Datasets:

Hothan

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OlympiadBench

like 11

[ "Suppose that Karuna and Jorge meet for the first time after $t_{1}$ seconds and for the second time after $t_{2}$ seconds.\n\nWhen they meet for the first time, Karuna has run partway from $A$ to $B$ and Jorge has run partway from $B$ to $A$.\n\n<img_3496>\n\nAt this instant, the sum of the distances that they have run equals the total distance from $A$ to $B$.\n\nSince Karuna runs at $6 \\mathrm{~m} / \\mathrm{s}$ for these $t_{1}$ seconds, she has run $6 t_{1} \\mathrm{~m}$.\n\nSince Jorge runs at $5 \\mathrm{~m} / \\mathrm{s}$ for these $t_{1}$ seconds, he has run $5 t_{1} \\mathrm{~m}$.\n\nTherefore, $6 t_{1}+5 t_{1}=297$ and so $11 t_{1}=297$ or $t_{1}=27$.\n\nWhen they meet for the second time, Karuna has run from $A$ to $B$ and is running back to $A$ and Jorge has run from $B$ to $A$ and is running back to $B$. This is because Jorge gets to $A$ halfway through his run before Karuna gets back to $A$ at the end of her run.\n\n<img_3542>\n\nSince they each finish running after 99 seconds, then each has $99-t_{2}$ seconds left to run. At this instant, the sum of the distances that they have left to run equals the total distance from $A$ to $B$.\n\nSince Karuna runs at $6 \\mathrm{~m} / \\mathrm{s}$ for these $\\left(99-t_{2}\\right)$ seconds, she has to run $6\\left(99-t_{2}\\right) \\mathrm{m}$.\n\nSince Jorge runs at $7.5 \\mathrm{~m} / \\mathrm{s}$ for these $\\left(99-t_{2}\\right)$ seconds, he has to run $7.5\\left(99-t_{2}\\right) \\mathrm{m}$.\n\nTherefore, $6\\left(99-t_{2}\\right)+7.5\\left(99-t_{2}\\right)=297$ and so $13.5\\left(99-t_{2}\\right)=297$ or $99-t_{2}=22$ and so $t_{2}=77$.\n\nAlternatively, to calculate the value of $t_{2}$, we note that when Karuna and Jorge meet for the second time, they have each run the distance from $A$ to $B$ one full time and are on their return trips.\n\nThis means that they have each run the full distance from $A$ to $B$ once and the distances that they have run on their return trip add up to another full distance from $A$ to $B$, for a total distance of $3 \\cdot 297 \\mathrm{~m}=891 \\mathrm{~m}$.\n\nKaruna has run at $6 \\mathrm{~m} / \\mathrm{s}$ for $t_{2}$ seconds, for a total distance of $6 t_{2} \\mathrm{~m}$.\n\nJorge ran the first $297 \\mathrm{~m}$ at $5 \\mathrm{~m} / \\mathrm{s}$, which took $\\frac{297}{5} \\mathrm{~s}$ and ran the remaining $\\left(t_{2}-\\frac{297}{5}\\right)$ seconds at $7.5 \\mathrm{~m} / \\mathrm{s}$, for a total distance of $\\left(297+7.5\\left(t_{2}-\\frac{297}{5}\\right)\\right) \\mathrm{m}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n6 t_{2}+297+7.5\\left(t_{2}-\\frac{297}{5}\\right) & =891 \\\\\n13.5 t_{2} & =891-297+7.5 \\cdot \\frac{297}{5} \\\\\n13.5 t_{2} & =1039.5 \\\\\nt_{2} & =77\n\\end{aligned}\n$$\n\nTherefore, Karuna and Jorge meet after 27 seconds and after 77 seconds." ]

[ "27, 77" ]

[ "Let $B C=x, P B=b$, and $B Q=a$.\n\nSince $B C=x$, then $A D=P S=Q R=x$.\n\nSince $B C=x$ and $B Q=a$, then $Q C=x-a$.\n\nSince $A B=718$ and $P B=b$, then $A P=718-b$.\n\nNote that $P Q=S R=250$.\n\nLet $\\angle B Q P=\\theta$.\n\nSince $\\triangle P B Q$ is right-angled at $B$, then $\\angle B P Q=90^{\\circ}-\\theta$.\n\nSince $B Q C$ is a straight angle and $\\angle P Q R=90^{\\circ}$, then $\\angle R Q C=180^{\\circ}-90^{\\circ}-\\theta=90^{\\circ}-\\theta$.\n\nSince $A P B$ is a straight angle and $\\angle S P Q=90^{\\circ}$, then $\\angle A P S=180^{\\circ}-90^{\\circ}-\\left(90^{\\circ}-\\theta\\right)=\\theta$.\n\nSince $\\triangle S A P$ and $\\triangle Q C R$ are each right-angled and have another angle in common with $\\triangle P B Q$, then these\n\n<img_3870>\nthree triangles are similar.\n\nContinuing in the same way, we can show that $\\triangle R D S$ is also similar to these three triangles.\n\nSince $R S=P Q$, then $\\triangle R D S$ is actually congruent to $\\triangle P B Q$ (angle-side-angle).\n\nSimilarly, $\\triangle S A P$ is congruent to $\\triangle Q C R$.\n\nIn particular, this means that $A S=x-a, S D=a, D R=b$, and $R C=718-b$.\n\nSince $\\triangle S A P$ and $\\triangle P B Q$ are similar, then $\\frac{S A}{P B}=\\frac{A P}{B Q}=\\frac{S P}{P Q}$.\n\nThus, $\\frac{x-a}{b}=\\frac{718-b}{a}=\\frac{x}{250}$.\n\nAlso, by the Pythagorean Theorem in $\\triangle P B Q$, we obtain $a^{2}+b^{2}=250^{2}$.\n\nBy the Pythagorean Theorem in $\\triangle S A P$,\n\n$$\n\\begin{aligned}\nx^{2} & =(x-a)^{2}+(718-b)^{2} \\\\\nx^{2} & =x^{2}-2 a x+a^{2}+(718-b)^{2} \\\\\n0 & =-2 a x+a^{2}+(718-b)^{2}\n\\end{aligned}\n$$\n\nSince $a^{2}+b^{2}=250^{2}$, then $a^{2}=250^{2}-b^{2}$.\n\nSince $\\frac{718-b}{a}=\\frac{x}{250}$, then $a x=250(718-b)$.\n\nTherefore, substituting into $(*)$, we obtain\n\n$$\n\\begin{aligned}\n0 & =-2(250)(718-b)+250^{2}-b^{2}+(718-b)^{2} \\\\\nb^{2} & =250^{2}-2(250)(718-b)+(718-b)^{2} \\\\\nb^{2} & =((718-b)-250)^{2} \\quad\\left(\\text { since } y^{2}-2 y z+z^{2}=(y-z)^{2}\\right) \\\\\nb^{2} & =(468-b)^{2} \\\\\nb & =468-b \\quad(\\text { since } b \\neq b-468) \\\\\n2 b & =468 \\\\\nb & =234\n\\end{aligned}\n$$\n\nTherefore, $a^{2}=250^{2}-b^{2}=250^{2}-234^{2}=(250+234)(250-234)=484 \\cdot 16=22^{2} \\cdot 4^{2}=88^{2}$ and so $a=88$.\n\nFinally, $x=\\frac{250(718-b)}{a}=\\frac{250 \\cdot 484}{88}=1375$. Therefore, $B C=1375$." ]

[ "1375" ]

[ "If we proceed by pattern recognition, we find after row 1 we have a total of 1 triangle, after two rows we have $2^{2}$ or 4 triangles. After ten rows we have $10^{2}$ or 100 triangles.\n\n<img_3802>", "This solution is based on the fact that the ratio of areas for similar triangles is the square of the ratio of corresponding sides. Thus the big triangle with side length ten times that of the smaller triangle has 100 times the area." ]

[ "100" ]

[ "If $P Q=\\frac{\\pi}{3}$, then by symmetry the coordinates of $R$\n\nare $\\left(\\frac{\\pi}{6}, k \\cos \\frac{\\pi}{6}\\right)$.\n\nArea of rectangle $P Q R S=\\frac{\\pi}{3}\\left(k \\cos \\frac{\\pi}{6}\\right)=\\frac{\\pi}{3}(k)\\left(\\frac{\\sqrt{3}}{2}\\right)$\n\nBut $\\frac{\\sqrt{3} k \\pi}{6}=\\frac{5 \\pi}{3} \\quad \\therefore k=\\frac{10}{\\sqrt{3}}$ or $\\frac{10}{3} \\sqrt{3}$.\n\n<img_3853>" ]

[ "$\\frac{10}{\\sqrt{3}}$,$\\frac{10}{3} \\sqrt{3}$" ]

[ "In $\\triangle B A N, \\angle B N A=25^{\\circ}$\n\nUsing the Sine Law in $\\triangle B A N$,\n\n$\\frac{N B}{\\sin 108^{\\circ}}=\\frac{100}{\\sin 25^{\\circ}}$\n\nTherefore $N B=\\frac{100 \\sin 108^{\\circ}}{\\sin 25^{\\circ}} \\approx 225.04$,\n\n<img_3946>\n\nNow in $\\triangle M N B, \\frac{M N}{N B}=\\tan 32^{\\circ}$\n\n$$\nM N=\\frac{100 \\sin 108^{\\circ}}{\\sin 25^{\\circ}} \\times \\tan 32^{\\circ} \\doteq 140.6\n$$\n\nThe tower is approximately $141 \\mathrm{~m}$ high." ]

[ "141" ]

[ "$Q(4,0), Q(0,4)$\n\n$Q(2,0), Q(0,2)$\n\n$Q(-2,2), Q(2,-2)$\n\n<img_3757>" ]

[ "$(4,0),(0,4),(2,0),(0,2),(-2,2),(2,-2)$" ]

[ "Let $[A B C]=K$. Then $[B C D]=\\frac{2}{3} \\cdot K$. Let $\\overline{D E}$ be the bisector of $\\angle B D C$, as shown below.\n\n<img_3399>\n\nNotice that $\\mathrm{m} \\angle D B A=\\mathrm{m} \\angle B D C-\\mathrm{m} \\angle A=\\mathrm{m} \\angle A$, so triangle $A D B$ is isosceles, and $B D=1$. (Alternately, notice that $\\overline{D E} \\| \\overline{A B}$, and by similar triangles, $[C D E]=\\frac{4}{9} \\cdot K$, which means $[B D E]=\\frac{2}{9} \\cdot K$. Because $[C D E]:[B D E]=2$ and $\\angle B D E \\cong \\angle C D E$, conclude that $\\frac{C D}{B D}=2$, thus $B D=1$.) Because $B C D$ is isosceles, it follows that $\\cos \\angle B D C=\\frac{1}{2} B D / C D=\\frac{1}{4}$. By the half-angle formula,\n\n$$\n\\sin A=\\sqrt{\\frac{1-\\cos \\angle B D C}{2}}=\\sqrt{\\frac{3}{8}}=\\frac{\\sqrt{6}}{\\mathbf{4}}\n$$" ]

[ "$\\frac{\\sqrt{6}}{4}$" ]

[ "First consider a regular octahedron of side length 1. To compute its volume, divide it into two square-based pyramids with edges of length 1 . Such a pyramid has slant height $\\frac{\\sqrt{3}}{2}$ and height $\\sqrt{\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}}=\\sqrt{\\frac{1}{2}}=\\frac{\\sqrt{2}}{2}$, so its volume is $\\frac{1}{3} \\cdot 1^{2} \\cdot \\frac{\\sqrt{2}}{2}=\\frac{\\sqrt{2}}{6}$. Thus the octahedron has volume twice that, or $\\frac{\\sqrt{2}}{3}$. The result of folding the net shown is actually the image of a regular octahedron after being stretched along an axis perpendicular to one face by a factor of $r$. Because the octahedron is only being stretched in one dimension, the volume changes by the same factor $r$. So the problem reduces to computing the factor $r$ and the edge length of the resulting octahedron.\n\nFor convenience, imagine that one face of the octahedron rests on a plane. Seen from above the plane, the octahedron appears as shown below.\n\n\n\n<img_3368>\n\nLet $P$ be the projection of $A$ onto the plane on which the octahedron rests, and let $Q$ be the foot of the perpendicular from $P$ to $\\overline{B C}$. Then $P Q=R_{C}-R_{I}$, where $R_{C}$ is the circumradius and $R_{I}$ the inradius of the equilateral triangle. Thus $P Q=\\frac{2}{3}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{1}{3}\\left(\\frac{\\sqrt{3}}{2}\\right)=\\frac{\\sqrt{3}}{6}$. Then $B P^{2}=P Q^{2}+B Q^{2}=\\frac{3}{36}+\\frac{1}{4}=\\frac{1}{3}$, so $A P^{2}=A B^{2}-B P^{2}=\\frac{2}{3}$, and $A P=\\frac{\\sqrt{6}}{3}$.\n\nNow let a vertical stretch take place along an axis parallel to $\\overleftrightarrow{A P}$. If the scale factor is $r$, then $A P=\\frac{r \\sqrt{6}}{3}$, and because the stretch occurs on an axis perpendicular to $\\overline{B P}$, the length $B P$ is unchanged, as can be seen below.\n\n<img_3791>\n\nThus $A B^{2}=\\frac{6 r^{2}}{9}+\\frac{1}{3}=\\frac{6 r^{2}+3}{9}$. It remains to compute $r$. But $r$ is simply the ratio of the new volume to the old volume:\n\n$$\nr=\\frac{6}{\\frac{\\sqrt{2}}{3}}=\\frac{18}{\\sqrt{2}}=9 \\sqrt{2}\n$$\n\nThus $A B^{2}=\\frac{6(9 \\sqrt{2})^{2}+3}{9}=\\frac{975}{9}=\\frac{325}{3}$, and $A B=\\frac{5 \\sqrt{39}}{3}$." ]

[ "$\\frac{5 \\sqrt{39}}{3}$" ]

[ "For each point of tangency of consecutive circles, drop a perpendicular from that point to $\\overline{A B}$. For each of the $T-2$ circles between the first and last circles, the distance between consecutive perpendiculars is $2 \\cdot 1=2$. Furthermore, the distance from $A$ to the first perpendicular equals 1 (i.e., the common radius of the circles), which also equals the distance from the last perpendicular to $B$. Thus $A B=1+(T-2) \\cdot 2+1=2(T-1)$. With $T=5$, it follows that $A B=2 \\cdot 4=8$." ]

[ "8" ]

[ "Note that $G H=A B-A H-B G=22-5-5=12$. Thus\n\n$$\n\\begin{aligned}\n{[A E F B C D] } & =[A B C D]+[E F G H]+[A E H]+[B F G] \\\\\n& =22^{2}+12^{2}+\\frac{1}{2} \\cdot 5 \\cdot 12+\\frac{1}{2} \\cdot 5 \\cdot 12 \\\\\n& =484+144+30+30 \\\\\n& =\\mathbf{6 8 8} .\n\\end{aligned}\n$$" ]

[ "688" ]

[ "Count the number of complete paths that pass through all vertices exactly once (such a path is called a Hamiltonian path). The set of vertices can be split into two rings:\n\n$$\n\\mathcal{I}=\\left\\{A_{1}, A_{2}, \\ldots, A_{6}\\right\\} \\text { (i.e., the inner ring), } \\quad \\mathcal{O}=\\left\\{B_{1}, B_{2}, \\ldots, B_{6}\\right\\} \\text { (i.e., the outer ring). }\n$$\n\nwhere $A_{1}=A$. The two rings are connected by the edges $E=\\left\\{A_{1} B_{1}, A_{2} B_{2}, \\ldots, A_{6} B_{6}\\right\\}$. Each vertex in the figure has exactly three edges joining it with the neighboring vertices. Also note that any closed loop must use exactly two edges (out of three) for each vertex.\n\nFurther note that a loop must use at least one edge from $E$ to move from one ring to the other. Consider two cases: the loop uses all six edges from $E$, or it uses some but not all of them.\n\nIf all edges from $E$ are used, there are two possible undirected loops. It is not possible to use both edges $A_{1} A_{2}$ and $B_{1} B_{2}$, so either $A_{1} A_{2}$ or $B_{1} B_{2}$ will be used. This choice determines how the entire loop is constructed.\n\nIf not all edges from $E$ are used, then there must be some $i$ for which the loop uses the edge $A_{i} B_{i}$ and does not use $A_{i+1} B_{i+1}$ (where $A_{j} B_{j}$ represents $A_{j-6} B_{j-6}$ if $7 \\leq j \\leq 12$ ). Because $A_{i+1}$ is only connected to three other vertices, the loop must use $A_{i} A_{i+1}$ and $A_{i+1} A_{i+2}$, and similarly must use $B_{i} B_{i+1}$ and $B_{i+1} B_{i+2}$. This also precludes using $A_{i+2} B_{i+2}$, because doing so would close the loop before it visits all 12 vertices. Therefore the loop must also use $A_{i+2} A_{i+3}$ and $B_{i+2} B_{i+3}$, which now precludes using $A_{i+3} B_{i+3}$. This continues to force the structure of the loop until it closes by using $A_{i+5} B_{i+5}=A_{i-1} B_{i-1}$. Hence the loop must use exactly two edges from $E$, and they must be consecutive: $A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose those two consecutive edges, so there are 6 possible undirected loops in this case.\n\nThe forced path is a loop, and the only way the given conditions are satisfied if $A_{i+k}=A_{i-1}$ and $B_{i+k}=B_{i-1}$. Hence the loop must use (precisely) two consecutive edges from $E: A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose two consecutive edges, so there are 6 possible undirected loops in this case.\n\nEach undirected loop can be traced in two ways, and thus the number of ways for Xena to trace the path is $(6+2) \\cdot 2=16$." ]

[ "16" ]

[ "Let $2 x$ be the side length of the squares. Then the intersection of $\\mathcal{S}_{1}$ and $\\mathcal{S}_{2}$ is a square of side length $x$, so its area is $x^{2}$. The area of the union of $\\mathcal{S}_{1}$ and $\\mathcal{S}_{2}$ is $(2 x)^{2}+(2 x)^{2}-x^{2}=7 x^{2}$. Thus the desired ratio of areas is $\\frac{7 x^{2}}{x^{2}}=7$ (independent of $T$ )." ]

[ "7" ]

[ "Using the Pythagorean Theorem, $I G=\\sqrt{(I L)^{2}-(L G)^{2}}=\\sqrt{41-25}=4$, and $G K=\\sqrt{(I K)^{2}-(I G)^{2}}=$ $\\sqrt{25-16}=3$." ]

[ "3" ]

[ "First we prove that for all maps $M, C(M)<h(M)$.\n\nProve. Let $n=h(M)$. The following strategy will always catch a Robber within two days using $n-1$ Cops, which proves that $C(M) \\leq n-1$. Choose any subset $\\mathcal{S}$ of $n-1$ hideouts and position $n-1$ Cops at the hideouts of $\\mathcal{S}$ for 2 days. If the Robber is not caught on the first day, he must have been at the hideout not in $\\mathcal{S}$, and therefore must move to a hideout in $\\mathcal{S}$ on the following day.\n\nso $C(M) \\leq 6$. To see that 6 is minimal, note that it is possible for the Robber to go from any hideout to any other hideout in a single night. Suppose that on a given day, the Robber successfully evades capture. Without loss of generality, assume that the Robber was at hideout $A_{1}$. Then the only valid conclusion the Cops can draw is that on the next day, the Robber will not be at $A_{1}$. Thus the only hideout the Cops can afford to leave open is $A_{1}$ itself. If they leave another hideout open, the Robber might have chosen to hide there, and the process will repeat. So we get $C(M)=6$." ]

[ "6" ]

[ "or the map $M$ from figure a, $W(M)=7$. The most efficient strategy is to use 7 Cops to blanket all the hideouts on the first day. Any strategy using fewer than 7 Cops would require 6 Cops on each of two consecutive days: given that any hideout can be reached from any other hideout, leaving more than one hideout unsearched on one day makes it is impossible to eliminate any hideouts the following day. So any other strategy would require a minimum of 12 Cop workdays.\n\nFor the map $M$ from figure b, $W(M)=8$. The strategy outlined in $2 \\mathrm{~b}$ used three Cops for a maximum of four workdays, yielding 12 Cop workdays. The most efficient strategy is to use 4 Cops, positioned at $\\{B, E, H, K\\}$ for 2 days each. The following argument demonstrates that 8 Cop workdays is in fact minimal. First, notice that there is no advantage to searching one of the hideouts between vertices of the square (for example, $C$ ) without searching the other hideout between the same vertices (for example, $D$ ). The Robber can reach $C$ on day $n$ if and only if he is at either $B$ or $E$ on day $n-1$, and in either case he could just as well go to $D$ instead of $C$. So there is no situation in which the Robber is certain to be caught at $C$ rather than at $D$. Additionally, the Robber's possible locations on day $n+1$ are the same whether he is at $C$ or $D$ on day $n$, so searching one rather than the other fails to rule out any locations for future days. So any successful strategy that involves searching $C$ should also involve searching $D$ on the same day, and similarly for $F$ and $G, I$ and $J$, and $L$ and $A$. On the other hand, if the Robber must be at one of $C$ and $D$ on day $n$, then he must be at either $B$ or $E$ on day $n+1$, because those are the only adjacent hideouts. So any strategy that involves searching both hideouts of one of the off-the-square pairs on day $n$ is equivalent to a strategy that searches the adjacent on-the-square hideouts on day $n+1$; the two strategies use the same number of Cops for the same number of workdays. Thus the optimal number of Cop workdays can be achieved using strategies that only search the \"corner\" hideouts $B, E, H, K$.\n\nRestricting the search to only those strategies searching corner hideouts $B, E, H, K$, a total of 8 workdays can be achieved by searching all four hideouts on two consecutive days: if the Robber is at one of the other eight hideouts the first day, he must move to one of the two adjacent corner hideouts the second day. But each of these corner hideouts is adjacent to two other corner hideouts. So if only one hideout is searched, for no matter how many consecutive days, the following day, the Robber could either be back at the previously-searched hideout or be at any other hideout: no possibilities are ruled out. If two adjacent corners are searched, the Cops do no better, as the following argument shows. Suppose that $B$ and $E$ are both searched for two consecutive days. Then the Cops can rule out $B, E, C$, and $D$ as possible locations, but if the Cops then switch to searching either $H$ or $K$ instead of $B$ or $E$, the Robber can go back to $C$ or $D$ within two days. So searching two adjacent corner hideouts for two days is fruitless and costs four Cop workdays. Searching diagonally opposite corner hideouts is even less fruitful, because doing so rules out none of the other hideouts as possible Robber locations. Using three Cops each day, it is easy to imagine scenarios in which the Robber evades capture for three days before being caught: for example, if $B, E, H$ are searched for two consecutive days, the Robber goes from $I$ or $J$ to $K$ to $L$. Therefore if three Cops are used, four days are required for a total of 12 Cop workdays.\n\nIf there are more than four Cops, the preceding arguments show that the number of Cops must be even to produce optimal results (because there is no advantage to searching one hideout between vertices of the square without searching the other). Using six Cops with four at corner hideouts yields no improvement, because the following day, the Robber could get to any of the four corner hideouts, requiring at least four Cops the second day, for ten Cop workdays. If two or fewer Cops are at corner hideouts, the situation is even worse, because if the Robber is not caught that day, he has at least nine possible hideouts the following day (depending on whether the unsearched corners are adjacent or diagonally opposite to each other). Using eight Cops (with four at corner vertices) could eliminate one corner vertex as a possible location for the second day (if the non-corner hideouts searched are on adjacent sides of the square), but eight Cop workdays have already been used on the first day. So 8 Cop workdays is minimal." ]

[ "12, 8" ]

[ "Because $A B=16, A C=B C=\\frac{16}{\\sqrt{2}}=8 \\sqrt{2}$. Then each of the large arcs has radius 8 , and the small arc has radius $8 \\sqrt{2}-8$. Each large arc has measure $45^{\\circ}$ and the small arc has measure $90^{\\circ}$. Therefore the area enclosed by each large arc is $\\frac{45}{360} \\cdot \\pi \\cdot 8^{2}=8 \\pi$, and the area enclosed by the small arc is $\\frac{90}{360} \\cdot \\pi \\cdot(8 \\sqrt{2}-8)^{2}=48 \\pi-32 \\pi \\sqrt{2}$. Thus the sum of the areas enclosed by the three arcs is $64 \\pi-32 \\pi \\sqrt{2}$. On the other hand, the area of the triangle is $\\frac{1}{2}(8 \\sqrt{2})^{2}=64$. So the area of the desired region is $64-64 \\pi+32 \\pi \\sqrt{2}$." ]

[ "$\\quad 64-64 \\pi+32 \\pi \\sqrt{2}$" ]

[ "Compute the desired area as $[E G F B]-[E H B]$. To compute the area of concave quadrilateral $E G F B$, draw segment $\\overline{B G}$, which divides the quadrilateral into three triangles, $\\triangle D E G, \\triangle B D G$, and $\\triangle B G F$. Then $[B D G]=[D E G]=18$ because the triangles have equal bases and heights. Because $D, G$, and $F$ are collinear, to compute $[B G F]$ it suffices to find the ratio $D G / G F$. Use Menelaus's Theorem on $\\triangle A D F$ with Menelaus Line $\\overline{E C}$ to obtain\n\n$$\n\\frac{A E}{E D} \\cdot \\frac{D G}{G F} \\cdot \\frac{F C}{C A}=1\n$$\n\n\n\nBecause $E$ and $F$ are the midpoints of $\\overline{A D}$ and $\\overline{C A}$ respectively, $A E / E D=1$ and $F C / C A=$ $1 / 2$. Therefore $D G / G F=2 / 1$, and $[B G F]=\\frac{1}{2}[B D G]=9$. Thus $[E G F B]=18+18+9=45$.\n\nTo compute $[E H B]$, consider that its base $\\overline{E B}$ is twice the base of $\\triangle D E G$. The ratio of their heights equals the ratio $E H / E G$ because the altitudes from $H$ and $G$ to $\\overleftrightarrow{B E}$ are parallel to each other. Use Menelaus's Theorem twice more on $\\triangle A E C$ to find these values:\n\n$$\n\\begin{gathered}\n\\frac{A D}{D E} \\cdot \\frac{E G}{G C} \\cdot \\frac{C F}{F A}=1 \\Rightarrow \\frac{E G}{G C}=\\frac{1}{2} \\Rightarrow E G=\\frac{1}{3} E C, \\text { and } \\\\\n\\frac{A B}{B E} \\cdot \\frac{E H}{H C} \\cdot \\frac{C F}{F A}=1 \\Rightarrow \\frac{E H}{H C}=\\frac{2}{3} \\Rightarrow E H=\\frac{2}{5} E C .\n\\end{gathered}\n$$\n\nTherefore $\\frac{E H}{E G}=\\frac{2 / 5}{1 / 3}=\\frac{6}{5}$. Thus $[E H B]=\\frac{6}{5} \\cdot 2 \\cdot[D E G]=\\frac{216}{5}$. Thus $[F G H]=45-\\frac{216}{5}=\\frac{9}{5}$.", "The method of mass points leads to the same results as Menelaus's Theorem, but corresponds to the physical intuition that masses on opposite sides of a fulcrum balance if and only if the products of the masses and their distances from the fulcrum are equal (in physics-speak, the net torque is zero). If a mass of weight 1 is placed at vertex $B$ and masses of weight 2 are placed at vertices $A$ and $C$, then $\\triangle A B C$ balances on the line $\\overleftrightarrow{B F}$ and also on the line $\\overleftrightarrow{C E}$. Thus it balances on the point $H$ where these two lines intersect. Replacing the masses at $A$ and $C$ with a single mass of weight 4 at their center of mass $F$, the triangle still balances at $H$. Thus $B H / H F=4$.\n\nNext, consider $\\triangle B E F$. Placing masses of weight 1 at the vertices $B$ and $E$ and a mass of weight 4 at $F$, the triangle balances at $G$. A similar argument shows that $D G / G F=2$ and that $E G / G H=5$. Because $\\triangle D E G$ and $\\triangle F H G$ have congruent (vertical) angles at $G$, it follows that $[D E G] /[F H G]=(D G / F G) \\cdot(E G / H G)=2 \\cdot 5=10$. Thus $[F G H]=[D E G] / 10=$ $\\frac{18}{10}=\\frac{9}{5}$." ]

[ "$\\frac{9}{5}$" ]

[ "Note that $D E S H$ is a trapezoid with height $\\frac{T}{2}$. Because $\\overline{A S}$ and $\\overline{B H}$ are midlines of triangles $E F M$ and $D G N$ respectively, it follows that $A S=B H=\\frac{T}{6}$. Thus $S H=T-2 \\cdot \\frac{T}{6}=\\frac{2 T}{3}$. Thus $[D E S H]=\\frac{1}{2}\\left(T+\\frac{2 T}{3}\\right) \\cdot \\frac{T}{2}=\\frac{5 T^{2}}{12}$. With $T=6$, the desired area is 15 ." ]

[ "15" ]

[ "Note that $\\overline{D Y}$ and $\\overline{K P}$ are both perpendicular to line $\\overleftrightarrow{Y P}$. Let $J$ be the foot of the perpendicular from $K$ to $\\overline{D Y}$. Then $P K J Y$ is a rectangle and $Y P=J K=\\sqrt{D K^{2}-D J^{2}}=$ $\\sqrt{(R+r)^{2}-(R-r)^{2}}=2 \\sqrt{R r}$. With $R=450$ and $r=\\frac{1}{3}$, the answer is $2 \\sqrt{150}=\\mathbf{1 0} \\sqrt{\\mathbf{6}}$." ]

[ "$10 \\sqrt{6}$" ]

[ "The first pair indicates an increase; the next three are decreases, and the last pair is an increase. So the 2-signature is $(12,21,21,21,12)$." ]

[ "$(12,21,21,21,12)$" ]

[ "12543,13542,14532,23541,24531,34521" ]

[ "12543,13542,14532,23541,24531,34521" ]

[ "The answer is $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$. The shape of this signature is a wedge: $n$ up steps followed by $n$ down steps. The wedge for $n=3$ is illustrated below:\n\n<img_3277>\n\nThe largest number in the label, $2 n+1$, must be placed at the peak in the center. If we choose the numbers to put in the first $n$ spaces, then they must be placed in increasing order. Likewise, the remaining $n$ numbers must be placed in decreasing order on the downward sloping piece of the shape. Thus there are exactly $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$ such labels.\n\n" ]

[ "$\\binom{2n}{n}$" ]

[ "The answer is 16 . We have a shape with two peaks and a valley in the middle. The 5 must go on one of the two peaks, so we place it on the first peak. By the shape's symmetry, we will double our answer at the end to account for the 5 -labels where the 5 is on the other peak.\n\n<img_3879>\n\nThe 4 can go to the left of the 5 or at the other peak. In the first case, shown below left, the 3 must go at the other peak and the 1 and 2 can go in either order. In the latter case, shown below right, the 1,2 , and 3 can go in any of 3 ! arrangements.\n<img_3481>\n\n\n\nSo there are $2 !+3 !=8$ possibilities. In all, there are 165 -labels (including the ones where the 5 is at the other peak)." ]

[ "16" ]

[ "The answer is 7936. The shape of this 2-signature has four peaks and three intermediate valleys:\n\n<img_3473>\n\nWe will solve this problem by building up from smaller examples. Let $f_{n}$ equal the number of $(2 n+1)$-labels whose 2 -signature consists of $n$ peaks and $n-1$ intermediate valleys. In part (b) we showed that $f_{2}=16$. In the case where we have one peak, $f_{1}=2$. For the trivial case (no peaks), we get $f_{0}=1$. These cases are shown below.\n\n1\n\n<img_3650>\n\n<img_3227>\n\nSuppose we know the peak on which the largest number, $2 n+1$, is placed. Then that splits our picture into two shapes with fewer peaks. Once we choose which numbers from $1,2, \\ldots, 2 n$ to place each shape, we can compute the number of arrangements of the numbers on each shape, and then take the product. For example, if we place the 9 at the second peak, as shown below, we get a 1-peak shape on the left and a 2-peak shape on the right.\n\n<img_4030>\n\nFor the above shape, there are $\\left(\\begin{array}{l}8 \\\\ 3\\end{array}\\right)$ ways to pick the three numbers to place on the left-hand side, $f_{1}=2$ ways to place them, and $f_{2}=16$ ways to place the remaining five numbers on the right.\n\nThis argument works for any $n>1$, so we have shown the following:\n\n$$\nf_{n}=\\sum_{k=1}^{n}\\left(\\begin{array}{c}\n2 n \\\\\n2 k-1\n\\end{array}\\right) f_{k-1} f_{n-k}\n$$\n\n\n\nSo we have:\n\n$$\n\\begin{aligned}\n& f_{1}=\\left(\\begin{array}{l}\n2 \\\\\n1\n\\end{array}\\right) f_{0}^{2}=2 \\\\\n& f_{2}=\\left(\\begin{array}{l}\n4 \\\\\n1\n\\end{array}\\right) f_{0} f_{1}+\\left(\\begin{array}{l}\n4 \\\\\n3\n\\end{array}\\right) f_{1} f_{0}=16 \\\\\n& f_{3}=\\left(\\begin{array}{l}\n6 \\\\\n1\n\\end{array}\\right) f_{0} f_{2}+\\left(\\begin{array}{l}\n6 \\\\\n3\n\\end{array}\\right) f_{1}^{2}+\\left(\\begin{array}{l}\n6 \\\\\n5\n\\end{array}\\right) f_{2} f_{0}=272 \\\\\n& f_{4}=\\left(\\begin{array}{l}\n8 \\\\\n1\n\\end{array}\\right) f_{0} f_{3}+\\left(\\begin{array}{l}\n8 \\\\\n3\n\\end{array}\\right) f_{1} f_{2}+\\left(\\begin{array}{l}\n8 \\\\\n5\n\\end{array}\\right) f_{2} f_{1}+\\left(\\begin{array}{l}\n8 \\\\\n7\n\\end{array}\\right) f_{3} f_{0}=7936 .\n\\end{aligned}\n$$" ]

[ "7936" ]

[ "The answer is $p ! \\cdot p^{n-p}$.\n\nCall two consecutive windows in a $p$-signature compatible if the last $p-1$ numbers in the first label and the first $p-1$ numbers in the second label (their \"overlap\") describe the same ordering. For example, in the $p$-signature $(. ., 2143,2431, \\ldots), 2143$ and 2431 are compatible. Notice that the last three digits of 2143 and the first three digits of 2431 can be described by the same 3-label, 132 .\n\nTheorem: A signature $\\sigma$ is possible if and only if every pair of consecutive windows is compatible.\n\nProof: $(\\Rightarrow)$ Consider a signature $\\sigma$ describing a $p$-label $L$. If some pair in $\\sigma$ is not compatible, then there is some string of $p-1$ numbers in our label $L$ that has two different $(p-1)$-signatures. This is impossible, since the $p$-signature is well-defined.\n\n$(\\Leftarrow)$ Now suppose $\\sigma$ is a $p$-signature such that that every pair of consecutive windows is compatible. We need to show that there is at least one label $L$ with $S_{p}[L]=\\sigma$. We do so by induction on the number of windows in $\\sigma$, using the results from $5(\\mathrm{~b})$.\n\nLet $\\sigma=\\left\\{\\omega_{1}, \\omega_{2}, \\ldots, \\omega_{k+1}\\right\\}$, and suppose $\\omega_{1}=a_{1}, a_{2}, \\ldots, a_{p}$. Set $L_{1}=\\omega_{1}$.\n\nSuppose that $L_{k}$ is a $(p+k-1)$-label such that $S_{p}\\left[L_{k}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k}\\right\\}$. We will construct $L_{k+1}$ for which $S_{p}\\left[L_{k+1}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k+1}\\right\\}$.\n\nAs in $5(\\mathrm{~b})$, denote by $L_{k}^{(j)}$ the label $L_{k}$ with a $j+0.5$ appended; we will eventually renumber the elements in the label to make them all integers. Appending $j+0.5$ does not affect any of the non-terminal windows of $S_{p}\\left[L_{k}\\right]$, and as $j$ varies from 0 to $p-k+1$ the final window of $S_{p}\\left[L_{k}^{(j)}\\right]$ varies over each of the $p$ windows compatible with $\\omega_{k}$. Since $\\omega_{k+1}$ is compatible with $\\omega_{k}$, there exists some $j$ for which $S_{p}\\left[L_{k}^{(j)}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k+1}\\right\\}$. Now we renumber as follows: set $L_{k+1}=S_{k+p}\\left[L_{k}^{(j)}\\right]$, which replaces $L_{k}^{(j)}$ with the integers 1 through $k+p$ and preserves the relative order of all integers in the label.\n\nBy continuing this process, we conclude that the $n$-label $L_{n-p+1}$ has $p$-signature $\\sigma$, so $\\sigma$ is possible.\n\nTo count the number of possible $p$-signatures, we choose the first window ( $p$ ! choices), then choose each of the remaining $n-p$ compatible windows ( $p$ choices each). In all, there are $p ! \\cdot p^{n-p}$ possible $p$-signatures." ]

[ "$p ! \\cdot p^{n-p}$" ]

[ "The answer is $n=7, p=5$.\n\nLet $P$ denote the probability that a randomly chosen $p$-signature is possible. We are\n\n\n\ngiven that $1-P=575$, so $P=\\frac{1}{576}$. We want to find $p$ and $n$ for which\n\n$$\n\\begin{aligned}\n\\frac{p ! \\cdot p^{n-p}}{(p !)^{n-p+1}} & =\\frac{1}{576} \\\\\n\\frac{p^{n-p}}{(p !)^{n-p}} & =\\frac{1}{576} \\\\\n((p-1) !)^{n-p} & =576\n\\end{aligned}\n$$\n\nThe only factorial that has 576 as an integer power is $4 !=\\sqrt{576}$. Thus $p=5$ and $n-p=2 \\Rightarrow n=7$." ]

[ "7,5" ]

[ "12345 and 54321 are the only ones." ]

[ "12345, 54321" ]

[ "The answer is $p=16$. To show this fact we will need to extend the idea from part 8(b) about \"linking\" inequalities forced by the various windows:\n\nTheorem: A $p$-signature for an $n$-label $L$ is unique if and only if for every $k<n, k$ and $k+1$ are in at least one window together. That is, the distance between them in the $n$-label is less than $p$.\n\nProof: Suppose that for some $k$, the distance between $k$ and $k+1$ is $p$ or greater. Then the label $L^{\\prime}$ obtained by swapping $k$ and $k+1$ has the same $p$-signature, because there are no numbers between $k$ and $k+1$ in any window and because the two numbers never appear in the same window.\n\nIf the distance between all such pairs is less than $k$, we need to show that $S_{p}[L]$ is unique. For $i=1,2, \\ldots, n$, let $r_{i}$ denote the position where $i$ appears in $L$. For example, if $L=4123$, then $r_{1}=2, r_{2}=3, r_{3}=4$, and $r_{4}=1$.\n\nLet $L=a_{1}, a_{2}, \\ldots, a_{n}$. Since 1 and 2 are in some window together, $a_{r_{1}}<a_{r_{2}}$. Similarly, for any $k$, since $k$ and $k+1$ are in some window together, $a_{r_{k}}<a_{r_{k+1}}$. We then get a linked inequality $a_{r_{1}}<a_{r_{2}}<\\cdots<a_{r_{n}}$, which can only be satisfied if $a_{r_{1}}=1, a_{r_{2}}=$ $2, \\ldots, a_{r_{n}}=n$. Therefore, $S_{p}[L]$ is unique.\n\nFrom the proof above, we know that the signature is unique if and only if every pair of consecutive integers coexists in at least one window. Therefore, we seek the largest distance between consecutive integers in $L$. That distance is 15 (from 8 to 9 , and from 17 to 18). Thus the smallest $p$ is 16 ." ]

[ "16" ]

[ "Because $A$ is on diagonal $\\overline{N Q}$, rectangles $N X A B$ and $A C Q Y$ are similar. Thus $\\frac{A B}{A X}=\\frac{Q Y}{Q C}=$ $\\frac{A C}{A Y} \\Rightarrow A B \\cdot A Y=A C \\cdot A X$. Therefore, we have $2009=[\\mathrm{I}]+2[\\mathrm{II}]+[\\mathrm{III}]$.\n\nLet the common ratio of the geometric progression be $\\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers ( $q$ may equal 1 ). Then [I] must be some integer multiple of $q^{2}$, which we will call $a q^{2}$. This gives $[\\mathrm{II}]=a p q$ and [III $=a p^{2}$. By factoring, we get\n\n$$\n2009=a q^{2}+2 a p q+a p^{2} \\Rightarrow 7^{2} \\cdot 41=a(p+q)^{2}\n$$\n\nThus we must have $p+q=7$ and $a=41$. Since $[\\mathrm{I}]=a q^{2}$ and $p, q>0$, the area is maximized when $\\frac{p}{q}=\\frac{1}{6}$, giving $[\\mathrm{I}]=41 \\cdot 36=\\mathbf{1 4 7 6}$. The areas of the other regions are 246,246, and 41 ." ]

[ "1476" ]

[ "We say that two numbers are neighbors if they occupy adjacent squares, and that $a$ is a friend of $b$ if $0<|a-b| \\leq 2$. Using this vocabulary, the problem's condition is that every pair of neighbors must be friends of each other. Each of the numbers 1 and 8 has two friends, and each number has at most four friends.\n\nIf there is no number written in the center square, then we must have one of the cycles in the figures below. For each cycle, there are 8 rotations. Thus there are 16 possible configurations with no number written in the center square.\n\n| 2 | 1 | 3 |\n| :--- | :--- | :--- |\n| 4 | - | 5 |\n| 6 | 8 | 7 |\n\n\n| 3 | 1 | 2 |\n| :--- | :--- | :--- |\n| 5 | - | 4 |\n| 7 | 8 | 6 |\n\nNow assume that the center square contains the number $n$. Because $n$ has at least three neighbors, $n \\neq 1$ and $n \\neq 8$. First we show that 1 must be in a corner. If 1 is a neighbor of $n$, then one of the corners neighboring 1 must be empty, because 1 has only two friends ( 2 and $3)$. If $c$ is in the other corner neighboring 1 , then $\\{n, c\\}=\\{2,3\\}$. But then $n$ must have three\n\n\n\nmore friends $\\left(n_{1}, n_{2}, n_{3}\\right)$ other than 1 and $c$, for a total of five friends, which is impossible, as illustrated below. Therefore 1 must be in a corner.\n\n| - | 1 | $c$ |\n| :--- | :--- | :--- |\n| $n_{1}$ | $n$ | $n_{2}$ |\n| | $n_{3}$ | |\n\nNow we show that 1 can only have one neighbor, i.e., one of the squares adjacent to 1 is empty. If 1 has two neighbors, then we have, up to a reflection and a rotation, the configuration shown below. Because 2 has only one more friend, the corner next to 2 is empty and $n=4$. Consequently, $m_{1}=5$ (refer to the figure below). Then 4 has one friend (namely 6) left with two neighbors $m_{2}$ and $m_{3}$, which is impossible. Thus 1 must have exactly one neighbor. An analogous argument shows that 8 must also be at a corner with exactly one neighbor.\n\n| 1 | 2 | - |\n| :--- | :--- | :--- |\n| 3 | $n$ | $m_{3}$ |\n| $m_{1}$ | $m_{2}$ | |\n\nTherefore, 8 and 1 must be in non-opposite corners, with the blank square between them. Thus, up to reflections and rotations, the only possible configuration is the one shown at left below.\n\n| 1 | - | 8 |\n| :--- | :--- | :--- |\n| $m$ | | |\n| | | |\n\n\n| 1 | - | 8 |\n| :---: | :---: | :---: |\n| $2 / 3$ | $4 / 5$ | $6 / 7$ |\n| $3 / 2$ | $5 / 4$ | $7 / 6$ |\n\nThere are two possible values for $m$, namely 2 and 3 . For each of the cases $m=2$ and $m=3$, the rest of the configuration is uniquely determined, as illustrated in the figure above right. We summarize our process: there are four corner positions for 1; two (non-opposite) corner positions for 8 (after 1 is placed); and two choices for the number in the square neighboring 1 but not neighboring 8 . This leads to $4 \\cdot 2 \\cdot 2=16$ distinct configurations with a number written in the center square.\n\nTherefore, there are 16 configurations in which the center square is blank and 16 configurations with a number in the center square, for a total of $\\mathbf{3 2}$ distinct configurations." ]

[ "32" ]

[ "Note that $\\mathrm{m} \\angle A E B=\\frac{1}{2}(\\mathrm{~m} \\widehat{A B}-m \\widehat{C D})=\\frac{1}{2}(\\mathrm{~m} \\widehat{A B}-\\mathrm{m} \\angle C O D)$. Also note that $\\mathrm{m} \\angle C O D=$ $360^{\\circ}-(\\mathrm{m} \\angle A O C+\\mathrm{m} \\angle B O D+\\mathrm{m} \\angle A O B)=360^{\\circ}-\\left(180^{\\circ}-2 \\mathrm{~m} \\angle O A C\\right)-\\left(180^{\\circ}-2 \\mathrm{~m} \\angle O B D\\right)-$ $\\mathrm{m} \\widehat{A B}=2(\\mathrm{~m} \\angle O A C+\\mathrm{m} \\angle O B D)-\\mathrm{m} \\widehat{A B}$. Thus $\\mathrm{m} \\angle A E B=\\mathrm{m} \\widehat{A B}-\\mathrm{m} \\angle O A C-\\mathrm{m} \\angle O B D=$ $\\frac{T}{4}-10^{\\circ}-5^{\\circ}$, and with $T=80$, the answer is 5 ." ]

[ "5" ]

[ "Let $\\mathrm{m} \\angle A=x$. Then $\\mathrm{m} \\angle P=\\mathrm{m} \\angle Q=T x$, and $(2 T+1) x=180^{\\circ}$, so $x=\\frac{180^{\\circ}}{2 T+1}$. Let $O$ be the center of the circle, as shown below.\n\n<img_3423>\n\nThen $\\mathrm{m} \\angle P O Q=2 \\mathrm{~m} \\angle P A Q=2\\left(\\frac{180^{\\circ}}{2 T+1}\\right)=\\frac{360^{\\circ}}{2 T+1}$. Because $\\mathrm{m} \\angle P O Q=\\frac{360^{\\circ}}{n}$, the denominators must be equal: $n=2 T+1$. Substitute $T=24$ to find $n=\\mathbf{4 9}$." ]

[ "49" ]

[ "The four \"removed\" circles have radii $\\frac{1}{2}, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{3}$ so the combined area of the six remaining curvilinear territories is:\n\n$$\n\\pi\\left(1^{2}-\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{3}\\right)^{2}-\\left(\\frac{1}{3}\\right)^{2}\\right)=\\frac{5 \\pi}{18}\n$$" ]

[ "$\\frac{5 \\pi}{18}$" ]

[ "At the beginning of day 2, there are six c-triangles, so six incircles are sold, dividing each of the six territories into three smaller curvilinear triangles. So a total of 18 curvilinear triangles exist at the start of day 3, each of which is itself divided into three pieces that day (by the sale of a total of 18 regions bounded by the territories' incircles). Therefore there are 54 regions at the end of day 3." ]

[ "54" ]

[ "The total number of plots sold up to and including day $n$ is\n\n$$\n\\begin{aligned}\n2+\\sum_{k=1}^{n} X_{k} & =2+2 \\sum_{k=1}^{n} 3^{k-1} \\\\\n& =2+2 \\cdot\\left(1+3+3^{2}+\\ldots+3^{n-1}\\right) \\\\\n& =3^{n}+1\n\\end{aligned}\n$$\n\nAlternatively, proceed by induction: on day 0 , there are $2=3^{0}+1$ plots sold, and for $n \\geq 0$,\n\n$$\n\\begin{aligned}\n\\left(3^{n}+1\\right)+X_{n+1} & =\\left(3^{n}+1\\right)+2 \\cdot 3^{n} \\\\\n& =3 \\cdot 3^{n}+1 \\\\\n& =3^{n+1}+1 .\n\\end{aligned}\n$$" ]

[ "$3^{n}+1$" ]

[ "Use Descartes' Circle Formula with $a=b=1$ and $c=\\frac{3}{2}$ to solve for $d$ :\n\n$$\n\\begin{aligned}\n2 \\cdot\\left(1^{2}+1^{2}+\\left(\\frac{3}{2}\\right)^{2}+d^{2}\\right) & =\\left(1+1+\\frac{3}{2}+d\\right)^{2} \\\\\n\\frac{17}{2}+2 d^{2} & =\\frac{49}{4}+7 d+d^{2} \\\\\nd^{2}-7 d-\\frac{15}{4} & =0\n\\end{aligned}\n$$\n\nfrom which $d=\\frac{15}{2}$ or $d=-\\frac{1}{2}$. These values correspond to radii of $\\frac{2}{15}$, a small circle nestled between the other three, or 2 , a large circle enclosing the other three.\n\nAlternatively, start by scaling the kingdom with the first four circles removed to match the situation given. Thus the three given circles are internally tangent to a circle of radius $r=2$ and curvature $d=-\\frac{1}{2}$. Descartes' Circle Formula gives a quadratic equation for $d$, and the sum of the roots is $2 \\cdot\\left(1+1+\\frac{3}{2}\\right)=7$, so the second root is $7+\\frac{1}{2}=\\frac{15}{2}$, corresponding to a circle of radius $r=\\frac{2}{15}$." ]

[ "$2$, $\\frac{2}{15}$" ]

[ "Day 3 begins with two circles of curvature 15 from the configuration $(2,2,3,15)$, and four circles of curvature 6 from the configuration $(-1,2,3,6)$. Consider the following two cases:\n\nCase 1: $(a, b, c, d)=(2,2,3,15), s=22$\n\n- $a=2: a^{\\prime}=2 s-3 a=\\mathbf{3 8}$\n- $b=2: b^{\\prime}=2 s-3 b=\\mathbf{3 8}$\n- $c=3: c^{\\prime}=2 s-3 c=\\mathbf{3 5}$\n- $d=15: d^{\\prime}=2 s-3 d=-1$, which is the configuration from day 1 .\n\nCase 2: $(a, b, c, d)=(-1,2,3,6), s=10$\n\n- $a=-1: a^{\\prime}=2 s-3 a=\\mathbf{2 3}$\n- $b=2: b^{\\prime}=2 s-3 b=\\mathbf{1 4}$\n- $c=3: c^{\\prime}=2 s-3 c=\\mathbf{1 1}$\n- $d=6: d^{\\prime}=2 s-3 d=2$, which is the configuration from day 1 .\n\n\n\nSo the areas of the plots removed on day 3 are:\n\n$$\n\\frac{\\pi}{38^{2}}, \\frac{\\pi}{35^{2}}, \\frac{\\pi}{23^{2}}, \\frac{\\pi}{14^{2}}, \\text { and } \\frac{\\pi}{11^{2}}\n$$\n\nThere are two circles with area $\\frac{\\pi}{35^{2}}$, and four circles with each of the other areas, for a total of 18 plots." ]

[ "$\\frac{\\pi}{38^{2}}, \\frac{\\pi}{35^{2}}, \\frac{\\pi}{23^{2}}, \\frac{\\pi}{14^{2}}, \\frac{\\pi}{11^{2}}$" ]

[ "Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n<img_3784>\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\nCompute $[A K M]$ as $\\frac{1}{2}$ base $\\cdot$ height, using base $\\overline{A M}$.\n\n<img_3921>\n\nBecause of (*), $\\triangle A Y O \\sim \\triangle A K L$. To compute $A M$, notice that in $\\triangle A Y O, A O=A M-r$, while in $\\triangle A K L$, the corresponding side $A L=A M+M L=A M+2$. Therefore:\n\n$$\n\\begin{aligned}\n\\frac{A O}{A L} & =\\frac{Y O}{K L} \\\\\n\\frac{A M-\\frac{5}{4}}{A M+2} & =\\frac{5 / 4}{3}\n\\end{aligned}\n$$\n\nfrom which $A M=\\frac{25}{7}$. Draw the altitude of $\\triangle A K M$ from vertex $K$, and let $h$ be its length. In right triangle $O K L, h$ is the altitude to the hypotenuse, so $\\frac{h}{3}=\\sin (\\angle K L O)=\\frac{r}{r+2}$. Hence $h=\\frac{15}{13}$. Therefore $[A K M]=\\frac{1}{2} \\cdot \\frac{25}{7} \\cdot \\frac{15}{13}=\\frac{375}{182}$.", "Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n<img_3784>\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\nBy the Power of the Point Theorem, $L K^{2}=L M \\cdot L R$, so\n\n$$\n\\begin{aligned}\nL R & =\\frac{9}{2} \\\\\nR M & =L R-L M=\\frac{5}{2} \\\\\nO L & =r+M L=\\frac{13}{4}\n\\end{aligned}\n$$\n\nFrom (*), we know that $\\triangle A Y O \\sim \\triangle A K L$. Hence by $(\\dagger)$,\n\n$$\n\\frac{A L}{A O}=\\frac{A L}{A L-O L}=\\frac{K L}{Y O}=\\frac{3}{5 / 4}=\\frac{12}{5}, \\quad \\text { thus } \\quad A L=\\frac{12}{7} \\cdot O L=\\frac{12}{7} \\cdot \\frac{13}{4}=\\frac{39}{7}\n$$\n\nHence $A M=A L-2=\\frac{25}{7}$. The ratio between the areas of triangles $A K M$ and $R K M$ is equal to\n\n$$\n\\frac{[A K M]}{[R K M]}=\\frac{A M}{R M}=\\frac{25 / 7}{5 / 2}=\\frac{10}{7}\n$$\n\nThus $[A K M]=\\frac{10}{7} \\cdot[R K M]$.\n\nBecause $\\angle K R L$ and $\\angle M K L$ both subtend $\\widehat{K M}, \\triangle K R L \\sim \\triangle M K L$. Therefore $\\frac{K R}{M K}=\\frac{L K}{L M}=$ $\\frac{3}{2}$. Thus let $K R=3 x$ and $M K=2 x$ for some positive real number $x$. Because $R M$ is a diameter of $\\omega$ (see left diagram below), $\\mathrm{m} \\angle R K M=90^{\\circ}$. Thus triangle $R K M$ is a right triangle with hypotenuse $\\overline{R M}$. In particular, $13 x^{2}=K R^{2}+M K^{2}=R M^{2}=\\frac{25}{4}$, so $x^{2}=\\frac{25}{52}$ and $[R K M]=\\frac{R K \\cdot K M}{2}=3 x^{2}$. Therefore\n\n$$\n[A K M]=\\frac{10}{7} \\cdot[R K M]=\\frac{10}{7} \\cdot 3 \\cdot \\frac{25}{52}=\\frac{\\mathbf{3 7 5}}{\\mathbf{1 8 2}}\n$$\n\n<img_3387>", "Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n<img_3784>\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\n\nLet $U$ and $V$ be the respective feet of the perpendiculars dropped from $A$ and $M$ to $\\overleftrightarrow{K L}$. From (*), $\\triangle A K L$ can be dissected into two infinite progressions of triangles: one progression of triangles similar to $\\triangle O K L$ and the other similar to $\\triangle Y O K$, as shown in the right diagram above. In both progressions, the corresponding sides of the triangles have common ratio equal to\n\n$$\n\\frac{Y O}{K L}=\\frac{5 / 4}{3}=\\frac{5}{12}\n$$\n\n\n\nThus\n\n$$\nA U=\\frac{5}{4}\\left(1+\\frac{5}{12}+\\left(\\frac{5}{12}\\right)^{2}+\\cdots\\right)=\\frac{5}{4} \\cdot \\frac{12}{7}=\\frac{15}{7}\n$$\n\nBecause $\\triangle L M V \\sim \\triangle L O K$, and because $L O=\\frac{13}{4}$ by $(\\dagger)$,\n\n$$\n\\frac{M V}{O K}=\\frac{L M}{L O}, \\quad \\text { thus } \\quad M V=\\frac{O K \\cdot L M}{L O}=\\frac{\\frac{5}{4} \\cdot 2}{\\frac{13}{4}}=\\frac{10}{13}\n$$\n\nFinally, note that $[A K M]=[A K L]-[K L M]$. Because $\\triangle A K L$ and $\\triangle K L M$ share base $\\overline{K L}$,\n\n$$\n[A K M]=\\frac{1}{2} \\cdot 3 \\cdot\\left(\\frac{15}{7}-\\frac{10}{13}\\right)=\\frac{\\mathbf{3 7 5}}{\\mathbf{1 8 2}}\n$$" ]

[ "$\\frac{375}{182}$" ]

[ "Let $r$ be the radius of the smaller circle. Then the conditions defining $Q$ imply that $P Q=$ $T<4 r$. With $T=8 \\sqrt{2}$, note that $r>2 \\sqrt{2} \\rightarrow 3 r>6 \\sqrt{2}=\\sqrt{72}$. The least integer greater than $\\sqrt{72}$ is 9 ." ]

[ "9" ]

[ "$$\n\\begin{aligned}\n& \\mathrm{Pa}(1,1)+\\mathrm{Pa}(2,1)+\\mathrm{Pa}(3,1)+\\mathrm{Pa}(4,1)+\\mathrm{Pa}(5,1)=1+2+3+4+5=\\mathbf{1 5} \\\\\n& \\mathrm{Pa}(2,2)+\\mathrm{Pa}(3,2)+\\mathrm{Pa}(4,2)+\\mathrm{Pa}(5,2)+\\mathrm{Pa}(6,2)=1+3+6+10+15=\\mathbf{3 5} \\\\\n& \\mathrm{Pa}(3,3)+\\mathrm{Pa}(4,3)+\\mathrm{Pa}(5,3)+\\mathrm{Pa}(6,3)+\\mathrm{Pa}(7,3)=1+4+10+20+35=\\mathbf{7 0} \\\\\n& \\mathrm{Pa}(4,4)+\\mathrm{Pa}(5,4)+\\mathrm{Pa}(6,4)+\\mathrm{Pa}(7,4)+\\mathrm{Pa}(8,4)=1+5+15+35+70=\\mathbf{1 2 6}\n\\end{aligned}\n$$" ]

[ "15, 35, 70, 126" ]

[ "Notice that $\\mathrm{Pa}(n, n)+\\operatorname{Pa}(n+1, n)+\\cdots+\\operatorname{Pa}(n+k, n)=\\mathrm{Pa}(n+k+1, n+1)$, so $m=n+k+1$ and $j=n+1$. (By symmetry, $j=k$ is also correct.) The equation is true for all $n$ when $k=0$, because the sum is simply $\\mathrm{Pa}(n, n)$ and the right side is $\\mathrm{Pa}(n+1, n+1)$, both of which are 1 . Proceed by induction on $k$. If $\\mathrm{Pa}(n, n)+\\mathrm{Pa}(n+1, n)+\\cdots+\\mathrm{Pa}(n+k, n)=\\mathrm{Pa}(n+k+1, n+1)$, then adding $\\mathrm{Pa}(n+k+1, n)$ to both sides yields $\\mathrm{Pa}(n+k+1, n)+\\mathrm{Pa}(n+k+1, n+1)=$ $\\mathrm{Pa}(n+k+2, n+1)$ by the recursive rule for $\\mathrm{Pa}$." ]

[ "$m=n+k+1$, $j=n+1$" ]

[ "Using the given values yields the system of equations below.\n\n$$\n\\left\\{\\begin{array}{l}\n\\mathrm{Cl}(1,1)=1=a(1)^{2}+b(1)+c \\\\\n\\mathrm{Cl}(2,1)=7=a(2)^{2}+b(2)+c \\\\\n\\mathrm{Cl}(3,1)=19=a(3)^{2}+b(3)+c\n\\end{array}\\right.\n$$\n\nSolving this system, $a=3, b=-3, c=1$." ]

[ "$3,-3,1$" ]

[ "$\\mathrm{Cl}(11,2)=1000$." ]

[ "1000" ]

[ "$\\mathrm{Cl}(11,3)=2025$." ]

[ "2025" ]

[ "$\\operatorname{Le}(17,1)=\\operatorname{Le}(16,0)-\\operatorname{Le}(17,0)=\\frac{1}{17}-\\frac{1}{18}=\\frac{1}{306}$." ]

[ "$\\frac{1}{306}$" ]

[ "$\\operatorname{Le}(17,2)=\\operatorname{Le}(16,1)-\\operatorname{Le}(17,1)=\\operatorname{Le}(15,0)-\\operatorname{Le}(16,0)-\\operatorname{Le}(17,1)=\\frac{1}{2448}$." ]

[ "$\\frac{1}{2448}$" ]

[ "Because $\\operatorname{Le}(n, 1)=\\frac{1}{n}-\\frac{1}{n+1}$,\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{2011} \\operatorname{Le}(i, 1) & =\\sum_{i=1}^{2011}\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right) \\\\\n& =\\left(\\frac{1}{1}-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\cdots+\\left(\\frac{1}{2010}-\\frac{1}{2011}\\right)+\\left(\\frac{1}{2011}-\\frac{1}{2012}\\right) \\\\\n& =1-\\frac{1}{2012} \\\\\n& =\\frac{2011}{2012} .\n\\end{aligned}\n$$" ]

[ "$\\frac{2011}{2012}$" ]

[ "Extending the result of $8 \\mathrm{~b}$ gives\n\n$$\n\\sum_{i=1}^{n} \\operatorname{Le}(i, 1)=\\frac{1}{1}-\\frac{1}{n}\n$$\n\nso as $n \\rightarrow \\infty, \\sum_{i=1}^{n} \\operatorname{Le}(i, 1) \\rightarrow 1$. This value appears as $\\operatorname{Le}(0,0)$, so $n=k=0$." ]

[ "$0,0$" ]

[ "$n=k=m-1$." ]

[ "$m-1,m-1$" ]

[ "The desired triangle is an isosceles triangle whose base vertices are the centers of the two smaller circles. The congruent sides of the triangle have length $T+\\frac{T}{2}$. Thus the altitude to the base has length $\\sqrt{\\left(\\frac{3 T}{2}\\right)^{2}-\\left(\\frac{T}{2}\\right)^{2}}=T \\sqrt{2}$. Thus the area of the triangle is $\\frac{1}{2} \\cdot\\left(\\frac{T}{2}+\\frac{T}{2}\\right) \\cdot T \\sqrt{2}=\\frac{T^{2} \\sqrt{2}}{2}$. With $T=12$, the area is $\\mathbf{7 2} \\sqrt{\\mathbf{2}}$." ]

[ "$72 \\sqrt{2}$" ]