x$.\n\nSince $\\triangle B C D$ is equilateral, $\\angle D B C=60^{\\circ}$.\n\nJoin $A$ to $D$.\n\n<img_3330>\n\nSince $A D$ and $B C$ are parallel, $\\angle A D B=\\angle D B C=60^{\\circ}$.\n\nConsider $\\triangle A D B$. This is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle since $\\angle A B D$ is a right angle.\n\nUsing ratios of side lengths, $\\frac{A B}{B D}=\\frac{\\sqrt{3}}{1}$ and so $A B=\\sqrt{3} B D=2 \\sqrt{3} x$" ]

[ "$2 \\sqrt{3} x$" ]

[ "We begin by determining the length of $A B$ in terms of $x$.\n\nSince $A B D E$ is a rectangle, $B D=A E=2 x$.\n\nSince $\\triangle B C D$ is equilateral, $\\angle D B C=60^{\\circ}$.\n\nJoin $A$ to $D$.\n\n<img_3330>\n\nSince $A D$ and $B C$ are parallel, $\\angle A D B=\\angle D B C=60^{\\circ}$.\n\nConsider $\\triangle A D B$. This is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle since $\\angle A B D$ is a right angle.\n\nUsing ratios of side lengths, $\\frac{A B}{B D}=\\frac{\\sqrt{3}}{1}$ and so $A B=\\sqrt{3} B D=2 \\sqrt{3} x$\n\nNext, we determine $\\frac{A C}{A D}$.\n\nNow, $\\frac{A D}{B D}=\\frac{2}{1}$ and so $A D=2 B D=4 x$.\n\nSuppose that $M$ is the midpoint of $A E$ and $N$ is the midpoint of $B D$.\n\nSince $A E=B D=2 x$, then $A M=M E=B N=N D=x$.\n\nJoin $M$ to $N$ and $N$ to $C$ and $A$ to $C$.\n\n<img_3952>\n\nSince $A B D E$ is a rectangle, then $M N$ is parallel to $A B$ and so $M N$ is perpendicular to both $A E$ and $B D$.\n\nAlso, $M N=A B=2 \\sqrt{3} x$.\n\nSince $\\triangle B C D$ is equilateral, its median $C N$ is perpendicular to $B D$.\n\nSince $M N$ and $N C$ are perpendicular to $B D, M N C$ is actually a straight line segment and so $M C=M N+N C$.\n\nNow $\\triangle B N C$ is also a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, and so $N C=\\sqrt{3} B N=\\sqrt{3} x$.\n\nThis means that $M C=2 \\sqrt{3} x+\\sqrt{3} x=3 \\sqrt{3} x$.\n\n\n\nFinally, $\\triangle A M C$ is right-angled at $M$ and so\n\n$$\nA C=\\sqrt{A M^{2}+M C^{2}}=\\sqrt{x^{2}+(3 \\sqrt{3} x)^{2}}=\\sqrt{x^{2}+27 x^{2}}=\\sqrt{28 x^{2}}=2 \\sqrt{7} x\n$$\n\nsince $x>0$.\n\nThis means that $\\frac{A C}{A D}=\\frac{2 \\sqrt{7} x}{4 x}=\\frac{\\sqrt{7}}{2}=\\sqrt{\\frac{7}{4}}$, which means that the integers $r=7$ and $s=4$ satisfy the conditions." ]

[ "7,4" ]

[ "Suppose that the integers in the first row are, in order, $a, b, c, d, e$.\n\nUsing these, we calculate the integer in each of the boxes below the top row in terms of these variables, using the rule that each integer is the product of the integers in the two boxes above:\n\n$a$\n\n| $b$ | $c$ | $c$ | | $d$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $a b^{2} c$ | $b c$ | $c d$ | $c$ | |\n| | $a b^{3} c^{3} d$ | $b c^{2} d$ | $c d^{2} e$ | |$\\quad d e$\n\nTherefore, $a b^{4} c^{6} d^{4} e=9953280000$.\n\n\n\nNext, we determine the prime factorization of the integer 9953280000 :\n\n$$\n\\begin{aligned}\n9953280000 & =10^{4} \\cdot 995328 \\\\\n& =2^{4} \\cdot 5^{4} \\cdot 2^{3} \\cdot 124416 \\\\\n& =2^{7} \\cdot 5^{4} \\cdot 2^{3} \\cdot 15552 \\\\\n& =2^{10} \\cdot 5^{4} \\cdot 2^{3} \\cdot 1944 \\\\\n& =2^{13} \\cdot 5^{4} \\cdot 2^{3} \\cdot 243 \\\\\n& =2^{16} \\cdot 5^{4} \\cdot 3^{5} \\\\\n& =2^{16} \\cdot 3^{5} \\cdot 5^{4}\n\\end{aligned}\n$$\n\nThus, $a b^{4} c^{6} d^{4} e=2^{16} \\cdot 3^{5} \\cdot 5^{4}$.\n\nSince the right side is not divisible by 7 , none of $a, b, c, d$, $e$ can equal 7 .\n\nThus, $a, b, c, d, e$ are five distinct integers chosen from $\\{1,2,3,4,5,6,8\\}$.\n\nThe only one of these integers divisible by 5 is 5 itself.\n\nSince $2^{16} \\cdot 3^{5} \\cdot 5^{4}$ includes exactly 4 factors of 5 , then either $b=5$ or $d=5$. No other placement of the 5 can give exactly 4 factors of 5 .\n\nCase 1: $b=5$\n\nHere, $a c^{6} d^{4} e=2^{16} \\cdot 3^{5}$ and $a, c, d, e$ are four distinct integers chosen from $\\{1,2,3,4,6,8\\}$. Since $a c^{6} d^{4} e$ includes exactly 5 factors of 3 and the possible values of $a, c, d$, e that are divisible by 3 are 3 and 6 , then either $d=3$ and one of $a$ and $e$ is 6 , or $d=6$ and one of $a$ and $e$ is 3 . No other placements of the multiples of 3 can give exactly 5 factors of 3 .\n\nCase 1a: $b=5, d=3, a=6$\n\nHere, $a \\cdot c^{6} \\cdot d^{4} \\cdot e=6 \\cdot c^{6} \\cdot 3^{4} \\cdot e=2 \\cdot 3^{5} \\cdot c^{6} \\cdot e$.\n\nThis gives $c^{6} e=2^{15}$ and $c$ and $e$ are distinct integers from $\\{1,2,4,8\\}$.\n\nTrying the four possible values of $c$ shows that $c=4$ and $e=8$ is the only solution in this case. Here, $(a, b, c, d, e)=(6,5,4,3,8)$.\n\nCase 1b: $b=5, d=3, e=6$ We obtain $(a, b, c, d, e)=(8,5,4,3,6)$.\n\nCase 1c: $b=5, d=6, a=3$\n\nHere, $a \\cdot c^{6} \\cdot d^{4} \\cdot e=3 \\cdot c^{6} \\cdot 6^{4} \\cdot e=2^{4} \\cdot 3^{5} \\cdot c^{6} \\cdot e$.\n\nThis gives $c^{6} e=2^{12}$ and $c$ and $e$ are distinct integers from $\\{1,2,4,8\\}$.\n\nTrying the four possible values of $c$ shows that $c=4$ and $e=1$ is the only solution in this case. Here, $(a, b, c, d, e)=(3,5,4,6,1)$.\n\nCase 1d: $b=5, d=6, e=3$ We obtain $(a, b, c, d, e)=(1,5,4,6,3)$.\n\nCase 2: $d=5$ : A similar analysis leads to 4 further quintuples $(a, b, c, d, e)$.\n\nTherefore, there are 8 ways in which the integers can be chosen and placed in the top row to obtain the desired integer in the bottom box." ]

[ "8" ]

[ "We label the centres of the outer circles, starting with the circle labelled $Z$ and proceeding clockwise, as $A, B, C, D, E, F, G, H, J$, and $K$, and the centre of the circle labelled $Y$ as $L$.\n\n<img_3893>\n\nJoin $L$ to each of $A, B, C, D, E, F, G, H, J$, and $K$. Join $A$ to $B, B$ to $C, C$ to $D, D$ to $E, E$ to $F, F$ to $G, G$ to $H, H$ to $J, J$ to $K$, and $K$ to $A$.\n\nWhen two circles are tangent, the distance between their centres equals the sum of their radii.\n\nThus,\n\n$$\n\\begin{array}{r}\nB C=C D=D E=E F=F G=G H=H J=J K=2+1=3 \\\\\nB L=D L=F L=H L=K L=2+4=6 \\\\\nC L=E L=G L=J L=1+4=5 \\\\\nA B=A K=r+2 \\\\\nA L=r+4\n\\end{array}\n$$\n\nBy side-side-side congruence, the following triangles are congruent:\n\n$$\n\\triangle B L C, \\triangle D L C, \\triangle D L E, \\triangle F L E, \\triangle F L G, \\triangle H L G, \\triangle H L J, \\triangle K L J\n$$\n\nSimilarly, $\\triangle A L B$ and $\\triangle A L K$ are congruent by side-side-side.\n\nLet $\\angle A L B=\\theta$ and let $\\angle B L C=\\alpha$.\n\n\n\nBy congruent triangles, $\\angle A L K=\\theta$ and\n\n$$\n\\angle B L C=\\angle D L C=\\angle D L E=\\angle F L E=\\angle F L G=\\angle H L G=\\angle H L J=\\angle K L J=\\alpha\n$$\n\nThe angles around $L$ add to $360^{\\circ}$ and so $2 \\theta+8 \\alpha=360^{\\circ}$ which gives $\\theta+4 \\alpha=180^{\\circ}$ and so $\\theta=180^{\\circ}-4 \\alpha$.\n\nSince $\\theta=180^{\\circ}-4 \\alpha$, then $\\cos \\theta=\\cos \\left(180^{\\circ}-4 \\alpha\\right)=-\\cos 4 \\alpha$.\n\nConsider $\\triangle A L B$ and $\\triangle B L C$.\n\n<img_3240>\n\nBy the cosine law in $\\triangle A L B$,\n\n$$\n\\begin{aligned}\nA B^{2} & =A L^{2}+B L^{2}-2 \\cdot A L \\cdot B L \\cdot \\cos \\theta \\\\\n(r+2)^{2} & =(r+4)^{2}+6^{2}-2(r+4)(6) \\cos \\theta \\\\\n12(r+4) \\cos \\theta & =r^{2}+8 r+16+36-r^{2}-4 r-4 \\\\\n\\cos \\theta & =\\frac{4 r+48}{12(r+4)} \\\\\n\\cos \\theta & =\\frac{r+12}{3 r+12}\n\\end{aligned}\n$$\n\nBy the cosine law in $\\triangle B L C$,\n\n$$\n\\begin{aligned}\nB C^{2} & =B L^{2}+C L^{2}-2 \\cdot B L \\cdot C L \\cdot \\cos \\alpha \\\\\n3^{2} & =6^{2}+5^{2}-2(6)(5) \\cos \\alpha \\\\\n60 \\cos \\alpha & =36+25-9 \\\\\n\\cos \\alpha & =\\frac{52}{60} \\\\\n\\cos \\alpha & =\\frac{13}{15}\n\\end{aligned}\n$$\n\nSince $\\cos \\alpha=\\frac{13}{15}$, then\n\n$$\n\\begin{aligned}\n\\cos 2 \\alpha & =2 \\cos ^{2} \\alpha-1 \\\\\n& =2 \\cdot \\frac{169}{225}-1 \\\\\n& =\\frac{338}{225}-\\frac{225}{225} \\\\\n& =\\frac{113}{225}\n\\end{aligned}\n$$\n\n\n\nand\n\n$$\n\\begin{aligned}\n\\cos 4 \\alpha & =2 \\cos ^{2} 2 \\alpha-1 \\\\\n& =2 \\cdot \\frac{113^{2}}{225^{2}}-1 \\\\\n& =\\frac{25538}{50625}-\\frac{50625}{50625} \\\\\n& =-\\frac{25087}{50625}\n\\end{aligned}\n$$\n\nFinally,\n\n$$\n\\begin{aligned}\n\\cos \\theta & =-\\cos 4 \\alpha \\\\\n\\frac{r+12}{3 r+12} & =\\frac{25087}{50625} \\\\\n\\frac{r+12}{r+4} & =\\frac{25087}{16875} \\\\\n\\frac{(r+4)+8}{r+4} & =\\frac{25087}{16875} \\\\\n1+\\frac{8}{r+4} & =\\frac{25087}{16875} \\\\\n\\frac{8}{r+4} & =\\frac{8212}{16875} \\\\\n\\frac{2}{r+4} & =\\frac{2053}{16875} \\\\\n\\frac{r+4}{2} & =\\frac{16875}{2053} \\\\\nr+4 & =\\frac{33750}{2053} \\\\\nr & =\\frac{25538}{2053}\n\\end{aligned}\n$$\n\nTherefore, the positive integers $s=25538$ and $t=2053$ satisfy the required conditions." ]

[ "$25538$,$2053$" ]

[ "We are told that when $a, b$ and $c$ are the numbers in consecutive sectors, then $b=a c$. This means that if $a$ and $b$ are the numbers in consecutive sectors, then the number in the next sector is $c=\\frac{b}{a}$. (That is, each number is equal to the previous number divided by the one before that.)\n\nStarting with the given 2 and 3 and proceeding clockwise, we obtain\n\n$$\n2,3, \\quad \\frac{3}{2}, \\frac{3 / 2}{3}=\\frac{1}{2}, \\frac{1 / 2}{3 / 2}=\\frac{1}{3}, \\frac{1 / 3}{1 / 2}=\\frac{2}{3}, \\frac{2 / 3}{1 / 3}=2, \\frac{2}{2 / 3}=3, \\quad \\frac{3}{2}, \\ldots\n$$\n\nAfter the first 6 terms, the first 2 terms ( 2 and 3) reappear, and so the first 6 terms will repeat again. (This is because each term comes from the previous two terms, so when two consecutive terms reappear, then the following terms are the same as when these two consecutive terms appeared earlier.)\n\nSince there are 36 terms in total, then the 6 terms repeat exactly $\\frac{36}{6}=6$ times.\n\nTherefore, the sum of the 36 numbers is $6\\left(2+3+\\frac{3}{2}+\\frac{1}{2}+\\frac{1}{3}+\\frac{2}{3}\\right)=6(2+3+2+1)=48$." ]

[ "48" ]

[ "Join $B E$.\n\n<img_3698>\n\nSince $\\triangle F B D$ is congruent to $\\triangle A E C$, then $F B=A E$.\n\nSince $\\triangle F A B$ and $\\triangle A F E$ are each right-angled, share a common side $A F$ and have equal hypotenuses $(F B=A E)$, then these triangles are congruent, and so $A B=F E$.\n\nNow $B A F E$ has two right angles at $A$ and $F$ (so $A B$ and $F E$ are parallel) and has equal sides $A B=F E$ so must be a rectangle.\n\nThis means that $B C D E$ is also a rectangle.\n\nNow the diagonals of a rectangle partition it into four triangles of equal area. (Diagonal $A E$ of the rectangle splits the rectangle into two congruent triangles, which have equal area. The diagonals bisect each other, so the four smaller triangles all have equal area.) Since $\\frac{1}{4}$ of rectangle $A B E F$ is shaded and $\\frac{1}{4}$ of rectangle $B C D E$ is shaded, then $\\frac{1}{4}$ of the total area is shaded. (If the area of $A B E F$ is $x$ and the area of $B C D E$ is $y$, then the total shaded area is $\\frac{1}{4} x+\\frac{1}{4} y$, which is $\\frac{1}{4}$ of the total area $x+y$.)\n\nSince $A C=200$ and $C D=50$, then the area of rectangle $A C D F$ is $200(50)=10000$, so the total shaded area is $\\frac{1}{4}(10000)=2500$." ]

[ "2500" ]

[ "Suppose that $M$ is the midpoint of $Y Z$.\n\nSuppose that the centre of the smaller circle is $O$ and the centre of the larger circle is $P$. Suppose that the smaller circle touches $X Y$ at $C$ and $X Z$ at $D$, and that the larger circle touches $X Y$ at $E$ and $X Z$ at $F$.\n\nJoin $O C, O D$ and $P E$.\n\nSince $O C$ and $P E$ are radii that join the centres of circles to points of tangency, then $O C$ and $P E$ are perpendicular to $X Y$.\n\nJoin $X M$. Since $\\triangle X Y Z$ is isosceles, then $X M$ (which is a median by construction) is an altitude (that is, $X M$ is perpendicular to $Y Z)$ and an angle bisector (that is, $\\angle M X Y=\\angle M X Z$ ).\n\nNow $X M$ passes through $O$ and $P$. (Since $X C$ and $X D$ are tangents from $X$ to the same circle, then $X C=X D$. This means that $\\triangle X C O$ is congruent to $\\triangle X D O$ by side-side-side. This means that $\\angle O X C=\\angle O X D$ and so $O$ lies on the angle bisector of $\\angle C X D$, and so $O$ lies on $X M$. Using a similar argument, $P$ lies on $X M$.)\n\n<img_3625>\nDraw a perpendicular from $O$ to $T$ on $P E$. Note that $O T$ is parallel to $X Y$ (since each is perpendicular to $P E$ ) and that $O C E T$ is a rectangle (since it has three right angles).\n\nConsider $\\triangle X M Y$ and $\\triangle O T P$.\n\nEach triangle is right-angled (at $M$ and at $T$ ).\n\nAlso, $\\angle Y X M=\\angle P O T$. (This is because $O T$ is parallel to $X Y$, since both are perpendicular to $P E$.)\n\nTherefore, $\\triangle X M Y$ is similar to $\\triangle O T P$.\n\nThus, $\\frac{X Y}{Y M}=\\frac{O P}{P T}$.\n\nNow $X Y=a$ and $Y M=\\frac{1}{2} b$.\n\nAlso, $O P$ is the line segment joining the centres of two tangent circles, so $O P=r+R$.\n\nLastly, $P T=P E-E T=R-r$, since $P E=R, E T=O C=r$, and $O C E T$ is a rectangle. Therefore,\n\n$$\n\\begin{aligned}\n\\frac{a}{b / 2} & =\\frac{R+r}{R-r} \\\\\n\\frac{2 a}{b} & =\\frac{R+r}{R-r} \\\\\n2 a(R-r) & =b(R+r) \\\\\n2 a R-b R & =2 a r+b r \\\\\nR(2 a-b) & =r(2 a+b) \\\\\n\\frac{R}{r} & =\\frac{2 a+b}{2 a-b} \\quad(\\text { since } 2 a>b \\text { so } 2 a-b \\neq 0, \\text { and } r>0)\n\\end{aligned}\n$$\n\nTherefore, $\\frac{R}{r}=\\frac{2 a+b}{2 a-b}$." ]

[ "$\\frac{2 a+b}{2 a-b}$" ]

[ "Because all of the angles in the figure are right angles, then $B C=D E=4$.\n\nThus, we can break up the figure into a 4 by 8 rectangle and a 4 by 4 square, by extending $B C$ to hit $F E$. Therefore, the area of the figure is $(8)(4)+(4)(4)=48$." ]

[ "48" ]

[ "By the Pythagorean Theorem in triangle $A B E$, $A B^{2}=15^{2}+20^{2}=625$, so $A B=25$.\n\nSince $A B C D$ is a rectangle, $C D=A B=25$, so by the Pythagorean Theorem in triangle $C F D$, we have $625=25^{2}=24^{2}+C F^{2}$, so $C F^{2}=625-576=49$, or $C F=7$." ]

[ "7" ]

[ "Since $A B C D$ is a square of side length 6 and each of $A E: E B, B F: F C, C G: G D$, and $D H: H A$ is equal to $1: 2$, then $A E=B F=C G=D H=2$ and $E B=F C=G D=H A=4$.\n\nThus, each of the triangles $H A E, E B F, F C G$, and $G D H$ is right-angled, with one leg of length 2 and the other of length 4.\n\nThen the area of $E F G H$ is equal to the area of square $A B C D$ minus the combined area of the four triangles, or $6^{2}-4\\left[\\frac{1}{2}(2)(4)\\right]=36-16=20$ square units.", "Since $A B C D$ is a square of side length 6 and each of $A E: E B, B F: F C, C G: G D$, and $D H: H A$ is equal to $1: 2$, then $A E=B F=C G=D H=2$ and $E B=F C=G D=H A=4$.\n\nThus, each of the triangles $H A E, E B F, F C G$, and $G D H$ is right-angled, with one leg of length 2 and the other of length 4.\n\nBy the Pythagorean Theorem,\n\n$E F=F G=G H=H E=\\sqrt{2^{2}+4^{2}}=\\sqrt{20}$.\n\nSince the two triangles $H A E$ and $E B F$ are congruent (we know the lengths of all three sides of each), then $\\angle A H E=\\angle B E F$. But $\\angle A H E+\\angle A E H=90^{\\circ}$, so $\\angle B E F+\\angle A E H=90^{\\circ}$, so $\\angle H E F=90^{\\circ}$.\n\nIn a similar way, we can show that each of the four angles of $E F G H$ is a right-angle, and so $E F G H$ is a square of side length $\\sqrt{20}$.\n\nTherefore, the area of $E F G H$ is $(\\sqrt{20})^{2}=20$ square units." ]

[ "20" ]

[ "When line $A$ with equation $y=2 x$ is reflected in the $y$-axis, the resulting line (line $B$ ) has equation $y=-2 x$. (Reflecting a line in the $y$-axis changes the sign of the slope.)\n\nSince the slope of line $B$ is -2 and line $C$ is perpendicular to line $B$, then the slope of line $C$ is $\\frac{1}{2}$ (the slopes of perpendicular lines are negative reciprocals)." ]

[ "$\\frac{1}{2}$" ]

[ "Consider the line through $O$ and $P$. To get from $O$ to $P$, we go right 2 and up 1. Since $B$ lies on this line and to get from $O$ to $B$ we go over 1, then we must go up $\\frac{1}{2}$, to keep the ratio constant.\n\nConsider the line through $O$ and $Q$. To get from $O$ to $Q$, we go right 3 and up 1. Since $A$ lies on this line and to get from $O$ to $A$ we go over 1, then we must go up $\\frac{1}{3}$, to keep the ratio constant.\n\nTherefore, since $A$ and $B$ lie on the same vertical line, then $A B=\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$.", "Since the line through $P$ passes through the origin, then its equation is of the form $y=m x$. Since it passes through the point $(2,1)$, then $1=2 m$, so the line has equation $y=\\frac{1}{2} x$. Since $B$ has $x$-coordinate 1, then $y=\\frac{1}{2}(1)=\\frac{1}{2}$, so $B$ has coordinates $\\left(1, \\frac{1}{2}\\right)$. Similarly, we can determine that the equation of the line through $Q$ is $y=\\frac{1}{3} x$, and so $A$ has coordinates $\\left(1, \\frac{1}{3}\\right)$.\n\nTherefore, since $A$ and $B$ lie on the same vertical line, then $A B=\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$." ]

[ "$\\frac{1}{6}$" ]

[ "Point $P$ is the point where the line $y=3 x+3$ crosses the $x$ axis, and so has coordinates $(-1,0)$.\n\nTherefore, one of the roots of the parabola $y=x^{2}+t x-2$ is $x=-1$, so\n\n$$\n\\begin{aligned}\n0 & =(-1)^{2}+t(-1)-2 \\\\\n0 & =1-t-2 \\\\\nt & =-1\n\\end{aligned}\n$$\n\nThe parabola now has equation $y=x^{2}-x-2=(x+1)(x-2)$ (we already knew one of the roots so this helped with the factoring) and so its two $x$-intercepts are -1 and 2 , ie. $P$ has coordinates $(-1,0)$ and $Q$ has coordinates $(2,0)$.\n\nWe now have to find the coordinates of the point $R$. We know that $R$ is one of the two points of intersection of the line and the parabola, so we equate their equations:\n\n$$\n\\begin{aligned}\n3 x+3 & =x^{2}-x-2 \\\\\n0 & =x^{2}-4 x-5 \\\\\n0 & =(x+1)(x-5)\n\\end{aligned}\n$$\n\n(Again, we already knew one of the solutions to this equation $(x=-1)$ so this made factoring easier.) Since $R$ does not have $x$-coordinate -1 , then $R$ has $x$-coordinate $x=5$. Since $R$ lies on the line, then $y=3(5)+3=18$, so $R$ has coordinates $(5,18)$.\n\nWe can now calculate the area of triangle $P Q R$. This triangle has base of length 3 (from $P$ to $Q$ ) and height of length 18 (from the $x$-axis to $R$ ), and so has area $\\frac{1}{2}(3)(18)=27$.\n\nThus, $t=-1$ and the area of triangle $P Q R$ is 27 ." ]

[ "-1,27" ]

[ "We first calculate the length of $A C$ using the cosine law:\n\n$$\n\\begin{aligned}\nA C^{2} & =7^{2}+8^{2}-2(7)(8) \\cos \\left(120^{\\circ}\\right) \\\\\nA C^{2} & =49+64-112\\left(-\\frac{1}{2}\\right) \\\\\nA C^{2} & =169 \\\\\nA C & =13\n\\end{aligned}\n$$\n\nSince triangle $A B C$ is right-angled and isosceles, then $x=A B=\\sqrt{2}(A C)=13 \\sqrt{2}$." ]

[ "$13 \\sqrt{2}$" ]

[ "First, we note that a triangle with one right angle and one angle with measure $45^{\\circ}$ is isosceles.\n\nThis is because the measure of the third angle equals $180^{\\circ}-90^{\\circ}-45^{\\circ}=45^{\\circ}$ which means that the triangle has two equal angles.\n\nIn particular, $\\triangle C D E$ is isosceles with $C D=D E$ and $\\triangle E F G$ is isosceles with $E F=F G$. Since $D E=E F=1 \\mathrm{~m}$, then $C D=F G=1 \\mathrm{~m}$.\n\nJoin $C$ to $G$.\n\n<img_3379>\n\nConsider quadrilateral $C D F G$. Since the angles at $D$ and $F$ are right angles and since $C D=G F$, it must be the case that $C D F G$ is a rectangle.\n\nThis means that $C G=D F=2 \\mathrm{~m}$ and that the angles at $C$ and $G$ are right angles.\n\nSince $\\angle C G F=90^{\\circ}$ and $\\angle D C G=90^{\\circ}$, then $\\angle B G C=180^{\\circ}-90^{\\circ}-45^{\\circ}=45^{\\circ}$ and $\\angle B C G=90^{\\circ}$.\n\nThis means that $\\triangle B C G$ is also isosceles with $B C=C G=2 \\mathrm{~m}$.\n\nFinally, $B D=B C+C D=2 \\mathrm{~m}+1 \\mathrm{~m}=3 \\mathrm{~m}$." ]

[ "3" ]

[ "After each round, each L shape is divided into 4 smaller $\\mathrm{L}$ shapes.\n\nThis means that the number of $\\mathrm{L}$ shapes increases by a factor of 4 after each round.\n\nAfter 1 round, there are $4 \\mathrm{~L}$ shapes.\n\nAfter 2 rounds, there are $4^{2}=16$ L's of the smallest size.\n\nAfter 3 rounds, there are $4^{3}=64$ L's of the smallest size." ]

[ "64" ]

[ "We note that $B D=B C+C D$ and that $B C=20 \\mathrm{~cm}$, so we need to determine $C D$.\n\nWe draw a line from $C$ to $P$ on $F D$ so that $C P$ is perpendicular to $D F$.\n\nSince $A C$ and $D F$ are parallel, then $C P$ is also perpendicular to $A C$.\n\nThe distance between $A C$ and $D F$ is $4 \\mathrm{~cm}$, so $C P=4 \\mathrm{~cm}$.\n\nSince $\\triangle A B C$ is isosceles and right-angled, then $\\angle A C B=45^{\\circ}$.\n\n<img_3723>\n\nThus, $\\angle P C D=180^{\\circ}-\\angle A C B-\\angle P C A=180^{\\circ}-45^{\\circ}-90^{\\circ}=45^{\\circ}$.\n\nSince $\\triangle C P D$ is right-angled at $P$ and $\\angle P C D=45^{\\circ}$, then $\\triangle C P D$ is also an isosceles right-angled triangle.\n\nTherefore, $C D=\\sqrt{2} C P=4 \\sqrt{2} \\mathrm{~cm}$.\n\nFinally, $B D=B C+C D=(20+4 \\sqrt{2}) \\mathrm{cm}$." ]

[ "$(20+4 \\sqrt{2})$" ]

[ "We use the cosine law in $\\triangle A B D$ to determine the length of $B D$ :\n\n$$\nB D^{2}=A B^{2}+A D^{2}-2(A B)(A D) \\cos (\\angle B A D)\n$$\n\nWe are given that $A B=75$ and $A D=20$, so we need to determine $\\cos (\\angle B A D)$.\n\nNow\n\n$$\n\\begin{aligned}\n\\cos (\\angle B A D) & =\\cos (\\angle B A C+\\angle E A D) \\\\\n& =\\cos (\\angle B A C) \\cos (\\angle E A D)-\\sin (\\angle B A C) \\sin (\\angle E A D) \\\\\n& =\\frac{A C}{A B} \\frac{A D}{A E}-\\frac{B C}{A B} \\frac{E D}{A E}\n\\end{aligned}\n$$\n\nsince $\\triangle A B C$ and $\\triangle A D E$ are right-angled.\n\nSince $A B=75$ and $B C=21$, then by the Pythagorean Theorem,\n\n$$\nA C=\\sqrt{A B^{2}-B C^{2}}=\\sqrt{75^{2}-21^{2}}=\\sqrt{5625-441}=\\sqrt{5184}=72\n$$\n\nsince $A C>0$.\n\nSince $A C=72$ and $C E=47$, then $A E=A C-C E=25$.\n\nSince $A E=25$ and $A D=20$, then by the Pythagorean Theorem,\n\n$$\nE D=\\sqrt{A E^{2}-A D^{2}}=\\sqrt{25^{2}-20^{2}}=\\sqrt{625-400}=\\sqrt{225}=15\n$$\n\nsince $E D>0$.\n\nTherefore,\n\n$$\n\\cos (\\angle B A D)=\\frac{A C}{A B} \\frac{A D}{A E}-\\frac{B C}{A B} \\frac{E D}{A E}=\\frac{72}{75} \\frac{20}{25}-\\frac{21}{75} \\frac{15}{25}=\\frac{1440-315}{75(25)}=\\frac{1125}{75(25)}=\\frac{45}{75}=\\frac{3}{5}\n$$\n\n\n\nFinally,\n\n$$\n\\begin{aligned}\nB D^{2} & =A B^{2}+A D^{2}-2(A B)(A D) \\cos (\\angle B A D) \\\\\n& =75^{2}+20^{2}-2(75)(20)\\left(\\frac{3}{5}\\right) \\\\\n& =5625+400-1800 \\\\\n& =4225\n\\end{aligned}\n$$\n\nSince $B D>0$, then $B D=\\sqrt{4225}=65$, as required." ]

[ "65" ]

[ "The centre of the circle is $(3,0)$ and the circle has a radius of 5.\n\nThus $\\sqrt{d^{2}+3^{2}}=5$\n\n$$\n\\begin{aligned}\n& d^{2}=5^{2}-3^{2} \\\\\n& d^{2}=16\n\\end{aligned}\n$$\n\nTherefore $d=4$, since $d>0$.", "Since $A B$ is a diameter of the circle, $\\angle A D B=90^{\\circ}$ and $\\angle A O D=90^{\\circ}$.\n\n$\\triangle A D O \\sim \\triangle D B O$\n\nTherefore, $\\frac{O D}{A O}=\\frac{B O}{O D}$\n\nand $d^{2}=2(8)$\n\n$$\n\\begin{aligned}\nd^{2} & =16 \\\\\nd & =4, \\text { since } d>0 .\n\\end{aligned}\n$$", "$\\angle A D B=\\angle A O D=\\angle B O D=90^{\\circ}$\n\nIn $\\triangle A O D, A D^{2}=4+d^{2}$.\n\nIn $\\triangle B O D, D B^{2}=64+d^{2}$.\n\nIn $\\Delta A D B,\\left(4+d^{2}\\right)+\\left(64+d^{2}\\right)=100$\n\n$$\n\\begin{aligned}\n2 d^{2} & =32 \\\\\nd & =4, d>0\n\\end{aligned}\n$$" ]

[ "4" ]

[ "Since the side length of the square is $x, T S=x-3$ and $V S=x-5$\n\nArea of triangle $A=\\frac{1}{2}(3)(x)$.\n\nArea of triangle $B=\\frac{1}{2}(5)(x)$\n\nArea of triangle $C=\\frac{1}{2}(x-5)(x-3)$.\n\nFrom the given information, $\\frac{1}{2}(3 x)+\\frac{1}{2}(5 x)=\\frac{1}{2}(x-5)(x-3)$. Labelled diagram\n\n$3 x+5 x=x^{2}-8 x+15$\n\n$x^{2}-16 x+15=0$\n\n<img_3393>\n\nThus $x=15$ or $x=1$.\n\nTherefore $x=15$ since $x=1$ is inadmissible." ]

[ "15" ]

[ "Let $D B=10$.\n\nTherefore, $D C=A D=6$.\n\nBy the theorem of Pythagoras, $B C^{2}=10^{2}-6^{2}=64$.\n\nTherefore, $B C=8$.\n\n\n\nThus, $\\tan \\angle A B C=\\frac{12}{8}=\\frac{3}{2}$." ]

[ "$\\frac{3}{2}$" ]

[ "$\\tan \\angle A O C=\\frac{3.39}{5.43}$\n\n$\\angle A O C=\\tan ^{-1}\\left(\\frac{3.39}{5.43}\\right)=31.97^{\\circ}$\n\nThe arc length $\\overparen{A C}=\\frac{31.97}{360^{\\circ}}[(2 \\pi)(6.40)]=3.57$ units\n\nThe distance is approximately $3570 \\mathrm{~km}$." ]

[ "3570" ]

[ "The parabola $y=-x^{2}+4$ has vertex $P(0,4)$ and intersects the $x$-axis at $A(-2,0)$ and $B(2,0)$. The intercept $B(2,0)$ has its pre-image, $B^{\\prime}$ on the parabola $y=-x^{2}+4$. To find $B^{\\prime}$, we find the point of intersection of the line passing through $B(2,0)$, with slope 1 , and the parabola $y=-x^{2}+4$.\n\nThe equation of the line is $y=x-2$.\n\nIntersection points, $x-2=-x^{2}+4$\n\n$$\n\\begin{array}{r}\nx^{2}+x-6=0 \\\\\n(x+3)(x-2)=0 .\n\\end{array}\n$$\n\nTherefore, $x=-3$ or $x=2$.\n\nFor $x=-3, y=-3-2=-5$. Thus $B^{\\prime}$ has coordinates $(-3,-5)$.\n\nIf $(-3,-5) \\rightarrow(2,0)$ then the required general translation mapping $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $(x, y) \\rightarrow(x+5, y+5)$.\n\nPossibility 1\n\nUsing the general translation, we find the coordinates of $Q$ to be, $P(0,4) \\rightarrow Q(0+5,4+5)=Q(5,9)$.\n\nIf $C$ is the reflection of $B$ in the axis of symmetry of the parabola, i.e. $x=5, C$ has coordinates $(8,0)$.\n\nPossibility 2\n\nIf $B^{\\prime}$ has coordinates $(-3,-5)$ then $C^{\\prime}$ is the reflection of $B^{\\prime}$ in the $y$-axis. Thus $C^{\\prime}$ has coordinates $(3,-5)$.\n\nIf we apply the general translation then $C$ has coordinates $(3+5,-5+5)$ or $(8,0)$.\n\nThus $C$ has coordinates $(8,0)$.\n\nPossibility 3\n\nUsing the general translation, we find the coordinates of $Q$ to be, $P(0,4) \\rightarrow Q(0+5,4+5)=Q(5,9)$.\n\nThe equation of the image parabola is $y=-(x-5)^{2}+9$.\n\n\n\nTo find its intercepts, $-(x-5)^{2}+9=0$\n\n$$\n\\begin{aligned}\n(x-5)^{2} & =9 \\\\\nx-5 & = \\pm 3 .\n\\end{aligned}\n$$\n\nTherefore $x=8$ or $x=2$.\n\nThus $C$ has coordinates $(8,0)$.", "The translation moving the parabola with equation $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $T(t, t)$ because the slope of the line $y=x+4$ is 1 .\n\nThe pre-image of $B^{\\prime}$ is $(2-t,-t)$.\n\nSince $B^{\\prime}$ is on the parabola with vertex $P$, we have\n\n$$\n\\begin{aligned}\n-t & =-(2-t)^{2}+4 \\\\\n-t & =-4+4 t-t^{2}+4 \\\\\nt^{2}-5 t & =0 \\\\\nt(t-5) & =0\n\\end{aligned}\n$$\n\nTherefore, $t=0$ or $t=5$.\n\nThus $B^{\\prime}$ is $(-3,-5)$.\n\nLet $C$ have coordinates $(c, 0)$.\n\nThe pre-image of $C$ is $(c-5,-5)$.\n\nTherefore, $-5=-(c-5)^{2}+4$.\n\nOr, $(c-5)^{2}=9$.\n\nTherefore $c-5=3$ or $c-5=-3$.\n\n$$\nc=8 \\text { or } \\quad c=2\n$$\n\nThus $C$ has coordinates $(8,0)$." ]

[ "$(8,0)$" ]

[ "Using the tangent properties of a circle, the lengths of line segments are as shown on the diagram.\n\nArea of trapezoid $A B C D=\\frac{1}{2}(8)(B C+A D)$\n\n$$\n\\begin{aligned}\n& =4(2 b+2 x-2 b) \\\\\n& =8 x .\n\\end{aligned}\n$$\n\n<img_3854>\n\nThus, $8 x=80$.\n\nTherefore, $x=10$." ]

[ "10" ]

[ "Since $\\angle O P B=90^{\\circ}$, then $O P$ and $P B$ are perpendicular, so the product of their slopes is -1 .\n\nThe slope of $O P$ is $\\frac{4-0}{p-0}=\\frac{4}{p}$ and the slope of $P B$ is $\\frac{4-0}{p-10}=\\frac{4}{p-10}$.\n\nTherefore, we need\n\n$$\n\\begin{aligned}\n\\frac{4}{p} \\cdot \\frac{4}{p-10} & =-1 \\\\\n16 & =-p(p-10) \\\\\np^{2}-10 p+16 & =0 \\\\\n(p-2)(p-8) & =0\n\\end{aligned}\n$$\n\nand so $p=2$ or $p=8$. Since each these steps is reversible, then $\\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$.", "Since $\\triangle O P B$ is right-angled at $P$, then $O P^{2}+P B^{2}=O B^{2}$ by the Pythagorean Theorem. Note that $O B=10$ since $O$ has coordinates $(0,0)$ and $B$ has coordinates $(10,0)$.\n\nAlso, $O P^{2}=(p-0)^{2}+(4-0)^{2}=p^{2}+16$ and $P B^{2}=(10-p)^{2}+(4-0)^{2}=p^{2}-20 p+116$. Therefore,\n\n$$\n\\begin{aligned}\n\\left(p^{2}+16\\right)+\\left(p^{2}-20 p+116\\right) & =10^{2} \\\\\n2 p^{2}-20 p+32 & =0 \\\\\np^{2}-10 p+16 & =0\n\\end{aligned}\n$$\n\n\n\nand so $(p-2)(p-8)=0$, or $p=2$ or $p=8$. Since each these steps is reversible, then $\\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$." ]

[ "2,8" ]

[ "In a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, the ratio of the side opposite the $90^{\\circ}$ to the side opposite the $60^{\\circ}$ angle is $2: \\sqrt{3}$.\n\nNote that each of $\\triangle A B C, \\triangle A C D, \\triangle A D E, \\triangle A E F, \\triangle A F G$, and $\\triangle A G H$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nTherefore, $\\frac{A H}{A G}=\\frac{A G}{A F}=\\frac{A F}{A E}=\\frac{A E}{A D}=\\frac{A D}{A C}=\\frac{A C}{A B}=\\frac{2}{\\sqrt{3}}$.\n\nThus, $A H=\\frac{2}{\\sqrt{3}} A G=\\left(\\frac{2}{\\sqrt{3}}\\right)^{2} A F=\\left(\\frac{2}{\\sqrt{3}}\\right)^{3} A E=\\left(\\frac{2}{\\sqrt{3}}\\right)^{4} A D=\\left(\\frac{2}{\\sqrt{3}}\\right)^{5} A C=\\left(\\frac{2}{\\sqrt{3}}\\right)^{6} A B$.\n\n(In other words, to get from $A B=1$ to the length of $A H$, we multiply by the \"scaling factor\" $\\frac{2}{\\sqrt{3}}$ six times.)\n\nTherefore, $A H=\\left(\\frac{2}{\\sqrt{3}}\\right)^{6}=\\frac{64}{27}$." ]

[ "$\\frac{64}{27}$" ]

[ "Since $\\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,\n\n$$\nA D=\\sqrt{A F^{2}+F D^{2}}=\\sqrt{4^{2}+2^{2}}=\\sqrt{20}=2 \\sqrt{5}\n$$\n\nsince $A D>0$.\n\nLet $\\angle F A D=\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B A F=90^{\\circ}-\\beta$.\n\nSince $\\triangle A F D$ is right-angled at $F$, then $\\angle A D F=90^{\\circ}-\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B D C=90^{\\circ}-\\left(90^{\\circ}-\\beta\\right)=\\beta$.\n\n\n\n<img_3979>\n\nTherefore, $\\triangle B F A, \\triangle A F D$, and $\\triangle D F E$ are all similar as each is right-angled and has either an angle of $\\beta$ or an angle of $90^{\\circ}-\\beta$ (and hence both of these angles).\n\nTherefore, $\\frac{A B}{A F}=\\frac{D A}{D F}$ and so $A B=\\frac{4(2 \\sqrt{5})}{2}=4 \\sqrt{5}$.\n\nAlso, $\\frac{F E}{F D}=\\frac{F D}{F A}$ and so $F E=\\frac{2(2)}{4}=1$.\n\nSince $A B C D$ is a rectangle, then $B C=A D=2 \\sqrt{5}$, and $D C=A B=4 \\sqrt{5}$.\n\nFinally, the area of quadrilateral $B C E F$ equals the area of $\\triangle D C B$ minus the area $\\triangle D F E$. Thus, the required area is\n\n$$\n\\frac{1}{2}(D C)(C B)-\\frac{1}{2}(D F)(F E)=\\frac{1}{2}(4 \\sqrt{5})(2 \\sqrt{5})-\\frac{1}{2}(2)(1)=20-1=19\n$$", "Since $\\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,\n\n$$\nA D=\\sqrt{A F^{2}+F D^{2}}=\\sqrt{4^{2}+2^{2}}=\\sqrt{20}=2 \\sqrt{5}\n$$\n\nsince $A D>0$.\n\nLet $\\angle F A D=\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B A F=90^{\\circ}-\\beta$. Since $\\triangle B A F$ is right-angled at $F$, then $\\angle A B F=\\beta$.\n\nSince $\\triangle A F D$ is right-angled at $F$, then $\\angle A D F=90^{\\circ}-\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B D C=90^{\\circ}-\\left(90^{\\circ}-\\beta\\right)=\\beta$.\n\n<img_3412>\n\nLooking at $\\triangle A F D$, we see that $\\sin \\beta=\\frac{F D}{A D}=\\frac{2}{2 \\sqrt{5}}=\\frac{1}{\\sqrt{5}}, \\cos \\beta=\\frac{A F}{A D}=\\frac{4}{2 \\sqrt{5}}=\\frac{2}{\\sqrt{5}}$, and $\\tan \\beta=\\frac{F D}{A F}=\\frac{2}{4}=\\frac{1}{2}$.\n\nSince $A F=4$ and $\\angle A B F=\\beta$, then $A B=\\frac{A F}{\\sin \\beta}=\\frac{4}{\\frac{1}{\\sqrt{5}}}=4 \\sqrt{5}$.\n\nSince $F D=2$ and $\\angle F D E=\\beta$, then $F E=F D \\tan \\beta=2 \\cdot \\frac{1}{2}=1$.\n\nSince $A B C D$ is a rectangle, then $B C=A D=2 \\sqrt{5}$, and $D C=A B=4 \\sqrt{5}$.\n\nFinally, the area of quadrilateral $E F B C$ equals the area of $\\triangle D C B$ minus the area $\\triangle D F E$. Thus, the required area is\n\n$$\n\\frac{1}{2}(D C)(C B)-\\frac{1}{2}(D F)(F E)=\\frac{1}{2}(4 \\sqrt{5})(2 \\sqrt{5})-\\frac{1}{2}(2)(1)=20-1=19\n$$" ]

[ "19" ]

[ "Suppose $\\angle P A R=x^{\\circ}$ and $\\angle Q D R=y^{\\circ}$.\n\n<img_3198>\n\nSince $P R$ and $P A$ are radii of the larger circle, then $\\triangle P A R$ is isosceles.\n\nThus, $\\angle P R A=\\angle P A R=x^{\\circ}$.\n\nSince $Q D$ and $Q R$ are radii of the smaller circle, then $\\triangle Q R D$ is isosceles.\n\nThus, $\\angle Q R D=\\angle Q D R=y^{\\circ}$.\n\nIn $\\triangle A R D$, the sum of the angles is $180^{\\circ}$, so $x^{\\circ}+\\left(x^{\\circ}+40^{\\circ}+y^{\\circ}\\right)+y^{\\circ}=180^{\\circ}$ or $2 x+2 y=140$ or $x+y=70$.\n\nTherefore, $\\angle C P D=x^{\\circ}+40^{\\circ}+y^{\\circ}=(x+y+40)^{\\circ}=110^{\\circ}$." ]

[ "$110$" ]

[ "To get from $P$ to $Q$, we move 3 units right and 4 units up.\n\nSince $P Q=Q R$ and $R$ lies on the line through $Q$, then we must use the same motion to get from $Q$ to $R$.\n\nTherefore, to get from $Q(0,4)$ to $R$, we move 3 units right and 4 units up, so the coordinates of $R$ are $(3,8)$.", "The line through $P(-3,0)$ and $Q(0,4)$ has slope $\\frac{4-0}{0-(-3)}=\\frac{4}{3}$ and $y$-intercept 4 , so has equation $y=\\frac{4}{3} x+4$.\n\nThus, $R$ has coordinates $\\left(a, \\frac{4}{3} a+4\\right)$ for some $a>0$.\n\nSince $P Q=Q R$, then $P Q^{2}=Q R^{2}$, so\n\n$$\n\\begin{aligned}\n(-3)^{2}+4^{2} & =a^{2}+\\left(\\frac{4}{3} a+4-4\\right)^{2} \\\\\n25 & =a^{2}+\\frac{16}{9} a^{2} \\\\\n\\frac{25}{9} a^{2} & =25 \\\\\na^{2} & =9\n\\end{aligned}\n$$\n\nso $a=3$ since $a>0$.\n\nThus, $R$ has coordinates $\\left(3, \\frac{4}{3}(3)+4\\right)=(3,8)$." ]

[ "(3,8)" ]

[ "Since $O P=9$, then the coordinates of $P$ are $(9,0)$.\n\nSince $O P=9$ and $O A=15$, then by the Pythagorean Theorem,\n\n$$\nA P^{2}=O A^{2}-O P^{2}=15^{2}-9^{2}=144\n$$\n\nso $A P=12$.\n\nSince $P$ has coordinates $(9,0)$ and $A$ is 12 units directly above $P$, then $A$ has coordinates $(9,12)$.\n\nSince $P B=4$, then $B$ has coordinates $(13,0)$.\n\nThe line through $A(9,12)$ and $B(13,0)$ has slope $\\frac{12-0}{9-13}=-3$ so, using the point-slope form, has equation $y-0=-3(x-13)$ or $y=-3 x+39$." ]

[ "$y=-3 x+39$" ]

[ "Since $\\cos (\\angle B A C)=\\frac{A B}{A C}$ and $\\cos (\\angle B A C)=\\frac{5}{13}$ and $A B=10$, then $A C=\\frac{13}{5} A B=26$.\n\nSince $\\triangle A B C$ is right-angled at $B$, then by the Pythagorean Theorem, $B C^{2}=A C^{2}-A B^{2}=26^{2}-10^{2}=576$ so $B C=24$ since $B C>0$.\n\nTherefore, $\\tan (\\angle A C B)=\\frac{A B}{B C}=\\frac{10}{24}=\\frac{5}{12}$." ]

[ "$\\frac{5}{12}$" ]

[ "Since $\\triangle A B C$ is isosceles and right-angled, then $\\angle B A C=45^{\\circ}$.\n\nAlso, $A C=\\sqrt{2} A B=\\sqrt{2}(2 \\sqrt{2})=4$.\n\nSince $\\angle E A B=75^{\\circ}$ and $\\angle B A C=45^{\\circ}$, then $\\angle C A E=\\angle E A B-\\angle B A C=30^{\\circ}$.\n\nSince $\\triangle A E C$ is right-angled and has a $30^{\\circ}$ angle, then $\\triangle A E C$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nThus, $E C=\\frac{1}{2} A C=2$ (since $E C$ is opposite the $30^{\\circ}$ angle) and $A E=\\frac{\\sqrt{3}}{2} A C=2 \\sqrt{3}$ (since $A E$ is opposite the $60^{\\circ}$ angle).\n\nIn $\\triangle C D E, E D=D C$ and $\\angle E D C=60^{\\circ}$, so $\\triangle C D E$ is equilateral.\n\nTherefore, $E D=C D=E C=2$.\n\nOverall, the perimeter of $A B C D E$ is\n\n$$\nA B+B C+C D+D E+E A=2 \\sqrt{2}+2 \\sqrt{2}+2+2+2 \\sqrt{3}=4+4 \\sqrt{2}+2 \\sqrt{3}\n$$" ]

[ "$4+4 \\sqrt{2}+2 \\sqrt{3}$" ]

[ "From the diagram, the parabola has $x$-intercepts $x=3$ and $x=-3$.\n\nTherefore, the equation of the parabola is of the form $y=a(x-3)(x+3)$ for some real number $a$.\n\nTriangle $A B C$ can be considered as having base $A B$ (of length $3-(-3)=6$ ) and height $O C$ (where $O$ is the origin).\n\nSuppose $C$ has coordinates $(0,-c)$. Then $O C=c$.\n\nThus, the area of $\\triangle A B C$ is $\\frac{1}{2}(A B)(O C)=3 c$. But we know that the area of $\\triangle A B C$ is 54 , so $3 c=54$ or $c=18$.\n\nSince the parabola passes through $C(0,-18)$, then this point must satisfy the equation of the parabola.\n\nTherefore, $-18=a(0-3)(0+3)$ or $-18=-9 a$ or $a=2$.\n\nThus, the equation of the parabola is $y=2(x-3)(x+3)=2 x^{2}-18$.", "Triangle $A B C$ can be considered as having base $A B$ (of length $3-(-3)=6$ ) and height $O C$ (where $O$ is the origin).\n\nSuppose $C$ has coordinates $(0,-c)$. Then $O C=c$.\n\nThus, the area of $\\triangle A B C$ is $\\frac{1}{2}(A B)(O C)=3 c$. But we know that the area of $\\triangle A B C$ is 54 , so $3 c=54$ or $c=18$.\n\nTherefore, the parabola has vertex $C(0,-18)$, so has equation $y=a(x-0)^{2}-18$.\n\n(The vertex of the parabola must lie on the $y$-axis since its roots are equally distant from the $y$-axis, so $C$ must be the vertex.)\n\nSince the parabola passes through $B(3,0)$, then these coordinates satisfy the equation, so $0=3^{2} a-18$ or $9 a=18$ or $a=2$.\n\nTherefore, the equation of the parabola is $y=2 x^{2}-18$." ]

[ "$y=2 x^{2}-18$" ]

[ "$\\triangle A O B$ is right-angled at $O$, so has area $\\frac{1}{2}(A O)(O B)=\\frac{1}{2} a(1)=\\frac{1}{2} a$.\n\nWe next need to calculate the area of $\\triangle B C D$.\n\nMethod 1: Completing the trapezoid\n\nDrop a perpendicular from $C$ to $P(3,0)$ on the $x$-axis.\n\n<img_3403>\n\nThen $D O P C$ is a trapezoid with parallel sides $D O$ of length 1 and $P C$ of length 2 and height $O P$ (which is indeed perpendicular to the parallel sides) of length 3.\n\nThe area of the trapezoid is thus $\\frac{1}{2}(D O+P C)(O P)=\\frac{1}{2}(1+2)(3)=\\frac{9}{2}$.\n\nBut the area of $\\triangle B C D$ equals the area of trapezoid $D O P C$ minus the areas of $\\triangle D O B$ and $\\triangle B P C$.\n\n$\\triangle D O B$ is right-angled at $O$, so has area $\\frac{1}{2}(D O)(O B)=\\frac{1}{2}(1)(1)=\\frac{1}{2}$.\n\n$\\triangle B P C$ is right-angled at $P$, so has area $\\frac{1}{2}(B P)(P C)=\\frac{1}{2}(2)(2)=2$.\n\nThus, the area of $\\triangle D B C$ is $\\frac{9}{2}-\\frac{1}{2}-2=2$.\n\n\n\n(A similar method for calculating the area of $\\triangle D B C$ would be to drop a perpendicular to $Q$ on the $y$-axis, creating a rectangle $Q O P C$.)\n\nMethod 2: $\\triangle D B C$ is right-angled\n\nThe slope of line segment $D B$ is $\\frac{1-0}{0-1}=-1$.\n\nThe slope of line segment $B C$ is $\\frac{2-0}{3-1}=1$.\n\nSince the product of these slopes is -1 (that is, their slopes are negative reciprocals), then $D B$ and $B C$ are perpendicular.\n\nTherefore, the area of $\\triangle D B C$ is $\\frac{1}{2}(D B)(B C)$.\n\nNow $D B=\\sqrt{(1-0)^{2}+(0-1)^{2}}=\\sqrt{2}$ and $B C=\\sqrt{(3-1)^{2}+(2-0)^{2}}=\\sqrt{8}$.\n\nThus, the area of $\\triangle D B C$ is $\\frac{1}{2} \\sqrt{2} \\sqrt{8}=2$.\n\nSince the area of $\\triangle A O B$ equals the area of $\\triangle D B C$, then $\\frac{1}{2} a=2$ or $a=4$." ]

[ "4" ]

[ "Suppose that $O$ is the centre of the planet, $H$ is the place where His Highness hovers in the helicopter, and $P$ is the furthest point on the surface of the planet that he can see.\n\n<img_3899>\n\nThen $H P$ must be a tangent to the surface of the planet (otherwise he could see further), so $O P$ (a radius) is perpendicular to $H P$ (a tangent).\n\nWe are told that $O P=24 \\mathrm{~km}$.\n\nSince the helicopter hovers at a height of $2 \\mathrm{~km}$, then $O H=24+2=26 \\mathrm{~km}$.\n\nTherefore, $H P^{2}=O H^{2}-O P^{2}=26^{2}-24^{2}=100$, so $H P=10 \\mathrm{~km}$.\n\nTherefore, the distance to the furthest point that he can see is $10 \\mathrm{~km}$." ]

[ "10" ]

[ "Since we know the measure of $\\angle A D B$, then to find the distance $A B$, it is enough to find the distances $A D$ and $B D$ and then apply the cosine law.\n\nIn $\\triangle D B E$, we have $\\angle D B E=180^{\\circ}-20^{\\circ}-70^{\\circ}=90^{\\circ}$, so $\\triangle D B E$ is right-angled, giving $B D=100 \\cos \\left(20^{\\circ}\\right) \\approx 93.969$.\n\nIn $\\triangle D A C$, we have $\\angle D A C=180^{\\circ}-50^{\\circ}-45^{\\circ}=85^{\\circ}$.\n\nUsing the sine law, $\\frac{A D}{\\sin \\left(50^{\\circ}\\right)}=\\frac{C D}{\\sin \\left(85^{\\circ}\\right)}$, so $A D=\\frac{150 \\sin \\left(50^{\\circ}\\right)}{\\sin \\left(85^{\\circ}\\right)} \\approx 115.346$.\n\n\n\nFinally, using the cosine law in $\\triangle A B D$, we get\n\n$$\n\\begin{aligned}\nA B^{2} & =A D^{2}+B D^{2}-2(A D)(B D) \\cos (\\angle A D B) \\\\\nA B^{2} & \\approx(115.346)^{2}+(93.969)^{2}-2(115.346)(93.969) \\cos \\left(35^{\\circ}\\right) \\\\\nA B^{2} & \\approx 4377.379 \\\\\nA B & \\approx 66.16\n\\end{aligned}\n$$\n\nTherefore, the distance from $A$ to $B$ is approximately $66 \\mathrm{~m}$." ]

[ "66" ]

[ "We consider placing the three coins individually.\n\nPlace one coin randomly on the grid.\n\nWhen the second coin is placed (in any one of 15 squares), 6 of the 15 squares will leave two coins in the same row or column and 9 of the 15 squares will leave the two coins in different rows and different columns.\n\n<img_3640>\n\nTherefore, the probability that the two coins are in different rows and different columns is $\\frac{9}{15}=\\frac{3}{5}$.\n\nThere are 14 possible squares in which the third coin can be placed.\n\n\n\nOf these 14 squares, 6 lie in the same row or column as the first coin and an additional 4 lie the same row or column as the second coin. Therefore, the probability that the third coin is placed in a different row and a different column than each of the first two coins is $\\frac{4}{14}=\\frac{2}{7}$.\n\nTherefore, the probability that all three coins are placed in different rows and different columns is $\\frac{3}{5} \\times \\frac{2}{7}=\\frac{6}{35}$." ]

[ "$\\frac{6}{35}$" ]

[ "Suppose that $A B=c, A C=b$ and $B C=a$.\n\nSince $D G$ is parallel to $A C, \\angle B D G=\\angle B A C$ and $\\angle D G B=\\angle A C B$, so $\\triangle D G B$ is similar to $\\triangle A C B$.\n\n(Similarly, $\\triangle A E D$ and $\\triangle E C F$ are also both similar to $\\triangle A B C$.)\n\nSuppose next that $D B=k c$, with $0<k<1$.\n\nThen the ratio of the side lengths of $\\triangle D G B$ to those of $\\triangle A C B$ will be $k: 1$, so $B G=k a$ and $D G=k b$.\n\nSince the ratio of the side lengths of $\\triangle D G B$ to $\\triangle A C B$ is $k: 1$, then the ratio of their areas will be $k^{2}: 1$, so the area of $\\triangle D G B$ is $k^{2}$ (since the area of $\\triangle A C B$ is 1 ).\n\nSince $A B=c$ and $D B=k c$, then $A D=(1-k) c$, so using similar triangles as before, $D E=(1-k) a$ and $A E=(1-k) b$. Also, the area of $\\triangle A D E$ is $(1-k)^{2}$.\n\nSince $A C=b$ and $A E=(1-k) b$, then $E C=k b$, so again using similar triangles, $E F=k c$, $F C=k a$ and the area of $\\triangle E C F$ is $k^{2}$.\n\nNow the area of trapezoid $D E F G$ is the area of the large triangle minus the combined areas of the small triangles, or $1-k^{2}-k^{2}-(1-k)^{2}=2 k-3 k^{2}$.\n\nWe know that $k \\geq 0$ by its definition. Also, since $G$ is to the left of $F$, then $B G+F C \\leq B C$ or $k a+k a \\leq a$ or $2 k a \\leq a$ or $k \\leq \\frac{1}{2}$.\n\nLet $f(k)=2 k-3 k^{2}$.\n\nSince $f(k)=-3 k^{2}+2 k+0$ is a parabola opening downwards, its maximum occurs at its vertex, whose $k$-coordinate is $k=-\\frac{2}{2(-3)}=\\frac{1}{3}$ (which lies in the admissible range for $k$ ).\n\nNote that $f\\left(\\frac{1}{3}\\right)=\\frac{2}{3}-3\\left(\\frac{1}{9}\\right)=\\frac{1}{3}$.\n\nTherefore, the maximum area of the trapezoid is $\\frac{1}{3}$." ]

[ "$\\frac{1}{3}$" ]

[ "Using the cosine law in $\\triangle P Q R$,\n\n$$\n\\begin{aligned}\nP R^{2} & =P Q^{2}+Q R^{2}-2 \\cdot P Q \\cdot Q R \\cdot \\cos (\\angle P Q R) \\\\\n21^{2} & =a^{2}+b^{2}-2 a b \\cos \\left(60^{\\circ}\\right) \\\\\n441 & =a^{2}+b^{2}-2 a b \\cdot \\frac{1}{2} \\\\\n441 & =a^{2}+b^{2}-a b\n\\end{aligned}\n$$\n\nUsing the sine law in $\\triangle S T U$, we obtain $\\frac{S T}{\\sin (\\angle T U S)}=\\frac{T U}{\\sin (\\angle T S U)}$ and so $\\frac{a}{4 / 5}=\\frac{b}{\\sin \\left(30^{\\circ}\\right)}$. Therefore, $\\frac{a}{4 / 5}=\\frac{b}{1 / 2}$ and so $a=\\frac{4}{5} \\cdot 2 b=\\frac{8}{5} b$.\n\nSubstituting into the previous equation,\n\n$$\n\\begin{aligned}\n& 441=\\left(\\frac{8}{5} b\\right)^{2}+b^{2}-\\left(\\frac{8}{5} b\\right) b \\\\\n& 441=\\frac{64}{25} b^{2}+b^{2}-\\frac{8}{5} b^{2} \\\\\n& 441=\\frac{64}{25} b^{2}+\\frac{25}{25} b^{2}-\\frac{40}{25} b^{2} \\\\\n& 441=\\frac{49}{25} b^{2} \\\\\n& 225=b^{2}\n\\end{aligned}\n$$\n\nSince $b>0$, then $b=15$ and so $a=\\frac{8}{5} b=\\frac{8}{5} \\cdot 15=24$." ]

[ "$24,15$" ]

[ "We make two copies of the given triangle, labelling them $\\triangle A B C$ and $\\triangle D E F$, as shown:\n<img_3909>\n\nThe combined area of these two triangles is $2 \\cdot 770 \\mathrm{~cm}^{2}=1540 \\mathrm{~cm}^{2}$, and the shaded area in each triangle is the same.\n\nNext, we rotate $\\triangle D E F$ by $180^{\\circ}$ :\n<img_3429>\n\nand join the two triangles together:\n\n<img_3440>\n\nWe note that $B C$ and $A E$ (which was $F E$ ) are equal in length (since they were copies of each other) and parallel (since they are $180^{\\circ}$ rotations of each other). The same is true for $A B$ and $E C$.\n\nTherefore, $A B C E$ is a parallelogram.\n\nFurther, $A B C E$ is divided into 11 identical parallelograms (6 shaded and 5 unshaded) by the horizontal lines. (Since the sections of the two triangles are equal in height, the horizontal lines on both sides of $A C$ align.)\n\nThe total area of parallelogram $A B C E$ is $1540 \\mathrm{~cm}^{2}$.\n\nThus, the shaded area of $A B C E$ is $\\frac{6}{11} \\cdot 1540 \\mathrm{~cm}^{2}=840 \\mathrm{~cm}^{2}$.\n\nSince this shaded area is equally divided between the two halves of the parallelogram, then the combined area of the shaded regions of $\\triangle A B C$ is $\\frac{1}{2} \\cdot 840 \\mathrm{~cm}^{2}=420 \\mathrm{~cm}^{2}$.", "We label the points where the horizontal lines touch $A B$ and $A C$ as shown:\n\n<img_3141>\n\nWe use the notation $|\\triangle A B C|$ to represent the area of $\\triangle A B C$ and use similar notation for the area of other triangles and quadrilaterals.\n\nLet $\\mathcal{A}$ be equal to the total area of the shaded regions.\n\nThus,\n\n$$\n\\mathcal{A}=\\left|\\triangle A B_{1} C_{1}\\right|+\\left|B_{2} B_{3} C_{3} C_{2}\\right|+\\left|B_{4} B_{5} C_{5} C_{4}\\right|+\\left|B_{6} B_{7} C_{7} C_{6}\\right|+\\left|B_{8} B_{9} C_{9} C_{8}\\right|+\\left|B_{10} B C C_{10}\\right|\n$$\n\nThe area of each of these quadrilaterals is equal to the difference of the area of two triangles. For example,\n\n$$\n\\left|B_{2} B_{3} C_{3} C_{2}\\right|=\\left|\\triangle A B_{3} C_{3}\\right|-\\left|\\triangle A B_{2} C_{2}\\right|=-\\left|\\triangle A B_{2} C_{2}\\right|+\\left|\\triangle A B_{3} C_{3}\\right|\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\mathcal{A}=\\mid & \\triangle A B_{1} C_{1}|-| \\triangle A B_{2} C_{2}|+| \\triangle A B_{3} C_{3}|-| \\triangle A B_{4} C_{4}|+| \\triangle A B_{5} C_{5} \\mid \\\\\n& \\quad-\\left|\\triangle A B_{6} C_{6}\\right|+\\left|\\triangle A B_{7} C_{7}\\right|-\\left|\\triangle A B_{8} C_{8}\\right|+\\left|\\triangle A B_{9} C_{9}\\right|-\\left|\\triangle A B_{10} C_{10}\\right|+|\\triangle A B C|\n\\end{aligned}\n$$\n\nEach of $\\triangle A B_{1} C_{1}, \\triangle A B_{2} C_{2}, \\ldots, \\triangle A B_{10} C_{10}$ is similar to $\\triangle A B C$ because their two base angles are equal due.\n\nSuppose that the height of $\\triangle A B C$ from $A$ to $B C$ is $h$.\n\nSince the height of each of the 11 regions is equal in height, then the height of $\\triangle A B_{1} C_{1}$ is $\\frac{1}{11} h$, the height of $\\triangle A B_{2} C_{2}$ is $\\frac{2}{11} h$, and so on.\n\nWhen two triangles are similar, their heights are in the same ratio as their side lengths:\n\nTo see this, suppose that $\\triangle P Q R$ is similar to $\\triangle S T U$ and that altitudes are drawn from $P$ and $S$ to $V$ and $W$.\n<img_3919>\n\nSince $\\angle P Q R=\\angle S T U$, then $\\triangle P Q V$ is similar to $\\triangle S T W$ (equal angle; right angle), which means that $\\frac{P Q}{S T}=\\frac{P V}{S W}$. In other words, the ratio of sides is equal to the ratio of heights.\n\nSince the height of $\\triangle A B_{1} C_{1}$ is $\\frac{1}{11} h$, then $B_{1} C_{1}=\\frac{1}{11} B C$.\n\nTherefore, $\\left|\\triangle A B_{1} C_{1}\\right|=\\frac{1}{2} \\cdot B_{1} C_{1} \\cdot \\frac{1}{11} h=\\frac{1}{2} \\cdot \\frac{1}{11} B C \\cdot \\frac{1}{11} h=\\frac{1^{2}}{11^{2}} \\cdot \\frac{1}{2} \\cdot B C \\cdot h=\\frac{1^{2}}{11^{2}}|\\triangle A B C|$.\n\nSimilarly, since the height of $\\triangle A B_{2} C_{2}$ is $\\frac{2}{11} h$, then $B_{2} C_{2}=\\frac{2}{11} B C$.\n\n\n\nTherefore, $\\left|\\triangle A B_{2} C_{2}\\right|=\\frac{1}{2} \\cdot B_{2} C_{2} \\cdot \\frac{2}{11} h=\\frac{1}{2} \\cdot \\frac{2}{11} B C \\cdot \\frac{2}{11} h=\\frac{2^{2}}{11^{2}} \\cdot \\frac{1}{2} \\cdot B C \\cdot h=\\frac{2^{2}}{11^{2}}|\\triangle A B C|$.\n\nThis result continues for each of the triangles.\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\mathcal{A} & =\\frac{1^{2}}{11^{2}}|\\triangle A B C|-\\frac{2^{2}}{11^{2}}|\\triangle A B C|+\\frac{3^{2}}{11^{2}}|\\triangle A B C|-\\frac{4^{2}}{11^{2}}|\\triangle A B C|+\\frac{5^{2}}{11^{2}}|\\triangle A B C| \\\\\n& \\quad-\\frac{6^{2}}{11^{2}}|\\triangle A B C|+\\frac{7^{2}}{11^{2}}|\\triangle A B C|-\\frac{8^{2}}{11^{2}}|\\triangle A B C|+\\frac{9^{2}}{11^{2}}|\\triangle A B C|-\\frac{10^{2}}{11^{2}}|\\triangle A B C|+\\frac{11^{2}}{11^{2}}|\\triangle A B C| \\\\\n& =\\frac{1}{11^{2}}|\\triangle A B C|\\left(11^{2}-10^{2}+9^{2}-8^{2}+7^{2}-6^{2}+5^{2}-4^{2}+3^{2}-2^{2}+1\\right) \\\\\n& =\\frac{1}{11^{2}}\\left(770 \\mathrm{~cm}^{2}\\right)((11+10)(11-10)+(9+8)(9-8)+\\cdots+(3+2)(3-2)+1) \\\\\n& =\\frac{1}{11^{2}}\\left(770 \\mathrm{~cm}^{2}\\right)(11+10+9+8+7+6+5+4+3+2+1) \\\\\n& =\\frac{1}{11}\\left(70 \\mathrm{~cm}^{2}\\right) \\cdot 66 \\\\\n& =420 \\mathrm{~cm}^{2}\n\\end{aligned}\n$$\n\nTherefore, the combined area of the shaded regions of $\\triangle A B C$ is $420 \\mathrm{~cm}^{2}$." ]

[ "$420$" ]

[ "We label five additional points in the diagram:\n\n<img_3827>\n\nSince $P Q=Q R=R S=1$, then $P S=3$ and $P R=2$.\n\nSince $\\angle P S T=90^{\\circ}$, then $P T=\\sqrt{P S^{2}+S T^{2}}=\\sqrt{3^{2}+1^{2}}=\\sqrt{10}$ by the Pythagorean Theorem.\n\nWe are told that $A B C D$ is a square.\n\nThus, $P T$ is perpendicular to $Q C$ and to $R B$.\n\nThus, $\\triangle P D Q$ is right-angled at $D$ and $\\triangle P A R$ is right-angled at $A$.\n\nSince $\\triangle P D Q, \\triangle P A R$ and $\\triangle P S T$ are all right-angled and all share an angle at $P$, then these three triangles are similar.\n\nThis tells us that $\\frac{P A}{P S}=\\frac{P R}{P T}$ and so $P A=\\frac{3 \\cdot 2}{\\sqrt{10}}$. Also, $\\frac{P D}{P S}=\\frac{P Q}{P T}$ and so $P D=\\frac{1 \\cdot 3}{\\sqrt{10}}$.\n\nTherefore,\n\n$$\nD A=P A-P D=\\frac{6}{\\sqrt{10}}-\\frac{3}{\\sqrt{10}}=\\frac{3}{\\sqrt{10}}\n$$\n\nThis means that the area of square $A B C D$ is equal to $D A^{2}=\\left(\\frac{3}{\\sqrt{10}}\\right)^{2}=\\frac{9}{10}$.", "We add coordinates to the diagram as shown:\n\n<img_3409>\n\nWe determine the side length of square $A B C D$ by determining the coordinates of $D$ and $A$ and then calculating the distance between these points.\n\nThe slope of the line through $(0,3)$ and $(3,2)$ is $\\frac{3-2}{0-3}=-\\frac{1}{3}$.\n\nThis equation of this line can be written as $y=-\\frac{1}{3} x+3$.\n\nThe slope of the line through $(0,0)$ and $(1,3)$ is 3.\n\nThe equation of this line can be written as $y=3 x$.\n\nThe slope of the line through $(1,0)$ and $(2,3)$ is also 3.\n\nThe equation of this line can be written as $y=3(x-1)=3 x-3$.\n\nPoint $D$ is the intersection point of the lines with equations $y=-\\frac{1}{3} x+3$ and $y=3 x$.\n\nEquating expressions for $y$, we obtain $-\\frac{1}{3} x+3=3 x$ and so $\\frac{10}{3} x=3$ which gives $x=\\frac{9}{10}$.\n\nSince $y=3 x$, we get $y=\\frac{27}{10}$ and so the coordinates of $D$ are $\\left(\\frac{9}{10}, \\frac{27}{10}\\right)$.\n\nPoint $A$ is the intersection point of the lines with equations $y=-\\frac{1}{3} x+3$ and $y=3 x-3$.\n\nEquating expressions for $y$, we obtain $-\\frac{1}{3} x+3=3 x-3$ and so $\\frac{10}{3} x=6$ which gives $x=\\frac{18}{10}$.\n\nSince $y=3 x-3$, we get $y=\\frac{24}{10}$ and so the coordinates of $A$ are $\\left(\\frac{18}{10}, \\frac{24}{10}\\right)$. (It is easier to not reduce these fractions.)\n\nTherefore,\n\n$$\nD A=\\sqrt{\\left(\\frac{9}{10}-\\frac{18}{10}\\right)^{2}+\\left(\\frac{27}{10}-\\frac{24}{10}\\right)^{2}}=\\sqrt{\\left(-\\frac{9}{10}\\right)^{2}+\\left(\\frac{3}{10}\\right)^{2}}=\\sqrt{\\frac{90}{100}}=\\sqrt{\\frac{9}{10}}\n$$\n\nThis means that the area of square $A B C D$ is equal to $D A^{2}=\\left(\\sqrt{\\frac{9}{10}}\\right)^{2}=\\frac{9}{10}$." ]

[ "$\\frac{9}{10}$" ]

[ "Suppose that $k$ is a positive integer.\n\nSuppose that $t$ people are seated at a table with $6 k+5$ chairs so that the table is full.\n\nWhen $t$ people are seated, there are $t$ gaps. Each gap consists of either 1 or 2 chairs. (A gap with 3 or more chairs can have an additional person seated in it, so the table is not full.)\n\nTherefore, there are between $t$ and $2 t$ empty chairs.\n\nThis means that the total number of chairs is between $t+t$ and $t+2 t$.\n\nIn other words, $2 t \\leq 6 k+5 \\leq 3 t$.\n\nSince $2 t \\leq 6 k+5$, then $t \\leq 3 k+\\frac{5}{2}$. Since $k$ and $t$ are integers, then $t \\leq 3 k+2$.\n\nWe note that it is possible to seat $3 k+2$ people around the table in seats\n\n$$\n\\{2,4,6, \\ldots, 6 k+2,6 k+4\\}\n$$\n\nThis table is full becase $3 k+1$ of the gaps consist of 1 chair and 1 gap consists of 2 chairs. Since $3 t \\geq 6 k+5$, then $t \\geq 2 k+\\frac{5}{3}$. Since $k$ and $t$ are integers, then $t \\geq 2 k+2$.\n\nWe note that it is possible to seat $2 k+2$ people around the table in seats\n\n$$\n\\{3,6,9, \\ldots, 6 k, 6 k+3,6 k+5\\}\n$$\n\nThis table is full becase $2 k+1$ of the gaps consist of 2 chairs and 1 gap consists of 1 chair.\n\nWe now know that, if there are $t$ people seated at a full table with $6 k+5$ chairs, then $2 k+2 \\leq t \\leq 3 k+2$.\n\nTo confirm that every such value of $t$ is possible, consider a table with $t$ people, $3 t-(6 k+5)$\n\n\n\ngaps of 1 chair, and $(6 k+5)-2 t$ gaps of 2 chairs.\n\nFrom the work above, we know that $3 t \\geq 6 k+5$ and so $3 t-(6 k+5) \\geq 0$, and that $2 t \\leq 6 k+5$ and so $(6 k+5)-2 t \\geq 0$.\n\nThe total number of gaps is $3 t-(6 k+5)+(6 k+5)-2 t=t$, since there are $t$ people seated.\n\nFinally, the total number of chairs is\n\n$$\nt+1 \\cdot(3 t-(6 k+5))+2 \\cdot((6 k+5)-2 t)=t+3 t-4 t-(6 k+5)+2(6 k+5)=6 k+5\n$$\n\nas expected.\n\nThis shows that every $t$ with $2 k+2 \\leq t \\leq 3 k+2$ can produce a full table.\n\nTherefore, the possible values of $t$ are those integers that satisfy $2 k+2 \\leq t \\leq 3 k+2$.\n\nThere are $(3 k+2)-(2 k+2)+1=k+1$ possible values of $t$." ]

[ "$k+1$" ]

[ "For each integer $n \\geq 3$, we define $f(n)$ to be the number of different full tables of size $n$. We can check that\n\n- $f(3)=3$ because the full tables when $n=3$ have people in chairs $\\{1\\},\\{2\\},\\{3\\}$,\n- $f(4)=2$ because the full tables when $n=4$ have people in chairs $\\{1,3\\},\\{2,4\\}$, and\n- $f(5)=5$ because the full tables when $n=4$ have people in chairs $\\{1,3\\},\\{2,4\\},\\{3,5\\}$, $\\{4,1\\},\\{5,2\\}$.\n\nIn the problem, we are told that $f(6)=5$ and in part (a), we determined that $f(8)=10$. This gives us the following table:\n\n| $n$ | $f(n)$ |\n| :---: | :---: |\n| 3 | 3 |\n| 4 | 2 |\n| 5 | 5 |\n| 6 | 5 |\n| 7 | $?$ |\n| 8 | 10 |\n\nBased on this information, we make the guess that for every integer $n \\geq 6$, we have $f(n)=f(n-2)+f(n-3)$.\n\nFor example, this would mean that $f(7)=f(5)+f(4)=5+2=7$ which we can verify is true.\n\nBased on this recurrence relation (which we have yet to prove), we deduce the values of $f(n)$ up to and including $n=19$ :\n\n| $n$ | $f(n)$ | $n$ | $f(n)$ |\n| :---: | :---: | :---: | :---: |\n| 3 | 3 | 11 | 22 |\n| 4 | 2 | 12 | 29 |\n| 5 | 5 | 13 | 39 |\n| 6 | 5 | 14 | 51 |\n| 7 | 7 | 15 | 68 |\n| 8 | 10 | 16 | 90 |\n| 9 | 12 | 17 | 119 |\n| 10 | 17 | 18 | 158 |\n| | | 19 | 209 |\n\nWe now need to prove that the equation $f(n)=f(n-2)+f(n-3)$ is true for all $n \\geq 6$.\n\n\n\nWe think about each full table as a string of 0 s and 1s, with 1 representing a chair that is occupied and 0 representing an empty chair.\n\nLet $a(n)$ be the number of full tables with someone in seat 1 (and thus nobody in seat 2). Let $b(n)$ be the number of full tables with someone in seat 2 (and thus nobody in seat 1). Let $c(n)$ be the number of full tables with nobody in seat 1 or in seat 2 .\n\nSince every full table must be in one of these categories, then $f(n)=a(n)+b(n)+c(n)$. A full table with $n$ seats $n \\geq 4$ must correspond to a string that starts with 10,01 or 00 . Since there cannot be more than two consecutive 0s, we can further specify this, namely to say that a full table with $n$ seats must correspond to a string that starts with 1010 or 1001 or 0100 or 0101 or 0010 . In each case, these are the first 4 characters of the string and correspond to full (1) and empty (0) chairs.\n\nConsider the full tables starting with 1010. Note that such strings end with 0 since the table is circular. Removing the 10 from positions 1 and 2 creates strings of length $n-2$ that begin 10. These strings will still correspond to a full table, and so there are $a(n-2)$ such strings. (We note that all possible strings starting 1010 of length $n$ will lead to all possible strings starting with 1010 of length $n-2$.)\n\nConsider the full tables starting with 1001. Note that such a string ends with 0 since the table is circular. Removing the 100 from positions 1, 2 and 3 creates strings of length $n-3$ that begin 10. (There must have been a 0 in position 5 after the 1 in position 4.) These strings will still correspond to full tables, and so there are $a(n-3)$ such strings. Consider the full tables starting with 0100 . Removing the 100 from positions 2,3 and 4 creates strings of length $n-3$ that begin 01 . (There must have been a 1 in position 5 after the 0 in position 4.) These strings will still correspond to full tables, and so there are $b(n-3)$ such strings.\n\nConsider the full tables starting with 0101. Removing the 01 from positions 3 and 4 creates strings of length $n-2$ that begin 01 . (The 1 in position 4 must have been followed by one or two 0s and so these strings maintains the desired properties.) These strings will still correspond to full tables, and so there are $b(n-2)$ such strings.\n\nConsider the full tables starting with 0010. These strings must begin with either 00100 or 00101.\n\nIf strings start 00100 , then they start 001001 and so we remove the 001 in positions 4,5 and 6 and obtain strings of length $n-3$ that start 001 (and thus start 00). There are $c(n-3)$ such strings.\n\nIf strings start 00101, we remove the 01 in positions 4 and 5 and obtain strings of length $n-2$ that start 001 (and thus start 00 ). There are $c(n-2)$ such strings.\n\nThese 6 cases and subcases count all strings counted by $f(n)$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nf(n) & =a(n-2)+a(n-3)+b(n-3)+b(n-2)+c(n-3)+c(n-2) \\\\\n& =a(n-2)+b(n-2)+c(n-2)+a(n-3)+b(n-3)+c(n-3) \\\\\n& =f(n-2)+f(n-3)\n\\end{aligned}\n$$\n\nas required, which means that the number of different full tables when $n=19$ is 209 .", "Extending our approach from (b), the number of people seated at a full table with 19 chairs is at least $\\frac{19}{3}=6 \\frac{1}{3}$ and at most $\\frac{19}{2}=9 \\frac{1}{2}$.\n\nSince the number of people is an integer, there must be 7,8 or 9 people at the table, which means that the number of empty chairs is 12,11 or 10 , respectively.\n\n\n\nSuppose that there are 9 people and 9 gaps with a total of 10 empty chairs.\n\nIn this case, there is 1 gap with 2 empty chairs and 8 gaps with 1 empty chair.\n\nThere are 19 pairs of chairs in which we can put 2 people with a gap of 2 in between: $\\{1,4\\},\\{2,5\\}, \\ldots,\\{19,3\\}$.\n\nOnce we choose one of these pairs, the seat choice for the remaining 8 people is completely determined by placing people in every other chair.\n\nTherefore, there are 19 different full tables with 9 people.\n\nSuppose that there are 8 people and 8 gaps with a total of 11 empty chairs.\n\nIn this case, there are 3 gaps with 2 empty chairs and 5 gaps with 1 empty chair. There are 7 different circular orderings in which these 8 gaps can be arranged:\n\n$$\n\\begin{array}{lllllll}\n22211111 & 22121111 & 22112111 & 22111211 & 22111121 & 21212111 & 21211211\n\\end{array}\n$$\n\nWe note that \"22211111\" would be the same as, for example, \"11222111\" since these gaps are arranged around a circle.\n\nIf the three gaps of length 2 are consecutive, there is only one configuration (22211111). If there are exactly 2 consecutive gaps of length 2 , there are 4 relative places in which the third gap of length 2 can be placed.\n\nIf there are no consecutive gaps of length 2 , these gaps can either be separated by 1 gap each (21212111) with 3 gaps on the far side, or can be separated by 1 gap, 2 gaps, and 2 gaps (21211211). There is only one configuration for the gaps in this last situation.\n\nThere are 7 different circular orderings for these 8 gaps.\n\nEach of these 7 different orderings can be placed around the circle of 19 chairs in 19 different ways, because each can be started in 19 different places. Because 19 is prime, none of these orderings overlap.\n\nTherefore, there are $7 \\cdot 19=133$ different full tables with 8 people.\n\nSuppose that there are 7 people and 7 gaps with a total of 12 empty chairs.\n\nIn this case, there are 2 gaps with 1 empty chair and 5 gaps with 2 empty chairs.\n\nThe 2 gaps with 1 empty chair can be separated by 0 gaps with 2 empty chairs, 1 gap with 2 empty chairs, or 2 gaps with 2 empty chairs. Because the chairs are around a circle, if there were 3 , 4 or 5 gaps with 2 empty chairs between them, there would be 2,1 or 0 gaps going the other way around the circle.\n\nThis means that there are 3 different configurations for the gaps.\n\nEach of these configurations can be placed in 19 different ways around the circle of chairs. Therefore, there are $3 \\cdot 19=57$ full tables with 7 people.\n\nIn total, there are $19+133+57=209$ full tables with 19 chairs.", "As in Solution 2, there must be 7, 8 or 9 people in chairs, and so there are 7,8 or 9 gaps. If there are 7 gaps, there are 2 gaps of 1 chair and 5 gaps of 2 chairs.\n\nIf there are 8 gaps, there are 5 gaps of 1 chair and 3 gaps of 2 chairs.\n\nIf there are 9 gaps, there are 8 gaps of 1 chair and 1 gap of 2 chairs.\n\nWe consider three mutually exclusive cases: (i) there is a person in chair 1 and not in chair 2 , (ii) there is a person in chair 2 and not in chair 1 , and (iii) there is nobody in chair 1 or in chair 2. Every full table fits into exactly one of these three cases.\n\nCase (i): there is a person in chair 1 and not in chair 2\n\nWe use the person in chair 1 to \"anchor\" the arrangement, by starting at chair 1 and\n\n\n\narranging the gaps (and thus the full chairs) clockwise around the table from chair 1.\n\nIf there are 7 gaps, we need to choose 2 of them to be of length 1 , and so there are $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)$ ways of arranging the gaps starting at chair 1.\n\nIf there are 8 gaps, we need to choose 3 of them to be of length 2 , and so there are $\\left(\\begin{array}{l}8 \\\\ 3\\end{array}\\right)$ ways of arranging the gaps starting at chair 1.\n\nIf there are 9 gaps, we need to choose 1 of them to be of length 2 , and so there are $\\left(\\begin{array}{l}9 \\\\ 1\\end{array}\\right)$ ways of arranging the gaps starting at chair 1.\n\nIn this case, there are a total of $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)+\\left(\\begin{array}{l}8 \\\\ 3\\end{array}\\right)+\\left(\\begin{array}{l}9 \\\\ 1\\end{array}\\right)=21+56+9=86$ full tables.\n\nCase (ii): there is a person in chair 2 and not in chair 1\n\nWe use the same reasoning starting with the person in chair 2 as the anchor.\n\nAgain, there are 86 full tables in this case.\n\nCase (iii): there is nobody in chair 1 or chair 2\n\nSince there is nobody in chair 1 or chair 2 , there must be a person in chair 3 and also in chair 19 , which fixes one gap of 2 chairs.\n\nHere, we use the person in chair 3 as the anchor.\n\nIf there are 7 gaps, there are 2 gaps of 1 chair and 4 gaps of 2 chairs left to place. There are $\\left(\\begin{array}{l}6 \\\\ 2\\end{array}\\right)$ ways of doing this.\n\nIf there are 8 gaps, there are 5 gaps of 1 chair and 2 gaps of 2 chairs left to place. There are $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)$ ways of doing this.\n\nIf there are 9 gaps, there are 8 gaps of 1 chair and 0 gaps of 2 chairs left to place. There is 1 way to do this.\n\nIn this case, there are a total of $\\left(\\begin{array}{l}6 \\\\ 2\\end{array}\\right)+\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)+1=15+21+1=37$ full tables.\n\nIn total, there are $86+86+37=209$ full tables with 19 chairs." ]

[ "209" ]

[ "The area of quadrilateral $A B C D$ is the sum of the areas of $\\triangle A B C$ and $\\triangle A C D$.\n\nSince $\\triangle A B C$ is right-angled at $B$, its area equals $\\frac{1}{2}(A B)(B C)=\\frac{1}{2}(3)(4)=6$.\n\nSince $\\triangle A B C$ is right-angled at $B$, then by the Pythagorean Theorem,\n\n$$\nA C=\\sqrt{A B^{2}+B C^{2}}=\\sqrt{3^{2}+4^{2}}=\\sqrt{25}=5\n$$\n\nbecause $A C>0$. (We could have also observed that $\\triangle A B C$ must be a \"3-4-5\" triangle.) Since $\\triangle A C D$ is right-angled at $A$, then by the Pythagorean Theorem,\n\n$$\nA D=\\sqrt{C D^{2}-A C^{2}}=\\sqrt{13^{2}-5^{2}}=\\sqrt{144}=12\n$$\n\nbecause $A D>0$. (We could have also observed that $\\triangle A C D$ must be a \" $5-12-13$ \" triangle.) Thus, the area of $\\triangle A C D$ equals $\\frac{1}{2}(A C)(A D)=\\frac{1}{2}(5)(12)=30$.\n\nFinally, the area of quadrilateral $A B C D$ is thus $6+30=36$." ]

[ "36" ]

[ "Let the width of each of the identical rectangles be $a$.\n\nIn other words, $Q P=R S=T W=W X=U V=V Y=a$.\n\nLet the height of each of the identical rectangles be $b$.\n\nIn other words, $Q R=P S=T U=W V=X Y=b$.\n\nThe perimeter of the whole shape equals\n\n$$\nQ P+P S+S X+X Y+V Y+U V+T U+T R+Q R\n$$\n\nSubstituting for known lengths, we obtain\n\n$$\na+b+S X+b+a+a+b+T R+b\n$$\n\nor $3 a+4 b+(S X+T R)$.\n\nBut $S X+T R=(T R+R S+S X)-R S=(T W+W X)-R S=a+a-a=a$.\n\nTherefore, the perimeter of the whole shape equals $4 a+4 b$.\n\nThe perimeter of one rectangle is $2 a+2 b$, which we are told equals $21 \\mathrm{~cm}$.\n\nFinally, the perimeter of the whole shape is thus $2(2 a+2 b)$ which equals $42 \\mathrm{~cm}$." ]

[ "42" ]

[ "Extend $C A$ and $D B$ downwards until they meet the horizontal through $O$ at $P$ and $Q$, respectively.\n\n<img_3828>\n\nSince $C A$ and $D B$ are vertical, then $\\angle C P O=\\angle D Q O=90^{\\circ}$.\n\nSince $O A=20 \\mathrm{~m}$, then $A P=O A \\sin 30^{\\circ}=(20 \\mathrm{~m}) \\cdot \\frac{1}{2}=10 \\mathrm{~m}$.\n\nSince $O B=20 \\mathrm{~m}$, then $B Q=O B \\sin 45^{\\circ}=(20 \\mathrm{~m}) \\cdot \\frac{1}{\\sqrt{2}}=10 \\sqrt{2} \\mathrm{~m}$.\n\nSince $A C=6 \\mathrm{~m}$, then $C P=A C+A P=16 \\mathrm{~m}$.\n\nFor $C D$ to be as short as possible and given that $C$ is fixed, then it must be the case that $C D$ is horizontal:\n\nIf $C D$ were not horizontal, then suppose that $X$ is on $D Q$, possibly extended, so that $C X$ is horizontal.\n\n<img_3255>\n\nThen $\\angle C X D=90^{\\circ}$ and so $\\triangle C X D$ is right-angled with hypotenuse $C D$.\n\nIn this case, $C D$ is longer than $C X$ or $X D$.\n\nIn particular, $C D>C X$, which means that if $D$ were at $X$, then $C D$ would be shorter.\n\nIn other words, a horizontal $C D$ makes $C D$ as short as possible.\n\nWhen $C D$ is horizontal, $C D Q P$ is a rectangle, since it has two vertical and two horizontal sides. Thus, $D Q=C P=16 \\mathrm{~m}$.\n\nFinally, this means that $B D=D Q-B Q=(16-10 \\sqrt{2}) \\mathrm{m}$." ]

[ "$(16-10 \\sqrt{2})$" ]

[ "Let $\\angle P E Q=\\theta$.\n\nJoin $P$ to $B$.\n\nWe 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.\n\nMore concretely, $\\angle D E P=\\angle P B E$ (using the chord $E P$ and the tangent through $E$ ) and $\\angle A B P=\\angle P E Q=\\theta$ (using the chord $B P$ and the tangent through $B$ ).\n\nNow $\\angle D E P$ is exterior to $\\triangle F E P$ and so $\\angle D E P=\\angle F P E+\\angle E F P=25^{\\circ}+30^{\\circ}$, and so $\\angle P B E=\\angle D E P=55^{\\circ}$.\n\nFurthermore, $\\angle A Q B$ is an exterior angle of $\\triangle P Q E$.\n\nThus, $\\angle A Q B=\\angle Q P E+\\angle P E Q=25^{\\circ}+\\theta$.\n\n<img_3991>\n\nIn $\\triangle A B Q$, we have $\\angle B A Q=35^{\\circ}, \\angle A B Q=\\theta+55^{\\circ}$, and $\\angle A Q B=25^{\\circ}+\\theta$.\n\nThus, $35^{\\circ}+\\left(\\theta+55^{\\circ}\\right)+\\left(25^{\\circ}+\\theta\\right)=180^{\\circ}$ or $115^{\\circ}+2 \\theta=180^{\\circ}$, and so $2 \\theta=65^{\\circ}$.\n\nTherefore $\\angle P E Q=\\theta=\\frac{1}{2}\\left(65^{\\circ}\\right)=32.5^{\\circ}$.\n\nAs 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.\n\nConsider a circle with centre $O$ and a chord $X Y$, with tangent $Z X$ meeting the circle at $X$. We prove that if $Z X$ is tangent to the circle, then $\\angle Z X Y$ equals $\\angle X W Y$ whenever $W$ is a point on the circle on the opposite side of $X Y$ as $X Z$ (that is, the angle subtended by $X Y$ on the opposite side of the circle).\n\nWe prove this in the case that $\\angle Z X Y$ is acute. The cases where $\\angle Z X Y$ is a right angle or an obtuse angle are similar.\n\nDraw diameter $X O V$ and join $V Y$.\n\n<img_3622>\n\nSince $\\angle Z X Y$ is acute, points $V$ and $W$ are on the same arc of chord $X Y$.\n\nThis means that $\\angle X V Y=\\angle X W Y$, since they are angles subtended by the same chord.\n\nSince $O X$ is a radius and $X Z$ is a tangent, then $\\angle O X Z=90^{\\circ}$.\n\nThus, $\\angle O X Y+\\angle Z X Y=90^{\\circ}$.\n\nSince $X V$ is a diameter, then $\\angle X Y V=90^{\\circ}$.\n\nFrom $\\triangle X Y V$, we see that $\\angle X V Y+\\angle V X Y=90^{\\circ}$.\n\nBut $\\angle O X Y+\\angle Z X Y=90^{\\circ}$ and $\\angle X V Y+\\angle V X Y=90^{\\circ}$ and $\\angle O X Y=\\angle V X Y$ tells us that $\\angle Z X Y=\\angle X V Y$.\n\nThis gives us that $\\angle Z X Y=\\angle X W Y$, as required." ]

[ "$32.5$" ]

[ "Draw a line segment through $M$ in the plane of $\\triangle P M N$ parallel to $P N$ 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.\n\nJoin $Q$ to $P$ and $A$. Join $R$ to $N$ and $B$.\n\n<img_3945>\n\nSo the volume of solid $A B C D P M N$ equals the volume of solid $A B C D P Q R N$ minus the volumes of solids $P M Q A$ and $N M R B$.\n\nSolid $A B C D P Q R N$ is a trapezoidal prism. This is because $N R$ and $B C$ are parallel (since they lie in parallel planes), which makes $N R B C$ a trapezoid. Similarly, $P Q A D$ is a trapezoid. Also, $P N, Q R, D C$, and $A B$ are all perpendicular to the planes of these trapezoids and equal in length, since they equal the side lengths of the squares.\n\nSolids $P M Q A$ and $N M R B$ are triangular-based pyramids. We can think of their bases as being $\\triangle P M Q$ and $\\triangle N M R$. 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.)\n\nThe volume of $A B C D P Q R N$ equals the area of trapezoid $N R B C$ times the width of the prism, which is 2.\n\nThat is, this volume equals $\\frac{1}{2}(N R+B C)(N C)(N P)=\\frac{1}{2}(N R+2)(2)(2)=2 \\cdot N R+4$.\n\nSo we need to find the length of $N R$.\n\nConsider quadrilateral $P N R Q$. This quadrilateral is a rectangle since $P N$ and $Q R$ are perpendicular to the two side planes of the original solid.\n\nThus, $N R$ equals the height of $\\triangle P M N$.\n\nJoin $M$ to the midpoint $T$ of $P N$.\n\nSince $\\triangle P M N$ is isosceles, then $M T$ is perpendicular to $P N$.\n\n<img_3305>\n\nSince $N T=\\frac{1}{2} P N=1$ and $\\angle P M N=90^{\\circ}$ and $\\angle T N M=45^{\\circ}$, then $\\triangle M T N$ is also right-angled and isosceles with $M T=T N=1$.\n\nTherefore, $N R=M T=1$ and so the volume of $A B C D P Q R N$ is $2 \\cdot 1+4=6$.\n\nThe volumes of solids $P M Q A$ and $N M R B$ are equal. Each has height 2 and their bases $\\triangle P M Q$ and $\\triangle N M R$ are congruent, because each is right-angled (at $Q$ and at $R$ ) with $P Q=N R=1$ and $Q M=M R=1$.\n\nThus, using the formula above, the volume of each is $\\frac{1}{3}\\left(\\frac{1}{2}(1)(1)\\right) 2=\\frac{1}{3}$.\n\nFinally, the volume of the original solid equals $6-2 \\cdot \\frac{1}{3}=\\frac{16}{3}$.", "We determine the volume of $A B C D P M N$ by splitting it into two solids: $A B C D P N$ and $A B N P M$ by slicing along the plane of $A B N P$.\n\nSolid $A B C D P N$ is a triangular prism, since $\\triangle B C N$ and $\\triangle A D P$ are each right-angled (at $C$ and $D$ ), $B C=C N=A D=D P=2$, and segments $P N, D C$ and $A B$ are perpendicular to each of the triangular faces and equal in length.\n\nThus, the volume of $A B C D P N$ equals the area of $\\triangle B C N$ times the length of $D C$, or $\\frac{1}{2}(B C)(C N)(D C)=\\frac{1}{2}(2)(2)(2)=4$. (This solid can also be viewed as \"half\" of a cube.)\n\nSolid $A B N P M$ is a pyramid with rectangular base $A B N P$. (Note that $P N$ and $A B$ are perpendicular to the planes of both of the side triangular faces of the original solid, that $P N=A B=2$ and $B N=A P=\\sqrt{2^{2}+2^{2}}=2 \\sqrt{2}$, by the Pythagorean Theorem.)\n\nTherefore, the volume of $A B N P M$ equals $\\frac{1}{3}(A B)(B N) h=\\frac{4 \\sqrt{2}}{3} h$, where $h$ is the height of the pyramid (that is, the distance that $M$ is above plane $A B N P$ ).\n\nSo we need to calculate $h$.\n\nJoin $M$ to the midpoint, $T$, of $P N$ and to the midpoint, $S$, of $A B$. Join $S$ and $T$. By symmetry, $M$ lies directly above $S T$. Since $A B N P$ is a rectangle and $S$ and $T$ are the midpoints of opposite sides, then $S T=A P=2 \\sqrt{2}$.\n\nSince $\\triangle P M N$ is right-angled and isosceles, then $M T$ is perpendicular to $P N$. Since $N T=\\frac{1}{2} P N=1$ and $\\angle T N M=45^{\\circ}$, then $\\triangle M T N$ is also right-angled and isosceles with $M T=T N=1$.\n\n<img_3858>\n\nAlso, $M S$ is the hypotenuse of the triangle formed by dropping a perpendicular from $M$ to $U$ in the plane of $A B C D$ (a distance of 2) and joining $U$ to $S$. Since $M$ is 1 unit horizontally from $P N$, then $U S=1$.\n\nThus, $M S=\\sqrt{2^{2}+1^{2}}=\\sqrt{5}$ by the Pythagorean Theorem.\n\n<img_3846>\n\nWe can now consider $\\triangle S M T . h$ is the height of this triangle, from $M$ to base $S T$.\n\n<img_3680>\n\nNow $h=M T \\sin (\\angle M T S)=\\sin (\\angle M T S)$.\n\nBy the cosine law in $\\triangle S M T$, we have\n\n$$\nM S^{2}=S T^{2}+M T^{2}-2(S T)(M T) \\cos (\\angle M T S)\n$$\n\nTherefore, $5=8+1-4 \\sqrt{2} \\cos (\\angle M T S)$ or $4 \\sqrt{2} \\cos (\\angle M T S)=4$.\n\nThus, $\\cos (\\angle M T S)=\\frac{1}{\\sqrt{2}}$ and so $\\angle M T S=45^{\\circ}$ which gives $h=\\sin (\\angle M T S)=\\frac{1}{\\sqrt{2}}$.\n\n(Alternatively, we note that the plane of $A B C D$ is parallel to the plane of $P M N$, and so since the angle between plane $A B C D$ and plane $P N B A$ is $45^{\\circ}$, then the angle between plane $P N B A$ and plane $P M N$ is also $45^{\\circ}$, and so $\\angle M T S=45^{\\circ}$.)\n\nFinally, this means that the volume of $A B N P M$ is $\\frac{4 \\sqrt{2}}{3} \\cdot \\frac{1}{\\sqrt{2}}=\\frac{4}{3}$, and so the volume of solid $A B C D P M N$ is $4+\\frac{4}{3}=\\frac{16}{3}$." ]

[ "$\\frac{16}{3}$" ]

[ "Since $\\sin C=\\frac{A B}{A C}$, then $A B=A C \\sin C=20\\left(\\frac{3}{5}\\right)=12$.\n\nBy Pythagoras, $B C^{2}=A C^{2}-A B^{2}=20^{2}-12^{2}=256$ or $B C=16$.", "Using the standard trigonometric ratios, $B C=A C \\cos C$.\n\nSince $\\sin C=\\frac{3}{5}$, then $\\cos ^{2} C=1-\\sin ^{2} C=1-\\frac{9}{25}=\\frac{16}{25}$ or $\\cos C=\\frac{4}{5}$. (Notice that $\\cos C$ is positive since angle $C$ is acute in triangle $A B C$.)\n\nTherefore, $B C=20\\left(\\frac{4}{5}\\right)=16$." ]

[ "16" ]

[ "Let $G$ be the point where the goat is standing, $H$ the position of the helicopter when the goat first measures the angle, $P$ the point directly below the helicopter at this time, $J$ the position of the helicopter one minute later, and $Q$ the point directly below the helicopter at this time.\n\n<img_3329>\n\nUsing the initial position of the helicopter, $\\tan \\left(6^{\\circ}\\right)=\\frac{H P}{P G}$ or $P G=\\frac{222}{\\tan \\left(6^{\\circ}\\right)} \\approx 2112.19 \\mathrm{~m}$.\n\nUsing the second position of the helicopter, $\\tan \\left(75^{\\circ}\\right)=\\frac{J Q}{Q G}$ or $Q G=\\frac{222}{\\tan \\left(75^{\\circ}\\right)} \\approx 59.48 \\mathrm{~m}$.\n\nSo in the one minute that has elapsed, the helicopter has travelled\n\n$2112.19 \\mathrm{~m}-59.48 \\mathrm{~m}=2052.71 \\mathrm{~m}$ or $2.0527 \\mathrm{~km}$.\n\nTherefore, in one hour, the helicopter will travel $60(2.0527)=123.162 \\mathrm{~km}$.\n\nThus, the helicopter is travelling $123 \\mathrm{~km} / \\mathrm{h}$." ]

[ "123" ]

[ "We label the vertices of the shaded hexagon $U, V, W, X$, $Y$, and $Z$.\n\nBy symmetry, all of the six triangles with two vertices on the inner hexagon and one on the outer hexagon (eg. triangle $U V A$ ) are congruent equilateral triangles. In order to determine the area of the inner hexagon, we determine the ratio of the side lengths of the two hexagons.\n\n<img_3663>\n\nLet the side length of the inner hexagon be $x$. Then $A U=U F=x$. Then triangle $A U F$ has a $120^{\\circ}$ between the two sides of length $x$. If we draw a perpendicular from $U$ to point $P$ on side $A F$, then $U P$ divides\n\n$\\triangle A U F$ into two $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangles. Thus, $F P=P A=\\frac{\\sqrt{3}}{2} x$ and so $A F=\\sqrt{3} x$.\n\nSo the ratio of the side lengths of the hexagons is $\\sqrt{3}: 1$, and so the ratio of\n\n<img_3205>\ntheir areas is $(\\sqrt{3})^{2}: 1=3: 1$.\n\nSince the area of the larger hexagon is 36 , then the area of the inner hexagon is 12.", "We label the vertices of the hexagon $U, V, W, X, Y$, and $Z$. By symmetry, all of the six triangles with two vertices on the inner hexagon and one on the outer hexagon (eg. triangle $U V A$ ) are congruent equilateral triangles.\n\nWe also join the opposite vertices of the inner hexagon, ie. we join $U$ to $X, V$ to $Y$, and $W$ to $Z$. (These 3 line segments all meet at a single point, say $O$.) This divides the inner hexagon into 6 small equilateral triangles identical to the\n\n<img_3385>\nsix earlier mentioned equilateral triangles.\n\nLet the area of one of these triangles be $a$. Then we can label the 12 small equilateral triangles as all having area $a$.\n\nBut triangle $A U F$ also has area $a$, because if we consider triangle $A F V$, then $A U$ is a median (since $F U=A U=U V$ by symmetry) and so divides triangle $A F V$ into two triangles of equal area. Since the area of\n\n<img_3160>\ntriangle $A U V$ is $a$, then the area of triangle $A U F$ is also $a$.\n\nTherefore, hexagon $A B C D E F$ is divided into 18 equal areas. Thus, $a=2$ since the area of the large hexagon is 36.\n\nSince the area of $U V W X Y Z$ is $6 a$, then its area is 12 ." ]

[ "12" ]

[ "We assign coordinates to the diagram, with the mouth of the cannon at the point $(0,0)$, with the positive $x$-axis in the horizontal direction towards the safety net from the cannon, and the positive $y$ axis upwards from $(0,0)$.\n\nSince Herc reaches his maximum height when his horizontal distance is $30 \\mathrm{~m}$, then the axis of symmetry of the parabola is the line $x=30$. Since the parabola has a root at $x=0$, then the other root must be at $x=60$.\n\nTherefore, the parabola has the form $y=\\operatorname{ax}(x-60)$.\n\nIn order to determine the value of $a$, we note that Herc passes through the point $(30,100)$, and so\n\n<img_3831>\n\n\n\n$$\n\\begin{aligned}\n100 & =30 a(-30) \\\\\na & =-\\frac{1}{9}\n\\end{aligned}\n$$\n\nThus, the equation of the parabola is $y=-\\frac{1}{9} x(x-60)$.\n\n(Alternatively, we could say that since the parabola has its maximum point at $(30,100)$, then it must be of the form $y=a(x-30)^{2}+100$.\n\nSince the parabola passes through $(0,0)$, then we have\n\n$$\n\\begin{aligned}\n& 0=a(0-30)^{2}+100 \\\\\n& 0=900 a+100 \\\\\n& a=-\\frac{1}{9}\n\\end{aligned}\n$$\n\nThus, the parabola has the equation $y=-\\frac{1}{9}(x-30)^{2}+100$.)\n\nWe would like to find the points on the parabola which have $y$-coordinate 64 , so we solve\n\n$$\n\\begin{aligned}\n64 & =-\\frac{1}{9} x(x-60) \\\\\n0 & =x^{2}-60 x+576 \\\\\n0 & =(x-12)(x-48)\n\\end{aligned}\n$$\n\nSince we want a point after Herc has passed his highest point, then $x=48$, ie. the horizontal distance from the cannon to the safety net is $48 \\mathrm{~m}$." ]

[ "48" ]

[ "First, we find the coordinates of $V$.\n\nTo do this, we use the given equation for the parabola and complete the square:\n\n$y=-x^{2}+4 x+1=-\\left(x^{2}-4 x-1\\right)=-\\left(x^{2}-4 x+2^{2}-2^{2}-1\\right)=-\\left((x-2)^{2}-5\\right)=-(x-2)^{2}+5$\n\nTherefore, the coordinates of the vertex $V$ are $(2,5)$.\n\nNext, we find the coordinates of $A$ and $B$.\n\nNote that $A$ and $B$ are the points of intersection of the line with equation $y=-x+1$ and the parabola with equation $y=-x^{2}+4 x+1$.\n\nWe equate $y$-values to obtain $-x+1=-x^{2}+4 x+1$ or $x^{2}-5 x=0$ or $x(x-5)=0$.\n\nTherefore, $x=0$ or $x=5$.\n\nIf $x=0$, then $y=-x+1=1$, and so $A$ (which is on the $y$-axis) has coordinates $(0,1)$.\n\nIf $x=5$, then $y=-x+1=-4$, and so $B$ has coordinates $(5,-4)$.\n\n\n\nWe now have the points $V(2,5), A(0,1), B(5,-4)$.\n\nThis gives\n\n$$\n\\begin{aligned}\nA V^{2} & =(0-2)^{2}+(1-5)^{2}=20 \\\\\nB V^{2} & =(5-2)^{2}+(-4-5)^{2}=90 \\\\\nA B^{2} & =(0-5)^{2}+(1-(-4))^{2}=50\n\\end{aligned}\n$$\n\nand so $A V^{2}+B V^{2}-A B^{2}=20+90-50=60$." ]

[ "60" ]

[ "Since $A B C$ is a quarter of a circular pizza with centre $A$ and radius $20 \\mathrm{~cm}$, then $A C=A B=20 \\mathrm{~cm}$.\n\nWe are also told that $\\angle C A B=90^{\\circ}$ (one-quarter of $360^{\\circ}$ ).\n\nSince $\\angle C A B=90^{\\circ}$ and $A, B$ and $C$ are all on the circumference of the circle, then $C B$ is a diameter of the pan. (This is a property of circles: if $X, Y$ and $Z$ are three points on a circle with $\\angle Z X Y=90^{\\circ}$, then $Y Z$ must be a diameter of the circle.)\n\nSince $\\triangle C A B$ is right-angled and isosceles, then $C B=\\sqrt{2} A C=20 \\sqrt{2} \\mathrm{~cm}$.\n\nTherefore, the radius of the circular plate is $\\frac{1}{2} C B$ or $10 \\sqrt{2} \\mathrm{~cm}$.\n\nThus, the area of the circular pan is $\\pi(10 \\sqrt{2} \\mathrm{~cm})^{2}=200 \\pi \\mathrm{cm}^{2}$.\n\nThe area of the slice of pizza is one-quarter of the area of a circle with radius $20 \\mathrm{~cm}$, or $\\frac{1}{4} \\pi(20 \\mathrm{~cm})^{2}=100 \\pi \\mathrm{cm}^{2}$.\n\nFinally, the fraction of the pan that is covered is the area of the slice of pizza divided by the area of the pan, or $\\frac{100 \\pi \\mathrm{cm}^{2}}{200 \\pi \\mathrm{cm}^{2}}=\\frac{1}{2}$." ]

[ "$\\frac{1}{2}$" ]

[ "Suppose that the length of $A F$ is $x \\mathrm{~m}$.\n\nSince the length of $A B$ is $8 \\mathrm{~m}$, then the length of $F B$ is $(8-x) \\mathrm{m}$.\n\nSince $\\triangle M A F$ is right-angled and has an angle of $60^{\\circ}$, then it is $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nTherefore, $M F=\\sqrt{3} A F$, since $M F$ is opposite the $60^{\\circ}$ angle and $A F$ is opposite the $30^{\\circ}$ angle.\n\nThus, $M F=\\sqrt{3} x \\mathrm{~m}$.\n\nSince $M P=2 \\mathrm{~m}$, then $P F=M F-M P=(\\sqrt{3} x-2) \\mathrm{m}$.\n\nWe can now look at $\\triangle B F P$ which is right-angled at $F$.\n\nWe have\n\n$$\n\\tan \\theta=\\frac{P F}{F B}=\\frac{(\\sqrt{3} x-2) \\mathrm{m}}{(8-x) \\mathrm{m}}=\\frac{\\sqrt{3} x-2}{8-x}\n$$\n\nTherefore, $(8-x) \\tan \\theta=\\sqrt{3} x-2$ or $8 \\tan \\theta+2=\\sqrt{3} x+(\\tan \\theta) x$.\n\nThis gives $8 \\tan \\theta+2=x(\\sqrt{3}+\\tan \\theta)$ or $x=\\frac{8 \\tan \\theta+2}{\\tan \\theta+\\sqrt{3}}$.\n\nFinally, $M F=\\sqrt{3} x=\\frac{8 \\sqrt{3} \\tan \\theta+2 \\sqrt{3}}{\\tan \\theta+\\sqrt{3}} \\mathrm{~m}$.\n\n" ]

[ "$\\frac{8 \\sqrt{3} \\tan \\theta+2 \\sqrt{3}}{\\tan \\theta+\\sqrt{3}}$" ]

[ "Since $M T$ is the perpendicular bisector of $B C$, then\n\n$B M=M C$, and $T M$ is perpendicular to $B C$.\n\nTherefore, $\\triangle C M T$ is similar to $\\triangle C B A$, since they share a common angle and each have a right angle.\n\n<img_3335>\n\nBut $\\frac{C M}{C B}=\\frac{1}{2}$ so $\\frac{C T}{C A}=\\frac{C M}{C B}=\\frac{1}{2}$, and thus $C T=A T=A B$, ie. $\\frac{A B}{A C}=\\frac{1}{2}$ or $\\sin (\\angle A C B)=\\frac{1}{2}$.\n\nTherefore, $\\angle A C B=30^{\\circ}$.", "Since $T M \\| A B$, and $C M=M B$, then $C T=T A=A B$.\n\nJoin $T$ to $B$.\n\nSince $\\angle A B C=90^{\\circ}$, then $A C$ is the diameter of a circle passing through $A, C$ and $B$, with $T$ as its centre.\n\n<img_3228>\n\nThus, $T A=A B=B T$ (all radii), and so $\\triangle A B T$ is equilateral. Therefore, $\\angle B A C=60^{\\circ}$, and so $\\angle A C B=30^{\\circ}$." ]

[ "$30^{\\circ}$" ]

[ "Extend $X A$ and $Z F$ to meet at point $T$.\n\nBy symmetry, $\\angle A X Z=\\angle F Z X=60^{\\circ}$ and $\\angle T A F=\\angle T F A=60^{\\circ}$, and so $\\triangle T A F$ and $\\triangle T X Z$ are both equilateral triangles.\n\nSince $A F=10$, then $T A=10$, which means\n\n$T X=10+5=15$, and so $X Z=T X=15$.\n\n<img_3285>", "We look at the quadrilateral $A X Z F$.\n\nSince $A B C D E F$ is a regular hexagon, then $\\angle F A X=\\angle A F Z=120^{\\circ}$.\n\nNote that $A F=10$, and also $A X=F Z=5$ since $X$ and $Z$ are midpoints of their respective sides.\n\n<img_3509>\n\nBy symmetry, $\\angle A X Z=\\angle F Z X=60^{\\circ}$, and so $A X Z F$ is a trapezoid.\n\nDrop perpendiculars from $A$ and $F$ to $P$ and $Q$, respectively, on $X Z$.\n\nBy symmetry again, $P X=Q Z$. Now, $P X=A X \\cos 60^{\\circ}=5\\left(\\frac{1}{2}\\right)=\\frac{5}{2}$.\n\nSince $A P Q F$ is a rectangle, then $P Q=10$.\n\nTherefore, $X Z=X P+P Q+Q Z=\\frac{5}{2}+10+\\frac{5}{2}=15$." ]

[ "15" ]

[ "Using a known formula for the area of a triangle, $A=\\frac{1}{2} a b \\sin C$,\n\n$$\n\\begin{aligned}\n18 & =\\frac{1}{2}(2 x+1)(2 x) \\sin 30^{\\circ} \\\\\n36 & =(2 x+1)(2 x)\\left(\\frac{1}{2}\\right) \\\\\n0 & =2 x^{2}+x-36 \\\\\n0 & =(2 x+9)(x-4)\n\\end{aligned}\n$$\n\nand so $x=4$ or $x=-\\frac{9}{2}$. Since $x$ is positive, then $x=4$.", "Draw a perpendicular from $A$ to $P$ on $B C$.\n\n<img_3390>\n\nUsing $\\triangle A P C, A P=A C \\sin 30^{\\circ}=2 x\\left(\\frac{1}{2}\\right)=x$.\n\nNow $A P$ is the height of $\\triangle A B C$, so Area $=\\frac{1}{2}(B C)(A P)$.\n\nThen\n\n$$\n\\begin{aligned}\n18 & =\\frac{1}{2}(2 x+1)(x) \\\\\n0 & =2 x^{2}+x-36 \\\\\n0 & =(2 x+9)(x-4)\n\\end{aligned}\n$$\n\nand so $x=4$ or $x=-\\frac{9}{2}$.\n\nSince $x$ is positive, then $x=4$." ]

[ "4" ]

[ "Let the length of the ladder be $L$.\n\nThen $A C=L \\cos 70^{\\circ}$ and $B C=L \\sin 70^{\\circ}$. Also, $A^{\\prime} C=L \\cos 55^{\\circ}$ and $B^{\\prime} C=L \\sin 55^{\\circ}$.\n\nSince $A^{\\prime} A=0.5$, then\n\n$$\n0.5=L \\cos 55^{\\circ}-L \\cos 70^{\\circ}\n$$\n\n$$\nL=\\frac{0.5}{\\cos 55^{\\circ}-\\cos 70^{\\circ}}\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\nB B^{\\prime} & =B C-B^{\\prime} C \\\\\n& =L \\sin 70^{\\circ}-L \\sin 55^{\\circ} \\\\\n& =L\\left(\\sin 70^{\\circ}-\\sin 55^{\\circ}\\right) \\\\\n& =\\frac{(0.5)\\left(\\sin 70^{\\circ}-\\sin 55^{\\circ}\\right)}{\\left(\\cos 55^{\\circ}-\\cos 70^{\\circ}\\right)} \\quad(\\text { from }(*)) \\\\\n& \\approx 0.2603 \\mathrm{~m}\n\\end{aligned}\n$$\n\n<img_3214>\n\nTherefore, to the nearest centimetre, the distance that the ladder slides down the wall is $26 \\mathrm{~cm}$." ]

[ "26" ]

[ "Draw perpendiculars from $P$ and $Q$ to $X$ and $Y$, respectively, on $S R$.\n\n<img_3755>\n\nSince $P Q$ is parallel to $S R$ (because $P Q R S$ is a trapezoid) and $P X$ and $Q Y$ are perpendicular to $S R$, then $P Q Y X$ is a rectangle.\n\nThus, $X Y=P Q=7$ and $P X=Q Y$.\n\nSince $\\triangle P X S$ and $\\triangle Q Y R$ are right-angled with $P S=Q R$ and $P X=Q Y$, then these triangles are congruent, and so $S X=Y R$.\n\nSince $X Y=7$ and $S R=15$, then $S X+7+Y R=15$ or $2 \\times S X=8$ and so $S X=4$.\n\nBy the Pythagorean Theorem in $\\triangle P X S$,\n\n$$\nP X^{2}=P S^{2}-S X^{2}=8^{2}-4^{2}=64-16=48\n$$\n\nNow $P R$ is the hypotenuse of right-angled $\\triangle P X R$.\n\nSince $P R>0$, then by the Pythagorean Theorem,\n\n$$\nP R=\\sqrt{P X^{2}+X R^{2}}=\\sqrt{48+(7+4)^{2}}=\\sqrt{48+11^{2}}=\\sqrt{48+121}=\\sqrt{169}=13\n$$\n\nTherefore, $P R=13$." ]

[ "13" ]

[ "Since $A B=A C$, then $\\triangle A B C$ is isosceles and $\\angle A B C=\\angle A C B$. Note that $\\angle B A C=\\theta$.\n\n<img_3938>\n\nThe angles in $\\triangle A B C$ add to $180^{\\circ}$, so $\\angle A B C+\\angle A C B+\\angle B A C=180^{\\circ}$.\n\nThus, $2 \\angle A C B+\\theta=180^{\\circ}$ or $\\angle A B C=\\angle A C B=\\frac{1}{2}\\left(180^{\\circ}-\\theta\\right)=90^{\\circ}-\\frac{1}{2} \\theta$.\n\nNow $\\triangle B C D$ is isosceles as well with $B C=B D$ and so $\\angle C D B=\\angle D C B=90^{\\circ}-\\frac{1}{2} \\theta$.\n\nSince the angles in $\\triangle B C D$ add to $180^{\\circ}$, then\n\n$$\n\\angle C B D=180^{\\circ}-\\angle D C B-\\angle C D B=180^{\\circ}-\\left(90^{\\circ}-\\frac{1}{2} \\theta\\right)-\\left(90^{\\circ}-\\frac{1}{2} \\theta\\right)=\\theta\n$$\n\nNow $\\angle E B D=\\angle A B C-\\angle D B C=\\left(90^{\\circ}-\\frac{1}{2} \\theta\\right)-\\theta=90^{\\circ}-\\frac{3}{2} \\theta$.\n\nSince $B E=E D$, then $\\angle E D B=\\angle E B D=90^{\\circ}-\\frac{3}{2} \\theta$.\n\nTherefore, $\\angle B E D=180^{\\circ}-\\angle E B D-\\angle E D B=180^{\\circ}-\\left(90^{\\circ}-\\frac{3}{2} \\theta\\right)-\\left(90^{\\circ}-\\frac{3}{2} \\theta\\right)=3 \\theta$." ]

[ "$3 \\theta$" ]

[ "Let $O$ be the centre of the ferris wheel and $B$ the lowest point on the wheel.\n\nSince the radius of the ferris wheel is $9 \\mathrm{~m}$ (half of the diameter of $18 \\mathrm{~m}$ ) and $B$ is $1 \\mathrm{~m}$ above the ground, then $O$ is $9+1=10 \\mathrm{~m}$ above the ground.\n\nLet $\\angle T O P=\\theta$.\n\n<img_3230>\n\nSince the ferris wheel rotates at a constant speed, then in 8 seconds, the angle through which the wheel rotates is twice the angle through which it rotates in 4 seconds. In other words, $\\angle T O Q=2 \\theta$.\n\nDraw a perpendicular from $P$ to $R$ on $T B$ and from $Q$ to $G$ on $T B$.\n\nSince $P$ is $16 \\mathrm{~m}$ above the ground and $O$ is $10 \\mathrm{~m}$ above the ground, then $O R=6 \\mathrm{~m}$.\n\nSince $O P$ is a radius of the circle, then $O P=9 \\mathrm{~m}$.\n\nLooking at right-angled $\\triangle O R P$, we see that $\\cos \\theta=\\frac{O R}{O P}=\\frac{6}{9}=\\frac{2}{3}$.\n\nSince $\\cos \\theta=\\frac{2}{3}<\\frac{1}{\\sqrt{2}}=\\cos \\left(45^{\\circ}\\right)$, then $\\theta>45^{\\circ}$.\n\nThis means that $2 \\theta>90^{\\circ}$, which means that $Q$ is below the horizontal diameter through $O$ and so $G$ is below $O$.\n\nSince $\\angle T O Q=2 \\theta$, then $\\angle Q O G=180^{\\circ}-2 \\theta$.\n\nKolapo's height above the ground at $Q$ equals $1 \\mathrm{~m}$ plus the length of $B G$.\n\nNow $B G=O B-O G$. We know that $O B=9 \\mathrm{~m}$.\n\nAlso, considering right-angled $\\triangle Q O G$, we have\n\n$$\nO G=O Q \\cos (\\angle Q O G)=9 \\cos \\left(180^{\\circ}-2 \\theta\\right)=-9 \\cos (2 \\theta)=-9\\left(2 \\cos ^{2} \\theta-1\\right)\n$$\n\nSince $\\cos \\theta=\\frac{2}{3}$, then $O G=-9\\left(2\\left(\\frac{2}{3}\\right)^{2}-1\\right)=-9\\left(\\frac{8}{9}-1\\right)=1 \\mathrm{~m}$.\n\nTherefore, $B G=9-1=8 \\mathrm{~m}$ and so $Q$ is $1+8=9 \\mathrm{~m}$ above the ground." ]

[ "9" ]

[ "The hour hand and minute hand both turn at constant rates. Since the hour hand moves $\\frac{1}{12}$ of the way around the clock in 1 hour and the minute hand moves all of the way around the clock in 1 hour, then the minute hand turns 12 times as quickly as the hour hand.\n<img_3418>\n\nSuppose also that the hour hand moves through an angle of $x^{\\circ}$ between Before and After. Therefore, the minute hand moves through an angle of $\\left(360^{\\circ}-x^{\\circ}\\right)$ between Before and After, since these two angles add to $360^{\\circ}$.\n\n\n\nSince the minute hand moves 12 times as quickly as the hour hand, then $\\frac{360^{\\circ}-x^{\\circ}}{x^{\\circ}}=12$ or $360-x=12 x$ and so $13 x=360$, or $x=\\frac{360}{13}$.\n\nIn one hour, the hour hand moves through $\\frac{1}{12} \\times 360^{\\circ}=30^{\\circ}$.\n\nSince the hour hand is moving for $t$ hours, then we have $30^{\\circ} t=\\left(\\frac{360}{13}\\right)^{\\circ}$ and so $t=\\frac{360}{30(13)}=\\frac{12}{13}$.", "Suppose that Jimmy starts painting $x$ hours after 9:00 a.m. and finishes painting $y$ hours after 10:00 a.m., where $0<x<1$ and $0<y<1$.\n\nSince $t$ is the amount of time in hours that he spends painting, then $t=(1-x)+y$, because he paints for $(1-x)$ hours until 10:00 a.m., and then for $y$ hours until his finishing time. The hour hand and minute hand both turn at constant rates.\n\nThe minute hand turns $360^{\\circ}$ in one hour and the hour hand turns $\\frac{1}{12} \\times 360^{\\circ}=30^{\\circ}$ in one hour.\n\nThus, in $x$ hours, where $0<x<1$, the minute hand turns $(360 x)^{\\circ}$ and the hour hand turns $(30 x)^{\\circ}$.\n\nIn the Before picture, the minute hand is $(360x)^{\\circ}$ clockwise from the 12 o'clock position.\n\nIn the After picture, the minute hand is $(360 y)^{\\circ}$ clockwise from the 12 o'clock position. The 9 is $9 \\times 30^{\\circ}=270^{\\circ}$ clockwise from the 12 o'clock position and the 10 is $10 \\times 30^{\\circ}=300^{\\circ}$ clockwise from the 12 o'clock position.\n\nTherefore, in the Before picture, the hour hand is $270^{\\circ}+(30 x)^{\\circ}$ clockwise from the 12 o'clock position, and in the After picture, the hour hand is $300^{\\circ}+(30y)^{\\circ}$ clockwirse from the 12 o'clock position.\n\nBecause the hour and minute hands have switched places from the Before to the After positions, then we can equate the corresponding positions to obtain $(360 x)^{\\circ}=300^{\\circ}+(30 y)^{\\circ}$ $($ or $360 x=300+30 y)$ and $(360 y)^{\\circ}=270^{\\circ}+(30 x)^{\\circ}($ or $360 y=270+30 x)$.\n\nDividing both equations by 30 , we obtain $12 x=10+y$ and $12 y=9+x$.\n\nSubtracting the second equation from the first, we obtain $12 x-12 y=10+y-9-x$ or $-1=13 y-13 x$.\n\nTherefore, $y-x=-\\frac{1}{13}$ and so $t=(1-x)+y=1+y-x=1-\\frac{1}{13}=\\frac{12}{13}$." ]

[ "$\\frac{12}{13}$" ]

[ "Since $\\angle A O B=90^{\\circ}, A B$ is a diameter of the circle.\n\nJoin $A B$.\n\n<img_3988>\n\nSince $C$ is the centre of the circle and $A B$ is a diameter, then $C$ is the midpoint of $A B$, so $A$ has coordinates $(0,2)$.\n\nTherefore, the area of the part of the circle inside the first quadrant is equal to the area of $\\triangle A O B$ plus the area of the semi-circle above $A B$.\n\nThe radius of the circle is equal to the distance from $C$ to $B$, or $\\sqrt{(1-2)^{2}+(1-0)^{2}}=\\sqrt{2}$, so the area of the semi-circle is $\\frac{1}{2} \\pi(\\sqrt{2})^{2}=\\pi$.\n\nThe area of $\\triangle A O B$ is $\\frac{1}{2}(O B)(A O)=\\frac{1}{2}(2)(2)=2$.\n\nThus, the area of the part of the circle inside the first quadrant is $\\pi+2$.", "Since $\\angle A O B=90^{\\circ}, A B$ is a diameter of the circle.\n\nJoin $A B$.\n\n<img_3957>\n\nSince $C$ is the centre of the circle and $A B$ is a diameter, then $C$ is the midpoint of $A B$, so $A$ has coordinates $(0,2)$.\n\nThus, $A O=B O$.\n\nWe \"complete the square\" by adding point $D(2,2)$, which is on the circle, by symmetry.\n\n\n\n<img_3605>\n\nThe area of the square is 4 .\n\nThe radius of the circle is equal to the distance from $C$ to $B$, or $\\sqrt{(1-2)^{2}+(1-0)^{2}}=\\sqrt{2}$, so the area of the circle is $\\pi(\\sqrt{2})^{2}=2 \\pi$.\n\nThe area of the portion of the circle outside the square is thus $2 \\pi-4$. This area is divided into four equal sections (each of area $\\frac{1}{4}(2 \\pi-4)=\\frac{1}{2} \\pi-1$ ), two of which are the only portions of the circle outside the first quadrant.\n\nTherefore, the area of the part of the circle inside the the first quadrant is $2 \\pi-2\\left(\\frac{1}{2} \\pi-1\\right)=$ $\\pi+2$.\n\nTwo additional ways to find the coordinates of $A$ :\n\n$*$ The length of $O C$ is $\\sqrt{1^{2}+1^{2}}=\\sqrt{2}$.\n\nSince $C$ is the centre of the circle and $O$ lies on the circle, then the circle has radius $\\sqrt{2}$.\n\nSince the circle has centre $(1,1)$ and radius $\\sqrt{2}$, its equation is $(x-1)^{2}+(y-1)^{2}=2$. To find the coordinates of $A$, we substitute $x=0$ to obtain $(0-1)^{2}+(y-1)^{2}=2$ or $(y-1)^{2}=1$, and so $y=0$ or $y=2$.\n\nSince $y=0$ gives us the point $O$, then $y=2$ gives us $A$, ie. $A$ has coordinates $(0,2)$.\n\n* Since $O$ and $A$ are both on the circle and each has a horizontal distance of 1 from $C$, then their vertical distances from $C$ must be same, ie. must each be 1.\n\nThus, $A$ has coordinates $(0,2)$." ]

[ "$(0,2),\\pi+2$" ]

[ "From the given information, $P C=P B$.\n\nIf we can calculate the length of $P C$, we can calculate the value of $h$, since we already know the length of $A C$.\n\nNow $\\triangle C P B$ is isosceles with $P C=P B, B C=2$ and $\\angle B P C=120^{\\circ}$.\n\nSince $\\triangle C P B$ is isosceles, $\\angle P C B=\\angle P B C=30^{\\circ}$.\n\n<img_3969>\n\nJoin $P$ to the midpoint, $M$, of $B C$.\n\nThen $P M$ is perpendicular to $B C$, since $\\triangle P C B$ is isosceles.\n\n\n\nTherefore, $\\triangle P M C$ is right-angled, has $\\angle P C M=30^{\\circ}$ and has $C M=1$.\n\nThus, $P C=\\frac{2}{\\sqrt{3}}$.\n\n(There are many other techniques that we can use to calculate the length of $P C$.)\n\nReturning to $\\triangle A P C$, we see $A P^{2}=A C^{2}-P C^{2}$ or $h^{2}=2^{2}-\\left(\\frac{2}{\\sqrt{3}}\\right)^{2}=4-\\frac{4}{3}=\\frac{8}{3}$, and so $h=\\sqrt{\\frac{8}{3}}=2 \\sqrt{\\frac{2}{3}}=\\frac{2 \\sqrt{6}}{3} \\approx 1.630$.\n\nTherefore, the height is approximately $1.63 \\mathrm{~m}$ or $163 \\mathrm{~cm}$." ]

[ "163" ]

[ "Since there are $N=7$ points from which the triangle's vertices can be chosen, there are $\\left(\\begin{array}{l}7 \\\\ 3\\end{array}\\right)=35$ triangles in total.\n\nWe compute the number of acute triangles.\n\nFix one of the vertices of such a triangle at $A_{1}$.\n\nWe construct the triangle by choosing the other two vertices in ascending subscript order. We choose the vertices by considering the arc length from the previous vertex - each of\n\n\n\nthese arc lengths must be smaller than half the total circumference of the circle.\n\nSince there are 7 equally spaced points on the circle, we assume the circumference is 7 , so the arc length formed by each side must be at most 3 .\n\nSince the first arc length is at most 3 , the second point can be only $A_{2}, A_{3}$ or $A_{4}$.\n\nIf the second point is $A_{2}$, then since the second and third arc lengths are each at most 3 , then the third point must be $A_{5}$. (Since the second arc length is at most 3, then the third point cannot be any further along than $A_{5}$. However, the arc length from $A_{5}$ around to $A_{1}$ is 3 , so it cannot be any closer than $A_{5}$.)\n\n<img_4024>\n\nIf the second point is $A_{3}$, the third point must be $A_{5}$ or $A_{6}$. If the second point is $A_{4}$, the third point must be $A_{5}$ or $A_{6}$ or $A_{7}$. Therefore, there are 6 acute triangles which include $A_{1}$ as one of its vertices.\n\nHow many acute triangles are there in total?\n\nWe can repeat the above process for each of the 6 other points, giving $7 \\times 6=42$ acute triangles.\n\nBut each triangle is counted three times here, as it has been counted once for each of its vertices.\n\nThus, there are $\\frac{7 \\times 6}{3}=14$ acute triangles.\n\nTherefore, the probability that a randomly chosen triangle is acute if $\\frac{14}{35}=\\frac{2}{5}$." ]

[ "$\\frac{2}{5}$" ]

[ "Since $\\triangle P Q S$ is right-angled at $P$, then by the Pythagorean Theorem,\n\n$$\nS Q^{2}=S P^{2}+P Q^{2}=(x+3)^{2}+x^{2}\n$$\n\nSince $\\triangle Q R S$ is right-angled at $Q$, then by the Pythagorean Theorem, we obtain\n\n$$\n\\begin{aligned}\nR S^{2} & =S Q^{2}+Q R^{2} \\\\\n(x+8)^{2} & =\\left((x+3)^{2}+x^{2}\\right)+8^{2} \\\\\nx^{2}+16 x+64 & =x^{2}+6 x+9+x^{2}+64 \\\\\n0 & =x^{2}-10 x+9 \\\\\n0 & =(x-1)(x-9)\n\\end{aligned}\n$$\n\nand so $x=1$ or $x=9$.\n\n(We can check that if $x=1, \\triangle P Q S$ has sides of lengths 4,1 and $\\sqrt{17}$ and $\\triangle Q R S$ has sides of lengths $\\sqrt{17}, 8$ and 9 , both of which are right-angled, and if $x=9, \\triangle P Q S$ has sides of lengths 12,9 and 15 and $\\triangle Q R S$ has sides of lengths 15,8 and 17 , both of which are right-angled.)\n\nIn terms of $x$, the perimeter of $P Q R S$ is $x+8+(x+8)+(x+3)=3 x+19$.\n\nThus, the possible perimeters of $P Q R S$ are 22 (when $x=1$ ) and 46 (when $x=9$ )." ]

[ "22,46" ]

[ "Since $\\frac{A C}{A D}=\\frac{3}{4}$, then we let $A C=3 t$ and $A D=4 t$ for some real number $t>0$.\n\n<img_3309>\n\nUsing the cosine law in $\\triangle A C D$, the following equations are equivalent:\n\n$$\n\\begin{aligned}\nA D^{2} & =A C^{2}+C D^{2}-2 \\cdot A C \\cdot C D \\cdot \\cos (\\angle A C D) \\\\\n(4 t)^{2} & =(3 t)^{2}+1^{2}-2(3 t)(1)\\left(-\\frac{3}{5}\\right) \\\\\n16 t^{2} & =9 t^{2}+1+\\frac{18}{5} t \\\\\n80 t^{2} & =45 t^{2}+5+18 t \\\\\n35 t^{2}-18 t-5 & =0 \\\\\n(7 t-5)(5 t+1) & =0\n\\end{aligned}\n$$\n\nSince $t>0$, then $t=\\frac{5}{7}$.\n\nThus, $A C=3 t=\\frac{15}{7}$.\n\nUsing the cosine law in $\\triangle A C B$ and noting that\n\n$$\n\\cos (\\angle A C B)=\\cos \\left(180^{\\circ}-\\angle A C D\\right)=-\\cos (\\angle A C D)=\\frac{3}{5}\n$$\n\n\n\nthe following equations are equivalent:\n\n$$\n\\begin{aligned}\nA B^{2} & =A C^{2}+B C^{2}-2 \\cdot A C \\cdot B C \\cdot \\cos (\\angle A C B) \\\\\n& =\\left(\\frac{15}{7}\\right)^{2}+2^{2}-2\\left(\\frac{15}{7}\\right)(2)\\left(\\frac{3}{5}\\right) \\\\\n& =\\frac{225}{49}+4-\\frac{36}{7} \\\\\n& =\\frac{225}{49}+\\frac{196}{49}-\\frac{252}{49} \\\\\n& =\\frac{169}{49}\n\\end{aligned}\n$$\n\nSince $A B>0$, then $A B=\\frac{13}{7}$." ]

[ "$\\frac{13}{7}$" ]

[ "The parabola with equation $y=a x^{2}+2$ is symmetric about the $y$-axis.\n\nThus, its vertex occurs when $x=0$ (which gives $y=a \\cdot 0^{2}+2=2$ ) and so $V$ has coordinates $(0,2)$.\n\nTo find the coordinates of $B$ and $C$, we use the equations of the parabola and line to obtain\n\n$$\n\\begin{aligned}\na x^{2}+2 & =-x+4 a \\\\\na x^{2}+x+(2-4 a) & =0\n\\end{aligned}\n$$\n\nUsing the quadratic formula,\n\n$$\nx=\\frac{-1 \\pm \\sqrt{1^{2}-4 a(2-4 a)}}{2 a}=\\frac{-1 \\pm \\sqrt{1-8 a+16 a^{2}}}{2 a}\n$$\n\nSince $1-8 a+16 a^{2}=(4 a-1)^{2}$ and $4 a-1>0\\left(\\right.$ since $\\left.a>\\frac{1}{2}\\right)$, then $\\sqrt{1-8 a+16 a^{2}}=4 a-1$ and so\n\n$$\nx=\\frac{-1 \\pm(4 a-1)}{2 a}\n$$\n\nwhich means that $x=\\frac{4 a-2}{2 a}=\\frac{2 a-1}{a}=2-\\frac{1}{a}$ or $x=\\frac{-4 a}{2 a}=-2$.\n\nWe can use the equation of the line to obtain the $y$-coordinates of $B$ and $C$.\n\nWhen $x=-2$ (corresponding to point $B$ ), we obtain $y=-(-2)+4 a=4 a+2$.\n\nWhen $x=2-\\frac{1}{a}$ (corresponding to point $C$ ), we obtain $y=-\\left(2-\\frac{1}{a}\\right)+4 a=4 a-2+\\frac{1}{a}$.\n\nLet $P$ and $Q$ be the points on the horizontal line through $V$ so that $B P$ and $C Q$ are perpendicular to $P Q$.\n\n<img_3382>\n\n\n\nThen the area of $\\triangle V B C$ is equal to the area of trapezoid $P B C Q$ minus the areas of right-angled $\\triangle B P V$ and right-angled $\\triangle C Q V$.\n\nSince $B$ has coordinates $(-2,4 a+2), P$ has coordinates $(-2,2), V$ has coordiantes $(0,2)$, $Q$ has coordinates $\\left(2-\\frac{1}{a}, 2\\right)$, and $C$ has coordinates $\\left(2-\\frac{1}{a}, 4 a-2+\\frac{1}{a}\\right)$, then\n\n$$\n\\begin{aligned}\nB P & =(4 a+2)-2=4 a \\\\\nC Q & =\\left(4 a-2+\\frac{1}{a}\\right)-2=4 a-4+\\frac{1}{a} \\\\\nP V & =0-(-2)=2 \\\\\nQ V & =2-\\frac{1}{a}-0=2-\\frac{1}{a} \\\\\nP Q & =P V+Q V=2+2-\\frac{1}{a}=4-\\frac{1}{a}\n\\end{aligned}\n$$\n\nTherefore, the area of trapezoid $P B C Q$ is\n\n$$\n\\frac{1}{2}(B P+C Q)(P Q)=\\frac{1}{2}\\left(4 a+4 a-4+\\frac{1}{a}\\right)\\left(4-\\frac{1}{a}\\right)=\\left(4 a-2+\\frac{1}{2 a}\\right)\\left(4-\\frac{1}{a}\\right)\n$$\n\nAlso, the area of $\\triangle B P V$ is $\\frac{1}{2} \\cdot B P \\cdot P V=\\frac{1}{2}(4 a)(2)=4 a$.\n\nFurthermore, the area of $\\triangle C Q V$ is\n\n$$\n\\frac{1}{2} \\cdot C Q \\cdot Q V=\\frac{1}{2}\\left(4 a-4+\\frac{1}{a}\\right)\\left(2-\\frac{1}{a}\\right)=\\left(2 a-2+\\frac{1}{2 a}\\right)\\left(2-\\frac{1}{a}\\right)\n$$\n\nFrom the given information,\n\n$$\n\\left(4 a-2+\\frac{1}{2 a}\\right)\\left(4-\\frac{1}{a}\\right)-4 a-\\left(2 a-2+\\frac{1}{2 a}\\right)\\left(2-\\frac{1}{a}\\right)=\\frac{72}{5}\n$$\n\nMultiplying both sides by $2 a^{2}$, which we distribute through the factors on the left side as $2 a \\cdot a$, we obtain\n\n$$\n\\left(8 a^{2}-4 a+1\\right)(4 a-1)-8 a^{3}-\\left(4 a^{2}-4 a+1\\right)(2 a-1)=\\frac{144}{5} a^{2}\n$$\n\nMultiplying both sides by 5 , we obtain\n\n$$\n5\\left(8 a^{2}-4 a+1\\right)(4 a-1)-40 a^{3}-5\\left(4 a^{2}-4 a+1\\right)(2 a-1)=144 a^{2}\n$$\n\nExpanding and simplifying, we obtain\n\n$$\n\\begin{aligned}\n\\left(160 a^{3}-120 a^{2}+40 a-5\\right)-40 a^{3}-\\left(40 a^{3}-60 a^{2}+30 a-5\\right) & =144 a^{2} \\\\\n80 a^{3}-204 a^{2}+10 a & =0 \\\\\n2 a\\left(40 a^{2}-102 a+5\\right) & =0 \\\\\n2 a(20 a-1)(2 a-5) & =0\n\\end{aligned}\n$$\n\nand so $a=0$ or $a=\\frac{1}{20}$ or $a=\\frac{5}{2}$. Since $a>\\frac{1}{2}$, then $a=\\frac{5}{2}$." ]

[ "$\\frac{5}{2}$" ]

[ "The $(4,2)$-sawtooth sequence consists of the terms\n\n$$\n1, \\quad 2,3,4,3,2,1, \\quad 2,3,4,3,2,1\n$$\n\nwhose sum is 31 ." ]

[ "31" ]

[ "Suppose that $m \\geq 2$.\n\nThe $(m, 3)$-sawtooth sequence consists of an initial 1 followed by 3 teeth, each of which goes from 2 to $m$ to 1 .\n\nConsider one of these teeth whose terms are\n\n$$\n2,3,4, \\ldots, m-1, m, m-1, m-2, m-3, \\ldots, 2,1\n$$\n\nWhen we write the ascending portion directly above the descending portion, we obtain\n\n$$\n\\begin{aligned}\n& 2, \\quad 3, \\quad 4, \\quad \\ldots, \\quad m-1, \\quad m \\\\\n& m-1, \\quad m-2, \\quad m-3, \\quad \\ldots, \\quad 2, \\quad 1\n\\end{aligned}\n$$\n\nFrom this presentation, we can see $m-1$ pairs of terms, the sum of each of which is $m+1$. $($ Note that $2+(m-1)=3+(m-2)=4+(m-3)=\\cdots=(m-1)+2=m+1$ and as we move from left to right, the terms on the top increase by 1 at each step and the terms on the bottom decrease by 1 at each step, so their sum is indeed constant.) Therefore, the sum of the numbers in one of the teeth is $(m-1)(m+1)=m^{2}-1$. This means that the sum of the terms in the $(m, 3)$-sawtooth sequence is $1+3\\left(m^{2}-1\\right)$, which equals $3 m^{2}-2$.", "Suppose that $m \\geq 2$.\n\nThe $(m, 3)$-sawtooth sequence consists of an initial 1 followed by 3 teeth, each of which goes from 2 to $m$ to 1 .\n\nConsider one of these teeth whose terms are\n\n$$\n2,3,4, \\ldots, m-1, m, m-1, m-2, m-3, \\ldots, 2,1\n$$\n\nThis tooth includes one 1 , two $2 \\mathrm{~s}$, two $3 \\mathrm{~s}$, and so on, until we reach two $(m-1) \\mathrm{s}$, and one $m$.\n\nThe sum of these numbers is\n\n$$\n1(1)+2(2)+2(3)+\\cdots+2(m-1)+m\n$$\n\nwhich can be rewritten as\n\n$$\n2(1+2+3+\\cdots+(m-1)+m)-1-m=2 \\cdot \\frac{1}{2} m(m+1)-m-1=m^{2}+m-m-1=m^{2}-1\n$$\n\nTherefore, the sum of the numbers in one of the teeth is $(m-1)(m+1)=m^{2}-1$.\n\nThis means that the sum of the terms in the $(m, 3)$-sawtooth sequence is $1+3\\left(m^{2}-1\\right)$, which equals $3 m^{2}-2$." ]

[ "$3 m^{2}-2$" ]

[ "Assume that the first topping is placed on the top half of the pizza. (We can rotate the pizza so that this is the case.)\n\nAssume that the second topping is placed on the half of the pizza that is above the horizontal diameter that makes an angle of $\\theta$ clockwise with the horizontal as shown. In other words, the topping covers the pizza from $\\theta$ to $\\theta+180^{\\circ}$.\n<img_3295>\n\nWe may assume that $0^{\\circ} \\leq \\theta \\leq 360^{\\circ}$.\n\nWhen $0^{\\circ} \\leq \\theta \\leq 90^{\\circ}$, the angle of the sector covered by both toppings is at least $90^{\\circ}$ (and so is at least a quarter of the circle).\n\nWhen $90^{\\circ}<\\theta \\leq 180^{\\circ}$, the angle of the sector covered by both toppings is less than $90^{\\circ}$ (and so is less than a quarter of the circle).\n\nWhen $\\theta$ moves past $180^{\\circ}$, the left-hand portion of the upper half circle starts to be covered with both toppings again. When $180^{\\circ} \\leq \\theta<270^{\\circ}$, the angle of the sector covered by both toppings is less than $90^{\\circ}$ (and so is less than a quarter of the circle).\n\nWhen $270^{\\circ} \\leq \\theta \\leq 360^{\\circ}$, the angle of the sector covered by both toppings at least $90^{\\circ}$ (and so is at least a quarter of the circle).\n\nTherefore, if $\\theta$ is chosen randomly between $0^{\\circ}$ and $360^{\\circ}$, the combined length of the intervals in which at least $\\frac{1}{4}$ of the pizza is covered with both toppings is $180^{\\circ}$.\n\nTherefore, the probability is $\\frac{180^{\\circ}}{360^{\\circ}}$, or $\\frac{1}{2}$." ]

[ "$\\frac{1}{2}$" ]

[ "Suppose that the first topping is placed on the top half of the pizza. (Again, we can rotate the pizza so that this is the case.)\n\nAssume that the second topping is placed on the half of the pizza that is above the diameter that makes an angle of $\\theta$ clockwise with the horizontal as shown. In other words, the topping covers the pizza from $\\theta$ to $\\theta+180^{\\circ}$.\n\nWe may assume that $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$. If $180^{\\circ} \\leq \\theta \\leq 360^{\\circ}$, the resulting pizza can be seen as a reflection of the one shown.\n<img_4026>\n\nConsider the third diameter added, shown dotted in the diagram above. Suppose that its angle with the horizontal is $\\alpha$. (In the diagram, $\\alpha<90^{\\circ}$.) We assume that the topping is added on the half pizza clockwise beginning at the angle of $\\alpha$, and that this topping stays in the same relative position as the diameter sweeps around the circle.\n\nFor what angles $\\alpha$ will there be a portion of the pizza covered with all three toppings?\n\nIf $0^{\\circ} \\leq \\alpha<180^{\\circ}$, there will be a portion covered with three toppings; this portion is above the right half of the horizontal diameter.\n\nIf $180^{\\circ} \\leq \\alpha<180^{\\circ}+\\theta$, the third diameter will pass through the two regions with angle $\\theta$ and the third topping will be below this diameter, so there will not be a region covered\n\n\n\nwith three toppings.\n\nIf $180^{\\circ}+\\theta \\leq \\alpha \\leq 360^{\\circ}$, the third topping starts to cover the leftmost part of the region currently covered with two toppings, and so a region is covered with three toppings.\n\nTherefore, for an angle $\\theta$ with $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$, a region of the pizza is covered with three toppings when $0^{\\circ} \\leq \\alpha<180^{\\circ}$ and when $180^{\\circ}+\\theta \\leq \\alpha \\leq 360^{\\circ}$.\n\nTo determine the desired probability, we graph points $(\\theta, \\alpha)$. A particular choice of diameters corresponds to a choice of angles $\\theta$ and $\\alpha$ with $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$ and $0^{\\circ} \\leq \\alpha \\leq 360^{\\circ}$, which corresponds to a point on the graph below.\n\nThe probability that we are looking for then equals the area of the region of this graph where three toppings are in a portion of the pizza divided by the total allowable area of the graph.\n\nThe shaded region of the graph corresponds to instances where a portion of the pizza will be covered by three toppings.\n\n<img_3744>\n\nThis shaded region consists of the entire portion of the graph where $0^{\\circ} \\leq \\alpha \\leq 180^{\\circ}$ (regardless of $\\theta$ ) as well as the region above the line with equation $\\alpha=\\theta+180^{\\circ}$ (that is, the region with $\\theta+180^{\\circ} \\leq \\alpha \\leq 360^{\\circ}$ ).\n\nSince the slope of the line is 1 , it divides the upper half of the region, which is a square, into two pieces of equal area.\n\nTherefore, $\\frac{3}{4}$ of the graph is shaded, which means that the probability that a region of the pizza is covered by all three toppings is $\\frac{3}{4}$." ]

[ "$\\frac{3}{4}$" ]

[ "The main idea of this solution is that the toppings all overlap exactly when there is one topping with the property that all other toppings \"begin\" somewhere in that toppings semi-circle. In the rest of this solution, we determine the probability using this fact and then justify this fact.\n\nSuppose that, for $1 \\leq j \\leq N$, topping $j$ is put on the semi-circle that starts at an angle of $\\theta_{j}$ clockwise from the horizontal left-hand radius and continues to an angle of $\\theta_{j}+180^{\\circ}$, where $0^{\\circ} \\leq \\theta_{j}<360^{\\circ}$. By establishing these variables and this convention, we are fixing both the angle of the diameter and the semi-circle defined by this diameter on which the topping is placed.\n\nSuppose that there is some region of the pizza with non-zero area that is covered by all $N$ toppings.\n\nThis region will be a sector with two bounding radii, each of which must be half of a diameter that defines one of the toppings.\n\nSuppose that the radius at the clockwise \"end\" of the sector is the end of the semi-circle where topping $X$ is placed, and that the radius at the counter-clockwise \"beginning\" of the sector is the start of the semi-circle where topping $Y$ is placed.\n\n\n\n<img_3683>\n\nThis means that each of the other $N-2$ toppings begins between (in the clockwise sense) the points where topping $X$ begins and where topping $Y$ begins.\n\nConsider the beginning angle for topping $X, \\theta_{X}$.\n\nTo say that the other $N-1$ toppings begin at some point before topping $X$ ends is the same as saying that each $\\theta_{j}$ with $j \\neq X$ is between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$.\n\nHere, we can allow for the possibility that $\\theta_{X}+180^{\\circ}$ is greater than $360^{\\circ}$ by saying that an angle equivalent to $\\theta_{j}$ (which is either $\\theta_{j}$ or $\\theta_{j}+360^{\\circ}$ ) is between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$. For each $j \\neq X$, the angle $\\theta_{j}$ is randomly, uniformly and independently chosen on the circle, so there is a probability of $\\frac{1}{2}$ that this angle (or its equivalent) will be in the semicircle between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$.\n\nSince there are $N-1$ such angles, the probability that all are between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$ is $\\frac{1}{2^{N-1}}$.\n\nSince there are $N$ possible selections for the first topping that can end the common sector, then the desired probability will be $\\frac{N}{2^{N-1}}$ as long as we can show that no set of angles can give two different sectors that are both covered with all toppings.\n\nTo show this last fact, we suppose without loss of generality that\n\n$$\n0^{\\circ}=\\theta_{1}<\\theta_{2}<\\theta_{3}<\\cdots<\\theta_{N-1}<\\theta_{N}<180^{\\circ}\n$$\n\n(We can relabel the toppings if necessary to obtain this order and rotate the pizza so that topping 1 begins at $0^{\\circ}$.)\n\nWe need to show that it is not possible to have a $Z$ for which $\\theta_{Z}, \\theta_{Z+1}, \\ldots, \\theta_{N}, \\theta_{1}, \\theta_{2}, \\ldots, \\theta_{Z-1}$ all lie in a semi-circle starting with $\\theta_{Z}$.\n\nSince $\\theta_{Z}<180^{\\circ}$ and $\\theta_{1}$ can be thought of as $360^{\\circ}$, then this is not possible as $\\theta_{1}$ and the angles after it are all not within $180^{\\circ}$ of $\\theta_{Z}$.\n\nTherefore, it is not possible to have two such regions with the same set of angles, and so the desired probability is $\\frac{N}{2^{N-1}}$." ]

[ "$\\frac{N}{2^{N-1}}$" ]

[ "Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.\n\nBy the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then\n\n$$\nB D=\\sqrt{30^{2}+40^{2}}=\\sqrt{900+1600}=\\sqrt{2500}=50\n$$\n\nWe calculate the area of $\\triangle A D B$ is two different ways.\n\nFirst, using $A B$ as base and $A D$ as height, we obtain an area of $\\frac{1}{2}(40)(30)=600$.\n\nNext, using $D B$ as base and $A F$ as height, we obtain an area of $\\frac{1}{2}(50) x=25 x$.\n\nWe must have $25 x=600$ and so $x=\\frac{600}{25}=24$.", "Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.\n\nBy the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then\n\n$$\nB D=\\sqrt{30^{2}+40^{2}}=\\sqrt{900+1600}=\\sqrt{2500}=50\n$$\n\nSince $\\triangle D A B$ is right-angled at $A$, then $\\sin (\\angle A D B)=\\frac{A B}{B D}=\\frac{40}{50}=\\frac{4}{5}$.\n\nBut $\\triangle A D F$ is right-angled at $F$ and $\\angle A D F=\\angle A D B$.\n\nTherefore, $\\sin (\\angle A D F)=\\frac{A F}{A D}=\\frac{x}{30}$.\n\nThus, $\\frac{x}{30}=\\frac{4}{5}$ and so $x=\\frac{4}{5}(30)=24$.\n\nSolution 3\n\nSince $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.\n\nBy the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then\n\n$$\nB D=\\sqrt{30^{2}+40^{2}}=\\sqrt{900+1600}=\\sqrt{2500}=50\n$$\n\nNote that $\\triangle B F A$ is similar to $\\triangle B A D$, since each is right-angled and they share a common angle at $B$.\n\nThus, $\\frac{A F}{A B}=\\frac{A D}{B D}$ and so $\\frac{x}{30}=\\frac{40}{50}$ which gives $x=\\frac{30(40)}{50}=24$." ]

[ "24" ]

[ "Since $\\triangle A B C$ is right-angled at $C$, then $\\sin B=\\cos A$.\n\nTherefore, $2 \\cos A=3 \\tan A=\\frac{3 \\sin A}{\\cos A}$ or $2 \\cos ^{2} A=3 \\sin A$.\n\nUsing the fact that $\\cos ^{2} A=1-\\sin ^{2} A$, this becomes $2-2 \\sin ^{2} A=3 \\sin A$\n\nor $2 \\sin ^{2} A+3 \\sin A-2=0$ or $(2 \\sin A-1)(\\sin A+2)=0$.\n\nSince $\\sin A$ is between -1 and 1 , then $\\sin A=\\frac{1}{2}$.\n\nSince $A$ is an acute angle, then $A=30^{\\circ}$.", "Since $\\triangle A B C$ is right-angled at $C$, then $\\sin B=\\frac{b}{c}$ and $\\tan A=\\frac{a}{b}$.\n\nThus, the given equation is $\\frac{2 b}{c}=\\frac{3 a}{b}$ or $2 b^{2}=3 a c$.\n\nUsing the Pythagorean Theorem, $b^{2}=c^{2}-a^{2}$ and so we obtain $2 c^{2}-2 a^{2}=3 a c$ or $2 c^{2}-3 a c-2 a^{2}=0$.\n\nFactoring, we obtain $(c-2 a)(2 c+a)=0$.\n\nSince $a$ and $c$ must both be positive, then $c=2 a$.\n\nSince $\\triangle A B C$ is right-angled, the relation $c=2 a$ means that $\\triangle A B C$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, with $A=30^{\\circ}$." ]

[ "$30^{\\circ}$" ]

[ "Since Alice drives at $60 \\mathrm{~km} / \\mathrm{h}$, then she drives $1 \\mathrm{~km}$ every minute.\n\nSince Alice drove from $G$ to $F$ in 45 minutes, then the distance from $G$ to $F$ is $45 \\mathrm{~km}$.\n\nLet the distance from $E$ to $G$ be $d \\mathrm{~km}$ and let Bob's speed be $B \\mathrm{~km} / \\mathrm{h}$.\n\nSince Bob drove from $G$ to $E$ in 20 minutes (or $\\frac{1}{3}$ of an hour), then $\\frac{d}{B}=\\frac{1}{3}$. Thus, $d=\\frac{1}{3} B$.\n\nThe time that it took Bob to drive from $F$ to $G$ was $\\frac{45}{B}$ hours.\n\nThe time that it took Alice to drive from $E$ to $G$ was $\\frac{d}{60}$ hours.\n\nSince the time that it took each of Alice and Bob to reach $G$ was the same, then $\\frac{d}{60}=\\frac{45}{B}$\n\nand so $B d=45(60)=2700$.\n\nThus, $B\\left(\\frac{1}{3} B\\right)=2700$ so $B^{2}=8100$ or $B=90$ since $B>0$.\n\nTherefore, Bob's speed was $90 \\mathrm{~km} / \\mathrm{h}$." ]

[ "90" ]

[ "First, we determine the coordinates of $A$.\n\nThe area of $\\triangle A B C$ is 4 . We can think of $A C$ as its base, and its height being the distance from $B$ to the $x$-axis.\n\nIf the coordinates of $A$ are $(a, 0)$, then the base has length $4-a$ and the height is 4 .\n\nThus, $\\frac{1}{2}(4-a)(4)=4$, so $4-a=2$ and so $a=2$.\n\nTherefore, the coordinates of $A$ are $(2,0)$.\n\nNext, we determine the equation of the parabola.\n\nThe parabola has $x$-intercepts 2 and 4 , so has equation $y=k(x-2)(x-4)$.\n\nSince the parabola passes through $(0,-4)$ as well, then $-4=k(-2)(-4)$ so $k=-\\frac{1}{2}$.\n\nTherefore, the parabola has equation $y=-\\frac{1}{2}(x-2)(x-4)$.\n\nNext, we determine the coordinates of $D$, the vertex of the parabola.\n\nSince the $x$-intercepts are 2 and 4 , then the $x$-coordinate of the vertex is the average of these, or 3.\n\n\n\nThe $y$-coordinate of $D$ can be obtained from the equation of the parabola; we obtain $y=-\\frac{1}{2}(3-2)(3-4)=-\\frac{1}{2}(1)(-1)=\\frac{1}{2}$.\n\nThus, the coordinates of $D$ are $\\left(3, \\frac{1}{2}\\right)$.\n\nLastly, we determine the area of $\\triangle B D C$, whose vertices have coordinates $B(0,-4)$, $D\\left(3, \\frac{1}{2}\\right)$, and $C(4,0)$.\n\nMethod 1 \n\nWe proceed be \"completing the rectangle\". That is, we draw the rectangle with horizontal sides along the lines $y=\\frac{1}{2}$ and $y=-4$ and vertical sides along the lines $x=0$ and $x=4$. We label this rectangle as $B P Q R$.\n\n<img_3586>\n\nThe area of $\\triangle B D C$ equals the area of the rectangle minus the areas of $\\triangle B P D, \\triangle D Q C$ and $\\triangle C R B$.\n\nRectangle $B P Q R$ has height $4+\\frac{1}{2}=\\frac{9}{2}$ and width 4 .\n\n$\\triangle B P D$ has height $\\frac{9}{2}$ and base 3 .\n\n$\\triangle D Q C$ has height $\\frac{1}{2}$ and base 1.\n\n$\\triangle C R B$ has height 4 and base 4.\n\nTherefore, the area of $\\triangle B D C$ is $4\\left(\\frac{9}{2}\\right)-\\frac{1}{2}\\left(\\frac{9}{2}\\right)(3)-\\frac{1}{2}\\left(\\frac{1}{2}\\right)(1)-\\frac{1}{2}(4)(4)=18-\\frac{27}{4}-\\frac{1}{4}-8=3$.\n\nMethod 2\n\nWe determine the coordinates of $E$, the point where $B D$ crosses the $x$-axis.\n\n<img_3686>\n\nOnce we have done this, then the area of $\\triangle B D C$ equals the sum of the areas of $\\triangle E C B$ and $\\triangle E C D$.\n\nSince $B$ has coordinates $(0,-4)$ and $D$ has coordinates $\\left(3, \\frac{1}{2}\\right)$, then the slope of $B D$ is $\\frac{\\frac{1}{2}-(-4)}{3-0}=\\frac{\\frac{9}{2}}{3}=\\frac{3}{2}$.\n\nSince $B$ is on the $y$-axis, then the equation of the line through $B$ and $D$ is $y=\\frac{3}{2} x-4$. To find the $x$-coordinate of $E$, we set $y=0$ to obtain $0=\\frac{3}{2} x-4$ or $\\frac{3}{2} x=4$ or $x=\\frac{8}{3}$.\n\nWe think of $E C$ as the base of each of the two smaller triangles. Note that $E C=4-\\frac{8}{3}=\\frac{4}{3}$. Thus, the area of $\\triangle E C D$ is $\\frac{1}{2}\\left(\\frac{4}{3}\\right)\\left(\\frac{1}{2}\\right)=\\frac{1}{3}$.\n\nAlso, the area of $\\triangle E C B$ is $\\frac{1}{2}\\left(\\frac{4}{3}\\right)(4)=\\frac{8}{3}$.\n\nTherefore, the area of $\\triangle B D C$ is $\\frac{1}{3}+\\frac{8}{3}=3$." ]

[ "3" ]

[ "Rotate a copy of $\\triangle P S U$ by $90^{\\circ}$ counterclockwise around $P$, forming a new triangle $P Q V$. Note that $V$ lies on the extension of $R Q$.\n\n<img_3428>\n\nThen $P V=P U$ by rotation.\n\nAlso, $\\angle V P T=\\angle V P Q+\\angle Q P T=\\angle U P S+\\angle Q P T=90^{\\circ}-\\angle U P T=90^{\\circ}-45^{\\circ}$.\n\nThis tells us that $\\triangle P T U$ is congruent to $\\triangle P T V$, by \"side-angle-side\".\n\nThus, the perimeter of $\\triangle R U T$ equals\n\n$$\n\\begin{aligned}\nU R+R T+U T & =U R+R T+T V \\\\\n& =U R+R T+T Q+Q V \\\\\n& =U R+R Q+S U \\\\\n& =S U+U R+R Q \\\\\n& =S R+R Q \\\\\n& =8\n\\end{aligned}\n$$\n\nThat is, the perimeter of $\\triangle R U T$ always equals 8 , so the maximum possible perimeter is 8 .", "Let $\\angle S P U=\\theta$. Note that $0^{\\circ} \\leq \\theta \\leq 45^{\\circ}$.\n\nThen $\\tan \\theta=\\frac{S U}{P S}$, so $S U=4 \\tan \\theta$.\n\nSince $S R=4$, then $U R=S R-S U=4-4 \\tan \\theta$.\n\nSince $\\angle U P T=45^{\\circ}$, then $\\angle Q P T=90^{\\circ}-45^{\\circ}-\\theta=45^{\\circ}-\\theta$.\n\nThus, $\\tan \\left(45^{\\circ}-\\theta\\right)=\\frac{Q T}{P Q}$ and so $Q T=4 \\tan \\left(45^{\\circ}-\\theta\\right)$.\n\nSince $Q R=4$, then $R T=4-4 \\tan \\left(45^{\\circ}-\\theta\\right)$.\n\nBut $\\tan (A-B)=\\frac{\\tan A-\\tan B}{1+\\tan A \\tan B}$, so $\\tan \\left(45^{\\circ}-\\theta\\right)=\\frac{\\tan \\left(45^{\\circ}\\right)-\\tan \\theta}{1+\\tan \\left(45^{\\circ}\\right) \\tan \\theta}=\\frac{1-\\tan \\theta}{1+\\tan \\theta}$, since $\\tan \\left(45^{\\circ}\\right)=1$.\n\nThis gives $R T=4-4\\left(\\frac{1-\\tan \\theta}{1+\\tan \\theta}\\right)=\\frac{4+4 \\tan \\theta}{1+\\tan \\theta}-\\frac{4-4 \\tan \\theta}{1+\\tan \\theta}=\\frac{8 \\tan \\theta}{1+\\tan \\theta}$.\n\nBy the Pythagorean Theorem in $\\triangle U R T$, we obtain\n\n$$\n\\begin{aligned}\nU T & =\\sqrt{U R^{2}+R T^{2}} \\\\\n& =\\sqrt{(4-4 \\tan \\theta)^{2}+\\left(\\frac{8 \\tan \\theta}{1+\\tan \\theta}\\right)^{2}} \\\\\n& =4 \\sqrt{(1-\\tan \\theta)^{2}+\\left(\\frac{2 \\tan \\theta}{1+\\tan \\theta}\\right)^{2}} \\\\\n& =4 \\sqrt{\\left(\\frac{1-\\tan ^{2} \\theta}{1+\\tan \\theta}\\right)^{2}+\\left(\\frac{2 \\tan \\theta}{1+\\tan \\theta}\\right)^{2}} \\\\\n& =\\sqrt[4]{\\frac{1-2 \\tan ^{2} \\theta+\\tan ^{4} \\theta+4 \\tan ^{2} \\theta}{\\left(1+\\tan ^{2}\\right)^{2}}} \\\\\n& =\\sqrt[4]{\\frac{1+2 \\tan ^{2} \\theta+\\tan ^{4} \\theta}{\\left(1+\\tan ^{2}\\right.}} \\\\\n& =4 \\sqrt{\\frac{\\left(1+\\tan ^{2} \\theta\\right)^{2}}{\\left(1+\\tan ^{2}\\right.}} \\\\\n& =4\\left(\\frac{1+\\tan ^{2} \\theta}{1+\\tan ^{2}}\\right)\n\\end{aligned}\n$$\n\nTherefore, the perimeter of $\\triangle U R T$ is\n\n$$\n\\begin{aligned}\nU R+R T+U T & =4-4 \\tan \\theta+\\frac{8 \\tan \\theta}{1+\\tan \\theta}+4\\left(\\frac{1+\\tan ^{2} \\theta}{1+\\tan \\theta}\\right) \\\\\n& =4\\left(\\frac{1-\\tan ^{2} \\theta}{1+\\tan \\theta}+\\frac{2 \\tan \\theta}{1+\\tan \\theta}+\\frac{1+\\tan ^{2} \\theta}{1+\\tan \\theta}\\right) \\\\\n& =4\\left(\\frac{2+2 \\tan \\theta}{1+\\tan \\theta}\\right) \\\\\n& =8\n\\end{aligned}\n$$\n\nThus, the perimeter is always 8 , regardless of the value of $\\theta$, so the maximum possible perimeter is 8 ." ]

[ "8" ]

[ "Suppose that $p_{1}=1$.\n\nThen $p_{2}=p_{7}=0$, so the string is of the form $10 p_{3} p_{4} p_{5} p_{6} 0$.\n\nSince $k=3$, then 2 of $p_{3}, p_{4}, p_{5}, p_{6}$ equal 1 , but in such a way that no two adjacent digits are both 1 .\n\nThe possible strings in this case are 1010100, 1010010 and 1001010.\n\nSuppose that $p_{1}=0$. Then $p_{2}$ can equal 1 or 0 .\n\nIf $p_{2}=1$, then $p_{3}=0$ as well. This means that the string is of the form $010 p_{4} p_{5} p_{6} p_{7}$, which is the same as the general string in the first case, but shifted by 1 position around the circle, so there are again 3 possibilities.\n\nIf $p_{2}=0$, then the string is of the form $00 p_{3} p_{4} p_{5} p_{6} p_{7}$ and 3 of the digits $p_{3}, p_{4}, p_{5}, p_{6}, p_{7}$ equal 1 in such a way that no 2 adjacent digits equal 1.\n\nThere is only 1 way in which this can happen: 0010101.\n\nOverall, this gives 7 possible configurations, so $f(7,3)=7$." ]

[ "7" ]

[ "We develop an explicit formula for $f(n, k)$ by building these strings.\n\nConsider the allowable strings of length $n$ that include $k$ 1's. Either $p_{n}=0$ or $p_{n}=1$.\n\nConsider first the case when $p_{n}=0$. (Here, $p_{1}$ can equal 0 or 1.)\n\nThese strings are all of the form $p_{1} p_{2} p_{3} \\cdots p_{n-1} 0$.\n\nIn this case, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nThus, these strings can all be built using $k 10$ blocks and $n-2 k 0$ blocks. This gives $k$ 1 's and $k+(n-2 k)=n-k 0$ 's. Note that any string built with these blocks will be allowable and will end with a 0 , and any such allowable string can be built in this way.\n\nThe number of ways of arranging $k$ blocks of one kind and $n-2 k$ blocks of another kind is $\\left(\\begin{array}{c}k+(n-2 k) \\\\ k\\end{array}\\right)$, which simplifies to $\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)$.\n\nConsider next the case when $p_{n}=1$.\n\nHere, we must have $p_{n-1}=p_{1}=0$, since these are the two digits adjacent to $p_{n}$.\n\nThus, these strings are all of the form $0 p_{2} p_{3} \\cdots 01$.\n\nConsider the strings formed by removing the first and last digits.\n\nThese strings are allowable, are of length $n-2$, include $k-11$ 's, end with 0 , and can begin with 0 or 1 .\n\nAgain, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nTranslating our method of counting from the first case, there are $\\left(\\begin{array}{c}(n-2)-(k-1) \\\\ k-1\\end{array}\\right)$ or\n\n\n\n$\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nThus, $f(n, k)=\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)+\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nIn order to look at divisibility, we need to first simplify the formula:\n\n$$\n\\begin{aligned}\nf(n, k) & =\\left(\\begin{array}{c}\nn-k \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right) \\\\\n& =\\frac{(n-k) !}{k !(n-k-k) !}+\\frac{(n-k-1) !}{(k-1) !((n-k-1)-(k-1)) !} \\\\\n& =\\frac{(n-k) !}{k !(n-2 k) !}+\\frac{(n-k-1) !}{(k-1) !(n-2 k) !} \\\\\n& =\\frac{(n-k-1) !(n-k)}{k !(n-2 k) !}+\\frac{(n-k-1) ! k}{k !(n-2 k) !} \\\\\n& =\\frac{(n-k-1) !(n-k+k)}{k !(n-2 k) !} \\\\\n& =\\frac{n(n-k-1) !}{k !(n-2 k) !} \\\\\n& =\\frac{n(n-k-1)(n-k-2) \\cdots(n-2 k+2)(n-2 k+1)}{k !}\n\\end{aligned}\n$$\n\nNow that we have written $f(n, k)$ as a product, it is significantly easier to look at divisibility.\n\nNote that $2009=41 \\times 49=7^{2} \\times 41$, so we need $f(n, k)$ to be divisible by 41 and by 7 twice. For this to be the case, the numerator of $f(n, k)$ must have at least one more factor of 41 and at least two more factors of 7 than the denominator.\n\nAlso, we want to minimize $n+k$, so we work to keep $n$ and $k$ as small as possible.\n\nIf $n=49$ and $k=5$, then\n\n$$\nf(49,5)=\\frac{49(43)(42)(41)(40)}{5 !}=\\frac{49(43)(42)(41)(40)}{5(4)(3)(2)(1)}=49(43)(14)\n$$\n\nwhich is divisible by 2009 .\n\nWe show that this pair minimizes the value of $n+k$ with a value of 54 .\n\nWe consider the possible cases by looking separately at the factors of 41 and 7 that must occur. We focus on the factor of 41 first.\n\nFor the numerator to contain a factor of 41 , either $n$ is divisible by 41 or one of the terms in the product $(n-k-1)(n-k-2) \\cdots(n-2 k+1)$ is divisible by 41 .\n\nCase 1: $n$ is divisible by 41\n\nWe already know that $n=82$ is too large, so we consider $n=41$. From the original interpretation of $f(n, k)$, we see that $k \\leq 20$, as there can be no more than 20 gifts placed on 41 plates.\n\nHere, the numerator becomes 41 times the product of $k-1$ consecutive integers, the largest of which is $40-k$.\n\nNow the numerator must also contain at least two factors of 7 more than the denominator. But the denominator is the product of $k$ consecutive integers. Since the numerator contains the product of $k-1$ consecutive integers and the denominator contains the product of $k$ consecutive integers, then the denominator will always include at least as many multiples of 7 as the numerator (since there are more consecutive integers in the product in the denominator). Thus, it is impossible for the numerator to contain even one more\n\n\n\nadditional factor of 7 than the denominator.\n\nTherefore, if $n=41$, then $f(n, k)$ cannot be divisible by 2009 .\n\nCase 2: $n$ is not divisible by 41\n\nThis means that the factor of 41 in the numerator must occur in the product\n\n$$\n(n-k-1)(n-k-2) \\cdots(n-2 k+1)\n$$\n\nIn this case, the integer 41 must occur in this product, since an occurrence of 82 would make $n$ greater than 82 , which does not minimize $n+k$.\n\nSo we try to find values of $n$ and $k$ that include the integer 41 in this list.\n\nNote that $n-k-1$ is the largest factor in the product and $n-2 k+1$ is the smallest.\n\nSince 41 is contained somewhere in the product, then $n-2 k+1 \\leq 41$ (giving $n \\leq 40+2 k$ ) and $41 \\leq n-k-1$ (giving $n \\geq 42+k$ ).\n\nCombining these restrictions, we get $42+k \\leq n \\leq 40+2 k$.\n\nNow, we focus on the factors of 7 .\n\nEither $n$ is not divisible by 7 or $n$ is divisible by 7 .\n\n* If $n$ is not divisible by 7 , then at least two factors of 7 must be included in the product\n\n$$\n(n-k-1)(n-k-2) \\cdots(n-2 k+1)\n$$\n\nwhich means that either $k \\geq 8$ (to give two multiples of 7 in this list of $k-1$ consecutive integers) or one of the factors is divisible by 49 .\n\n- If $k \\geq 8$, then $n \\geq 42+k \\geq 50$ so $n+k \\geq 58$, which is not minimal.\n- If one of the factors is a multiple of 49 , then 49 must be included in the list so $n-2 k+1 \\leq 49$ (giving $n \\leq 48+2 k$ ) and $49 \\leq n-k-1$ (giving $n \\geq 50+k$ ). In this case, we already know that $42+k \\leq n \\leq 40+2 k$ and now we also have $50+k \\leq n \\leq 48+2 k$.\n\nFor these ranges to overlap, we need $50+k \\leq 40+2 k$ and so $k \\geq 10$, which means that $n \\geq 50+k \\geq 60$, and so $n+k \\geq 70$, which is not minimal.\n\n* Next, we consider the case where $n$ is a multiple of 7 .\n\nHere, $42+k \\leq n \\leq 40+2 k$ (to include 41 in the product) and $n$ is a multiple of 7 .\n\nSince $k$ is at least 2 by definition, then $n \\geq 42+k \\geq 44$, so $n$ is at least 49 .\n\nIf $n$ was 56 or more, we do not get a minimal value for $n+k$.\n\nThus, we need to have $n=49$. In this case, we do not need to look for another factor of 7 in the list.\n\nTo complete this case, we need to find the smallest value of $k$ for which 49 is in the range from $42+k$ to $40+2 k$ because we need to have $42+k \\leq n \\leq 40+2 k$.\n\nThis value of $k$ is $k=5$, which gives $n+k=49+5=54$.\n\nSince $f(49,5)$ is divisible by 2009 , as determined above, then this is the case that minimizes $n+k$, giving a value of 54 ." ]

[ "54" ]

[ "The angles in a polygon with $n$ sides have a sum of $(n-2) \\cdot 180^{\\circ}$.\n\nThis means that the angles in a pentagon have a sum of $3 \\cdot 180^{\\circ}$ or $540^{\\circ}$, which means that each interior angle in a regular pentagon equals $\\frac{1}{5} \\cdot 540^{\\circ}$ or $108^{\\circ}$.\n\nAlso, each interior angle in a regular polygon with $n$ sides equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$. (This is the general version of the statement in the previous sentence.)\n\nConsider the portion of the regular polygon with $n$ sides that lies outside the pentagon and join the points from which the angles that measure $a^{\\circ}$ and $b^{\\circ}$ emanate to form a hexagon.\n\n<img_3516>\n\nThis polygon has 6 sides, and so the sum of its 6 angles is $4 \\cdot 180^{\\circ}$.\n\nFour of its angles are the original angles from the $n$-sided polygon, so each equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$.\n\nThe remaining two angles have measures $a^{\\circ}+c^{\\circ}$ and $b^{\\circ}+d^{\\circ}$.\n\nWe are told that $a^{\\circ}+b^{\\circ}=88^{\\circ}$.\n\nAlso, the angles that measure $c^{\\circ}$ and $d^{\\circ}$ are two angles in a triangle whose third angle is $108^{\\circ}$.\n\nThus, $c^{\\circ}+d^{\\circ}=180^{\\circ}-108^{\\circ}=72^{\\circ}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n4 \\cdot \\frac{n-2}{n} \\cdot 180^{\\circ}+88^{\\circ}+72^{\\circ} & =4 \\cdot 180^{\\circ} \\\\\n160^{\\circ} & =\\left(4-\\frac{4(n-2)}{n}\\right) \\cdot 180^{\\circ} \\\\\n160^{\\circ} & =\\frac{4 n-(4 n-8)}{n} \\cdot 180^{\\circ} \\\\\n\\frac{160^{\\circ}}{180^{\\circ}} & =\\frac{8}{n} \\\\\n\\frac{8}{9} & =\\frac{8}{n}\n\\end{aligned}\n$$\n\nand so the value of $n$ is 9 .", "The angles in a polygon with $n$ sides have a sum of $(n-2) \\cdot 180^{\\circ}$.\n\nThis means that the angles in a pentagon have a sum of $3 \\cdot 180^{\\circ}$ or $540^{\\circ}$, which means that each interior angle in a regular pentagon equals $\\frac{1}{5} \\cdot 540^{\\circ}$ or $108^{\\circ}$.\n\nAlso, each interior angle in a regular polygon with $n$ sides equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$. (This is the general version of the statement in the previous sentence.)\n\nConsider the portion of the regular polygon with $n$ sides that lies outside the pentagon.\n\n<img_3345>\n\nThis polygon has 7 sides, and so the sum of its 7 angles is $5 \\cdot 180^{\\circ}$.\n\nFour of its angles are the original angles from the $n$-sided polygon, so each equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$.\n\nTwo of its angles are the angles equal to $a^{\\circ}$ and $b^{\\circ}$, whose sum is $88^{\\circ}$.\n\nIts seventh angle is the reflex angle corresponding to the pentagon's angle of $108^{\\circ}$, which equals $360^{\\circ}-108^{\\circ}$ or $252^{\\circ}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n4 \\cdot \\frac{n-2}{n} \\cdot 180^{\\circ}+88^{\\circ}+252^{\\circ} & =5 \\cdot 180^{\\circ} \\\\\n340^{\\circ} & =\\left(5-\\frac{4(n-2)}{n}\\right) \\cdot 180^{\\circ} \\\\\n340^{\\circ} & =\\frac{5 n-(4 n-8)}{n} \\cdot 180^{\\circ} \\\\\n\\frac{340^{\\circ}}{180^{\\circ}} & =\\frac{n+8}{n} \\\\\n\\frac{17}{9} & =\\frac{n+8}{n} \\\\\n17 n & =9(n+8) \\\\\n17 n & =9 n+72 \\\\\n8 n & =72\n\\end{aligned}\n$$\n\nand so the value of $n$ is 9 ." ]

[ "9" ]

[ "Since $\\angle A O B=\\angle B O C=\\angle C O D=\\angle D O A$ and these angles form a complete circle around $O$, then $\\angle A O B=\\angle B O C=\\angle C O D=\\angle D O A=\\frac{1}{4} \\cdot 360^{\\circ}=90^{\\circ}$.\n\nJoin point $O$ to $P, B, Q, C, S, D, T$, and $A$.\n\n<img_3604>\n\nSince $P, Q, S$, and $T$ are points of tangency, then the radii meet the sides of $A B C D$ at right angles at these points.\n\nSince $A O=3$ and $O T=1$ and $\\angle O T A=90^{\\circ}$, then by the Pythagorean Theorem, $A T=\\sqrt{A O^{2}-O T^{2}}=\\sqrt{8}=2 \\sqrt{2}$.\n\nSince $\\triangle O T A$ is right-angled at $T$, then $\\angle T A O+\\angle A O T=90^{\\circ}$.\n\nSince $\\angle D O A=90^{\\circ}$, then $\\angle A O T+\\angle D O T=90^{\\circ}$.\n\nThus, $\\angle T A O=\\angle D O T$.\n\nThis means that $\\triangle A T O$ is similar to $\\triangle O T D$.\n\nThus, $\\frac{D T}{O T}=\\frac{O T}{A T}$ and so $D T=\\frac{O T^{2}}{A T}=\\frac{1}{2 \\sqrt{2}}$.\n\nSince $D S$ and $D T$ are tangents to the circle from the same point, then $D S=D T=\\frac{1}{2 \\sqrt{2}}$." ]

[ "$\\frac{1}{2 \\sqrt{2}}$" ]

[ "By the cosine law in $\\triangle C B A$,\n\n$$\n\\begin{aligned}\nC A^{2} & =C B^{2}+B A^{2}-2(C B)(B A) \\cos (\\angle C B A) \\\\\nC A^{2} & =16^{2}+21^{2}-2(16)(21) \\cos \\left(60^{\\circ}\\right) \\\\\nC A^{2} & =256+441-2(16)(21)\\left(\\frac{1}{2}\\right) \\\\\nC A^{2} & =256+441-(16)(21) \\\\\nC A^{2} & =361 \\\\\nC A & =\\sqrt{361}=19 \\quad(\\text { since } C A>0)\n\\end{aligned}\n$$\n\nIn $\\triangle C A D, \\angle C D A=180^{\\circ}-\\angle D C A-\\angle D A C=180^{\\circ}-45^{\\circ}-30^{\\circ}=105^{\\circ}$.\n\nBy the sine law in $\\triangle C D A$,\n\n$$\n\\begin{aligned}\n\\frac{C D}{\\sin (\\angle D A C)} & =\\frac{C A}{\\sin (\\angle C D A)} \\\\\nC D & =\\frac{19 \\sin \\left(30^{\\circ}\\right)}{\\sin \\left(105^{\\circ}\\right)} \\\\\nC D & =\\frac{19\\left(\\frac{1}{2}\\right)}{\\sin \\left(105^{\\circ}\\right)} \\\\\nC D & =\\frac{19}{2 \\sin \\left(105^{\\circ}\\right)} \\\\\nC D & \\approx 9.835\n\\end{aligned}\n$$\n\nso, to the nearest tenth, $C D$ equals 9.8 .\n\n(Note that we could have used\n\n$$\n\\begin{aligned}\n\\sin \\left(105^{\\circ}\\right) & =\\sin \\left(60^{\\circ}+45^{\\circ}\\right)=\\sin \\left(60^{\\circ}\\right) \\cos \\left(45^{\\circ}\\right)+\\cos \\left(60^{\\circ}\\right) \\sin \\left(45^{\\circ}\\right) \\\\\n& =\\frac{\\sqrt{3}}{2} \\cdot \\frac{1}{\\sqrt{2}}+\\frac{1}{2} \\cdot \\frac{1}{\\sqrt{2}}=\\frac{\\sqrt{3}+1}{2 \\sqrt{2}}\n\\end{aligned}\n$$\n\nto say that $C D=\\frac{19}{2\\left(\\frac{\\sqrt{3}+1}{2 \\sqrt{2}}\\right)}=\\frac{19 \\sqrt{2}}{\\sqrt{3}+1}$ exactly, and then evaluated this expression.)" ]

[ "$\\frac{19 \\sqrt{2}}{\\sqrt{3}+1}$" ]

[ "Consider $P$ on $A B$ with $C P$ perpendicular to $A B$. Note that $C P=12$.\n\nSince the small circle with centre $A$ is tangent to the large circle with centre $C$, then $A C$ equals the sum of the radii of these circles, or $A C=4+9=13$. Similarly, $B C=13$.\n\nThis tells us that $\\triangle A P C$ is congruent to $\\triangle B P C$ (they have equal hypotenuses and each is right-angled and has a common side), so $B P=A P$.\n\nBy the Pythagorean Theorem in $\\triangle A P C$,\n\n$$\nA P^{2}=A C^{2}-P C^{2}=13^{2}-12^{2}=169-144=25\n$$\n\nso $A P=5$ (since $A P>0)$.\n\nTherefore, $B P=A P=5$ and so $A B=10$.\n\nSince it takes the bug 5 seconds to walk this distance, then in 1 second, the bug walks a distance of $\\frac{10}{5}=2$." ]

[ "2" ]

[ "Let $A P=x$ and $Q P=h$.\n\nSince $Q P$ is parallel to $C B$, then $Q P$ is perpendicular to $B A$.\n\nConsider trapezoid $C B P Q$. We can think of this as having parallel bases of lengths 4 and $h$ and height 5 . Thus, its area is $\\frac{1}{2}(4+h)(5)$.\n\nHowever, we can also compute its area by adding the areas of $\\triangle C B R$ (which is $\\left.\\frac{1}{2}(4)(3)\\right)$, $\\triangle C R Q$ (which is given as 5), and $\\triangle R P Q$ (which is $\\left.\\frac{1}{2}(2)(h)\\right)$.\n\nThus,\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(4+h)(5) & =\\frac{1}{2}(4)(3)+5+\\frac{1}{2}(2)(h) \\\\\n20+5 h & =12+10+2 h \\\\\n3 h & =2 \\\\\nh & =\\frac{2}{3}\n\\end{aligned}\n$$\n\nNow, $\\triangle A P Q$ is similar to $\\triangle A B C$, as each has a right angle and they share a common angle at $A$. Thus,\n\n$$\n\\begin{aligned}\n\\frac{A P}{P Q} & =\\frac{A B}{B C} \\\\\n(A P)(B C) & =(P Q)(A B) \\\\\n4 x & =\\frac{2}{3}(x+5) \\\\\n4 x & =\\frac{2}{3} x+\\frac{10}{3} \\\\\n\\frac{10}{3} x & =\\frac{10}{3} \\\\\nx & =1\n\\end{aligned}\n$$\n\n\n\nTherefore, $A P=x=1$.", "Let $A P=x$ and $Q P=h$.\n\nSince $Q P$ is parallel to $C B$, then $Q P$ is perpendicular to $B A$.\n\nSince $\\triangle A B C$ is right-angled at $B$, its area is $\\frac{1}{2}(4)(5+x)=10+2 x$.\n\nHowever, we can look at the area of the $\\triangle A B C$ in terms of its four triangular pieces:\n\n$\\triangle C B R$ (which has area $\\left.\\frac{1}{2}(4)(3)\\right), \\triangle C R Q$ (which has area 5), $\\triangle Q P R$ (which has area $\\left.\\frac{1}{2} h(2)\\right)$, and $\\triangle Q P A$ (which has area $\\frac{1}{2} x h$ ).\n\nTherefore, $10+2 x=6+5+h+\\frac{1}{2} x h$ so $x h-4 x+2 h+2=0$.\n\nNow, $\\triangle A P Q$ is similar to $\\triangle A B C$, as each has a right angle and they share a common angle at $A$. Thus,\n\n$$\n\\begin{aligned}\n\\frac{A P}{P Q} & =\\frac{A B}{B C} \\\\\n(A P)(B C) & =(P Q)(A B) \\\\\nx(4) & =h(x+5) \\\\\n4 x & =h x+5 h \\\\\n-5 h & =h x-4 x\n\\end{aligned}\n$$\n\nSubstituting this into the equation above, $x h+2 h-4 x+2=0$ becomes $-5 h+2 h+2=0$ or $3 h=2$ or $h=\\frac{2}{3}$.\n\nLastly, we solve for $x$ by subsituting our value for $h$ : $-5\\left(\\frac{2}{3}\\right)=\\frac{2}{3} x-4 x$ or $-\\frac{10}{3}=-\\frac{10}{3} x$ and so $x=1$.\n\nTherefore, $A P=x=1$." ]

[ "1" ]

[ "Since sector $A O B$ is $\\frac{1}{6}$ of a circle with radius 18 , its area is $\\frac{1}{6}\\left(\\pi \\cdot 18^{2}\\right)$ or $54 \\pi$.\n\nFor the line $A P$ to divide this sector into two pieces of equal area, each piece has area $\\frac{1}{2}(54 \\pi)$ or $27 \\pi$.\n\nWe determine the length of $O P$ so that the area of $\\triangle P O A$ is $27 \\pi$.\n\nSince sector $A O B$ is $\\frac{1}{6}$ of a circle, then $\\angle A O B=\\frac{1}{6}\\left(360^{\\circ}\\right)=60^{\\circ}$.\n\nDrop a perpendicular from $A$ to $T$ on $O B$.\n\n<img_3642>\n\nThe area of $\\triangle P O A$ is $\\frac{1}{2}(O P)(A T)$.\n\n$\\triangle A O T$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nSince $A O=18$, then $A T=\\frac{\\sqrt{3}}{2}(A O)=9 \\sqrt{3}$.\n\nFor the area of $\\triangle P O A$ to equal $27 \\pi$, we have $\\frac{1}{2}(O P)(9 \\sqrt{3})=27 \\pi$ which gives $O P=\\frac{54 \\pi}{9 \\sqrt{3}}=\\frac{6 \\pi}{\\sqrt{3}}=2 \\sqrt{3} \\pi$.\n\n(Alternatively, we could have used the fact that the area of $\\triangle P O A$ is $\\frac{1}{2}(O A)(O P) \\sin (\\angle P O A)$.)" ]

[ "$2 \\sqrt{3} \\pi$" ]