Datasets:

qq8933

/

AOPS

like 0

[ "Take a force-overlaid inversion about $A$ and note $D$ and $E$ map to each other. As $DE$ was originally the diameter of $\\gamma$ $DE$ is still the diameter of $\\gamma$ . Thus $\\gamma$ is preserved. Note that the midpoint $M$ of $BC$ lies on $\\gamma$ , and $BC$ and $\\omega$ are swapped. Thus points $F$ and $M$ map to each other, and are isogonal. It follows that $AF$ is a symmedian of $\\triangle{ABC}$ , or that $ABFC$ is harmonic. Then $(AB)(FC)=(BF)(CA)$ , and thus we can let $BF=5x, CF=3x$ for some $x$ . By the LoC, it is easy to see $\\angle{BAC}=120^\\circ$ so $(5x)^2+(3x)^2-2\\cos{60^\\circ}(5x)(3x)=49$ . Solving gives $x^2=\\frac{49}{19}$ , from which by Ptolemy's we see $AF=\\frac{30}{\\sqrt{19}}$ . We conclude the answer is $900+19=\\boxed{919}$", "Use the angle bisector theorem to find $CD=\\tfrac{21}{8}$ $BD=\\tfrac{35}{8}$ , and use Stewart's Theorem to find $AD=\\tfrac{15}{8}$ . Use Power of Point $D$ to find $DE=\\tfrac{49}{8}$ , and so $AE=8$ . Use law of cosines to find $\\angle CAD = \\tfrac{\\pi} {3}$ , hence $\\angle BAD = \\tfrac{\\pi}{3}$ as well, and $\\triangle BCE$ is equilateral, so $BC=CE=BE=7$ In triangle $AEF$ , let $X$ be the foot of the altitude from $A$ ; then $EF=EX+XF$ , where we use signed lengths. Writing $EX=AE \\cdot \\cos \\angle AEF$ and $XF=AF \\cdot \\cos \\angle AFE$ , we get \\begin{align}\\tag{1} EF = AE \\cdot \\cos \\angle AEF + AF \\cdot \\cos \\angle AFE. \\end{align} Note $\\angle AFE = \\angle ACE$ , and the Law of Cosines in $\\triangle ACE$ gives $\\cos \\angle ACE = -\\tfrac 17$ . \nAlso, $\\angle AEF = \\angle DEF$ , and $\\angle DFE = \\tfrac{\\pi}{2}$ $DE$ is a diameter), so $\\cos \\angle AEF = \\tfrac{EF}{DE} = \\tfrac{8}{49}\\cdot EF$\nPlugging in all our values into equation $(1)$ , we get: \\[EF = \\tfrac{64}{49} EF -\\tfrac{1}{7} AF \\quad \\Longrightarrow \\quad EF = \\tfrac{7}{15} AF.\\] The Law of Cosines in $\\triangle AEF$ , with $EF=\\tfrac 7{15}AF$ and $\\cos\\angle AFE = -\\tfrac 17$ gives \\[8^2 = AF^2 + \\tfrac{49}{225} AF^2 + \\tfrac 2{15} AF^2 = \\tfrac{225+49+30}{225}\\cdot AF^2\\] Thus $AF^2 = \\frac{900}{19}$ . The answer is $\\boxed{919}$", "Let $a = BC$ $b = CA$ $c = AB$ for convenience. Let $M$ be the midpoint of segment $BC$ . We claim that $\\angle MAD=\\angle DAF$\n$\\textit{Proof}$ . Since $AE$ is the angle bisector, it follows that $EB = EC$ and consequently $EM\\perp BC$ . Therefore, $M\\in \\gamma$ . Now let $X = FD\\cap \\omega$ . Since $\\angle EFX=90^\\circ$ $EX$ is a diameter, so $X$ lies on the perpendicular bisector of $BC$ ; hence $E$ $M$ $X$ are collinear. From $\\angle DAG = \\angle DMX = 90^\\circ$ , quadrilateral $ADMX$ is cyclic. Therefore, $\\angle MAD = \\angle MXD$ . But $\\angle MXD$ and $\\angle EAF$ are both subtended by arc $EF$ in $\\omega$ , so they are equal. Thus $\\angle MAD=\\angle DAF$ , as claimed. As a result, $\\angle CAM = \\angle FAB$ . Combined with $\\angle BFA=\\angle MCA$ , we get $\\triangle ABF\\sim\\triangle AMC$ and therefore \\[\\frac c{AM}=\\frac {AF}b\\qquad \\Longrightarrow \\qquad AF^2=\\frac{b^2c^2}{AM^2} = \\frac{15^2}{AM^2}\\] By Stewart's Theorem on $\\triangle ABC$ (with cevian $AM$ ), we get \\[AM^2 = \\tfrac 12 (b^2+c^2)-\\tfrac 14 a^2 = \\tfrac{19}{4},\\] so $AF^2 = \\tfrac{900}{19}$ , so the answer is $900+19=\\boxed{919}$", "Use the angle bisector theorem to find $CD=\\tfrac{21}{8}$ $BD=\\tfrac{35}{8}$ , and use Stewart's Theorem to find $AD=\\tfrac{15}{8}$ . Use Power of Point $D$ to find $DE=\\tfrac{49}{8}$ , and so $AE=8$ . Then use the Extended Law of Sine to find that the length of the circumradius of $\\triangle ABC$ is $\\tfrac{7\\sqrt{3}}{3}$ Since $DE$ is the diameter of circle $\\gamma$ $\\angle DFE$ is $90^\\circ$ . Extending $DF$ to intersect circle $\\omega$ at $X$ , we find that $XE$ is the diameter of $\\omega$ (since $\\angle DFE$ is $90^\\circ$ ). Therefore, $XE=\\tfrac{14\\sqrt{3}}{3}$\nLet $EF=x$ $XD=u$ , and $DF=v$ . Then $XE^2-XF^2=EF^2=DE^2-DF^2$ , so we get \\[(u+v)^2-v^2=\\frac{196}{3}-\\frac{2401}{64}\\] which simplifies to \\[u^2+2uv = \\frac{5341}{192}.\\] By Power of Point $D$ $uv=BD \\cdot DC=735/64$ . Combining with above, we get \\[XD^2=u^2=\\frac{931}{192}.\\] Note that $\\triangle XDE\\sim \\triangle ADF$ and the ratio of similarity is $\\rho = AD : XD = \\tfrac{15}{8}:u$ . Then $AF=\\rho\\cdot XE = \\tfrac{15}{8u}\\cdot R$ and \\[AF^2 = \\frac{225}{64}\\cdot \\frac{R^2}{u^2} = \\frac{900}{19}.\\] The answer is $900+19=\\boxed{919}$", "Denote $AB = c, BC = a, AC = b, \\angle A = 2 \\alpha.$ Let M be midpoint BC. Let $\\theta$ be the circle centered at $A$ with radius $\\sqrt{AB \\cdot AC} =\\sqrt{bc}.$\nWe calculate the length of some segments.\nThe median $AM = \\sqrt{\\frac {b^2}{2} + \\frac {c^2}{2} - \\frac {a^2}{4}}.$ The bisector $AD = \\frac {2 b c \\cos \\alpha}{b+c}.$ One can use Stewart's Theorem in both cases.\n$AD$ is bisector of $\\angle A \\implies BD = \\frac {a c}{b + c}, CD = \\frac {a b}{b + c} \\implies$ \\[BD \\cdot CD = \\frac {a^2 bc }{(b+c)^2}.\\] We use Power of Point $D$ and get $AD \\cdot DE = BD \\cdot CD.$ \\[AE = AD + DE = AD + \\frac {BD \\cdot CD}{AD},\\] \\[AE =\\frac {2 b c \\cos \\alpha}{b+c} + \\frac {a^2 bc \\cdot (b+c) }{(b+c)^2 \\cdot 2 b c \\cos \\alpha} =\\] \\[= \\frac {b c \\cos^2 \\alpha + a^2}{2(b+c)\\cos \\alpha} =\\frac {4bc \\cos^2 \\alpha + b^2 +c^2 -2 b c \\cos 2\\alpha}{2(b+c) \\cos \\alpha} = \\frac {b+c}{2} \\implies AD \\cdot AE = 2 bc \\cos \\alpha.\\] We consider the inversion with respect $\\theta.$\n$B$ swap $B' \\implies AB' = AC, B' \\in AB \\implies B'$ is symmetric to $C$ with respect to $AE.$\n$C$ swap $C' \\implies AC' = AB, C'$ lies on line $AC \\implies C'$ is symmetric to $B$ with respect to $AE.$\n$BC^2 = AB^2 + AC^2 + AB \\cdot BC \\implies \\alpha = 60^\\circ \\implies AD \\cdot AE = bc \\implies D$ swap $E.$\nPoints $D$ and $E$ lies on $\\Gamma \\implies \\Gamma$ swap $\\Gamma.$\n$DE$ is diameter $\\Gamma, \\angle DME = 90^\\circ \\implies M \\in \\Gamma.$ Therefore $M$ is crosspoint of $BC$ and $\\Gamma.$\nLet $\\Omega$ be circumcircle $AB'C'. \\Omega$ is image of line $BC.$ Point $M$ maps into $M' \\implies M' = \\Gamma \\cap \\Omega.$\nPoints $A, B',$ and $C'$ are symmetric to $A, C,$ and $B,$ respectively.\nPoint $M'$ lies on $\\Gamma$ which is symmetric with respect to $AE$ and on $\\Omega$ which is symmetric to $\\omega$ with respect to $AE \\implies$\n$M'$ is symmetric $F$ with respect to $AE \\implies AM' = AF.$\nWe use Power of Point $A$ and get \\[AF = AM' = \\frac {AD \\cdot AE}{AM} = \\frac {4b c}{\\sqrt{2 b^2 + 2 c^2 – a^2}} = \\frac {4 \\cdot 3 \\cdot 5}{\\sqrt{ 50 + 18 – 49}} = \\frac {30}{\\sqrt{19}} \\implies \\boxed{919}.\\]" ]

[ "Let $x=\\angle CAM$ , so $3x=\\angle CDM$ . Then, $\\frac{\\tan 3x}{\\tan x}=\\frac{CM/1}{CM/11}=11$ . Expanding $\\tan 3x$ using the angle sum identity gives \\[\\tan 3x=\\tan(2x+x)=\\frac{3\\tan x-\\tan^3x}{1-3\\tan^2x}.\\] Thus, $\\frac{3-\\tan^2x}{1-3\\tan^2x}=11$ . Solving, we get $\\tan x= \\frac 12$ . Hence, $CM=\\frac{11}2$ and $AC= \\frac{11\\sqrt{5}}2$ by the Pythagorean Theorem . The total perimeter is $2(AC + CM) = \\sqrt{605}+11$ . The answer is thus $a+b=\\boxed{616}$", "In a similar fashion, we encode the angles as complex numbers, so if $BM=x$ , then $\\angle BAD=\\text{Arg}(11+xi)$ and $\\angle BDM=\\text{Arg}(1+xi)$ . So we need only find $x$ such that $\\text{Arg}((11+xi)^3)=\\text{Arg}(1331-33x^2+(363x-x^3)i)=\\text{Arg}(1+xi)$ . This will happen when $\\frac{363x-x^3}{1331-33x^2}=x$ , which simplifies to $121x-4x^3=0$ . Therefore, $x=\\frac{11}{2}$ . By the Pythagorean Theorem, $AB=\\frac{11\\sqrt{5}}{2}$ , so the perimeter is $11+11\\sqrt{5}=11+\\sqrt{605}$ , giving us our answer, $\\boxed{616}$", "Let $\\angle BAD=\\alpha$ , so $\\angle BDM=3\\alpha$ $\\angle BDA=180-3\\alpha$ , and thus $\\angle ABD=2\\alpha.$ We can then draw the angle bisector of $\\angle ABD$ , and let it intersect $\\overline{AM}$ at $E.$ Since $\\angle BAE=\\angle ABE$ $AE=BE.$ Let $AE=x$ . Then we see by the Pythagorean Theorem, $BM=\\sqrt{BE^2-ME^2}=\\sqrt{x^2-(11-x)^2}=\\sqrt{22x-121}$ $BD=\\sqrt{BM^2+1}=\\sqrt{22x-120}$ $BA=\\sqrt{BM^2+121}=\\sqrt{22x}$ , and $DE=10-x.$ By the angle bisector theorem, $BA/BD=EA/ED.$ Substituting in what we know for the lengths of those segments, we see that \\[\\frac{\\sqrt{22x}}{\\sqrt{22x-120}}=\\frac{x}{10-x}.\\] multiplying by both denominators and squaring both sides yields \\[22x(10-x)^2=x^2(22x-120)\\] which simplifies to $x=\\frac{55}{8}.$ Substituting this in for x in the equations for $BA$ and $BM$ yields $BA=\\frac{\\sqrt{605}}{2}$ and $BM=\\frac{11}{2}.$ Thus the perimeter is $11+\\sqrt{605}$ , and the answer is $\\boxed{616}$", "The triangle is symmetrical so we can split it in half ( $\\triangle ABM$ and $\\triangle ACM$ ).\nLet $\\angle BAM = y$ and $\\angle BDM = 3y$ . By the Law of Sines on triangle $BAD$ $\\frac{10}{\\sin 2y} = \\frac{BD}{\\sin y}$ . Using $\\sin 2y = 2\\sin y\\cos y$ we can get $BD = \\frac{5}{\\cos y}$ . We can use this information to relate $BD$ to $DM$ by using the Law of Sines on triangle $BMD$\n\\[\\frac{\\frac{5}{\\cos y}}{\\sin BMD} = \\frac{1}{\\sin 90^\\circ - 3y}\\]\n$\\sin BMD = 1$ (as $\\angle BMD$ is a right angle), so $\\frac{1}{\\sin 90^\\circ - 3y} = \\frac{5}{\\cos y}$ . Using the identity $\\sin 90^\\circ - x = \\cos x$ , we can turn the equation into::\n\\[\\frac{1}{\\cos 3y} = \\frac{5}{\\cos y}\\]\n\\[5\\cos 3y = \\cos y\\]\n\\[5(4\\cos ^3 y - 3\\cos y) = \\cos y\\]\n\\[20\\cos ^3 y = 16 \\cos y\\]\n\\[5\\cos ^3 y = 4\\cos y\\]\n\\[5\\cos ^2 y = 4\\]\n\\[\\cos ^2 y = \\frac{4}{5}\\]\nNow that we've found $\\cos y$ , we can look at the side lengths of $BM$ and $AB$ (since they are symmetrical, the perimeter of $\\triangle ABC$ is $2(BM + AB)$\nWe note that $BM = 11\\tan y$ and $AB = 11\\sec y$\n\\[\\sin ^2 y = 1 - \\cos ^2 y\\]\n\\[\\sin ^2 y = \\frac{1}{5}\\]\n\\[\\tan ^2 y = \\frac{1}{4}\\]\n\\[\\tan y = \\frac{1}{2}\\]\n(Note it is positive since $BM > 0$ ).\n\\[\\sec ^2 y = \\frac{5}{4}\\]\n\\[\\sec y = \\frac{\\sqrt{5}}{2}\\]\n\\[BM + AB = 11\\frac{\\sqrt{5}+1}{2}\\]\n\\[2(BM + AB) = 11(\\sqrt{5} + 1)\\]\n\\[2(BM + AB) = 11\\sqrt{5} + 11\\]\n\\[2(BM + AB) = \\sqrt{605} + 11\\]\nThe answer is $\\boxed{616}$", "Suppose $\\angle BAM=\\angle CAM =x$ , since $\\angle BDC=3\\angle BAC$ , we have $\\angle BDM=\\angle MDC = 3x$ . Therefore, $\\angle DBC=\\angle DCB = 90^\\circ -3x$ and $\\angle ABD=\\angle DCA=2x$ . As a result, $\\triangle KAC$ is isosceles, $KC=KA$\nLet $H$ be a point on the extension of $CD$ through $D$ such that $\\overline{HB}\\perp\\overline{BC}$ and denote the intersection of $\\overline{HC}$ and $\\overline{AB}$ as $K$ . Then, $BH=2DM=2, \\overline{HB}\\parallel\\overline{DM}$ , and $HD=DC$ by the Midpoint Theorem. So, $\\angle HBA=x$ and $\\angle CDM=\\angle CHB=\\angle HDA= 3x$\nConsequently, $\\triangle HBK\\sim \\triangle DAK$ \\[\\frac{BK}{KA}=\\frac{HK}{KD}=\\frac{1}{5}\\] Assume $BK=a$ and $HK=b$ , then $KA=5a$ and $KD = 5b$ . Since $KC=KA, KC=5a$ , and since $HD=DC$ $KC=11b$ . Therefore, $a=\\frac{11}{5}b$\nIn $\\triangle BDM$ , by the Pythagorean Theorem $BM=\\sqrt{36b^2-1}$ . Similarly in $\\triangle BAM$ $BM=\\sqrt{36a^2-121}$ . So \\[\\sqrt{36a^2-121}=\\sqrt{36b^2-1}\\] Since $a=\\frac{11}{5}b$ , we have $b=\\frac{5\\sqrt{5}}{12}$ and $a=\\frac{11\\sqrt{5}}{12}$ . Consequently, $BM=\\frac{11}{2}$ and $AB=\\frac{11\\sqrt{5}}{2}$ . Thus, the perimeter of $\\triangle ABC$ is $11+\\sqrt{605}$ , and the answer is $\\boxed{616}$" ]

[ "Take point $N$ inside $\\triangle ABC$ such that $\\angle CBN = 7^\\circ$ and $\\angle BCN = 23^\\circ$\n$\\angle MCN = 106^\\circ - 2\\cdot 23^\\circ = 60^\\circ$ . Also, since $\\triangle AMC$ and $\\triangle BNC$ are congruent (by ASA), $CM = CN$ . Hence $\\triangle CMN$ is an equilateral triangle , so $\\angle CNM = 60^\\circ$\nThen $\\angle MNB = 360^\\circ - \\angle CNM - \\angle CNB = 360^\\circ - 60^\\circ - 150^\\circ = 150^\\circ$ . We now see that $\\triangle MNB$ and $\\triangle CNB$ are congruent. Therefore, $CB = MB$ , so $\\angle CMB = \\angle MCB = \\boxed{83}$", "From the givens, we have the following angle measures $m\\angle AMC = 150^\\circ$ $m\\angle MCB = 83^\\circ$ . If we define $m\\angle CMB = \\theta$ then we also have $m\\angle CBM = 97^\\circ - \\theta$ . Then apply the Law of Sines to triangles $\\triangle AMC$ and $\\triangle BMC$ to get\n\\[\\frac{\\sin 150^\\circ}{\\sin 7^\\circ} = \\frac{AC}{CM} = \\frac{BC}{CM} = \\frac{\\sin \\theta}{\\sin (97^\\circ - \\theta)}\\]\nClearing denominators , evaluating $\\sin 150^\\circ = \\frac 12$ and applying one of our trigonometric identities to the result gives\n\\[\\frac{1}{2} \\cos (7^\\circ - \\theta )= \\sin 7^\\circ \\sin \\theta\\]\nand multiplying through by 2 and applying the double angle formulas gives\n\\[\\cos 7^\\circ\\cos\\theta + \\sin7^\\circ\\sin\\theta = 2 \\sin7^\\circ \\sin\\theta\\]\nand so $\\cos 7^\\circ \\cos \\theta = \\sin 7^\\circ \\sin\\theta \\Longleftrightarrow \\tan 7^{\\circ} = \\cot \\theta$ ; since $0^\\circ < \\theta < 180^\\circ$ , we must have $\\theta = 83^\\circ$ , so the answer is $\\boxed{83}$", "Without loss of generality , let $AC = BC = 1$ . Then, using the Law of Sines in triangle $AMC$ , we get $\\frac {1}{\\sin 150} = \\frac {MC}{\\sin 7}$ , and using the sine addition formula to evaluate $\\sin 150 = \\sin (90 + 60)$ , we get $MC = 2 \\sin 7$\nThen, using the Law of Cosines in triangle $MCB$ , we get $MB^2 = 4\\sin^2 7 + 1 - 4\\sin 7(\\cos 83) = 1$ , since $\\cos 83 = \\sin 7$ . So triangle $MCB$ is isosceles, and $\\angle CMB = \\boxed{83}$", "Note: A diagram would be much appreciated; I cannot make one since I'm bad at asymptote. Also, please make this less cluttered :) ~tauros\nFirst, take point $E$ outside of $\\triangle{ABC}$ so that $\\triangle{CEB}$ is equilateral. Then, connect $A$ $C$ , and $M$ to $E$ . Also, let $ME$ intersect $AB$ at $F$ $\\angle{MCE} = 83^\\circ - 60^\\circ = 23^\\circ$ $CE = AB$ , and (trivially) $CM = CM$ , so $\\triangle{MCE} \\cong \\triangle{MCA}$ by SAS congruence. Also, $\\angle{CMA} = \\angle{CME} = 150^\\circ$ , so $\\angle{AME} = 60^\\circ$ , and $AM = ME$ ,\nmaking $\\triangle{AME}$ also equilateral. (it is isosceles with a $60^\\circ$ angle) $\\triangle{MAF} \\cong \\triangle{EAF}$ by SAS ( $MA = AE$ $AF = AF$ , and $m\\angle{MAF} = m\\angle{EAF} = 30^\\circ$ ), and $\\triangle{MAB} \\cong \\triangle{EAB}$ by SAS ( $MA = AE$ $AB = AB$ , and $m\\angle{MAB} = m\\angle{EAB} = 30^\\circ$ ). Thus, $\\triangle{BME}$ is isosceles, with $m\\angle{BME} = m\\angle{BEM} = 60^\\circ + 7^\\circ = 67^\\circ$ . Also, $\\angle{EMB} + \\angle {CMB} = \\angle{CME} = 150^\\circ$ , so $\\angle{CME} = 150^\\circ - 67^\\circ = \\boxed{83}$", "Noticing that we have three concurrent cevians, we apply Ceva's theorem:\n\\[(\\sin \\angle ACM)(\\sin \\angle BAM)(\\sin \\angle CBM) = (\\sin \\angle CAM)(\\sin \\angle ABM)(\\sin \\angle BCM)\\] \\[(\\sin 23)(\\sin 30)(\\sin x) = (\\sin 7)(\\sin 37-x)(\\sin 83)\\]\nusing the fact that $\\sin 83 = \\cos 7$ and $(\\sin 7)(\\cos 7) = 1/2 (\\sin 14)$ we have:\n\\[(\\sin 23)(\\sin x) = (\\sin 14)(\\sin 37-x)\\]\nBy inspection, $x=14^\\circ$ works, so the answer is $180-83-14= \\boxed{083}$" ]

[ "The midpoint $M$ of line segment $\\overline{BC}$ is $\\left(\\frac{35}{2}, \\frac{39}{2}\\right)$ . The equation of the median can be found by $-5 = \\frac{q - \\frac{39}{2}}{p - \\frac{35}{2}}$ . Cross multiply and simplify to yield that $-5p + \\frac{35 \\cdot 5}{2} = q - \\frac{39}{2}$ , so $q = -5p + 107$\nUse determinants to find that the area of $\\triangle ABC$ is $\\frac{1}{2} \\begin{vmatrix}p & 12 & 23 \\\\ q & 19 & 20 \\\\ 1 & 1 & 1\\end{vmatrix} = 70$ (note that there is a missing absolute value ; we will assume that the other solution for the triangle will give a smaller value of $p+q$ , which is provable by following these steps over again) (alternatively, we could use the Shoelace Theorem ). We can calculate this determinant to become $140 = \\begin{vmatrix} 12 & 23 \\\\ 19 & 20 \\end{vmatrix} - \\begin{vmatrix} p & q \\\\ 23 & 20 \\end{vmatrix} + \\begin{vmatrix} p & q \\\\ 12 & 19 \\end{vmatrix}$ $\\Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q$ $= -197 - p + 11q$ . Thus, $q = \\frac{1}{11}p + \\frac{337}{11}$\nSetting this equation equal to the equation of the median, we get that $\\frac{1}{11}p + \\frac{337}{11} = -5p + 107$ , so $\\frac{56}{11}p = \\frac{107 \\cdot 11 - 337}{11}$ . Solving produces that $p = 15$ Substituting backwards yields that $q = 32$ ; the solution is $p + q = \\boxed{047}$" ]

[ "Let $BC = a$ $BG = x$ $GC = y$ , and the length of the perpendicular from $BC$ through $A$ be $h$ . By angle bisector theorem, we have that \\[\\frac{50}{x} = \\frac{10}{y},\\] where $y = -x+a$ . Therefore substituting we have that $BG=\\frac{5a}{6}$ . By similar triangles, we have that $DF=\\frac{5a}{12}$ , and the height of this trapezoid is $\\frac{h}{2}$ . Then, we have that $\\frac{ah}{2}=120$ . We wish to compute $\\frac{5a}{8}\\cdot\\frac{h}{2}$ , and we have that it is $\\boxed{75}$ by substituting.", "For this problem, we have $\\triangle{ADE}\\sim\\triangle{ABC}$ because of SAS and $DE = \\frac{BC}{2}$ . Therefore, $\\bigtriangleup ADE$ is a quarter of the area of $\\bigtriangleup ABC$ , which is $30$ . Subsequently, we can compute the area of quadrilateral $BDEC$ to be $120 - 30 = 90$ . Using the angle bisector theorem in the same fashion as the previous problem, we get that $\\overline{BG}$ is $5$ times the length of $\\overline{GC}$ . We want the larger piece, as described by the problem. Because the heights are identical, one area is $5$ times the other, and $\\frac{5}{6} \\cdot 90 = \\boxed{75}$", "The ratio of the $\\overline{BG}$ to $\\overline{GC}$ is $5:1$ by the Angle Bisector Theorem, so area of $\\bigtriangleup ABG$ to the area of $\\bigtriangleup ACG$ is also $5:1$ (They have the same height). Therefore, the area of $\\bigtriangleup ABG$ is $\\frac{5}{5+1}\\times120=100$ . Since $\\overline{DE}$ is the midsegment of $\\bigtriangleup ABC$ , so $\\overline{DF}$ is the midsegment of $\\bigtriangleup ABG$ . Thus, the ratio of the area of $\\bigtriangleup ADF$ to the area of $\\bigtriangleup ABG$ is $1:4$ , so the area of $\\bigtriangleup ACG$ is $\\frac{1}{4}\\times100=25$ . Therefore, the area of quadrilateral $FDBG$ is $[ABG]-[ADF]=100-25=\\boxed{75}$", "The area of quadrilateral $FDBG$ is the area of $\\bigtriangleup ABG$ minus the area of $\\bigtriangleup ADF$ . Notice, $\\overline{DE} || \\overline{BC}$ , so $\\bigtriangleup ABG \\sim \\bigtriangleup ADF$ , and since $\\overline{AD}:\\overline{AB}=1:2$ , the area of $\\bigtriangleup ADF:\\bigtriangleup ABG=(1:2)^2=1:4$ . Given that the area of $\\bigtriangleup ABC$ is $120$ , using $\\frac{bh}{2}$ on side $AB$ yields $\\frac{50h}{2}=120\\implies h=\\frac{240}{50}=\\frac{24}{5}$ . Using the Angle Bisector Theorem, $\\overline{BG}:\\overline{BC}=50:(10+50)=5:6$ , so the height of $\\bigtriangleup ABG: \\bigtriangleup ACB=5:6$ . Therefore our answer is $\\big[ FDBG\\big] = \\big[ABG\\big]-\\big[ ADF\\big] = \\big[ ABG\\big]\\big(1-\\frac{1}{4}\\big)=\\frac{3}{4}\\cdot \\frac{bh}{2}=\\frac{3}{8}\\cdot 50\\cdot \\frac{5}{6}\\cdot \\frac{24}{5}=\\frac{3}{8}\\cdot 200=\\boxed{75}$", "We try to find the area of quadrilateral $FDBG$ by subtracting the area outside the quadrilateral but inside triangle $ABC$ . Note that the area of $\\triangle ADE$ is equal to $\\frac{1}{2} \\cdot 25 \\cdot 5 \\cdot \\sin{A}$ and the area of triangle $ABC$ is equal to $\\frac{1}{2} \\cdot 50 \\cdot 10 \\cdot \\sin A$ . The ratio $\\frac{[ADE]}{[ABC]}$ is thus equal to $\\frac{1}{4}$ and the area of triangle $ADE$ is $\\frac{1}{4} \\cdot 120 = 30$ . Let side $BC$ be equal to $6x$ , then $BG = 5x, GC = x$ by the angle bisector theorem. Similarly, we find the area of triangle $AGC$ to be $\\frac{1}{2} \\cdot 10 \\cdot x \\cdot \\sin C$ and the area of triangle $ABC$ to be $\\frac{1}{2} \\cdot 6x \\cdot 10 \\cdot \\sin C$ . A ratio between these two triangles yields $\\frac{[ACG]}{[ABC]} = \\frac{x}{6x} = \\frac{1}{6}$ , so $[AGC] = 20$ . Now we just need to find the area of triangle $AFE$ and subtract it from the combined areas of $[ADE]$ and $[ACG]$ , since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$ , so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$ , and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$ , and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$ , and we are done." ]

[ "Let $BC = a$ $BG = x$ $GC = y$ , and the length of the perpendicular from $BC$ through $A$ be $h$ . By angle bisector theorem, we have that \\[\\frac{50}{x} = \\frac{10}{y},\\] where $y = -x+a$ . Therefore substituting we have that $BG=\\frac{5a}{6}$ . By similar triangles, we have that $DF=\\frac{5a}{12}$ , and the height of this trapezoid is $\\frac{h}{2}$ . Then, we have that $\\frac{ah}{2}=120$ . We wish to compute $\\frac{5a}{8}\\cdot\\frac{h}{2}$ , and we have that it is $\\boxed{75}$ by substituting.", "For this problem, we have $\\triangle{ADE}\\sim\\triangle{ABC}$ because of SAS and $DE = \\frac{BC}{2}$ . Therefore, $\\bigtriangleup ADE$ is a quarter of the area of $\\bigtriangleup ABC$ , which is $30$ . Subsequently, we can compute the area of quadrilateral $BDEC$ to be $120 - 30 = 90$ . Using the angle bisector theorem in the same fashion as the previous problem, we get that $\\overline{BG}$ is $5$ times the length of $\\overline{GC}$ . We want the larger piece, as described by the problem. Because the heights are identical, one area is $5$ times the other, and $\\frac{5}{6} \\cdot 90 = \\boxed{75}$", "The ratio of the $\\overline{BG}$ to $\\overline{GC}$ is $5:1$ by the Angle Bisector Theorem, so area of $\\bigtriangleup ABG$ to the area of $\\bigtriangleup ACG$ is also $5:1$ (They have the same height). Therefore, the area of $\\bigtriangleup ABG$ is $\\frac{5}{5+1}\\times120=100$ . Since $\\overline{DE}$ is the midsegment of $\\bigtriangleup ABC$ , so $\\overline{DF}$ is the midsegment of $\\bigtriangleup ABG$ . Thus, the ratio of the area of $\\bigtriangleup ADF$ to the area of $\\bigtriangleup ABG$ is $1:4$ , so the area of $\\bigtriangleup ACG$ is $\\frac{1}{4}\\times100=25$ . Therefore, the area of quadrilateral $FDBG$ is $[ABG]-[ADF]=100-25=\\boxed{75}$", "The area of quadrilateral $FDBG$ is the area of $\\bigtriangleup ABG$ minus the area of $\\bigtriangleup ADF$ . Notice, $\\overline{DE} || \\overline{BC}$ , so $\\bigtriangleup ABG \\sim \\bigtriangleup ADF$ , and since $\\overline{AD}:\\overline{AB}=1:2$ , the area of $\\bigtriangleup ADF:\\bigtriangleup ABG=(1:2)^2=1:4$ . Given that the area of $\\bigtriangleup ABC$ is $120$ , using $\\frac{bh}{2}$ on side $AB$ yields $\\frac{50h}{2}=120\\implies h=\\frac{240}{50}=\\frac{24}{5}$ . Using the Angle Bisector Theorem, $\\overline{BG}:\\overline{BC}=50:(10+50)=5:6$ , so the height of $\\bigtriangleup ABG: \\bigtriangleup ACB=5:6$ . Therefore our answer is $\\big[ FDBG\\big] = \\big[ABG\\big]-\\big[ ADF\\big] = \\big[ ABG\\big]\\big(1-\\frac{1}{4}\\big)=\\frac{3}{4}\\cdot \\frac{bh}{2}=\\frac{3}{8}\\cdot 50\\cdot \\frac{5}{6}\\cdot \\frac{24}{5}=\\frac{3}{8}\\cdot 200=\\boxed{75}$", "We try to find the area of quadrilateral $FDBG$ by subtracting the area outside the quadrilateral but inside triangle $ABC$ . Note that the area of $\\triangle ADE$ is equal to $\\frac{1}{2} \\cdot 25 \\cdot 5 \\cdot \\sin{A}$ and the area of triangle $ABC$ is equal to $\\frac{1}{2} \\cdot 50 \\cdot 10 \\cdot \\sin A$ . The ratio $\\frac{[ADE]}{[ABC]}$ is thus equal to $\\frac{1}{4}$ and the area of triangle $ADE$ is $\\frac{1}{4} \\cdot 120 = 30$ . Let side $BC$ be equal to $6x$ , then $BG = 5x, GC = x$ by the angle bisector theorem. Similarly, we find the area of triangle $AGC$ to be $\\frac{1}{2} \\cdot 10 \\cdot x \\cdot \\sin C$ and the area of triangle $ABC$ to be $\\frac{1}{2} \\cdot 6x \\cdot 10 \\cdot \\sin C$ . A ratio between these two triangles yields $\\frac{[ACG]}{[ABC]} = \\frac{x}{6x} = \\frac{1}{6}$ , so $[AGC] = 20$ . Now we just need to find the area of triangle $AFE$ and subtract it from the combined areas of $[ADE]$ and $[ACG]$ , since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$ , so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$ , and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$ , and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$ , and we are done." ]

[ "Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$ . It now follows that $\\angle{DOA} = 2\\angle ACP = \\angle{APC} = \\angle{DPB}$ . Hence $ODP$ is isosceles and $OD = DP = 2$\nDenote $E$ the projection of $O$ onto $CD$ . Now $CD = CP + DP = 3$ . By the Pythagorean Theorem $OE = \\sqrt {2^2 - \\frac {3^2}{2^2}} = \\sqrt {\\frac {7}{4}}$ . Now note that $EP = \\frac {1}{2}$ . By the Pythagorean Theorem, $OP = \\sqrt {\\frac {7}{4} + \\frac {1^2}{2^2}} = \\sqrt {2}$ . Hence it now follows that,\n\\[\\frac {AP}{BP} = \\frac {AO + OP}{BO - OP} = \\frac {2 + \\sqrt {2}}{2 - \\sqrt {2}} = 3 + 2\\sqrt {2}\\]\nThis gives that the answer is $\\boxed{007}$", "Let $\\angle{ACP}$ be equal to $x$ . Then by Law of Sines, $PB = -\\frac{\\cos{x}}{\\cos{3x}}$ and $AP = \\frac{\\sin{x}}{\\sin{3x}}$ . We then obtain $\\cos{3x} = 4\\cos^3{x} - 3\\cos{x}$ and $\\sin{3x} = 3\\sin{x} - 4\\sin^3{x}$ . Solving, we determine that $\\sin^2{x} = \\frac{4 \\pm \\sqrt{2}}{8}$ . Plugging this in gives that $\\frac{AP}{PB} = \\frac{\\sqrt{2}+1}{\\sqrt{2}-1} = 3 + 2\\sqrt{2}$ . The answer is $\\boxed{007}$", "Let $\\alpha=\\angle{ACP}$ $\\beta=\\angle{ABC}$ , and $x=BP$ . By Law of Sines,\n$\\frac{1}{\\sin(\\beta)}=\\frac{x}{\\sin(90-\\alpha)}\\implies \\sin(\\beta)=\\frac{\\cos(\\alpha)}{x}$ (1), and\n$\\frac{4-x}{\\sin(\\alpha)}=\\frac{4\\sin(\\beta)}{\\sin(2\\alpha)} \\implies 4-x=\\frac{2\\sin(\\beta)}{\\cos(\\alpha)}$ . (2)\nThen, substituting (1) into (2), we get\n$4-x=\\frac{2}{x} \\implies x^2-4x+2=0 \\implies x=2-\\sqrt{2} \\implies \\frac{4-x}{x}=\\frac{2+\\sqrt{2}}{2-\\sqrt{2}}=3+2\\sqrt{2}$\nThe answer is $\\boxed{007}$ .\n~Rowechen", "Let $\\angle{ACP}=x$ . Then, $\\angle{APC}=2x$ and $\\angle{A}=180-3x$ . Let the foot of the angle bisector of $\\angle{APC}$ on side $AC$ be $D$ . Then,\n$CD=DP$ and $\\triangle{DAP}\\sim{\\triangle{APC}}$ due to the angles of these triangles.\nLet $CD=a$ . By the Angle Bisector Theorem, $\\frac{1}{a}=\\frac{AP}{AD}$ , so $AD=a\\cdot{AP}$ . Moreover, since $CD=DP=a$ , by similar triangle ratios, $\\frac{AP}{a+a\\cdot{AP}}=a$ . Therefore, $AP = \\frac{a^2}{1-a^2}$\nConstruct the perpendicular from $D$ to $AP$ and denote it as $F$ . Denote the midpoint of $CP$ as $M$ . Since $PD$ is an angle bisector, $PF$ is congruent to $PM$ , so $PF=\\frac{1}{2}$\nAlso, $\\triangle{DFA}\\sim{\\triangle{BCA}}$ . Thus, $\\frac{FA}{AC}=\\frac{AD}{AB}\\Longrightarrow\\frac{\\frac{a^2}{1-a^2}-\\frac{1}{2}}{a+\\frac{a^3}{1-a^2}}=\\frac{\\frac{a^3}{1-a^3}}{4}$ . After some major cancellation, we have $7a^4-8a^2+2=0$ , which is a quadratic in $a^2$ . Thus, $a^2 = \\frac{4\\pm\\sqrt{2}}{7}$\nTaking the negative root implies $AP<BP$ , contradiction. Thus, we take the positive root to find that $AP=2+\\sqrt{2}$ . Thus, $BP=2-\\sqrt{2}$ , and our desired ratio is $\\frac{2+\\sqrt{2}}{2-\\sqrt{2}}\\implies{3+2\\sqrt{2}}$\nThe answer is $\\boxed{007}$", "Let $O$ be the circumcenter of $\\triangle ABC$ . Since $\\triangle ABC$ is a right triangle, $O$ will be on $\\overline{AB}$ and $\\overline{AO} \\cong \\overline{OB} \\cong \\overline{OC} = 2$ . Let $\\overline{OP} = x$\nNext, let's do some angle chasing. Label $\\angle ACP = \\theta^{\\circ}$ , and $\\angle APC = 2\\theta^{\\circ}$ . Thus, $\\angle PAC = (180-3\\theta)^{\\circ}$ , and by isosceles triangles, $\\angle ACO = (180-3\\theta)^{\\circ}$ . Then, by angle subtraction, $\\angle OCP = (\\theta - (180-3\\theta))^{\\circ} = (4\\theta - 180)^{\\circ}$\nUsing the Law of Sines: \\[\\frac{x}{\\sin (4\\theta-180)^{\\circ}}=\\frac{2}{\\sin (2\\theta)^{\\circ}}\\] Using trigonometric identies, $\\sin (4\\theta-180)^{\\circ}=-\\sin (4\\theta)^{\\circ}=-2\\sin (2\\theta)^{\\circ}\\cos (2\\theta)^{\\circ}$ . Plugging this back into the Law of Sines formula gives us: \\[\\frac{x}{-2\\sin (2\\theta)^{\\circ}\\cos (2\\theta)^{\\circ}}=\\frac{2}{\\sin (2\\theta)^{\\circ}}\\]\n\\[-4\\sin (2\\theta)^{\\circ}\\cos (2\\theta)^{\\circ}=x\\sin (2\\theta)^{\\circ}\\] \\[-4\\cos (2\\theta)^{\\circ}=x\\] \\[\\cos(2\\theta)^{\\circ}=\\frac{-x}4\\]\nNext, using the Law of Cosines: \\[2^2=1^2+x^2-2\\cdot 1\\cdot x\\cdot \\cos (2\\theta)^{\\circ}\\] Substituting $\\cos(2\\theta)^{\\circ}=\\frac{-x}4$ gives us: \\[2^2=1^2+x^2-2\\cdot 1\\cdot x\\cdot \\frac{-x}4\\] \\[4=1+x^2+\\frac{x^2}{2}\\]\nSolving for x gives $x=\\sqrt 2$\nFinally: $\\frac{\\overline{AP}}{\\overline{BP}}=\\frac{\\overline{AO}+\\overline{OP}}{\\overline{BO}-\\overline{OP}}=\\frac{2+\\sqrt 2}{2-\\sqrt 2}=3+2\\sqrt2$ , which gives us an answer of $3+2+2=\\boxed{007}$ . ~adyj" ]

[ "Do note that by counting the area in 2 ways, the first altitude is $x = \\frac{ab}{c}$ . By similar triangles, the common ratio is $\\rho = \\frac{a}{c}$ for each height, so by the geometric series formula, we have \\begin{align} 6p=\\frac{x}{1-\\rho} = \\frac{ab}{c-a}. \\end{align} Writing $p=a+b+c$ and clearing denominators, we get \\[13a=6p .\\] Thus $p=13q$ $a=6q$ , and $b+c=7q$ , i.e. $c=7q-b$ . Plugging these into $(1)$ , we get $78q(q-b)=6bq$ , i.e., $14b=13q$ . Thus $q=14r$ and $p=182r$ $b=13r$ $a=84r$ $c=85r$ . Taking $r=1$ (since $a,b,c$ are relatively prime) we get $p=\\boxed{182}$", "Note that by counting the area in 2 ways, the first altitude is $\\dfrac{ab}{c}$ . By similar triangles, the common ratio is $\\dfrac{a}{c}$ for reach height, so by the geometric series formula, we have $6p=\\dfrac{\\dfrac{ab}{c}}{1-\\dfrac{a}{c}}$ . Multiplying by the denominator and expanding, the equation becomes $\\dfrac{ab}{c}=6a+6b+6c-\\dfrac{6a^2}{c}-\\dfrac{6ab}{c}-6a$ . Cancelling $6a$ and multiplying by $c$ yields $ab=6bc+6c^2-6a^2-6ab$ , so $7ab = 6bc+6b^2$ and $7a=6b+6c$ . Checking for Pythagorean triples gives $13,84,$ and $85$ , so $p=13+84+85=\\boxed{182}$", "We start by splitting the sum of all $C_{n-2}C_{n-1}$ into two parts: those where $n-2$ is odd and those where $n-2$ is even.\nFirst consider the sum of the lengths of the segments for which $n-2$ is odd for each $n\\geq2$ . The perimeters of these triangles can be expressed using $p$ and ratios that result because of similar triangles. Considering triangles where $n-2$ is odd, we find that the perimeter for each such $n$ is $p\\left(\\frac{C_{n-1}C_{n}}{C_{0}B}\\right)$ . Thus,\n$p\\sum_{n=1}^{\\infty}\\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B$\nSimplifying,\n$\\sum_{n=1}^{\\infty}C_{2n-1}C_{2n}=6C_{0}B + \\frac{(C_{0}B)^2}{p}=C_{0}B\\left(6+\\frac{C_{0}B}{p}\\right)$ . (1)\nContinuing with a similar process for the sum of the lengths of the segments for which $n-2$ is even,\n$p\\sum_{n=1}^{\\infty}\\frac{C_{2n-2}C_{2n-1}}{C_{0}B}=6p+C_{0}A+AB=7p-C_{0}B$\nSimplifying,\n$\\sum_{n=1}^{\\infty}C_{2n-2}C_{2n-1}=C_{0}B\\left(7-\\frac{C_{0}B}{p}\\right)$ . (2)\nAdding (1) and (2) together, we find that\n$6p=13C_{0}B \\Rightarrow p=\\frac{13C_{0}B}{6}=C_{0}B+C_{0}A+AB \\Rightarrow \\frac{7C_{0}B}{6}=C_{0}A+AB \\Rightarrow 7C_{0}B=6C_{0}A + 6AB$\nSetting $a=C_{0}B$ $b=C_{0}A$ , and $c=AB$ , we can now proceed as in Shaddoll's solution, and our answer is $p=13+84+85=\\boxed{182}$", "\nLet $a = BC_0$ $b = AC_0$ , and $c = AB$ .\nNote that the total length of the red segments in the figure above is equal to the length of the blue segment times $\\frac{a+c}{b}$\nThe desired sum is equal to the total length of the infinite path $C_0 C_1 C_2 C_3 \\cdots$ , shown in red in the figure below.\nSince each of the triangles $\\triangle C_0 C_1 C_2, \\triangle C_2 C_3 C_4, \\dots$ on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the blue segment times $\\frac{a+c}{b}$ .\nIn other words, we have that $a\\left(\\frac{a+c}{b}\\right) = 6p$\nGuessing and checking Pythagorean triples reveals that $a = 84$ $b=13$ $c = 85$ , and $p = a + b + c = \\boxed{182}$ satisfies this equation.", "This solution proceeds from $\\frac{\\frac{ab}{c}}{1-\\frac{a}{c}} = \\frac{\\frac{ab}{c}}{\\frac{c-a}{c}} = \\frac{ab}{c-a} = 6(a+b+c)$ . Note the general from for a primitive pythagorean triple, $m^2-n^2, 2mn, m^2+n^2$ and after substitution, letting $a = m^2-n^2, b = 2mn, c = m^2+n^2$ into the previous equation simplifies down very nicely into $m = 13n$ . Thus $a = 168n^2, b = 26n^2, c = 170n^2$ . Since we know all three side lengths are relatively prime, we must divide each by 2 and let n = 1 giving $a = 84, b = 13, c = 85$ yielding $p = a + b + c = \\boxed{182}$", "For this problem, first notice that its an infinite geometric series of $6(a+b+c)=\\frac{ab}{c-b}$ if $c$ is the hypotenuse. WLOG $a<b$ , we can generalize a pythagorean triple of $x^2-y^2, 2xy, x^2+y^2$ . Let $b=2xy$ , then this generalization gives $6(a+b+c)(c-b)=ab$ \\[(x^2-y^2)2xy=6(2x^2+2xy)(x-y)^2\\] \\[(x+y)xy=6(x^2+xy)(x-y)\\] \\[xy=6x(x-y)\\] \\[7xy=6x^2\\] \\[y=\\frac{6}{7}x\\]\nNow this is just clear. Let $x=7m$ and $y=6m$ for $m$ to be a positive integer, the pythagorean triple is $13-84-85$ which yields $\\boxed{182}$" ]

[ "Note that every $B_nC_n$ is parallel to each other for any nonnegative $n$ . Also, the area we seek is simply the ratio $k=\\frac{[B_0B_1C_1]}{[B_0B_1C_1]+[C_1C_0B_0]}$ , because it repeats in smaller and smaller units. Note that the area of the triangle, by Heron's formula, is 90.\nFor ease, all ratios I will use to solve this problem are with respect to the area of $[AB_0C_0]$ . For example, if I say some area has ratio $\\frac{1}{2}$ , that means its area is 45.\nNow note that $k=$ 1 minus ratio of $[B_1C_1A]$ minus ratio $[B_0C_0C_1]$ . We see by similar triangles given that ratio $[B_0C_0C_1]$ is $\\frac{17^2}{25^2}$ . Ratio $[B_1C_1A]$ is $(\\frac{336}{625})^2$ , after seeing that $C_1C_0 = \\frac{289}{625}$ , . Now it suffices to find 90 times ratio $[B_0B_1C_1]$ , which is given by 1 minus the two aforementioned ratios. Substituting these ratios to find $k$ and clearing out the $5^8$ , we see that the answer is $90\\cdot \\frac{5^8-336^2-17^2\\cdot 5^4}{5^8-336^2}$ , which gives $q= \\boxed{961}$", "Let $k$ be the coefficient of the similarity of triangles \\[\\triangle B_0 C_1 C_0 \\sim \\triangle AB_0 C_0 \\implies k = \\frac {B_0 C_0}{AC_0} = \\frac {17}{25}.\\] Then area $\\frac {[B_0 C_1 C_0]}{[AB_0 C_0 ]} = k^2 \\implies \\frac {[AB_0 C_1]}{[AB_0 C_0]} = 1 – k^2.$\nThe height of triangles $\\triangle B_0C_1A$ and $\\triangle AB_0C_0$ from $B_0$ is the same $\\implies \\frac {AC_1}{AC_0} = 1 – k^2.$\nThe coefficient of the similarity of triangles $\\triangle AB_1C_1 \\sim \\triangle AB_0C_0$ is $\\frac {AC_1}{AC_0} = 1 – k^2 \\implies \\frac {[B_1C_1C_2 ]}{[AB_0C_0 ]} = k^2 (1 – k^2)^2.$\nAnalogically the coefficient of the similarity of triangles $\\triangle AB_2C_2 \\sim \\triangle AB_0C_0$ is $(1 – k^2)^2 \\implies \\frac {[B_2C_2C_3]}{[AB_0C_0 ]} = k^2 (1 – k^2)^4$ and so on.\nThe yellow area $[Y]$ is $\\frac {[Y]}{[AB_0C_0 ]} = k^2 + k^2 (1 – k^2)^2 + k^2 (1 – k^2)^4 +.. = \\frac {k^2}{1 – (1 – k^2)^2} = \\frac{1}{2 – k^2}.$\nThe required area is $[AB_0C_0 ] – [Y] = [AB_0C_0 ] \\cdot (1 – \\frac{1}{2 – k^2}) = [AB_0C_0 ] \\cdot \\frac {1 – k^2}{2 – k^2} = [AB_0C_0 ] \\cdot \\frac {25^2 – 17^2} {2 \\cdot 25^2 – 17^2} = [AB_0C_0 ] \\cdot \\frac {336}{961}.$\nThe number $961$ is prime, $[AB_0C_0]$ is integer but not $961,$ therefore the answer is $\\boxed{961}$" ]

[ "Since quadrilateral $UVCM$ has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that $\\triangle AUV$ has $\\frac 14$ the area of triangle $AMC$ by similarity, so $[UVCM]=\\frac 34\\cdot [AMC].$ Thus, \\[\\frac 12 \\cdot 12\\cdot 12=\\frac 34 \\cdot [AMC]\\] \\[72=\\frac 34\\cdot [AMC]\\] \\[[AMC]=96\\rightarrow \\boxed{96}.\\]", "\nWe know that $\\triangle AUV \\sim \\triangle AMC$ , and since the ratios of its sides are $\\frac{1}{2}$ , the ratio of of their areas is $\\left(\\frac{1}{2}\\right)^2=\\frac{1}{4}$\nIf $\\triangle AUV$ is $\\frac{1}{4}$ the area of $\\triangle AMC$ , then trapezoid $MUVC$ is $\\frac{3}{4}$ the area of $\\triangle AMC$\nLet's call the intersection of $\\overline{UC}$ and $\\overline{MV}$ $P$ . Let $\\overline{UP}=x$ . Then $\\overline{PC}=12-x$ . Since $\\overline{UC} \\perp \\overline{MV}$ $\\overline{UP}$ and $\\overline{CP}$ are heights of triangles $\\triangle MUV$ and $\\triangle MCV$ , respectively. Both of these triangles have base $12$\nArea of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$\nArea of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$\nAdding these two gives us the area of trapezoid $MUVC$ , which is $6x+(72-6x)=72$\nThis is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{96}$ ~quacker88, diagram by programjames1", "Draw median $\\overline{AB}$ \nSince we know that all medians of a triangle intersect at the centroid, we know that $\\overline{AB}$ passes through point $P$ . We also know that medians of a triangle divide each other into segments of ratio $2:1$ . Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$ , and since the two segments sum to $12$ $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$ , respectively.\nFinally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\\triangle PUM$ is enough. $\\overline{PC} = \\overline{MP} = 8$\nThe area of $\\triangle PUM = \\frac{4\\cdot8}{2}=16$ . Multiplying this by $6$ gives us $6\\cdot16=\\boxed{96}$", " We know that $AU = UM$ $AV = VC$ , so $UV = \\frac{1}{2} MC$\nAs $\\angle UPM = \\angle VPC = 90$ , we can see that $\\triangle UPM \\cong \\triangle VPC$ and $\\triangle UVP \\sim \\triangle MPC$ with a side ratio of $1 : 2$\nSo $UP = VP = 4$ $MP = PC = 8$\nWith that, we can see that $[\\triangle UPM] = 16$ , and the area of trapezoid $MUVC$ is 72.\nAs said in solution 1, $[\\triangle AMC] = 72 / \\frac{3}{4} = \\boxed{96}$", "\nLet $AB$ be the height. Since medians divide each other into a $2:1$ ratio, and the medians have length 12, we have $PC=MP=8$ and $UP=VP=4$ . From right triangle $\\triangle{MUP}$ \\[MU^2=MP^2+UP^2=8^2+4^2=80,\\] so $MU=\\sqrt{80}=4\\sqrt{5}$ . Since $CU$ is a median, $AM=8\\sqrt{5}$ . From right triangle $\\triangle{MPC}$ \\[MC^2=MP^2+PC^2=8^2+8^2=128,\\] which implies $MC=\\sqrt{128}=8\\sqrt{2}$ . By symmetry $MB=\\dfrac{8\\sqrt{2}}{2}=4\\sqrt{2}$\nApplying the Pythagorean Theorem to right triangle $\\triangle{MAB}$ gives $AB^2=AM^2-MB^2=8\\sqrt{5}^2-4\\sqrt{2}^2=288$ , so $AB=\\sqrt{288}=12\\sqrt{2}$ . Then the area of $\\triangle{AMC}$ is \\[\\dfrac{AB \\cdot MC}{2}=\\dfrac{8\\sqrt{2} \\cdot 12\\sqrt{2}}{2}=\\dfrac{96 \\cdot 2}{2}=\\boxed{96}\\]", "By similarity, the area of $AUV$ is equal to $\\frac{1}{4}$\nThe area of $UVCM$ is equal to 72.\nAssuming the total area of the triangle is S, the equation will be : $\\frac{3}{4}$ S = 72.\nS = $\\boxed{96}$", "Given a triangle with perpendicular medians with lengths $x$ and $y$ , the area will be $\\frac{2xy}{3}=\\boxed{96}$", "Connect the line segment $UV$ and it's easy to see quadrilateral $UVMC$ has an area of the product of its diagonals divided by $2$ which is $72$ . Now, solving for triangle $AUV$ could be an option, but the drawing shows the area of $AUV$ will be less than the quadrilateral meaning the the area of $AMC$ is less than $72*2$ but greater than $72$ , leaving only one possible answer choice, $\\boxed{96}$", "\nConnect $AP$ , and let $B$ be the point where $AP$ intersects $MC$ $MB=CB$ because all medians of a triangle intersect at one point, which in this case is $P$ $MP:PV=2:1$ because the point at which all medians intersect divides the medians into segments of ratio $2:1$ , so $MP=8$ and similarly $CP=8$ . We apply the Pythagorean Theorem to triangle $MPC$ and get $MC=\\sqrt{128}=8\\sqrt{2}$ . The area of triangle $MPC$ is $\\dfrac{MP\\cdot CP}{2}=32$ , and that must equal to $\\dfrac{MC\\cdot BP}{2}$ , so $BP=\\dfrac{8}{\\sqrt{2}}=4\\cdot\\sqrt{2}$ $BP=\\dfrac{1}{3}BA$ , so $BA=12\\sqrt{2}$ . The area of triangle $AMC$ is equal to $\\dfrac{MC\\cdot BA}{2}=\\dfrac{8 \\cdot \\sqrt{2} \\cdot 12 \\cdot \\sqrt{2}}{2}=\\boxed{96}$", "$[\\square MUVC] = 72$ . Let intersection of line $AP$ and base $MC$ be $B$ \\[[AUV]=[MUB]=[UVB]=[BVC] \\implies \\left[\\frac{\\triangle AMC}{4}\\right] = \\left[\\frac{\\square MUVC}{3}\\right] \\implies [AMC] = \\boxed{96}\\]", "\nSince $\\overline{MV}$ and $\\overline{CU}$ intersect at a right angle, this means $MUVC$ is a kite. Hence, the area of this kite is $\\frac{12 \\cdot 12}{2} = 72$\nAlso, notice that $\\triangle AUV \\sim \\triangle AMC$ by AAA Similarity. Since the ratios of its sides is $\\frac{1}{2}$ , the ratios of the area is $\\left(\\frac{1}{2}\\right)^2=\\frac{1}{4}$ . Therefore,\n\\[[AMC] = [MUVC] + \\frac{1}{4} \\cdot [AMC]\\]\nSimplifying gives us $\\frac{3}{4} \\cdot [AMC] = 72$ , so $[AMC] = 72 \\cdot \\frac{4}{3} = \\boxed{96}$", "Horizontally translate line $\\overline{UC}$ until point $U$ is at point $V$ , with $C$ subsequently at $C'$ , and then connect up $C'$ and $C$ to create $\\triangle MVC'$ , which is a right triangle.\n\nNotice that $\\triangle MVC'$ $12 \\cdot 12 \\cdot \\frac{1}{2} = 72$ , and $\\triangle MVC'$ $\\triangle MVC + \\triangle MUV$ (since the latter has the same base and height as the sub-triangle $\\triangle CVC'$ inside $\\triangle MVC'$ ).\nFrom this, we can deduce that $\\textbf{(B)}$ cannot be true, since an incomplete part of $\\triangle AMC$ is equal to it. We can also deduce that $\\textbf{(D)}$ also cannot be true, since the unknown triangle $\\triangle AUV = \\triangle MUV$ , and $\\triangle MUV = \\triangle CVC' < \\triangle MVC'$ . Therefore, the answer must be between $72$ and $144$ , leaving $\\boxed{96}$ as the only possible correct answer." ]

[ "First, from triangle $ABO$ $\\angle AOB = 180^\\circ - \\angle BAO - \\angle ABO$ . Note that $AO$ bisects $\\angle BAT$ (to see this, draw radii from $O$ to $AB$ and $AT,$ creating two congruent right triangles), so $\\angle BAO = \\angle BAT/2$ . Similarly, $\\angle ABO = \\angle ABR/2$\nAlso, $\\angle BAT = 180^\\circ - \\angle BAP$ , and $\\angle ABR = 180^\\circ - \\angle ABP$ . Hence,\n$\\angle AOB = 180^\\circ - \\angle BAO - \\angle ABO = 180^\\circ - \\frac{\\angle BAT}{2} - \\frac{\\angle ABR}{2} = 180^\\circ - \\frac{180^\\circ - \\angle BAP}{2} - \\frac{180^\\circ - \\angle ABP}{2}= \\frac{\\angle BAP + \\angle ABP}{2}.$\nFinally, from triangle $ABP$ $\\angle BAP + \\angle ABP = 180^\\circ - \\angle APB = 180^\\circ - 40^\\circ = 140^\\circ$ , so \\[\\angle AOB = \\frac{\\angle BAP + \\angle ABP}{2} = \\frac{140^\\circ}{2} = \\boxed{70}.\\]" ]

[ "The large triangle $ABC$ has sides of length $4$ . The medium triangle has sides of length $2$ . The small triangle has sides of length $1$ . There are $3$ segment sizes, and all segments depicted are one of these lengths.\nStarting at $A$ and going clockwise, the perimeter is:\n$AB + BC + CD + DE + EF + FG + GA$\n$4 + 4 + 2 + 2 + 1 + 1 + 1$\n$15$ , thus the answer is $\\boxed{15}$", "The perimeter of $ABCDEFG$ is the perimeter of the three triangles, minus segments $AD$ and $EG$ , which are on the interior of the figure. Because each of these segments is on two triangles, each segment must be subtracted two times.\nAs in solution 1, the sides of the triangles are $4, 2,$ and $1$ , and the perimeters of the triangles are thus $12, 6,$ and $3$\nThe perimeter of the three triangles is $12 + 6 + 3 = 21$ . Subtracting the two segments $AD$ and $EG$ two times, the perimeter of $ABCDEFG$ is $21 - 2 - 1 - 2 - 1 = 15$ , and the answer is $\\boxed{15}$" ]

[ "After sketching, it is clear a $90^{\\circ}$ rotation is done about $(x,y)$ . Looking between $A$ and $A'$ $x+y=18$ . Thus $90+18=\\boxed{108}$ .\n~mn28407", "\nWe first draw a diagram with the correct Cartesian coordinates and a center of rotation $P$ . Note that $PC=PC'$ because $P$ lies on the perpendicular bisector of $CC'$ (it must be equidistant from $C$ and $C'$ by properties of a rotation).\nSince $AB$ is vertical while $A'B'$ is horizontal, we have that the angle of rotation must be $90^{\\circ}$ , and therefore $\\angle P = 90^{\\circ}$ . Therefore, $CPC'$ is a 45-45-90 right triangle, and $CD=DP$\nWe calculate $D$ to be $(20,1)$ . Since we translate $4$ right and $1$ up to get from point $C$ to point $D$ , we must translate $1$ right and $4$ down to get to point $P$ . This gives us $P(21,-3)$ . Our answer is then $90+21-3=\\boxed{108}$ . ~Lopkiloinm & samrocksnature", "For the above reasons, the transformation is simply a $90^\\circ$ rotation. Proceed with complex numbers on the points $C$ and $C'$ . Let $(x, y)$ be the origin. Thus, $C \\rightarrow (16-x)+(-y)i$ and $C' \\rightarrow (24-x)+(2-y)i$ . The transformation from $C'$ to $C$ is a multiplication of $i$ , which yields $(16-x)+(-y)i=(y-2)+(24-x)i$ . Equating the real and complex terms results in the equations $16-x=y-2$ and $-y=24-x$ . Solving, $(x, y) : (21, -3) \\rightarrow 90+21-3=\\boxed{108}$", "We know that the rotation point $P$ has to be equidistant from both $A$ and $A'$ so it has to lie on the line that is on the midpoint of the segment $AA'$ and also the line has to be perpendicular to $AA'$ . Solving, we get the line is $y=\\frac{-4}{3}x+25$ . Doing the same for $B$ and $B'$ , we get that $y=-6x+123$ . Since the point $P$ of rotation must lie on both of these lines, we set them equal, solve and get: $x=21$ $y=-3$ . We can also easily see that the degree of rotation is $90$ since $AB$ is initially vertical, and now it is horizontal. Also, we can just sketch this on a coordinate plane and easily realize the same. Hence, the answer is $21-3+90 = \\boxed{108}$" ]

[ "If you started backwards you would get: \\[0\\Rightarrow (+40)=40 , \\Rightarrow \\left(\\frac{1}{2}\\right)=20 , \\Rightarrow (+40)=60 , \\Rightarrow \\left(\\frac{1}{2}\\right)=30 , \\Rightarrow (+40)=70 , \\Rightarrow \\left(\\frac{1}{2}\\right)=\\boxed{35}\\]", "If you have $x$ as the amount of money Foolish Fox started with we have $2(2(2x-40)-40)-40=0.$ Solving this we get $\\boxed{35}$" ]

[ "We wish to find the radius of one circle, so that we can find the total area.\nNotice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle. Now consider two adjacent smaller circles. This means that the line connecting the radii is a segment of length $2r$ that is tangent to the larger circle at the midpoint of the two centers. Thus, we have essentially have a regular dodecagon whose vertices are the centers of the smaller triangles circumscribed about a circle of radius $1$\nWe thus know that the apothem of the dodecagon is equal to $1$ . To find the side length, we make a triangle consisting of a vertex, the midpoint of a side, and the center of the dodecagon, which we denote $A, M,$ and $O$ respectively. Notice that $OM=1$ , and that $\\triangle OMA$ is a right triangle with hypotenuse $OA$ and $m \\angle MOA = 15^\\circ$ . Thus $AM = (1) \\tan{15^\\circ} = 2 - \\sqrt {3}$ , which is the radius of one of the circles. The area of one circle is thus $\\pi(2 - \\sqrt {3})^{2} = \\pi (7 - 4 \\sqrt {3})$ , so the area of all $12$ circles is $\\pi (84 - 48 \\sqrt {3})$ , giving an answer of $84 + 48 + 3 = \\boxed{135}$" ]

[ "Set up the proportion $\\frac{12\\ \\text{meals}}{18\\ \\text{people}}=\\frac{x\\ \\text{meals}}{12\\ \\text{people}}$ . Solving for $x$ gives us $x= \\boxed{8}$" ]

[ "Let $a$ $b$ , and $c$ be the vertex of a triangle that satisfies this problem, where $a > b > c$ \\[a - b = p_1\\] \\[b - c = p_2\\] \\[a - c = p_3\\]\n$p_3 = a - c = a - b + b - c = p_1 + p_2$ . Because $p_3$ is the sum of two primes, $p_1$ and $p_2$ $p_1$ or $p_2$ must be $2$ . Let $p_1 = 2$ , then $p_3 = p_2 + 2$ . There are only $8$ primes less than $20$ $2, 3, 5, 7, 11, 13, 17, 19$ . Only $3, 5, 11, 17$ plus $2$ equals another prime. $p_2 \\in \\{ 3, 5, 11, 17 \\}$\nOnce $a$ is determined, $a = b+2$ and $b = c + p_2$ . There are $18$ values of $a$ where $a+2 \\le 20$ , and $4$ values of $p_2$ . Therefore the answer is $18 \\cdot 4 = \\boxed{072}$", "As above, we must deduce that the sum of two primes must be equal to the third prime. Then, we can finish the solution using casework.\nIf the primes are $2,3,5$ , then the smallest number can range between $1$ and $15$ .\nIf the primes are $2,5,7$ , then the smallest number can range between $1$ and $13$ .\nIf the primes are $2,11,13$ , then the smallest number can range between $1$ and $7$ . \nIf the primes are $2,17,19$ , then the smallest number can only be $1$\nAdding all cases gets $15+13+7+1=36$ . However, due to the commutative property, we must multiply this by 2. For example, in the $2,17,19$ case the numbers can be $1,3,20$ or $1,18,20$ . Therefore the answer is $36\\cdot2=\\boxed{072}$" ]

[ "We can use complementary counting , by finding the probability that none of the three knights are sitting next to each other and subtracting it from $1$\nImagine that the $22$ other (indistinguishable) people are already seated, and fixed into place.\nWe will place $A$ $B$ , and $C$ with and without the restriction.\nThere are $22$ places to put $A$ , followed by $21$ places to put $B$ , and $20$ places to put $C$ after $A$ and $B$ . Hence, there are $22\\cdot21\\cdot20$ ways to place $A, B, C$ in between these people with restrictions.\nWithout restrictions, there are $22$ places to put $A$ , followed by $23$ places to put $B$ , and $24$ places to put $C$ after $A$ and $B$ . Hence, there are $22\\cdot23\\cdot24$ ways to place $A,B,C$ in between these people without restrictions.\nThus, the desired probability is $1-\\frac{22\\cdot21\\cdot20}{22\\cdot23\\cdot24}=1-\\frac{420}{552}=1-\\frac{35}{46}=\\frac{11}{46}$ , and the answer is $11+46=\\boxed{057}$", "There are $(25-1)! = 24!$ configurations for the knights about the table (since configurations that are derived from each other simply by a rotation are really the same, and should not be counted multiple times).\nThere are ${3\\choose 2} = 3$ ways to pick a pair of knights from the trio, and there are $2! = 2$ ways to determine in which order they are seated. Since these two knights must be together, we let them be a single entity, so there are $(24-1)! = 23!$ configurations for the entities.\nHowever, this overcounts the instances in which the trio sits together; when all three knights sit together, then two of the pairs from the previous case are counted. However, we only want to count this as one case, so we need to subtract the number of instances in which the trio sits together (as a single entity). There are $3! = 6$ ways to determine their order, and there are $(23-1)! = 22!$ configurations.\nThus, the required probability is $\\frac{2 \\times 3 \\times 23! - 6 \\times 22!}{24!} = \\frac{11}{46}$ , and the answer is $\\boxed{057}$", "Number the knights around the table from $1$ to $25$ . The total number of ways to pick the knights is \\[{25\\choose 3} = \\frac{25\\cdot24\\cdot23}{3\\cdot2\\cdot1} = 25\\cdot23\\cdot4\\] There are two possibilities: either all three sit next to each other, or two sit next to each other and one is not sitting next to the other two.\nCase 1: All three sit next to each other. In this case, you are picking $(1,2,3)$ $(2,3,4)$ $(3,4,5)$ $(4,5,6)$ , ..., $(23,24,25)$ $(24,25,1)$ $(25,1,2)$ . This makes $25$ combinations.\nCase 2: Like above, there are $25$ ways to pick the pair of knights sitting next to each other. Once a pair is picked, you cannot pick either of the two adjacent knights. (For example, if you pick $(5,6)$ , you may not pick 4 or 7.) Thus, there are $25-4=21$ ways to pick the third knight, for a total of $25\\cdot21$ combinations.\nThus, you have a total of $25 + (25\\cdot21) = 25\\cdot22$ allowable ways to pick the knights.\nThe probability is $\\frac{25\\cdot22}{25\\cdot23\\cdot4} = \\frac{11}{46}$ , and the answer is $\\boxed{057}$", "Pick an arbitrary spot for the first knight. Then pick spots for the next two knights in order.\nCase 1: The second knight sits next to the first knight. There are $2$ possible places for this out of $24$ , so the probability of this is $\\frac{1}{12}$ . We do not need to consider the third knight.\nCase 2: The second knight sits two spaces apart from the first knight. There are $2$ possible places for this out of $24$ , so the probability is $\\frac{1}{12}$ . Then there are $3$ places out of a remaining $23$ for the third knight to sit, so the total probability for this case is $\\frac{1}{12} \\times \\frac{3}{23}$\nCase 3: The second knight sits three or more spaces apart from the first knight. There are $20$ possible places for this out of $24$ , so the probability is $\\frac{5}{6}$ . Then there are $4$ places to put the last knight out of $23$ , so the total probability for this case is $\\frac{5}{6}\\times\\frac{4}{23}$\nNow we add the probabilities to get the total:\n\\[\\frac{1}{12}+\\frac{1}{12}\\times\\frac{3}{23}+\\frac{5}{6}\\times\\frac{4}{23}=\\frac{1}{12}\\times\\frac{1}{23}\\left(23+3+40\\right)=\\frac{66}{12\\times 23}\\] \\[=\\frac{6\\times 11}{6\\times 2 \\times 23}=\\frac{11}{46}\\] so the answer is $\\boxed{057}$", "We simplify this problem by using complementary counting and fixing one knight in place. Then, either a knight can sit two spaces apart from the fixed knight, or a knight can sit more than two spaces apart from the fixed knight. The probability is then $\\frac{24\\left(23\\right)-\\left[2\\left(20\\right)+20\\left(19\\right)\\right]}{24\\left(23\\right)}=\\frac{11}{46}$ , so the answer is $11+46=\\boxed{057}$", "Let $K_1, K_2, K_3$ be the knights in clockwise order. Let $A$ be the distance between $K_1$ and $K_2$ $B$ the distance between $K_2$ and $K_3$ and $C$ the distance between $K_3$ and $K_1$ $A + B + C = 25$ and $A, B, C \\geq 1$ . In order to use stars and bars the numbers must be greater than or equal to 0 instead of 1, so we define $A_1 = A - 1, B_1 = B - 1, C_1 = C - 1$ $A_1 + B_1 + C_1 = 22$ , so by stars and bars there are $\\binom{22 + 3 - 1}{3 - 1} = 276$ possibilities.\nThe condition is not satisfied if $A, B, C \\geq 2$ , so we can use complementary counting. Let $A_2 = A - 2, B_2 = B - 2, C_2 = C - 2$ $A_2 + B_2 + C_2 = 19$ , and by stars and bars there are $\\binom{19 + 3 - 1}{3 - 1} = 210$ possibilities. This means there are $276 - 210 = 66$ possibilities where the condition is satisfied, so the probability is $\\frac{11}{46}$ , resulting in $\\boxed{057}$", "There are $\\binom{25}{3}=\\frac{25 \\cdot 24 \\cdot 23}{3 \\cdot 2}=2300$ ways to chose $3$ knights out of $25$ knights.\nTo ensure at least $2$ adjacent knights are chosen, first choose $1$ of the $25$ pairs of adjacent knights. After choosing the adjacent pair of knights, there are $23$ ways to choose the third knight. There are $25$ choices of $3$ knights sitting consecutively, which are counted twice. For example, choosing $(1,2)$ first, then choosing $3$ is the same as choosing $(2,3)$ first, then choosing $1$ . Therefore, there are $25 \\cdot 23 -25=550$ ways to choose $3$ knights where at least $2$ of the $3$ had been sitting next to each other. \\[P=\\frac{550}{2300}=\\frac{11}{46}\\] The answer is $11+46=\\boxed{057}$", "(I need to get this out)\nDenominator: $\\binom{25}{3}$ ways to choose three knights\nNumerator: $\\implies$ Case 1: Two knights are next to each other, third is lonely :(\n$\\implies$ Case 2: All three knights are next to each other\nOur answer is $\\frac{21 \\cdot 25+25}{\\binom{25}{3}}$ Simplifies to: $\\frac{22 \\cdot 25 \\cdot 6}{25 \\cdot 24 \\cdot 3}$ $\\implies$ $\\frac{11}{46}$ , and our answer is $11+46=\\boxed{057}$" ]

[ "Twenty percent less than 60 is $\\frac 45 \\cdot 60 = 48$ . One-third more than a number is $\\frac 43n$ . Therefore $\\frac 43n = 48$ and the number is $\\boxed{36}$" ]

[ "Twenty percent less than 60 is $\\frac 45 \\cdot 60 = 48$ . One-third more than a number is $\\frac 43n$ . Therefore $\\frac 43n = 48$ and the number is $\\boxed{36}$" ]

[ "2005 I AIME-9.png\nWe can consider the orientation of each of the individual cubes independently.\nThe unit cube at the center of our large cube has no exterior faces, so all of its orientations work.\nFor the six unit cubes and the centers of the faces of the large cube, we need that they show an orange face. This happens in $\\frac{4}{6} = \\frac{2}{3}$ of all orientations, so from these cubes we gain a factor of $\\left(\\frac{2}{3}\\right)^6$\nThe twelve unit cubes along the edges of the large cube have two faces showing, and these two faces are joined along an edge. Thus, we need to know the number of such pairs that are both painted orange. We have a pair for each edge, and 7 edges border one of the unpainted faces while only 5 border two painted faces. Thus, the probability that two orange faces show for one of these cubes is $\\frac{5}{12}$ , so from all of these cubes we gain a factor of $\\left(\\frac{5}{12}\\right)^{12} = \\frac{5^{12}}{2^{24}3^{12}}$\nFinally, we need to orient the eight corner cubes. Each such cube has 3 faces showing, and these three faces share a common vertex. Thus, we need to know the number of vertices for which all three adjacent faces are painted orange. There are six vertices which are a vertex of one of the unpainted faces and two vertices which have our desired property, so each corner cube contributes a probability of $\\frac{2}{8} = \\frac{1}{4}$ and all the corner cubes together contribute a probability of $\\left(\\frac{1}{4}\\right)^8 = \\frac{1}{2^{16}}$\nSince these probabilities are independent, the overall probability is just their product, $\\frac{2^6}{3^6} \\cdot \\frac{5^{12}}{2^{24}3^{12}} \\cdot \\frac{1}{2^{16}} = \\frac{5^{12}}{2^{34}\\cdot 3^{18}}$ and so the answer is $2 + 3 + 5 + 12 + 34 + 18 = \\boxed{074}$" ]

[ "There are 3 cases: where $x^\\circ$ is a base angle with the $70^\\circ$ as the other angle, where $x^\\circ$ is a base angle with $70^\\circ$ as the vertex angle, and where $x^\\circ$ is the vertex angle with $70^\\circ$ as a base angle.\nCase 1: $x^\\circ$ is a base angle with the $70^\\circ$ as the other angle:\nHere, $x=70$ , since base angles are congruent.\nCase 2: $x^\\circ$ is a base angle with $70^\\circ$ as the vertex angle:\nHere, the 2 base angles are both $x^\\circ$ , so we can use the equation $2x+70=180$ , which simplifies to $x=55$\nCase 3: $x^\\circ$ is the vertex angle with $70^\\circ$ as a base angle:\nHere, both base angles are $70^\\circ$ , since base angles are congruent. Thus, we can use the equation $x+140=180$ , which simplifies to $x=40$\nAdding up all the cases, we get $70+55+40=165$ , so the answer is $\\boxed{165}$" ]

[ "Let the speed of boy $A$ be $a$ , and the speed of boy $B$ be $b$ . Notice that $A$ travels $4$ miles per hour slower than boy $B$ , so we can replace $b$ with $a+4$\nNow let us see the distances that the boys each travel. Boy $A$ travels $60-12=48$ miles, and boy $B$ travels $60+12=72$ miles. Now, we can use $d=rt$ to make an equation, where we set the time to be equal: \\[\\frac{48}{a}=\\frac{72}{a+4}\\] Cross-multiplying gives $48a+192=72a$ . Isolating the variable $a$ , we get the equation $24a=192$ , so $a=\\boxed{8}$" ]

[ "Let $d$ be the length of the track in feet and $x$ be the number of laps that one of the boys did, so time one of the boys traveled before the two finish is $\\tfrac{dx}{5}$ . Since the time elapsed for both boys is equal, one boy ran $5(\\tfrac{dx}{5})$ feet while the other boy ran $9(\\tfrac{dx}{5})$ feet. Because both finished at the starting point, both ran an integral number of laps, so $5(\\tfrac{dx}{5})$ and $9(\\tfrac{dx}{5})$ are multiples of $d$ . Because both stopped when both met at the start for the first time, $x = 5$\nNote that between the time a runner finishes a lap and a runner (can be same) finishes a lap, both runners must meet each other. When $0 < x \\le 5$ and either $5(\\tfrac{dx}{5})$ or $9(\\tfrac{dx}{5})$ is a multiple of $d$ , one of the runners completed a lap. This is achieved when $x = \\tfrac59, 1, \\tfrac{10}{9}, \\tfrac{15}{9}, 2, \\tfrac{20}{9}, \\tfrac{25}{9}, 3, \\tfrac{30}{9}, \\tfrac{35}{9}, 4, \\tfrac{40}{9}, 5$ , so the two meet each other (excluding start and finish) a total of $\\boxed{13}$ times." ]

[ "There are $2 \\times 90 = 180$ minutes of total playing time. Divided equally among the five children, each child gets $180/5 = \\boxed{36}$ minutes." ]

[ "A good diagram is very helpful.\nThe first circle is in red, the second in blue.\nWith this diagram, we can see that the first circle is inscribed in equilateral triangle $GBA$ while the second circle is inscribed in $GKJ$ .\nFrom this, it's evident that the ratio of the blue area to the red area is equal to the ratio of the areas $\\triangle GKJ$ to $\\triangle GBA$\nSince the ratio of areas is equal to the square of the ratio of lengths, we know our final answer is $\\left(\\frac{GK}{GB}\\right)^2$ .\nFrom the diagram, we can see that this is $9^2=\\boxed{81}$", "As above, we note that the first circle is inscribed in an equilateral triangle of sidelength 1 (if we assume, WLOG, that the regular hexagon has sidelength 1). The inradius of an equilateral triangle with sidelength 1 is equal to $\\frac{\\sqrt{3}}{6}$ . Therefore, the area of the first circle is $(\\frac{\\sqrt{3}}{6})^2 \\cdot \\pi =\\frac{\\pi}{12}$\nCall the center of the second circle $O$ . Now we drop a perpendicular from $O$ to Circle O's point of tangency with $GK$ and draw another line connecting O to G. Note that because triangle $BGA$ is equilateral, $\\angle BGA=60^{\\circ}$ $OG$ bisects $\\angle BGA$ , so we have a 30-60-90 triangle.\nCall the radius of Circle O $r$ $OG=2r= \\text{height of equilateral triangle} + \\text{height of regular hexagon} + r$\nThe height of an equilateral triangle of sidelength 1 is $\\frac{\\sqrt{3}}{2}$ . The height of a regular hexagon of sidelength 1 is $\\sqrt{3}$ . Therefore, $OG=\\frac{\\sqrt{3}}{2} + \\sqrt{3} + r$\nWe can now set up the following equation:\n$\\frac{\\sqrt{3}}{2} + \\sqrt{3} + r=2r$\n$\\frac{\\sqrt{3}}{2} + \\sqrt{3}=r$\n$\\frac{3\\sqrt{3}}{2}=r$\nThe area of Circle O equals $\\pi r^2=\\frac{27}{4} \\pi$\nTherefore, the ratio of the areas is $\\frac{\\frac{27}{4}}{\\frac{1}{12}}=\\boxed{81}$" ]

[ "\nLet the center of the surrounding circle be $X$ . The circle that is tangent at point $A$ will have point $Y$ as the center. Similarly, the circle that is tangent at point $B$ will have point $Z$ as the center. Connect $AB$ $YZ$ $XA$ , and $XB$ . Now observe that $\\triangle XYZ$ is similar to $\\triangle XAB$ by SAS.\nWriting out the ratios, we get \\[\\frac{XY}{XA}=\\frac{YZ}{AB} \\Rightarrow \\frac{13-5}{13}=\\frac{5+5}{AB} \\Rightarrow \\frac{8}{13}=\\frac{10}{AB} \\Rightarrow AB=\\frac{65}{4}.\\] Therefore, our answer is $65+4= \\boxed{69}$" ]

[ "Let the center of the two circles be $O$ . Now pick an arbitrary point $A$ on the boundary of the circle with radius $2$ . We want to find the range of possible places for the second point, $A'$ , such that $AA'$ passes through the circle of radius $1$ . To do this, first draw the tangents from $A$ to the circle of radius $1$ . Let the intersection points of the tangents (when extended) with circle of radius $2$ be $B$ and $C$ . Let $H$ be the foot of the altitude from $O$ to $\\overline{BC}$ . Then we have the following diagram.\n\nWe want to find $\\angle BOC$ , as the range of desired points $A'$ is the set of points on minor arc $\\overarc{BC}$ . This is because $B$ and $C$ are part of the tangents, which \"set the boundaries\" for $A'$ . Since $OH = 1$ and $OB = 2$ as shown in the diagram, $\\triangle OHB$ is a $30-60-90$ triangle with $\\angle BOH = 60^\\circ$ . Thus, $\\angle BOC = 120^\\circ$ , and the probability $A'$ lies on the minor arc $\\overarc{BC}$ is thus $\\dfrac{120}{360} = \\boxed{13}$" ]

[ "Observe that $\\triangle{EAB}$ is equilateral. Therefore, $m\\angle{AEB}=m\\angle{EAB}=m\\angle{EBA} = 60^{\\circ}$ . Since $CD$ is a straight line, we conclude that $m\\angle{EBD} = 180^{\\circ}-60^{\\circ}=120^{\\circ}$ . Since $BE=BD$ (both are radii of the same circle), $\\triangle{BED}$ is isosceles, meaning that $m\\angle{BED}=m\\angle{BDE}=30^{\\circ}$ . Similarly, $m\\angle{AEC}=m\\angle{ACE}=30^{\\circ}$\nNow, $\\angle{CED}=m\\angle{AEC}+m\\angle{AEB}+m\\angle{BED} = 30^{\\circ}+60^{\\circ}+30^{\\circ} = 120^{\\circ}$ . Therefore, the answer is $\\boxed{120}$", "We know that $\\triangle{EAB}$ is equilateral, because all of its sides are congruent radii. Because point $A$ is the center of a circle, $C$ is at the border of a circle, and $E$ and $B$ are points on the edge of that circle, $m\\angle{ECB}=\\frac{1}{2}\\cdot m\\angle{EAB}=\\frac{1}{2}\\cdot60^{\\circ}=30^{\\circ}$ . Since $\\triangle{CED}$ is isosceles, angle $\\angle{CED}=180^{\\circ}-2\\cdot30^{\\circ}=\\boxed{120}$ degrees -SweetMango77." ]

[ "Using the diagram above, we notice that the desired length is simply the distance between the point $C$ and $\\overline{AB}$ . We can mark $C$ as $(3,3)$ since it is $3$ units away from each of the bases. Point $B$ is $(8,3)$ . Thus, line $\\overline{AB}$ is $y = \\frac{3}{8}x \\Rightarrow 3x + 8y = 0$ . We can use the distance from point to line formula $\\frac{|Ax_0 + By_0 + C|}{\\sqrt{A^2 + B^2}}$ , where $x_0$ and $y_0$ are the coordinates of the point, and A, B, and C are the coefficients of the line in form $Ax + By + C = 0$ . Plugging everything in, we get \\[\\frac{|3(3) - 8(3)|}{\\sqrt{8^2 + 3^2}} = \\frac{15}{\\sqrt{73}} \\Rightarrow \\frac{225}{73} \\Rightarrow \\boxed{298}\\]" ]

[ "The overlap length is $2(15)-25=5$ , so the shaded area is $5 \\cdot 15 =75$ . The area of the whole shape is $25 \\cdot 15 = 375$ . The fraction $\\dfrac{75}{375}$ reduces to $\\dfrac{1}{5}$ or 20%. Therefore, the answer is $\\boxed{20}$" ]

[ "It is easier to think of the dice as $21$ sided dice with $6$ sixes, $5$ fives, etc. Then there are $21^2=441$ possible rolls. There are $2\\cdot(1\\cdot 6+2\\cdot 5+3\\cdot 4)=56$ rolls that will result in a seven. The odds are therefore $\\frac{56}{441}=\\frac{8}{63}$ . The answer is $8+63=\\boxed{071}$", "Since the probability of rolling any number is 1, and the problem tells us the dice are unfair, we can assign probabilities to the individual faces. The probability of rolling $n$ is $\\frac{n}{21}$ because $21=\\frac{6 \\cdot 7}{2}$ Next, we notice that 7 can be rolled by getting individual results of 1 and 6, 2 and 5, or 3 and 4 on the separate dice. \nThe probability that 7 is rolled is now $2(\\frac{1}{21} \\cdot \\frac{6}{21}+\\frac{2}{21} \\cdot \\frac{5}{21} + \\frac{3}{21} \\cdot \\frac{4}{21})$ which is equal to $\\frac{56}{441}=\\frac{8}{63}$ . Therefore the answer is $8+63=\\boxed{071}$ ~PEKKA", "Since the probability of rolling a $1$ is $\\frac{1}{21}$ , the probability of rolling a $2$ is $\\frac{2}{21}$ the probability of rolling a $3$ is $\\frac{3}{21}$ and so on, we can make a chart of probabilities and add them together. Note that we only need the probabilities of $1$ and $6$ $2$ and $5$ , and $3$ and $4$ , and the rest is symmetry and the others are irrelevant.\nWe have: $2 \\cdot (\\frac{1}{21} \\cdot \\frac{2}{7}$ $+$ $\\frac{2}{21} \\cdot \\frac{5}{21}$ $+$ $\\frac{1}{7} \\cdot \\frac{4}{21})$ $=$ $2 \\cdot (\\frac{2}{147} + \\frac{10}{441} + \\frac{4}{147})$ $2 \\cdot \\frac{4}{63} = \\frac{8}{63}$ . Therefore, the answer is $8 + 63$ $\\boxed{071}$" ]

[ "The product will be even if at least one selected number is even, and odd if none are. Using complementary counting, the chance that both numbers are odd is $\\frac{\\tbinom32}{\\tbinom52}=\\frac3{10}$ , so the answer is $1-0.3$ which is $\\boxed{0.7}$", "There are $2$ cases to get an even number. Case 1: $\\text{Even} \\times \\text{Even}$ and Case 2: $\\text{Odd} \\times \\text{Even}$ . Thus, to get an $\\text{Even} \\times \\text{Even}$ , you get $\\frac {\\binom {2}{2}}{\\binom {5}{2}}= \\frac {1}{10}$ . And to get $\\text{Odd} \\times \\text{Even}$ , you get $\\frac {\\binom {3}{1}}{\\binom {5}{2}}= \\frac {6}{10}$ $\\frac {1}{10}+\\frac {6}{10}=\\frac {7}{10}$ which is $0.7$ and the answer is $\\boxed{0.7}$", "Note that we have three cases to get an even number: even $\\times$ even, odd $\\times$ even and even $\\times$ odd.\nThe probability of case 1 is $\\dfrac{2}{5}\\cdot\\dfrac{1}{4}$ , the probability of case 2 is $\\dfrac{2}{5}\\cdot\\dfrac{3}{4}$ and the probability of case 3 is $\\dfrac{3}{5}\\cdot\\dfrac12$\nAdding these up we get $\\dfrac{1}{10}+\\dfrac{3}{10}+\\dfrac{3}{10} = \\boxed{0.7}.$" ]

[ " - Diagram by Brendanb4321\nExtend $AB$ to form a right triangle with legs $6$ and $8$ such that $AD$ is the hypotenuse and connect the points $CD$ so \nthat you have a rectangle. (We know that $\\triangle ADE$ is a $6-8-10$ , since $\\triangle DEB$ is an $8-15-17$ .) The base $CD$ of the rectangle will be $9+6+6=21$ . Now, let $O$ be the intersection of $BD$ and $AC$ . This means that $\\triangle ABO$ and $\\triangle DCO$ are with ratio $\\frac{21}{9}=\\frac73$ . Set up a proportion, knowing that the two heights add up to 8. We will let $y$ be the height from $O$ to $DC$ , and $x$ be the height of $\\triangle ABO$ \\[\\frac{7}{3}=\\frac{y}{x}\\] \\[\\frac{7}{3}=\\frac{8-x}{x}\\] \\[7x=24-3x\\] \\[10x=24\\] \\[x=\\frac{12}{5}\\]\nThis means that the area is $A=\\tfrac{1}{2}(9)(\\tfrac{12}{5})=\\tfrac{54}{5}$ . This gets us $54+5=\\boxed{059}.$", "Using the diagram in Solution 1, let $E$ be the intersection of $BD$ and $AC$ . We can see that angle $C$ is in both $\\triangle BCE$ and $\\triangle ABC$ . Since $\\triangle BCE$ and $\\triangle ADE$ are congruent by AAS, we can then state $AE=BE$ and $DE=CE$ . It follows that $BE=AE$ and $CE=17-BE$ . We can now state that the area of $\\triangle ABE$ is the area of $\\triangle ABC-$ the area of $\\triangle BCE$ . Using Heron's formula, we compute the area of $\\triangle ABC=36$ . Using the Law of Cosines on angle $C$ , we obtain\n\\[9^2=17^2+10^2-2(17)(10)cosC\\] \\[-308=-340cosC\\] \\[cosC=\\frac{308}{340}\\] (For convenience, we're not going to simplify.)\nApplying the Law of Cosines on $\\triangle BCE$ yields \\[BE^2=10^2+(17-BE)^2-2(10)(17-BE)cosC\\] \\[BE^2=389-34BE+BE^2-20(17-BE)(\\frac{308}{340})\\] \\[0=389-34BE-(340-20BE)(\\frac{308}{340})\\] \\[0=389-34BE+\\frac{308BE}{17}\\] \\[0=81-\\frac{270BE}{17}\\] \\[81=\\frac{270BE}{17}\\] \\[BE=\\frac{51}{10}\\] This means $CE=17-BE=17-\\frac{51}{10}=\\frac{119}{10}$ . Next, apply Heron's formula to get the area of $\\triangle BCE$ , which equals $\\frac{126}{5}$ after simplifying. Subtracting the area of $\\triangle BCE$ from the area of $\\triangle ABC$ yields the area of $\\triangle ABE$ , which is $\\frac{54}{5}$ , giving us our answer, which is $54+5=\\boxed{059}.$ -Solution by flobszemathguy", " - Diagram by Brendanb4321 extended by Duoquinquagintillion\nBegin with the first step of solution 1, seeing $AD$ is the hypotenuse of a $6-8-10$ triangle and calling the intersection of $DB$ and $AC$ point $E$ . Next, notice $DB$ is the hypotenuse of an $8-15-17$ triangle. Drop an altitude from $E$ with length $h$ , so the other leg of the new triangle formed has length $4.5$ . Notice we have formed similar triangles, and we can solve for $h$\n\\[\\frac{h}{4.5} = \\frac{8}{15}\\] \\[h = \\frac{36}{15} = \\frac{12}{5}\\]\nSo $\\triangle ABE$ has area \\[\\frac{ \\frac{12}{5} \\cdot 9}{2} = \\frac{54}{5}\\] And $54+5=\\boxed{059}.$ - Solution by Duoquinquagintillion", "Let $a = \\angle{CAB}$ . By Law of Cosines, \\[\\cos a = \\frac{17^2+9^2-10^2}{2*9*17} = \\frac{15}{17}\\] \\[\\sin a = \\sqrt{1-\\cos^2 a} = \\frac{8}{17}\\] \\[\\tan a = \\frac{8}{15}\\] \\[A = \\frac{1}{2}* 9*\\frac{9}{2}\\tan a = \\frac{54}{5}\\] And $54+5=\\boxed{059}.$", "Let $A=(0,0), B=(9,0)$ . Now consider $C$ , and if we find the coordinates of $C$ , by symmetry about $x=4.5$ , we can find the coordinates of D.\nSo let $C=(a,b)$ . So the following equations hold:\n$\\sqrt{(a-9)^2+(b)^2}=17$\n$\\sqrt{a^2+b^2}=10$\nSolving by squaring both equations and then subtracting one from the other to eliminate $b^2$ , we get $C=(-6,8)$ because $C$ is in the second quadrant.\nNow by symmetry, $D=(16, 8)$\nSo now you can proceed by finding the intersection and then calculating the area directly. We get $\\boxed{059}$", "\nSince $\\triangle ABC \\cong \\triangle BAD$ $\\angle ADB = \\angle BCA$ . Thus, $A$ $B$ $C$ $D$ are concyclic.\nBy Ptolemy's Theorem on $ABCD$ \\[(AD)(BC) + (AB)(DC) = (BD)(AC)\\] \\[10^2 + 9(DC) = 17^2\\] \\[DC = 21\\]\nThe altitudes dropped from $C$ and $D$ onto the extension of AB are equal, meaning that $DC \\parallel AB$ . Therefore, $\\triangle DCO \\sim \\triangle ABO$ . It follows that \\[\\frac{OB}{17 - OB} = \\frac{AB}{DC} = \\frac{9}{21} = \\frac{3}{7}\\] Solving yields $OB = \\frac{51}{10}$\nIn $\\triangle ABD$ , drop an altitude from $A$ to $BD$ . Call the intersection of this altitude and $BD$ $K$\nThe area of $\\triangle ABD$ is $\\frac{1}{2}(AB)(DE) = 36$ . Thus, $\\frac{1}{2}(AK)(BD) = 36$ , and $AK = \\frac{72}{17}$\nTherefore, the area of $\\triangle AOB$ is $\\frac{1}{2}(OB)(AK) = \\frac{1}{2}(\\frac{51}{10})(\\frac{72}{17}) = \\frac{54}{5}$\nThe requested answer is $54 + 5 = \\boxed{59}$" ]

[ "Each of the numbers $a$ and $b$ is a solution to $\\left| x - \\frac 1x \\right| = 1$\nHence it is either a solution to $x - \\frac 1x = 1$ , or to $\\frac 1x - x = 1$ . Then it must be a solution either to $x^2 - x - 1 = 0$ , or to $x^2 + x - 1 = 0$\nThere are in total four such values of $x$ , namely $\\frac{\\pm 1 \\pm \\sqrt 5}2$\nOut of these, two are positive: $\\frac{-1+\\sqrt 5}2$ and $\\frac{1+\\sqrt 5}2$ . We can easily check that both of them indeed have the required property, and their sum is $\\frac{-1+\\sqrt 5}2 + \\frac{1+\\sqrt 5}2 = \\boxed{5}$" ]

[ "Any two prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate A, B, and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is $(13)(17)-(13+17) = 221 - 30 = 191$ . Thus, we can eliminate E. So, the answer must be $\\boxed{119}$", "Let the two primes be $p$ and $q$ . We wish to obtain the value of $pq-(p+q)$ , or $pq-p-q$ . Using Simon's Favorite Factoring Trick , we can rewrite this expression as $(1-p)(1-q) -1$ or $(p-1)(q-1) -1$ . Noticing that $(13-1)(11-1) - 1 = 120-1 = 119$ , we see that the answer is $\\boxed{119}$" ]

[ "Any two prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate A, B, and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is $(13)(17)-(13+17) = 221 - 30 = 191$ . Thus, we can eliminate E. So, the answer must be $\\boxed{119}$", "Let the two primes be $p$ and $q$ . We wish to obtain the value of $pq-(p+q)$ , or $pq-p-q$ . Using Simon's Favorite Factoring Trick , we can rewrite this expression as $(1-p)(1-q) -1$ or $(p-1)(q-1) -1$ . Noticing that $(13-1)(11-1) - 1 = 120-1 = 119$ , we see that the answer is $\\boxed{119}$" ]

[ "Let the area of the shaded region be $S$ , the area of the unshaded region be $U$ , and the acute angle that is formed by the two lines be $\\theta$ . We can set up two equations between $S$ and $U$\n$S+U=9\\pi$\n$S=\\dfrac{8}{13}U$\nThus $\\dfrac{21}{13}U=9\\pi$ , and $U=\\dfrac{39\\pi}{7}$ , and thus $S=\\dfrac{8}{13}\\cdot \\dfrac{39\\pi}{7}=\\dfrac{24\\pi}{7}$\nNow we can make a formula for the area of the shaded region in terms of $\\theta$\n$\\dfrac{2\\theta}{2\\pi} \\cdot \\pi +\\dfrac{2(\\pi-\\theta)}{2\\pi} \\cdot (4\\pi-\\pi)+\\dfrac{2\\theta}{2\\pi}(9\\pi-4\\pi)=\\theta +3\\pi-3\\theta+5\\theta=3\\theta+3\\pi=\\dfrac{24\\pi}{7}$\nThus $3\\theta=\\dfrac{3\\pi}{7}\\Rightarrow \\theta=\\dfrac{\\pi}{7}\\Rightarrow\\boxed{7}$", "As mentioned in Solution #1, we can make an equation for the area of the shaded region in terms of $\\theta$\n$\\implies\\dfrac{2\\theta}{2\\pi} \\cdot \\pi +\\dfrac{2(\\pi-\\theta)}{2\\pi} \\cdot (4\\pi-\\pi)+\\dfrac{2\\theta}{2\\pi}(9\\pi-4\\pi)=\\theta +3\\pi-3\\theta+5\\theta=3\\theta+3\\pi$\nSo, the shaded region is $3\\theta+3\\pi$ . This means that the unshaded region is $9\\pi-(3\\theta+3\\pi)$\nAlso, the shaded region is $\\frac{8}{13}$ of the unshaded region. Hence, we can now make an equation and solve for $\\theta$\n$3\\theta+3\\pi=\\frac{8}{13}(9\\pi-(3\\theta+3\\pi)\\implies 39\\theta+39\\pi=8(6\\pi-3\\theta)\\implies 39\\theta+39\\pi=48\\pi-24\\theta$\nSimplifying, we get $63\\theta=9\\pi\\implies \\theta=\\boxed{7}$" ]

[ "The sum of the first $n$ integers is given by $\\frac{n(n+1)}{2}$ , so $\\frac{37(37+1)}{2}=703$\nTherefore, $703-x-y=xy$\nRearranging, $xy+x+y=703$ . We can factor this equation by SFFT to get\n$(x+1)(y+1)=704$\nLooking at the possible divisors of $704 = 2^6\\cdot11$ $22$ and $32$ are within the constraints of $0 < x \\leq y \\leq 37$ so we try those:\n$(x+1)(y+1) = 22\\cdot32$\n$x+1=22, y+1 = 32$\n$x = 21, y = 31$\nTherefore, the difference $y-x=31-21=\\boxed{10}$" ]

[ "Let the second term of each series be $x$ . Then, the common ratio is $\\frac{1}{8x}$ , and the first term is $8x^2$\nSo, the sum is $\\frac{8x^2}{1-\\frac{1}{8x}}=1$ . Thus, $64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \\Rightarrow x = \\frac{1}{4}, \\frac{-1 \\pm \\sqrt{5}}{8}$\nThe only solution in the appropriate form is $x = \\frac{\\sqrt{5}-1}{8}$ . Therefore, $100m+10n+p = \\boxed{518}$", "Let's ignore the \"two distinct, real, infinite geometric series\" part for now and focus on what it means to be a geometric series.\nLet the first term of the series with the third term equal to $\\frac18$ be $a,$ and the common ratio be $r.$ Then, we get that $\\frac{a}{1-r} = 1 \\implies a = 1-r,$ and $ar^2 = \\frac18 \\implies (1-r)(r^2) = \\frac18.$\nWe see that this cubic is equivalent to $r^3 - r^2 + \\frac18 = 0.$ Through experimenting, we find that one of the solutions is $r = \\frac12.$ Using synthetic division leads to the quadratic $4x^2 - 2x - 1 = 0.$ This has roots $\\dfrac{2 \\pm \\sqrt{4 - 4(4)(-1)}}{8},$ or, when reduced, $\\dfrac{1 \\pm \\sqrt{5}}{4}.$\nIt becomes clear that the two geometric series have common ratio $\\frac{1 + \\sqrt{5}}{4}$ and $\\frac{1 - \\sqrt{5}}{4}.$ Let $\\frac{1 + \\sqrt{5}}{4}$ be the ratio that we are inspecting. We see that in this case, $a = \\dfrac{3 - \\sqrt{5}}{4}.$\nSince the second term in the series is $ar,$ we compute this and have that \\[ar = \\left(\\dfrac{3 - \\sqrt{5}}{4} \\right)\\left(\\dfrac{1+\\sqrt{5}}{4}\\right) = \\dfrac{\\sqrt{5} - 1}{8},\\] for our answer of $100 \\cdot 5 + 1 \\cdot 10 + 8 = \\boxed{518}.$" ]

[ "Two congruent coplanar circles will either be tangent to one another (resulting in $3$ common tangents), intersect one another (resulting in $2$ common tangents), or be separate from one another (resulting in $4$ common tangents).\nHaving only $\\boxed{1}$ common tangent is impossible, unless the circles are non-congruent and internally tangent." ]

[ "First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$ . Let points $A'$ and $B'$ be the reflections of $A$ and $B$ , respectively, about the perpendicular bisector of $\\overline{O_1O_2}$ . Then quadrilaterals $ABO_1O_2$ and $B'A'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $A'B'O_1CDO_2$ have the same area. Furthermore, triangles $DO_2A'$ and $B'O_1C$ are congruent, so $A'D = B'C$ and quadrilateral $A'B'CD$ is an isosceles trapezoid. Next, remark that $B'O_1 = DO_2$ , so quadrilateral $B'O_1DO_2$ is also an isosceles trapezoid; in turn, $B'D = O_1O_2 = 15$ , and similarly $A'C = 15$ . Thus, Ptolmey's theorem on $A'B'CD$ yields $A'D\\cdot B'C + 2\\cdot 16 = 15^2$ , whence $A'D = B'C = \\sqrt{193}$ . Let $\\alpha = \\angle A'B'D$ . The Law of Cosines on triangle $A'B'D$ yields \\[\\cos\\alpha = \\frac{15^2 + 2^2 - (\\sqrt{193})^2}{2\\cdot 2\\cdot 15} = \\frac{36}{60} = \\frac 35,\\] and hence $\\sin\\alpha = \\tfrac 45$ . Thus the distance between bases $A’B’$ and $CD$ is $12$ (in fact, $\\triangle A'B'D$ is a $9-12-15$ triangle with a $7-12-\\sqrt{193}$ triangle removed), which implies the area of $A'B'CD$ is $\\tfrac12\\cdot 12\\cdot(2+16) = 108$\nNow let $O_1C = O_2A' = r_1$ and $O_2D = O_1B' = r_2$ ; the tangency of circles $\\omega_1$ and $\\omega_2$ implies $r_1 + r_2 = 15$ . Furthermore, angles $A'O_2D$ and $A'B'D$ are opposite angles in cyclic quadrilateral $B'A'O_2D$ , which implies the measure of angle $A'O_2D$ is $180^\\circ - \\alpha$ . Therefore, the Law of Cosines applied to triangle $\\triangle A'O_2D$ yields \\begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \\tfrac45r_1r_2\\\\ &= (r_1+r_2)^2 - \\tfrac45 r_1r_2 = 225 - \\tfrac45r_1r_2. \\end{align*}\nThus $r_1r_2 = 40$ , and so the area of triangle $A'O_2D$ is $\\tfrac12r_1r_2\\sin\\alpha = 16$\nThus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\\cdot 16 = \\boxed{140}$", "Denote by $O$ the center of $\\Omega$ .\nDenote by $r$ the radius of $\\Omega$\nWe have $O_1$ $O_2$ $A$ $B$ $C$ $D$ are all on circle $\\Omega$\nDenote $\\angle O_1 O O_2 = 2 \\theta$ .\nDenote $\\angle O_1 O B = \\alpha$ .\nDenote $\\angle O_2 O A = \\beta$\nBecause $B$ and $C$ are on circles $\\omega_1$ and $\\Omega$ $BC$ is a perpendicular bisector of $O_1 O$ . Hence, $\\angle O_1 O C = \\alpha$\nBecause $A$ and $D$ are on circles $\\omega_2$ and $\\Omega$ $AD$ is a perpendicular bisector of $O_2 O$ . Hence, $\\angle O_2 O D = \\beta$\nIn $\\triangle O O_1 O_2$ \\[ O_1 O_2 = 2 r \\sin \\theta . \\]\nHence, \\[ 2 r \\sin \\theta = 15 . \\]\nIn $\\triangle O AB$ \\begin{align*} AB & = 2 r \\sin \\frac{2 \\theta - \\alpha - \\beta}{2} \\\\ & = 2 r \\sin \\theta \\cos \\frac{\\alpha + \\beta}{2} - 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} \\\\ & = 15 \\cos \\frac{\\alpha + \\beta}{2} - 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} . \\end{align*}\nHence, \\[ 15 \\cos \\frac{\\alpha + \\beta}{2} - 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} = 2 . \\hspace{1cm} (1) \\]\nIn $\\triangle O CD$ \\begin{align*} CD & = 2 r \\sin \\frac{360^\\circ - 2 \\theta - \\alpha - \\beta}{2} \\\\ & = 2 r \\sin \\left( \\theta + \\frac{\\alpha + \\beta}{2} \\right) \\\\ & = 2 r \\sin \\theta \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} \\\\ & = 15 \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} . \\end{align*}\nHence, \\[ 15 \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} = 16 . \\hspace{1cm} (2) \\]\nTaking $\\frac{(1) + (2)}{30}$ , we get $\\cos \\frac{\\alpha + \\beta}{2} = \\frac{3}{5}$ .\nThus, $\\sin \\frac{\\alpha + \\beta}{2} = \\frac{4}{5}$\nTaking these into (1), we get $2 r \\cos \\theta = \\frac{35}{4}$ .\nHence, \\begin{align*} 2 r & = \\sqrt{ \\left( 2 r \\sin \\theta \\right)^2 + \\left( 2 r \\cos \\theta \\right)^2} \\\\ & = \\frac{5}{4} \\sqrt{193} . \\end{align*}\nHence, $\\cos \\theta = \\frac{7}{\\sqrt{193}}$\nIn $\\triangle O O_1 B$ \\[ O_1 B = 2 r \\sin \\frac{\\alpha}{2} . \\]\nIn $\\triangle O O_2 A$ , by applying the law of sines, we get \\[ O_2 A = 2 r \\sin \\frac{\\beta}{2} . \\]\nBecause circles $\\omega_1$ and $\\omega_2$ are externally tangent, $B$ is on circle $\\omega_1$ $A$ is on circle $\\omega_2$ \\begin{align*} O_1 O_2 & = O_1 B + O_2 A \\\\ & = 2 r \\sin \\frac{\\alpha}{2} + 2 r \\sin \\frac{\\beta}{2} \\\\ & = 2 r \\left( \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} \\right) . \\end{align*}\nThus, $\\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} = \\frac{12}{\\sqrt{193}}$\nNow, we compute $\\sin \\alpha$ and $\\sin \\beta$\nRecall $\\cos \\frac{\\alpha + \\beta}{2} = \\frac{3}{5}$ and $\\sin \\frac{\\alpha + \\beta}{2} = \\frac{4}{5}$ .\nThus, $e^{i \\frac{\\alpha}{2}} e^{i \\frac{\\beta}{2}} = e^{i \\frac{\\alpha + \\beta}{2}} = \\frac{3}{5} + i \\frac{4}{5}$\nWe also have \\begin{align*} \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} & = \\frac{1}{2i} \\left( e^{i \\frac{\\alpha}{2}} - e^{-i \\frac{\\alpha}{2}} + e^{i \\frac{\\beta}{2}} - e^{-i \\frac{\\beta}{2}} \\right) \\\\ & = \\frac{1}{2i} \\left( 1 - \\frac{1}{e^{i \\frac{\\alpha + \\beta}{2}} } \\right) \\left( e^{i \\frac{\\alpha}{2}} + e^{i \\frac{\\beta}{2}} \\right) . \\end{align*}\nThus, \\begin{align*} \\sin \\alpha + \\sin \\beta & = \\frac{1}{2i} \\left( e^{i \\alpha} - e^{-i \\alpha} + e^{i \\beta} - e^{-i \\beta} \\right) \\\\ & = \\frac{1}{2i} \\left( 1 - \\frac{1}{e^{i \\left( \\alpha + \\beta \\right)}} \\right) \\left( e^{i \\alpha} + e^{i \\beta} \\right) \\\\ & = \\frac{1}{2i} \\left( 1 - \\frac{1}{e^{i \\left( \\alpha + \\beta \\right)}} \\right) \\left( \\left( e^{i \\frac{\\alpha}{2}} + e^{i \\frac{\\beta}{2}} \\right)^2 - 2 e^{i \\frac{\\alpha + \\beta}{2}} \\right) \\\\ & = \\frac{1}{2i} \\left( 1 - \\frac{1}{e^{i \\left( \\alpha + \\beta \\right)}} \\right) \\left( \\left( \\frac{2 i \\left( \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} \\right)}{1 - \\frac{1}{e^{i \\frac{\\alpha + \\beta}{2}} }} \\right)^2 - 2 e^{i \\frac{\\alpha + \\beta}{2}} \\right) \\\\ & = - \\frac{1}{i} \\left( e^{i \\frac{\\alpha + \\beta}{2}} - e^{-i \\frac{\\alpha + \\beta}{2}} \\right) \\left( \\frac{2 \\left( \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} \\right)^2} {e^{i \\frac{\\alpha + \\beta}{2}} + e^{-i \\frac{\\alpha + \\beta}{2}} - 2 } + 1 \\right) \\\\ & = \\frac{167 \\cdot 8}{193 \\cdot 5 } . \\end{align*}\nTherefore, \\begin{align*} {\\rm Area} \\ ABO_1CDO_2 & = {\\rm Area} \\ \\triangle O_3 AB + {\\rm Area} \\ \\triangle O_3 BO_1 + {\\rm Area} \\ \\triangle O_3 O_1 C \\\\ & \\quad + {\\rm Area} \\ \\triangle O_3 C D + {\\rm Area} \\ \\triangle O_3 D O_2 + {\\rm Area} \\ \\triangle O_3 O_2 A \\\\ & = \\frac{1}{2} r^2 \\left( \\sin \\left( 2 \\theta - \\alpha - \\beta \\right) + \\sin \\alpha + \\sin \\alpha + \\sin \\left( 360^\\circ - 2 \\theta - \\alpha - \\beta \\right) + \\sin \\beta + \\sin \\beta \\right) \\\\ & = \\frac{1}{2} r^2 \\left( \\sin \\left( 2 \\theta - \\alpha - \\beta \\right) - \\sin \\left( 2 \\theta + \\alpha + \\beta \\right) + 2 \\sin \\alpha + 2 \\sin \\beta \\right) \\\\ & = r^2 \\left( - \\cos 2 \\theta \\sin \\left( \\alpha + \\beta \\right) + \\sin \\alpha + \\sin \\beta \\right) \\\\ & = r^2 \\left( \\left( 1 - 2 \\cos^2 \\theta \\right) 2 \\sin \\frac{\\alpha + \\beta}{2} \\cos \\frac{\\alpha + \\beta}{2} + \\sin \\alpha + \\sin \\beta \\right) \\\\ & = \\boxed{140}", "Let points $A'$ and $B'$ be the reflections of $A$ and $B,$ respectively, about the perpendicular bisector of $O_1 O_2.$ \\[B'O_2 = BO_1 = O_1 P = O_1 C,\\] \\[A'O_1 = AO_2 = O_2 P = O_2 D.\\] We establish the equality of the arcs and conclude that the corresponding chords are equal \\[\\overset{\\Large\\frown} {CO_1} + \\overset{\\Large\\frown} {A'O_1} +\\overset{\\Large\\frown} {A'B'} = \\overset{\\Large\\frown} {B'O_2} +\\overset{\\Large\\frown} {A'O_1} +\\overset{\\Large\\frown} {A'B'} =\\overset{\\Large\\frown} {B'O_2} +\\overset{\\Large\\frown} {DO_2} +\\overset{\\Large\\frown} {A'B'}\\] \\[\\implies A'D = B'C = O_1 O_2 = 15.\\] Similarly $A'C = B'D \\implies \\triangle A'CO_1 = \\triangle B'DO_2.$\nPtolemy's theorem on $A'CDB'$ yields \\[B'D \\cdot A'C + A'B' \\cdot CD = A'D \\cdot B'C \\implies\\] \\[B'D^2 + 2 \\cdot 16 = 15^2 \\implies B'D = A'C = \\sqrt{193}.\\] The area of the trapezoid $A'CDB'$ is equal to the area of an isosceles triangle with sides $A'D = B'C = 15$ and $A'B' + CD = 18.$\nThe height of this triangle is $\\sqrt{15^2-9^2} = 12.$ The area of $A'CDB'$ is $108.$\n\\[\\sin \\angle B'CD = \\frac{12}{15} = \\frac{4}{5},\\] \\[\\angle B'CD + \\angle B'O_2 D = 180^o \\implies \\sin \\angle B'O_2 D = \\frac{4}{5}.\\]\nDenote $\\angle B'O_2 D = 2\\alpha.$ $\\angle B'O_2 D > \\frac{\\pi}{2},$ hence $\\cos \\angle B'O_2 D = \\cos 2\\alpha = -\\frac{3}{5}.$ \\[\\tan \\alpha =\\frac { \\sin 2 \\alpha}{1+\\cos 2 \\alpha} = \\frac {4/5}{1 - 3/5}=2.\\]\nSemiperimeter of $\\triangle B'O_2 D$ is $s = \\frac {15 + \\sqrt{193}}{2}.$\nThe distance from the vertex $O_2$ to the tangent points of the inscribed circle of the triangle $B'O_2 D$ is equal $s – B'D = \\frac{15 – \\sqrt{193}}{2}.$\nThe radius of the inscribed circle is $r = (s – B'D) \\tan \\alpha.$\nThe area of triangle $B'O_2 D$ is $[B'O_2 D] = sr = s (s – B'D) \\tan \\alpha = \\frac {15^2 – 193}{2} = 16.$\nThe hexagon $ABO_1 CDO_2$ has the same area as hexagon $B'A'O_1 CDO_2.$\nThe area of hexagon $B'A'O_1 CDO_2$ is equal to the sum of the area of the trapezoid $A'CDB'$ and the areas of two equal triangles $B'O_2 D$ and $A'O_1 C,$ so the area of the hexagon $ABO_1 CDO_2$ is \\[108 + 16 + 16 = \\boxed{140}.\\]" ]

[ "Suppose the dice have $a$ and $b$ faces, and WLOG $a\\geq{b}$ . Since each die has at least $6$ faces, there will always be $6$ ways to sum to $7$ . As a result, there must be $\\tfrac{4}{3}\\cdot6=8$ ways to sum to $10$ . There are at most nine distinct ways to get a sum of $10$ , which are possible whenever $a,b\\geq{9}$ . To achieve exactly eight ways, $b$ must have $8$ faces, and $a\\geq9$ . Let $n$ be the number of ways to obtain a sum of $12$ , then $\\tfrac{n}{8a}=\\tfrac{1}{12}\\implies n=\\tfrac{2}{3}a$ . Since $b=8$ $n\\leq8\\implies a\\leq{12}$ . In addition to $3\\mid{a}$ , we only have to test $a=9,12$ , of which both work. Taking the smaller one, our answer becomes $a+b=9+8=\\boxed{17}$", "Suppose the dice have $a$ and $b$ faces, and WLOG $a\\geq{b}$ . Note that if $a+b=12$ since they are both $6$ , there is one way to make $12$ , and incrementing $a$ or $b$ by one will add another way. This gives us the probability of making a 12 as \\[\\frac{a+b-11}{ab}=\\frac{1}{12}\\] Cross-multiplying, we get \\[12a+12b-132=ab\\] Simon's Favorite Factoring Trick now gives \\[(a-12)(b-12)=12\\] This narrows the possibilities down to 3 ordered pairs of $(a,b)$ , which are $(13,24)$ $(6,10)$ , and $(8,9)$ . We can obviously ignore the first pair and test the next two straightforwardly. The last pair yields the answer: \\[\\frac{6}{72}=\\frac{3}{4}\\left(\\frac{9+8-9}{72}\\right)\\] The answer is then $a+b=8+9=\\boxed{17}$" ]

[ "Call the common ratio $r.$ Now since the $n$ th term of a geometric sequence with first term $x$ and common ratio $y$ is $xy^{n-1},$ we see that $a_1 \\cdot r^{14} = b_1 \\cdot r^{10} \\implies r^4 = \\frac{99}{27} = \\frac{11}{3}.$ But $a_9$ equals $a_1 \\cdot r^8 = a_1 \\cdot (r^4)^2=27\\cdot {\\left(\\frac{11}{3}\\right)}^2=27\\cdot \\frac{121} 9=\\boxed{363}$" ]

[ "The desired average can be found by dividing the total number of points earned by the total number of students. There are $20\\cdot 80+30\\cdot 70=3700$ points earned and $20+30=50$ students. Thus, our answer is $\\frac{3700}{50}$ , or $\\boxed{74}$" ]

[ "To double the range, we must find the current range, which is $28 - 3 = 25$ , to then double to: $2(25) = 50$ . Since we do not want to change the median, we need to get a value less than $8$ (as $8$ would change the mode) for the smaller, making $53$ fixed for the larger. Remember, anything less than $3$ is not beneficial to the optimization because you want to get the largest range without changing the mode. So, taking our optimal values of $7$ and $53$ , we have an answer of $7 + 53 = \\boxed{60}$" ]

[ "Since, $x + y = 26$ $x$ can equal $15$ , and $y$ can equal $11$ , so no even integers are required to make 26. To get to $41$ , we have to add $41 - 26 = 15$ . If $a+b=15$ , at least one of $a$ and $b$ must be even because two odd numbers sum to an even number. Therefore, one even integer is required when transitioning from $26$ to $41$ . Finally, we have the last transition is $57-41=16$ . If $m+n=16$ $m$ and $n$ can both be odd because two odd numbers sum to an even number, meaning only $1$ even integer is required. The answer is $\\boxed{1}$ . ~Extremelysupercooldude (Latex, grammar, and solution edits)", "Just worded and formatted a little differently than above.\nThe first two integers sum up to $26$ . Since $26$ is even, in order to minimize the number of even integers, we make both of the first two odd.\nThe second two integers sum up to $41-26=15$ . Since $15$ is odd, we must have at least one even integer in these next two.\nFinally, $57-41=16$ , and once again, $16$ is an even number so both of these integers can be odd.\nTherefore, we have a total of one even integer and our answer is $\\boxed{1}$" ]

[ "Out of the first two integers, it's possible for both to be even: for example, $10 + 16 = 26.$ But the next two integers, when added, increase the sum by $15,$ which is odd, so one of them must be odd and the other must be even: for example, $3 + 12 = 15.$ Finally, the next two integers increase the sum by $16,$ which is even, so we can have both be even: for example, $2 + 14 = 16.$ Therefore, $\\boxed{1}$ is the minimum number of integers that must be odd." ]

[ "Call the number of marbles in each jar $x$ (because the problem specifies that they each contain the same number). Thus, $\\frac{x}{10}$ is the number of green marbles in Jar $1$ , and $\\frac{x}{9}$ is the number of green marbles in Jar $2$ . Since $\\frac{x}{9}+\\frac{x}{10}=\\frac{19x}{90}$ , we have $\\frac{19x}{90}=95$ , so there are $x=450$ marbles in each jar.\nBecause $\\frac{9x}{10}$ is the number of blue marbles in Jar $1$ , and $\\frac{8x}{9}$ is the number of blue marbles in Jar $2$ , there are $\\frac{9x}{10}-\\frac{8x}{9}=\\frac{x}{90} = 5$ more marbles in Jar $1$ than Jar $2$ . This means the answer is $\\boxed{5}$", "Let $b_1$ $g_1$ $b_2$ $g_2$ , represent the amount of blue marbles in jar 1, the amount of green marbles in jar 1, the \nthe amount of blue marbles in jar 2, and the amount of green marbles in jar 2, respectively. We now have the equations, $\\frac{b_1}{g_1} = \\frac{9}{1}$ $\\frac{b_2}{g_2} = \\frac{8}{1}$ $g_1 + g_2 =95$ , and $b_1 + g_1 = b_2 + g_2$ .\nSince $b_1 = 9g_1$ and $b_2 = 8g_2$ , we substitute that in to obtain $10g_1 = 9g_2$ . \nCoupled with our third equation, we find that $g_1 = 45$ , and that $g_2 = 50$ . We now use this information to find $b_1 = 405$ and $b_2 = 400$\nTherefore, $b_1 - b_2 = 5$ so our answer is $\\boxed{5}$ .\n~Binderclips1" ]

[ "Writing out to ratios, we have $9:1$ in jar $1$ and $8:1$ in jar $2$ . Since the jar must have to same amount of marbles, let's make a variable $a$ and $b$ for each of the ratios to be multiplied by. Now we would have $9a + a = 8b + b \\rightarrow 10a = 9b$ . We can take the most obvious values of $a$ and $b$ and then scale it from there. We should be able to see that $a$ and $b$ could be $9$ and $10$ respectively. Now remember that there are $95$ green marbles or $x(a + b) = 95$ for some integer $x$ to scale it. Substituting and dividing, we find $x = 5$ . Thus to find the difference of the blue marbles we must do \\begin{align*} x(9a - 8b) &= \\\\ 5(81 - 80) &= \\\\ 5(1) &= \\boxed{5}" ]

[ "Let's first work out the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=\\frac{x}{2} + b$ implies $2=\\frac{2}{2} +b=1+b$ so $b=1$ , while $y=2x + c$ implies $2= 2 \\cdot 2+c=4+c$ so $c=-2$ . Also, $x+y=10$ implies $y=-x+10$ . Thus the lines are $y=\\frac{x}{2} +1, y=2x-2,$ and $y=-x+10$ . \nNow we find the intersection points between each of the lines with $y=-x+10$ , which are $(6,4)$ and $(4,6)$ . Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base $2\\sqrt{2}$ and height $3\\sqrt{2}$ , whose area is $\\boxed{6}$", "Like in Solution 1, we determine the coordinates of the three vertices of the triangle. The coordinates that we get are: $(2,2)$ $(6,4)$ $(4,6)$ . Now, using the Shoelace Theorem , we can directly find that the area is $\\boxed{6}$", "Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at $(4, 6)$ and $(6, 4)$ . Then apply Heron's Formula: the semi-perimeter will be $s = \\sqrt{2} + \\sqrt{20}$ , so the area reduces nicely to a difference of squares, making it $\\boxed{6}$", "Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . We can now draw the bounding square with vertices $(2, 2)$ $(2, 6)$ $(6, 6)$ and $(6, 2)$ , and deduce that the triangle's area is $16-4-2-4=\\boxed{6}$", "Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . Using graph paper, we can see that this triangle has $6$ boundary lattice points and $4$ interior lattice points. By Pick's Theorem, the area is $\\frac62 + 4 - 1 = \\boxed{6}$", "Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\\sqrt5$ and $BC = 2\\sqrt2$ . By the Law of Cosines, \\[\\cos A = \\frac{(2\\sqrt5)^2 + (2\\sqrt5)^2 - (2\\sqrt2)^2}{2 \\cdot 2\\sqrt5 \\cdot 2\\sqrt5} = \\frac{20 + 20 - 8}{40} = \\frac{32}{40} = \\frac45\\] Therefore, $\\sin A = \\sqrt{1 - \\cos^2 A} = \\frac35$ , so the area is $\\frac12 bc \\sin A = \\frac12 \\cdot 2\\sqrt5 \\cdot 2\\sqrt5 \\cdot \\frac35 = \\boxed{6}$", "Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . The area of the triangle is half the absolute value of the determinant of the matrix determined by these points. \\[\\frac12\\begin{Vmatrix} 2&2&1\\\\ 4&6&1\\\\ 6&4&1\\\\ \\end{Vmatrix} = \\frac12|-12| = \\frac12 \\cdot 12 = \\boxed{6}\\]", "Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . Then vectors $\\overrightarrow{AB} = \\langle 2, 4 \\rangle$ and $\\overrightarrow{AC} = \\langle 4, 2 \\rangle$ . The area of the triangle is half the magnitude of the cross product of these two vectors. \\[\\frac12\\begin{Vmatrix} i&j&k\\\\ 2&4&0\\\\ 4&2&0\\\\ \\end{Vmatrix} = \\frac12|-12k| = \\frac12 \\cdot 12 = \\boxed{6}\\]", "Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . By the Pythagorean theorem, this is an isosceles triangle with base $2\\sqrt2$ and equal length $2\\sqrt5$ . The area of an isosceles triangle with base $b$ and equal length $l$ is $\\frac{b\\sqrt{4l^2-b^2}}{4}$ . Plugging in $b = 2\\sqrt2$ and $l = 2\\sqrt5$ \\[\\frac{2\\sqrt2 \\cdot \\sqrt{80-8}}{4} = \\frac{\\sqrt{576}}{4} = \\frac{24}{4} = \\boxed{6}\\]", "Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\\sqrt5$ and $BC = 2\\sqrt2$ . By the Law of Cosines, \\[\\cos A = \\frac{(2\\sqrt5)^2 + (2\\sqrt5)^2 - (2\\sqrt2)^2}{2 \\cdot 2\\sqrt5 \\cdot 2\\sqrt5} = \\frac{20 + 20 - 8}{40} = \\frac{32}{40} = \\frac45\\] Therefore, $\\sin A = \\sqrt{1 - \\cos^2 A} = \\frac35$ . By the extended Law of Sines, \\[2R = \\frac{a}{\\sin A} = \\frac{2\\sqrt2}{\\frac35} = \\frac{10\\sqrt2}{3}\\] \\[R = \\frac{5\\sqrt2}{3}\\] Then the area is $\\frac{abc}{4R} = \\frac{2\\sqrt2 \\cdot 2\\sqrt5^2}{4 \\cdot \\frac{5\\sqrt2}{3}} = \\boxed{6}$", "The area of a triangle formed by three lines, \\[a_1x + a_2y + a_3 = 0\\] \\[b_1x + b_2y + b_3 = 0\\] \\[c_1x + c_2y + c_3 = 0\\] is the absolute value of \\[\\frac12 \\cdot \\frac{1}{(b_1c_2-b_2c_1)(a_1c_2-a_2c_1)(a_1b_2-a_2b_1)} \\cdot \\begin{vmatrix} a_1&a_2&a_3\\\\ b_1&b_2&b_3\\\\ c_1&c_2&c_3\\\\ \\end{vmatrix}^2\\] Plugging in the three lines, \\[-x + 2y - 2 = 0\\] \\[-2x + y + 2 = 0\\] \\[x + y - 10 = 0\\] the area is the absolute value of \\[\\frac12 \\cdot \\frac{1}{(-2-1)(-1-2)(-1+4)} \\cdot \\begin{vmatrix} -1&2&-2\\\\ -2&1&2\\\\ 1&1&-10\\\\ \\end{vmatrix}^2 = \\frac12 \\cdot \\frac{1}{27} \\cdot 18^2 = \\boxed{6}\\] Source: Orrick, Michael L. “THE AREA OF A TRIANGLE FORMED BY THREE LINES.” Pi Mu Epsilon Journal, vol. 7, no. 5, 1981, pp. 294–298. JSTOR, www.jstor.org/stable/24336991.", "Like in other solutions, we find that our triangle is isosceles with legs of $2\\sqrt5$ and base $2\\sqrt2$ . Then, the semi - perimeter of our triangle is, \\[\\frac{4\\sqrt5+2\\sqrt2}{2} = 2\\sqrt5 + \\sqrt2.\\] Applying Heron's formula, we find that the area of this triangle is equivalent to \\[\\sqrt{{(2\\sqrt5+\\sqrt2)}{(2\\sqrt5-\\sqrt2)}{(2)}} = \\sqrt{{(20-2)}{(2)}} = \\boxed{6}.\\]" ]

[ "Let's first work out the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=\\frac{x}{2} + b$ implies $2=\\frac{2}{2} +b=1+b$ so $b=1$ , while $y=2x + c$ implies $2= 2 \\cdot 2+c=4+c$ so $c=-2$ . Also, $x+y=10$ implies $y=-x+10$ . Thus the lines are $y=\\frac{x}{2} +1, y=2x-2,$ and $y=-x+10$ . \nNow we find the intersection points between each of the lines with $y=-x+10$ , which are $(6,4)$ and $(4,6)$ . Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base $2\\sqrt{2}$ and height $3\\sqrt{2}$ , whose area is $\\boxed{6}$", "Like in Solution 1, we determine the coordinates of the three vertices of the triangle. The coordinates that we get are: $(2,2)$ $(6,4)$ $(4,6)$ . Now, using the Shoelace Theorem , we can directly find that the area is $\\boxed{6}$", "Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at $(4, 6)$ and $(6, 4)$ . Then apply Heron's Formula: the semi-perimeter will be $s = \\sqrt{2} + \\sqrt{20}$ , so the area reduces nicely to a difference of squares, making it $\\boxed{6}$", "Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . We can now draw the bounding square with vertices $(2, 2)$ $(2, 6)$ $(6, 6)$ and $(6, 2)$ , and deduce that the triangle's area is $16-4-2-4=\\boxed{6}$", "Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . Using graph paper, we can see that this triangle has $6$ boundary lattice points and $4$ interior lattice points. By Pick's Theorem, the area is $\\frac62 + 4 - 1 = \\boxed{6}$", "Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\\sqrt5$ and $BC = 2\\sqrt2$ . By the Law of Cosines, \\[\\cos A = \\frac{(2\\sqrt5)^2 + (2\\sqrt5)^2 - (2\\sqrt2)^2}{2 \\cdot 2\\sqrt5 \\cdot 2\\sqrt5} = \\frac{20 + 20 - 8}{40} = \\frac{32}{40} = \\frac45\\] Therefore, $\\sin A = \\sqrt{1 - \\cos^2 A} = \\frac35$ , so the area is $\\frac12 bc \\sin A = \\frac12 \\cdot 2\\sqrt5 \\cdot 2\\sqrt5 \\cdot \\frac35 = \\boxed{6}$", "Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . The area of the triangle is half the absolute value of the determinant of the matrix determined by these points. \\[\\frac12\\begin{Vmatrix} 2&2&1\\\\ 4&6&1\\\\ 6&4&1\\\\ \\end{Vmatrix} = \\frac12|-12| = \\frac12 \\cdot 12 = \\boxed{6}\\]", "Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . Then vectors $\\overrightarrow{AB} = \\langle 2, 4 \\rangle$ and $\\overrightarrow{AC} = \\langle 4, 2 \\rangle$ . The area of the triangle is half the magnitude of the cross product of these two vectors. \\[\\frac12\\begin{Vmatrix} i&j&k\\\\ 2&4&0\\\\ 4&2&0\\\\ \\end{Vmatrix} = \\frac12|-12k| = \\frac12 \\cdot 12 = \\boxed{6}\\]", "Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . By the Pythagorean theorem, this is an isosceles triangle with base $2\\sqrt2$ and equal length $2\\sqrt5$ . The area of an isosceles triangle with base $b$ and equal length $l$ is $\\frac{b\\sqrt{4l^2-b^2}}{4}$ . Plugging in $b = 2\\sqrt2$ and $l = 2\\sqrt5$ \\[\\frac{2\\sqrt2 \\cdot \\sqrt{80-8}}{4} = \\frac{\\sqrt{576}}{4} = \\frac{24}{4} = \\boxed{6}\\]", "Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\\sqrt5$ and $BC = 2\\sqrt2$ . By the Law of Cosines, \\[\\cos A = \\frac{(2\\sqrt5)^2 + (2\\sqrt5)^2 - (2\\sqrt2)^2}{2 \\cdot 2\\sqrt5 \\cdot 2\\sqrt5} = \\frac{20 + 20 - 8}{40} = \\frac{32}{40} = \\frac45\\] Therefore, $\\sin A = \\sqrt{1 - \\cos^2 A} = \\frac35$ . By the extended Law of Sines, \\[2R = \\frac{a}{\\sin A} = \\frac{2\\sqrt2}{\\frac35} = \\frac{10\\sqrt2}{3}\\] \\[R = \\frac{5\\sqrt2}{3}\\] Then the area is $\\frac{abc}{4R} = \\frac{2\\sqrt2 \\cdot 2\\sqrt5^2}{4 \\cdot \\frac{5\\sqrt2}{3}} = \\boxed{6}$", "The area of a triangle formed by three lines, \\[a_1x + a_2y + a_3 = 0\\] \\[b_1x + b_2y + b_3 = 0\\] \\[c_1x + c_2y + c_3 = 0\\] is the absolute value of \\[\\frac12 \\cdot \\frac{1}{(b_1c_2-b_2c_1)(a_1c_2-a_2c_1)(a_1b_2-a_2b_1)} \\cdot \\begin{vmatrix} a_1&a_2&a_3\\\\ b_1&b_2&b_3\\\\ c_1&c_2&c_3\\\\ \\end{vmatrix}^2\\] Plugging in the three lines, \\[-x + 2y - 2 = 0\\] \\[-2x + y + 2 = 0\\] \\[x + y - 10 = 0\\] the area is the absolute value of \\[\\frac12 \\cdot \\frac{1}{(-2-1)(-1-2)(-1+4)} \\cdot \\begin{vmatrix} -1&2&-2\\\\ -2&1&2\\\\ 1&1&-10\\\\ \\end{vmatrix}^2 = \\frac12 \\cdot \\frac{1}{27} \\cdot 18^2 = \\boxed{6}\\] Source: Orrick, Michael L. “THE AREA OF A TRIANGLE FORMED BY THREE LINES.” Pi Mu Epsilon Journal, vol. 7, no. 5, 1981, pp. 294–298. JSTOR, www.jstor.org/stable/24336991.", "Like in other solutions, we find that our triangle is isosceles with legs of $2\\sqrt5$ and base $2\\sqrt2$ . Then, the semi - perimeter of our triangle is, \\[\\frac{4\\sqrt5+2\\sqrt2}{2} = 2\\sqrt5 + \\sqrt2.\\] Applying Heron's formula, we find that the area of this triangle is equivalent to \\[\\sqrt{{(2\\sqrt5+\\sqrt2)}{(2\\sqrt5-\\sqrt2)}{(2)}} = \\sqrt{{(20-2)}{(2)}} = \\boxed{6}.\\]" ]

[ "2007 AIME II-11.png\nIf it weren’t for the small tube, the larger tube would travel $144\\pi$ . Consider the distance from which the larger tube first contacts the smaller tube, until when it completely loses contact with the smaller tube.\nDrawing the radii as shown in the diagram, notice that the hypotenuse of the right triangle in the diagram has a length of $72 + 24 = 96$ . The horizontal line divides the radius of the larger circle into $72 - 24 = 48$ on the top half, which indicates that the right triangle has leg of 48 and hypotenuse of 96, a $30-60-90 \\triangle$\nFind the length of the purple arc in the diagram (the distance the tube rolled, but not the horizontal distance). The sixty degree central angle indicates to take $\\frac{60}{360} = \\frac 16$ of the circumference of the larger circle (twice), while the $180 - 2(30) = 120^{\\circ}$ central angle in the smaller circle indicates to take $\\frac{120}{360} = \\frac 13$ of the circumference. This adds up to $2 \\cdot \\frac 16 144\\pi + \\frac 13 48\\pi = 64\\pi$\nThe actual horizontal distance it takes can be found by using the $30-60-90 \\triangle$ s. The missing leg is equal in length to $48\\sqrt{3}$ . Thus, the total horizontal distance covered is $96\\sqrt{3}$\nThus, we get $144\\pi - 64\\pi + 96\\sqrt{3} = 80\\pi + 96\\sqrt{3}$ , and our answer is $\\boxed{179}$" ]

[ "First, find the number of hours it takes for the two to meet together. After $h$ hours, the person at $R$ walks $4.5h$ miles. In the same amount of time, the person at $S$ has been walking at $3.25+0.5(h-1)$ mph for the past hour, so the person walks $\\frac{h(6.5+0.5(h-1))}{2}$ miles.\nIn order for both to meet, the sum of both of the distances walked must total $76$ miles, so \\[4.5h + \\frac{h(6+0.5h)}{2} = 76\\] \\[4.5h + 3h + 0.25h^2 = 76\\] \\[0.25h^2 + 7.5h - 76 = 0\\] \\[h^2 + 30h - 304 = 0\\] \\[(h + 38)(h - 8) = 0\\] Since $h$ must be positive, $h = 8$ . Because it takes $8$ hours to meet, the person from $R$ traveled $36$ miles while the person from $S$ traveled $40$ miles. Thus, they are $4$ miles closer to $R$ than $S$ , so the answer is $\\boxed{4}$" ]

[ "Let the first two terms of the first sequence be $x_{1}$ and $x_{2}$ and the first two of the second sequence be $y_{1}$ and $y_{2}$ . Computing the seventh term, we see that $5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}$ . Note that this means that $x_{1}$ and $y_{1}$ must have the same value modulo $8$ . To minimize, let one of them be $0$ WLOG , assume that $x_{1} = 0$ . Thus, the smallest possible value of $y_{1}$ is $8$ ; and since the sequences are non-decreasing we get $y_{2} \\ge 8$ . To minimize, let $y_{2} = 8$ . Thus, $5y_{1} + 8y_{2} = 40 + 64 = \\boxed{104}$", "WLOG, let $a_i$ $b_i$ be the sequences with $a_1<b_1$ . Then \\[N=5a_1+8a_2=5b_1+8b_2\\] or \\[5a_1+8a_2=5(a_1+c)+8(a_2-d)\\] for some natural numbers $c$ $d$ . Thus $5c=8d$ . To minimize $c$ and $d$ , we have $(c,d)=(8,5)$ , or \\[5a_1+8a_2=5(a_1+8)+8(a_2-5).\\] To minimize $a_1$ and $b_1$ , we have $(a_1,b_1)=(0,0+c)=(0,8)$ . Using the same method, since $b_2\\ge b_1$ , we have $b_2\\ge8$\nThus the minimum $N=5b_1+8b_2=104\\boxed{104}$" ]

[ "Let the first two terms of the first sequence be $x_{1}$ and $x_{2}$ and the first two of the second sequence be $y_{1}$ and $y_{2}$ . Computing the seventh term, we see that $5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}$ . Note that this means that $x_{1}$ and $y_{1}$ must have the same value modulo $8$ . To minimize, let one of them be $0$ WLOG , assume that $x_{1} = 0$ . Thus, the smallest possible value of $y_{1}$ is $8$ ; and since the sequences are non-decreasing we get $y_{2} \\ge 8$ . To minimize, let $y_{2} = 8$ . Thus, $5y_{1} + 8y_{2} = 40 + 64 = \\boxed{104}$", "WLOG, let $a_i$ $b_i$ be the sequences with $a_1<b_1$ . Then \\[N=5a_1+8a_2=5b_1+8b_2\\] or \\[5a_1+8a_2=5(a_1+c)+8(a_2-d)\\] for some natural numbers $c$ $d$ . Thus $5c=8d$ . To minimize $c$ and $d$ , we have $(c,d)=(8,5)$ , or \\[5a_1+8a_2=5(a_1+8)+8(a_2-5).\\] To minimize $a_1$ and $b_1$ , we have $(a_1,b_1)=(0,0+c)=(0,8)$ . Using the same method, since $b_2\\ge b_1$ , we have $b_2\\ge8$\nThus the minimum $N=5b_1+8b_2=104\\boxed{104}$" ]

[ "Let $s$ be the semiperimeter of the two triangles. Also, let the base of the longer triangle be $16x$ and the base of the shorter triangle be $14x$ for some arbitrary factor $x$ . Then, the dimensions of the two triangles must be $s-8x,s-8x,16x$ and $s-7x,s-7x,14x$ . By Heron's Formula, we have\nSince $15$ and $338$ are coprime, to minimize, we must have $s=338$ and $x=15$ . However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by $2$ , which gives us a final answer of $\\boxed{676}$", "Let the first triangle have sides $16n,a,a$ , so the second has sides $14n,a+n,a+n$ . The height of the first triangle is $\\frac{7}{8}$ the height of the second triangle. Therefore, we have \\[a^2-64n^2=\\frac{49}{64}((a+n)^2-49n^2).\\] Multiplying this, we get \\[64a^2-4096n^2=49a^2+98an-2352n^2,\\] which simplifies to \\[15a^2-98an-1744n^2=0.\\] Solving this for $a$ , we get $a=n\\cdot\\frac{218}{15}$ , so $n=15$ and $a=218$ and the perimeter is $15\\cdot16+218+218=\\boxed{676}$" ]

[ "Intersect at the origin and select a point on each line to define vectors $\\mathbf{v}_{i}=(x_{i},y_{i})$ .\nNote that $\\theta=45^{\\circ}$ gives equal magnitudes of the vector products \\[\\mathbf{v}_1\\cdot\\mathbf{v}_2 = v_{1}v_{2}\\cos\\theta \\quad\\mathrm{and}\\quad |\\mathbf{v}_1\\times\\mathbf{v}_2| = v_{1}v_{2}\\sin\\theta .\\]\nSubstituting coordinate expressions for vector products, we find \\[\\mathbf{v}_1\\cdot\\mathbf{v}_2 = |\\mathbf{v}_1\\times\\mathbf{v}_2| \\ \\implies\\ x_{1}x_{2}+y_{1}y_{2} = x_{1}y_{2}-x_{2}y_{1} .\\] Divide this equation by $x_{1}x_{2}$ to obtain \\[1+m_{1}m_{2} = m_{2}-m_{1} ,\\] where $m_{i}=y_{i}/x_{i}$ is the slope of line $i$ .\nTaking $m_{2}=6m_{1}$ , we obtain \\[6m_{1}^{2}-5m_{1}+1 = 0 \\ \\implies\\ m_{1} \\in \\{\\frac{1}{3},\\frac{1}{2}\\} .\\] The latter solution gives the largest product of slopes $m_{1}m_{2} = 6m_{1}^2 = \\frac{3}{2} . \\quad \\boxed{32}$", "Place on coordinate plane.\nLines are $y=mx, y=6mx.$ The intersection point at the origin.\nGoes through $(0,0),(1,m),(1,6m),(1,0).$ So by law of sines, $\\frac{5m}{\\sin{45^{\\circ}}} = \\frac{\\sqrt{1+m^2}}{1/(\\sqrt{1+36m^2})},$ lettin $a=m^2$ we want $6a.$ Simplifying gives $50a = (1+a)(1+36a),$ so $36a^2-13a+1=0 \\implies 36(a-1/4)(a-1/9)=0,$ so max $a=1/4,$ and $6a=3/2 \\quad \\boxed{32}.$", "Let one of the lines have equation $y=ax$ . Let $\\theta$ be the angle that line makes with the x-axis, so $\\tan(\\theta)=a$ . The other line will have a slope of $\\tan(45^{\\circ}+\\theta)=\\frac{\\tan(45^{\\circ})+\\tan(\\theta)}{1-\\tan(45^{\\circ})\\tan(\\theta)} = \\frac{1+a}{1-a}$ . Since the slope of one line is $6$ times the other, and $a$ is the smaller slope, $6a = \\frac{1+a}{1-a} \\implies 6a-6a^2=1+a \\implies 6a^2-5a+1=0 \\implies a=\\frac{1}{2},\\frac{1}{3}$ . If $a = \\frac{1}{2}$ , the other line will have slope $\\frac{1+\\frac{1}{2}}{1-\\frac{1}{2}} = 3$ . If $a = \\frac{1}{3}$ , the other line will have slope $\\frac{1+\\frac{1}{3}}{1-\\frac{1}{3}} = 2$ . The first case gives the bigger product of $\\frac{3}{2}$ , so our answer is $\\boxed{32}$", "Let the smaller slope be $m$ , then the larger slope is $6m$ . Since we want the greatest product we begin checking each answer choice, starting with (E).\n$6m^2=6$\n$m^2=1$\nThis gives $m=1$ and $6m=6$ . Checking with a protractor we see that this does not form a 45 degree angle.\nUsing this same method for the other answer choices, we eventually find that the answer is $\\boxed{32}$ since our slopes are $\\frac12$ and $3$ which forms a perfect 45 degree angle.", "If you have this formula memorized then it will be very easy to do: If $\\theta$ is the angle between two lines with slopes $m_1$ and $m_2$ , then $\\tan(\\theta)=\\frac{m_1-m_2}{1+m_1m_2}$\nNow let the smaller slope be $m$ , thus the other slope is $6m$ . Using our formula above: \\[\\tan(45^\\circ)=\\frac{6m-m}{1+6m^2} \\implies 6m^2-5m+1=0 \\implies (2m-1)(3m-1)=0.\\] Therefore the two possible values for $m$ are $\\tfrac{1}{2}$ and $\\tfrac{1}{3}$ . We choose the larger one and thus our answer is \\[6m^2=6 \\cdot \\frac{1}{4} = \\frac{3}{2} \\implies \\boxed{32}.\\]" ]

[ "Set the two numbers as $x$ and $y$ . Therefore, $x+y=7(x-y), xy=24(x-y)$ , and $24(x+y)=7xy$ . Simplifying the first equation gives $y=\\frac{3}{4}x$ . Substituting for $y$ in the second equation gives $\\frac{3}{4}x^2=6x.$ Solving yields $x=8$ or $x=0$ . Substituting $x=0$ back into the first equation yields $1=-7$ which is false, so $x=0$ is not valid and $x=8$ . Substituting into $y=\\frac{3}{4}x$ gives $y=6$ and $xy=\\boxed{48}$" ]

[ "Letting $x$ be the third side, then by the triangle inequality, $20-15 < x < 20+15$ , or $5 < x < 35$ . Therefore the perimeter must be greater than 40 but less than 70. 72 is not in this range, so $\\boxed{72}$ is our answer." ]

[ "Call the two integers $b$ and $b+60$ , so we have $\\sqrt{b}+\\sqrt{b+60}=\\sqrt{c}$ . Square both sides to get $2b+60+2\\sqrt{b^2+60b}=c$ . Thus, $b^2+60b$ must be a square, so we have $b^2+60b=n^2$ , and $(b+n+30)(b-n+30)=900$ . The sum of these two factors is $2b+60$ , so they must both be even. To maximize $b$ , we want to maximixe $b+n+30$ , so we let it equal $450$ and the other factor $2$ , but solving gives $b=196$ , which is already a perfect square, so we have to keep going. In order to keep both factors even, we let the larger one equal $150$ and the other $6$ , which gives $b=48$ . This checks, so the solution is $48+108=\\boxed{156}$" ]

[ "It should first be noted that given any quadrilateral of fixed side lengths, there is exactly one way to manipulate the angles so that the quadrilateral becomes cyclic.\nProof. Given a quadrilateral $ABCD$ where all sides are fixed (in a certain order), we can construct the diagonal $\\overline{BD}$ . When $BD$ is the minimum allowed by the triangle inequality, one of the angles $\\angle DAB$ or $\\angle BCD$ will be degenerate and measure $0^\\circ$ , so opposite angles will sum to less than $180^\\circ$ . When $BD$ is the maximum allowed, one of the angles will be degenerate and measure $180^\\circ$ , so opposite angles will sum to more than $180^\\circ$ . Thus, since the sum of opposite angles increases continuously as $BD$ is lengthened from the minimum to the maximum values, there is a unique value of $BD$ somewhere in the middle such that the sum of opposite angles is exactly $180^\\circ$\nDenote $a$ $b$ $c$ , and $d$ as the integer side lengths of the quadrilateral. Without loss of generality, let $a\\ge b \\ge c \\ge d$\nSince $a+b+c+d = 32$ , the Triangle Inequality implies that $a \\le 15$\nWe will now split into $5$ cases.\nCase $1$ $a = b = c = d$ $4$ side lengths are equal)\nClearly there is only $1$ way to select the side lengths $(8,8,8,8)$ , and no matter how the sides are rearranged only $1$ unique quadrilateral can be formed.\nCase $2$ $a = b = c > d$ or $a > b = c = d$ $3$ side lengths are equal)\nIf $3$ side lengths are equal, then each of those side lengths can only be integers from $6$ to $10$ except for $8$ (because that is counted in the first case). Obviously there is still only $1$ unique quadrilateral that can be formed from one set of side lengths, resulting in a total of $4$ quadrilaterals.\nCase $3$ $a = b > c = d$ $2$ pairs of side lengths are equal)\n$a$ and $b$ can be any integer from $9$ to $15$ , and likewise $c$ and $d$ can be any integer from $1$ to $7$ . However, a single set of side lengths can form $2$ different cyclic quadrilaterals (a rectangle and a kite), so the total number of quadrilaterals for this case is $7\\cdot{2} = 14$\nCase $4$ $a = b > c > d$ or $a > b = c > d$ or $a > b > c = d$ $2$ side lengths are equal)\nIf the $2$ equal side lengths are each $1$ , then the other $2$ sides must each be $15$ , which we have already counted in an earlier case. If the equal side lengths are each $2$ , there is $1$ possible set of side lengths. Likewise, for side lengths of $3$ there are $2$ sets. Continuing this pattern, we find a total of $1+2+3+4+4+5+7+5+4+4+3+2+1 = 45$ sets of side lengths. (Be VERY careful when adding up the total for this case!) For each set of side lengths, there are $3$ possible quadrilaterals that can be formed, so the total number of quadrilaterals for this case is $3\\cdot{45} = 135$\nCase $5$ $a > b > c > d$ (no side lengths are equal)\nUsing the same counting principles starting from $a = 15$ and eventually reaching $a = 9$ , we find that the total number of possible side lengths is $69$ . There are $4!$ ways to arrange the $4$ side lengths, but there is only $1$ unique quadrilateral for $4$ rotations, so the number of quadrilaterals for each set of side lengths is $\\frac{4!}{4} = 6$ . The total number of quadrilaterals is $6\\cdot{69} = 414$\nAnd so, the total number of quadrilaterals that can be made is $414 + 135 + 14 + 4 + 1 = \\boxed{568}$", "As with solution $1$ we would like to note that given any quadrilateral we can change its angles to make a cyclic one.\nLet $a \\ge b \\ge c\\ge d$ be the sides of the quadrilateral.\nThere are $\\binom{31}{3}$ ways to partition $32$ . However, some of these will not be quadrilaterals since they would have one side bigger than the sum of the other three. This occurs when $a \\ge 16$ . For $a=16$ $b+c+d=16$ . There are $\\binom{15}{2}$ ways to partition $16$ . Since $a$ could be any of the four sides, we have counted $4\\binom{15}{2}$ degenerate quadrilaterals. Similarly, there are $4\\binom{14}{2}$ $4\\binom{13}{2} \\cdots 4\\binom{2}{2}$ for other values of $a$ . Thus, there are $\\binom{31}{3} - 4\\left(\\binom{15}{2}+\\binom{14}{2}+\\cdots+\\binom{2}{2}\\right) = \\binom{31}{3} - 4\\binom{16}{3} = 2255$ non-degenerate partitions of $32$ by the hockey stick theorem. We then account for symmetry. If all sides are congruent (meaning the quadrilateral is a square), the quadrilateral will be counted once. If the quadrilateral is a rectangle (and not a square), it will be counted twice. In all other cases, it will be counted 4 times. Since there is $1$ square case, and $7$ rectangle cases, there are $2255-1-2\\cdot7=2240$ quadrilaterals counted 4 times. Thus there are $1+7+\\frac{2240}{4} = \\boxed{568}$ total quadrilaterals.", "As with solution $1$ we find that there are $2255$ ways to form a quadrilateral if we don't account for rotations. We now apply Burnside's_Lemma . There are four types of actions in the group acting on the set of quadrilaterals. We will consider each individually:\nIdentity: maps a quadrilateral with sides $a,b,c,d$ in that order to $a,b,c,d$ . Obviously all members of the set of quadrilaterals are fixed points, for a total of $2255$ .\nRotation by one: maps a quadrilateral from $a,b,c,d$ to $b,c,d,a$ . For this to have a fixed point we need $a=b,b=c,c=d,d=a$ , so the only quadrilateral that is a fixed point is the square with side length $8$ , for a total of $1$ .\nRotation by two: maps a quadrilateral from $a,b,c,d$ to $c,d,a,b$ . For this to be a fixed point we need $a=c$ and $b=d$ . Thus the quadrilateral is of the form $x,y,x,y$ —a rectangle. We can count that there are $15$ rectangle cases, namely $(a, b, c, d) = \\{ (1, 15, 1, 15), (2, 14, 2, 14), \\cdots, (15, 1, 15, 1) \\}$ .\nRoration by three: maps a quadrilateral from $a,b,c,d$ to $d,a,b,c$ . Similarly to the rotation by one case, there is one fixed point here.\nSumming up, we get that the total number of groups is $2255+1+15+1=2272$ . Since there are $4$ members of the group our final answer is $\\frac{2272}{4}=\\boxed{568}$" ]

[ "Noting that $\\angle OQP$ and $\\angle ORP$ are right angles, we realize that we can draw a semicircle with diameter $\\overline{OP}$ and points $Q$ and $R$ on the semicircle. Since the radius of the semicircle is $100$ , if $\\overline{QR} \\leq 100$ , then $\\overarc{QR}$ must be less than or equal to $60^{\\circ}$\nThis simplifies the problem greatly. Since the degree measure of an angle on a circle is simply half the degree measure of its subtended arc, the problem is simply asking:\nGiven $a, b$ such that $0<a, b<75$ , what is the probability that $|a-b| \\leq 30$ ? \nThrough simple geometric probability, we get that $P = \\frac{16}{25}$\nThe answer is $16+25=\\boxed{041}$", "Put $\\triangle POQ$ and $\\triangle POR$ with $O$ on the origin and the triangles on the $1^{st}$ quadrant.\nThe coordinates of $Q$ and $P$ is $(200 \\cos^{2}a,200 \\cos a\\sin a )$ $(200\\cos^{2}b,200\\cos(b)\\sin b)$ . So $PQ^{2}$ $(200 \\cos^{2} a - 200 \\cos^{2} b)^{2} +(200 \\cos a \\sin a - 200 \\cos b \\sin b)^{2}$ , which we want to be less then $100^{2}$ .\nSo $(200 \\cos^{2} a - 200 \\cos^{2} b)^{2} +(200 \\cos a \\sin a - 200 \\cos b \\sin b)^{2} <= 100^{2}$ \\[(\\cos^{2} a - \\cos^{2} b)^{2} +(\\cos a \\sin a - \\cos b \\sin b)^{2} \\le \\frac{1}{4}\\] \\[\\cos^{4} a + \\cos^{4} b - 2\\cos^{2} a \\cos^{2} b +\\cos^{2}a \\sin^{2} a + \\cos^{2} b \\sin^{2} b - 2 \\cos a \\sin a \\cos b \\sin b \\le \\frac{1}{4}\\] \\[\\cos^{2} a(\\cos^{2} a + \\sin^{2} a)+\\cos^{2} b(\\cos^{2} b+\\sin^{2} b) - 2\\cos^{2} a \\cos^{2} b- 2 \\cos a \\sin a \\cos b \\sin b \\le \\frac{1}{4}\\] \\[\\cos^{2} a(1-\\cos^{2} b)+\\cos^{2} b(1-\\cos^{2} a) - 2 \\cos a \\sin a \\cos b \\sin b \\le \\frac{1}{4}\\] \\[(\\cos a\\sin b)^{2} +(\\cos b\\sin a)^{2} - 2 (\\cos a \\sin b)(\\cos b \\sin a)\\le \\frac{1}{4}\\] \\[(\\cos a\\sin b-\\cos b\\sin a)^{2}\\le \\frac{1}{4}\\] \\[\\sin^{2} (b-a) \\le \\frac{1}{4}\\] So we want $-\\frac{1}{2} \\le \\sin (b-a) \\le \\frac{1}{2}$ , which is equivalent to $-30 \\le b-a \\le 30$ or $150 \\le b-a \\le 210$ . The second inequality is impossible so we only consider what the first inequality does to our $75$ by $75$ box in the $ab$ plane. This cuts off two isosceles right triangles from opposite corners with side lengths $45$ from the $75$ by $75$ box. Hence the probability is $1-\\frac{45^2}{75^2} = 1- \\frac{9}{25}=\\frac{16}{25}$ and the answer is $16+25 = \\boxed{41}$", " Let $QR=x.$ Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: $OQ = 200 \\cos a, PQ = 200 \\sin a, PR = 200 \\sin b, OR = 200 \\cos b.$ Now observe that quadrilateral $OQRP$ is a cyclic quadrilateral . Thus, we are able to apply Ptolemy's Theorem to it: \\[200 x + (200 \\cos a) (200 \\sin b) = (200 \\sin a) (200 \\cos b),\\] \\[x + 200 (\\cos a \\sin b) = 200 (\\sin a \\cos b),\\] \\[x = 200(\\sin a \\cos b - \\sin b \\cos a),\\] \\[x = 200 \\sin(a-b).\\] We want $|x| \\le 100$ (the absolute value comes from the fact that $a$ is not necessarily greater than $b,$ so we cannot assume that $Q$ is to the right of $R$ as in the diagram), so we substitute: \\[|200 \\sin(a-b)| \\le 100,\\] \\[|\\sin(a-b)| \\le \\frac{1}{2},\\] \\[|a-b| \\le 30 ^\\circ,\\] \\[-30 \\le a-b \\le 30.\\] By simple geometric probability (see Solution 2 for complete explanation), $\\frac{m}{n} = 1 - \\frac{2025}{5625} = 1 - \\frac{9}{25} = \\frac{16}{25},$ so $m+n = \\boxed{041}.$", "Impose a coordinate system as follows:\nLet the midpoint of $\\overline{OP}$ be the origin, and let $\\overline{OP}$ be the x-axis. We construct a circle with center at the origin with radius 100. Since $\\angle OQP$ and $\\angle ORP$ are both right angles, points $Q$ and $R$ are on our circle. Place $Q$ and $R$ in the first quadrant of the Cartesian Plane. Suppose we construct $Q'$ and $R'$ such that they are clockwise rotations of $Q$ and $R$ , respectively by an angle of $2b$ degrees. Thus, we see that $\\overline{QR}=100\\sqrt{2}\\sqrt{\\cos(2|a-b|)}$ . We want this quantity to be less than $100$ . This happens when $\\cos(2|a-b|) \\ge 1/2,$ or when $|a-b|\\le 30^{\\circ}$ . The probability that the last inequality is satisfied is $16/25$ . Therefore, the probability that $QR$ is less than $100$ is $16/25$ . Hence, $m+n=\\boxed{41}$", "WLOG, let $b\\ge a$ . It does not actually matter, but it is necessary for this particular setup. It should be apparent that $\\Delta RAQ\\sim\\Delta OAP$ . We write the equation\n\\[\\dfrac{RA}{AO}=\\dfrac{RQ}{OP}.\\]\nIf we examine right triangle $\\Delta ROA$ , we can see that $\\sin(b-a)=\\dfrac{RA}{AO}$ . Also, we are given $OP=200$ , so now we have\n\\[\\sin(b-a)=\\dfrac{QR}{200}.\\]\nWe want $QR$ to be less than or equal to $100$ ; this is equivalent to $\\dfrac{QR}{200}\\le\\dfrac12.$ We solve from there:\n\\begin{align*}\n\\dfrac{QR}{200}&\\le\\dfrac12 \\\\\n\\sin(b-a)&\\le\\dfrac12 \\\\\n\\arcsin(\\sin(b-a))&\\le\\arcsin\\left(\\dfrac12\\right) \\\\\nb-a&\\le30^\\circ. \\\\\n\\end{align*}\n(Notice that if $a>b$ , then this would become $a-b\\le30^circ.$ As in Solution 1, we can write $|a-b|\\le30$ .) One can now proceed as in Solution 1, but let us tackle the geometric probability for completeness.\nWe now have transformed this problem into another problem asking for the probability of two uniformly, randomly, and independently chosen real numbers between $0$ and $75$ being no more than $30$ from each other.\nIf the first number (let this be $x$ ) is between $30$ and $45$ , then the other number can be from $x-30$ to $x+30$ - a range of $60$ . Thus, the probability that this contributes is $\\dfrac{45-30}{75}\\cdot\\dfrac{60}{75}=\\dfrac4{25}$\nIf $x$ is between $0$ and $30$ or $45$ to $75$ (these two cases are equivalent), the chance is the same as that of the average value since the ranges are uniform. For $x=15$ (the average), the second number can be from $0$ to $15+30=45$ - a range of $45$ . The total range is $(30-0)+(75-45)=30+30=60$ . Thus, this case contributes $\\dfrac{45}{75}\\cdot\\dfrac{60}{75}=\\dfrac{12}{25}$\nAdding the two, we get $\\dfrac{16}{25}$ for an answer of $16+25=\\boxed{041}$" ]

[ "We will use complementary counting. The probability that the product is negative can be found by finding the probability that one number is positive and the other number is negative. The probability of a positive number being selected is $\\frac13$ , and the probability of a negative number being selected is $\\frac23$ . So, we see that the probability of the product being negative is \\[2 \\cdot \\frac23 \\cdot \\frac13 = \\frac49.\\]\nWe multiply by $2$ because you can either pick the negative or positive number first.\nThus, the probability of the product being positive is \\[1-\\frac49 = \\boxed{59}\\]" ]

[ "The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii $3$ and $6$ and infinitely large height. Then the base area of the wide cylinder is $4$ times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise $\\boxed{4}$ times as much." ]

[ "The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii $3$ and $6$ and infinitely large height. Then the base area of the wide cylinder is $4$ times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise $\\boxed{4}$ times as much." ]

[ "Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$ . Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$ . We are told \\[r_2=\\frac{11r_1}{10}\\] \\[\\pi r_1^2h_1=\\pi r_2^2h_2\\] Substituting the first equation into the second and dividing both sides by $\\pi$ , we get \\[r_1^2h_1=\\frac{121r_1^2}{100}h_2\\implies h_1=\\frac{121h_2}{100}.\\] Therefore, $\\boxed{21}$" ]

[ "Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$ . Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$ . We are told \\[r_2=\\frac{11r_1}{10}\\] \\[\\pi r_1^2h_1=\\pi r_2^2h_2\\] Substituting the first equation into the second and dividing both sides by $\\pi$ , we get \\[r_1^2h_1=\\frac{121r_1^2}{100}h_2\\implies h_1=\\frac{121h_2}{100}.\\] Therefore, $\\boxed{21}$" ]

[ "The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes,\ntherefore the length of the side perpendicular to that altitude will be between $10$ and $15$ . The only answer choice that meets this requirement is $\\boxed{12}$", "Let the height to the side of length $15$ be $h_{1}$ , the height to the side of length 10 be $h_{2}$ , the area be $A$ , and the height to the unknown side be $h_{3}$\nBecause the area of a triangle is $\\frac{bh}{2}$ , we get that $15(h_{1}) = 2A$ and $10(h_{2}) = 2A$ , so, setting them equal, $h_{2} = \\frac{3h_{1}}{2}$ . From the problem, we know that $2h_{3} = h_{1} + h_{2}$ . Substituting, we get that \\[h_{3} = 1.25h_{1}.\\] Thus, the side length is going to be $\\frac{2A}{1.25h_{1}} = \\frac{15}{\\frac{5}{4}} = \\boxed{12}$" ]

[ "Let $x$ be the height of triangle when the base is $10$ and $y$ is the height of the triangles when the base is $15$ . This means the height for when the triangles has the $3$ rd side length, the height would be $\\dfrac{(x+y)}{2}$ giving us the following equation:\n$\\dfrac{10x}{2}=\\dfrac{15y}{2}=\\dfrac{n(x+y)}{2}$ For $n$ , we can plug in the answer choices and check when the ratio of $\\dfrac{x}{y}$ for $\\dfrac{10x}{2}=\\dfrac{n(x+y)}{2}$ and $\\dfrac{15y}{2}=\\dfrac{n(x+y)}{2}$ is the same because we can observe that only for one value, the ratio remains constant. From brute force we get that $\\boxed{12}$ meaning that $12$ is the $3$ rd base.\n~ Batmanstark" ]

[ "Label the point of intersection as $C$ . Since $d = rt$ $AC = 8t$ and $BC = 7t$ . According to the law of cosines\n\\begin{align*}(7t)^2 &= (8t)^2 + 100^2 - 2 \\cdot 8t \\cdot 100 \\cdot \\cos 60^\\circ\\\\ 0 &= 15t^2 - 800t + 10000 = 3t^2 - 160t + 2000\\\\ t &= \\frac{160 \\pm \\sqrt{160^2 - 4\\cdot 3 \\cdot 2000}}{6} = 20, \\frac{100}{3}.\\end{align*}\nSince we are looking for the earliest possible intersection, $20$ seconds are needed. Thus, $8 \\cdot 20 = \\boxed{160}$ meters is the solution.", "Let $P$ be the point of intersection between the skaters, Allie and Billie. We can draw a line that goes through $P$ and is parallel to $\\overline{AB}$ . Letting this line be the $x$ -axis, we can reflect $B$ over the $x$ -axis to get $B'$ . As reflections preserve length, $B'X = XB$\nWe then draw lines $BB'$ and $PB'$ . We can let the foot of the perpendicular from $P$ to $BB'$ be $X$ , and we can let the foot of the perpendicular from $P$ to $AB$ be $Y$ . In doing so, we have constructed rectangle $PXBY$\nBy $d=rt$ , we have $AP = 8t$ and $PB = PB' = 7t$ , where $t$ is the number of seconds it takes the skaters to meet. Furthermore, we have $30-60-90$ triangle $PAY$ , so $AY = 4t$ , and $PY = 4t\\sqrt{3}$ . Since we have $PY = XB = B'X$ $B'X = 4t\\sqrt{3}$ . By Pythagoras, $PX = t$\nAs $PXBY$ is a rectangle, $PX = YB$ . Thus $AY + YB = AB \\Rightarrow AY + PX = AB$ , so we get $4t + t = 100$ . Solving for $t$ , we find $t = 20$\nOur answer, $AP$ , is equivalent to $8t$ . Thus, $AP = 8 \\cdot 20 = \\boxed{160}$", "We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$ . Then we can produce the following diagram:\n\nNow, if we drop an altitude from point $M$ , we get :\n\nWe know this from the $30-60-90$ triangle that is formed. From this we get that: \\[(7x)^2 = (4 \\sqrt{3} x)^2 + (100-4x)^2\\]\n\\[\\Longrightarrow 49x^2 - 48x^2 = x^2 = (100-4x)^2\\]\n\\[\\Longrightarrow 0=(100-4x)^2 - x^2 = (100-3x)(100-5x)\\]\nTherefore, we get that $x = \\frac{100}{3}$ or $x = 20$ . Since $20< \\frac{100}{3}$ , we have that $x=20$ (since the problem asks for the quickest possible meeting point), so the distance Allie travels before meeting Billie would be $8 \\cdot x = 8 \\cdot 20 = \\boxed{160}$ meters." ]

[ "This solution refers to the Diagram section.\nAs shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ has radius $13$ ) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\\mathcal{P}.$ Let $\\mathcal{R}$ be the plane that is determined by $O_1,O_2,$ and $O_3.$ Suppose $A$ is the foot of the perpendicular from $O_3$ to line $\\ell,$ so $\\overleftrightarrow{O_3A}$ is the perpendicular bisector of $\\overline{O_1O_2}.$ We wish to find $T_3A.$ Note that:\nNow, we focus on cross-sections $O_1O_3T_3T_1$ and $\\mathcal{R}:$\nWe have the following diagram: In cross-section $O_1O_3T_3T_1,$ since $\\overline{O_1T_1}\\parallel\\overline{O_3T_3}$ as discussed, we obtain $\\triangle O_1T_1B\\sim\\triangle O_3T_3B$ by AA, with the ratio of similitude $\\frac{O_1T_1}{O_3T_3}=\\frac{36}{13}.$ Therefore, we get $\\frac{O_1B}{O_3B}=\\frac{49+O_3B}{O_3B}=\\frac{36}{13},$ or $O_3B=\\frac{637}{23}.$\nIn cross-section $\\mathcal{R},$ note that $O_1O_3=49$ and $DO_3=\\frac{O_1O_2}{2}=36.$ Applying the Pythagorean Theorem to right $\\triangle O_1DO_3,$ we have $O_1D=\\sqrt{1105}.$ Moreover, since $\\ell\\perp\\overline{O_1C}$ and $\\overline{DO_3}\\perp\\overline{O_1C},$ we obtain $\\ell\\parallel\\overline{DO_3}$ so that $\\triangle O_1CB\\sim\\triangle O_1DO_3$ by AA, with the ratio of similitude $\\frac{O_1B}{O_1O_3}=\\frac{49+\\frac{637}{23}}{49}.$ Therefore, we get $\\frac{O_1C}{O_1D}=\\frac{\\sqrt{1105}+DC}{\\sqrt{1105}}=\\frac{49+\\frac{637}{23}}{49},$ or $DC=\\frac{13\\sqrt{1105}}{23}.$\nFinally, note that $\\overline{O_3T_3}\\perp\\overline{T_3A}$ and $O_3T_3=13.$ Since quadrilateral $DCAO_3$ is a rectangle, we have $O_3A=DC=\\frac{13\\sqrt{1105}}{23}.$ Applying the Pythagorean Theorem to right $\\triangle O_3T_3A$ gives $T_3A=\\frac{312}{23},$ from which the answer is $312+23=\\boxed{335}.$", "The centers of the three spheres form a $49$ $49$ $72$ triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the $72$ side of this triangle. Take its midpoint $M$ , which is $36$ away from the midpoint $A$ of the $72$ side, and connect these two midpoints.\nNow consider the point at which the plane is tangent to the small sphere, and connect $M$ with the small sphere's tangent point $B$ . Extend $\\overline{MB}$ through $B$ until it hits the ray from $A$ through the center of the small sphere (convince yourself that these two intersect). Call this intersection $D$ , the center of the small sphere $C$ , we want to find $BD$\nBy Pythagoras, $AC=\\sqrt{49^2-36^2}=\\sqrt{1105}$ , and we know that $MA=36$ and $BC=13$ . We know that $\\overline{MA}$ and $\\overline{BC}$ must be parallel, using ratios we realize that $CD=\\frac{13}{23}\\sqrt{1105}$ . Apply the Pythagorean theorem to $\\triangle BCD$ $BD=\\frac{312}{23}$ , so $312 + 23 = \\boxed{335}$", "This solution refers to the Diagram section. The isosceles triangle of centers $O_1 O_2 O$ $O$ is the center of sphere of radii $13$ ) has sides $O_1 O = O_2 O = 36 + 13 = 49,$ and $O_1 O_2 = 36 + 36 = 72.$\nLet $N$ be the midpoint $O_1 O_2$\nThe isosceles triangle of points of tangency $T_1 T_2 T$ has sides $T_1 T = T_2 T = 2 \\sqrt{13 \\cdot 36} = 12 \\sqrt{13}$ and $T_1 T_2 = 72.$\nLet $M$ be the midpoint $T_1 T_2.$\nThe height $TM$ is $\\sqrt {12^2 \\cdot 13 - 36^2} = 12 \\sqrt {13-9} = 24.$\nThe tangents of the half-angle between the planes is $\\frac {TO}{AT} = \\frac {MN - TO}{TM},$ so $\\frac {13}{AT} = \\frac {36 - 13}{24},$ \\[AT = \\frac{24\\cdot 13}{23} = \\frac {312}{23} \\implies 312 + 23 = \\boxed{335}.\\] vladimir.shelomovskii@gmail.com, vvsss" ]

[ "There are ${49 \\choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards.\nNote that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $\\frac{49-1}{2}=24$ such pairs. There are then ${49 \\choose 2}-24$ pairs that yield $4$ distinct boards upon rotation; in other words, for each of the ${49 \\choose 2}-24$ pairs, there are three other pairs that yield an equivalent board.\nThus, the number of inequivalent boards is $\\frac{{49 \\choose 2} - 24}{4} + \\frac{24}{2} = \\boxed{300}$ . For a $(2n+1) \\times (2n+1)$ board, this argument generalizes to $n(n+1)(2n^2+2n+1)$ inequivalent configurations.", "There are 4 cases: \n1. The center square is occupied, in which there are $12$ cases.\n2. The center square isn't occupied and the two squares that are opposite to each other with respect to the center square, in which there are $12$ cases.\n3. The center square isn't occupied and the two squares can rotate to each other with a $90^{\\circ}$ rotation with each other and with respect to the center square, in which case there are $12$ cases.\n4. And finally, the two squares can't rotate to each other with respect to the center square in which case there are $\\dbinom{12}{2} \\cdot \\frac{16}{4} = 264$ cases. \nAdd up all the values for each case to get $\\boxed{300}$ as your answer." ]

[ "First, choose the two letters to be repeated in each set. $\\dbinom{5}{2}=10$ . Now we have three remaining elements that we wish to place into two separate subsets. There are $2^3 = 8$ ways to do so because each of the three remaining letters can be placed either into the first or second subset. Both of those subsets contain the two chosen elements, so their intersection is the two chosen elements). Unfortunately, we have over-counted (Take for example $S_{1} = \\{a,b,c,d \\}$ and $S_{2} = \\{a,b,e \\}$ ). Notice how $S_{1}$ and $S_{2}$ are interchangeable. Division by two will fix this problem. Thus we have:\n$\\dfrac{10 \\times 8}{2} = 40 \\implies \\boxed{40}$", "Another way of looking at this problem is to break it down into cases.\nFirst, our two subsets can have 2 and 5 elements. The 5-element subset (aka the set) will contain the 2-element subset. There are $\\dbinom{5}{2}=10$ ways to choose the 2-element subset. Thus, there are $10$ ways to create these sets.\nSecond, the subsets can have 3 and 4 elements. $3+4=7$ non-distinct elements. $7-5=2$ elements in the intersection. There are $\\dbinom{5}{3}=10$ ways to choose the 3-element subset. For the 4-element subset, two of the elements must be the remaining elements (not in the 3-element subset). The other two have to be a subset of the 3-element subset. There are $\\dbinom{3}{2}=3$ ways to choose these two elements, which means there are 3 ways to choose the 4-element subset. Therefore, there are $10\\cdot3=30$ ways to choose these sets.\nThis leads us to the answer:\n$10+30=40 \\implies \\boxed{40}$", "There are $\\dbinom{5}{2}=10$ ways to choose the 2 shared elements. We now must place the 3 remaining elements into the subsets. Using stars and bars, we can notate this as: $I I I X \\rightarrow \\dbinom{4}{3}=4$ . Thus, $4*10=\\boxed{40}$" ]

[ "First, note that it will take $30$ seconds for the first swimmer to reach the other side and $45$ seconds for the second swimmer to reach the other side. Also, note that after $180$ seconds (or $3$ minutes), both swimmers will complete an even number of laps, essentially returning to their starting point.\n\nAt this point, find the number of meeting points in the first $3$ minutes, then multiply by four to get the answer. From the graph (where the x-axis is the time in seconds and the y-axis is distance from one side of the pool), there are five meeting points, so the two swimmers will pass each other $20$ times, which is answer choice $\\boxed{20}$" ]

[ "Since $Q$ can be anywhere on the circle between $A$ and $B$ , it can basically be \"on top\" of $A$ . Then $R$ will be at the same point as $A$ , so $APR$ form a degenerate triable with side length $AB=20$ . So its perimeter will be $40$ . Since $BP=PQ$ and $QR=CR$ by power of a point, as $AP$ and $AR$ decrease in length, $PR=PQ+QR$ will \"grow\" to compensate, so the perimeter will stay constant with a value of $\\boxed{40}$" ]

[ "In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).\nIn order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d.\nSetting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.\nFinally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:\n1/2 (216°-144°) = 1/2 (72°) $=\\boxed{36}.$", "Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is $\\frac{3x-2x}2 = \\frac {x} 2$ by the arc length formula. Also note that $3x + 2x = 360$ , which simplifies to $x= 72.$ Hence the angle formed by the tangents is equal to $\\boxed{36}$", "Let the center of the circle be $O$ . The radii extending from $O$ to the points of tangency $B$ and $C$ are both perpendicular to lines $AB$ and $AC$ . Thus $\\angle ABO = \\angle ACO = 90^{\\circ}$ , and quadrilateral $ABOC$ is cyclic, since the opposite angles add to $180^{\\circ}$\nSince the ratio of the arc lengths is $2 : 3$ , the shorter arc (the arc cut off by $\\angle BOC$ ) is $144^{\\circ}$\n$\\angle BOC$ and $\\angle BAC$ must add up to $180^{\\circ}$ since quadrilateral $ABOC$ is cyclic, so $\\angle BAC = 180 - 144 = \\boxed{36}$" ]

[ "The label $1993$ will occur on the $\\frac12(1993)(1994) \\pmod{2000}$ th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $\\frac12(n)(n + 1)\\equiv \\frac12(1993)(1994) \\pmod{2000}$\nSimplifying this expression, we see that $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\\equiv 0\\pmod{2000}$ . Therefore, one of $1993 - n$ or $1994 + n$ is odd, and each of them must be a multiple of $125$ or $16$\nFor $1993 - n$ to be a multiple of $125$ and $1994 + n$ to be a multiple of $16$ $n \\equiv 118 \\pmod {125}$ and $n\\equiv 6 \\pmod {16}$ . The smallest $n$ for this case is $118$\nIn order for $1993 - n$ to be a multiple of $16$ and $1994 + n$ to be a multiple of $125$ $n\\equiv 9\\pmod{16}$ and $n\\equiv 6\\pmod{125}$ . The smallest $n$ for this case is larger than $118$ , so $\\boxed{118}$ is our answer.", "Two labels $a$ and $b$ occur on the same point if $\\ a(a+1)/2\\equiv \\ b(b+1)/2\\pmod{2000}$ . If we assume the final answer be $n$ , then we have $\\frac12(n)(n + 1)\\equiv \\frac12(1993)(1994) \\pmod{2000}$\nMultiply $2$ on both side we have $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\\equiv 0\\pmod{4000}$ . As they have different parities, the even one must be divisible by $32$ . As $(1993 - n)+(1994 + n)\\equiv 2\\pmod{5}$ , one of them is divisible by $5$ , which indicates it's divisible by $125$\nWhich leads to four different cases: $1993-n\\equiv 0\\pmod{4000}$ $1994+n\\equiv 0\\pmod{4000}$ $1993-n\\equiv 0\\pmod{32}$ and $1994+n\\equiv 0\\pmod{125}$ $1993-n\\equiv 0\\pmod{125}$ and $1994+n\\equiv 0\\pmod{32}$ . Which leads to $n\\equiv 1993,2006,3881$ and $118\\pmod{4000}$ respectively, and only $n=118$ satisfied.Therefore answer is $\\boxed{118}$ .(by ZJY)" ]

[ "Let us make a chart of values in alphabetical order, where $P_a,\\ P_b$ are the probabilities that each string comes from $aaa$ and $bbb$ multiplied by $27$ , and $S_b$ denotes the partial sums of $P_b$ (in other words, $S_b = \\sum_{n=1}^{b} P_b$ ): \\[\\begin{array}{|r||r|r|r|} \\hline \\text{String}&P_a&P_b&S_b\\\\ \\hline aaa & 8 & 1 & 1 \\\\ aab & 4 & 2 & 3 \\\\ aba & 4 & 2 & 5 \\\\ abb & 2 & 4 & 9 \\\\ baa & 4 & 2 & 11 \\\\ bab & 2 & 4 & 15 \\\\ bba & 2 & 4 & 19 \\\\ bbb & 1 & 8 & 27 \\\\ \\hline \\end{array}\\]\nThe probability is $p=\\sum P_a \\cdot (27 - S_b)$ , so the answer turns out to be $\\frac{8\\cdot 26 + 4 \\cdot 24 \\ldots 2 \\cdot 8 + 1 \\cdot 0}{27^2} = \\frac{532}{729}$ , and the solution is $\\boxed{532}$", "Let $S(a,n)$ be the $n$ th letter of string $S(a)$ .\nCompare the first letter of the string $S(a)$ to the first letter of the string $S(b)$ .\nThere is a $(2/3)^2=4/9$ chance that $S(a,1)$ comes before $S(b,1)$ .\nThere is a $2(1/3)(2/3)=4/9$ that $S(a,1)$ is the same as $S(b,1)$\nIf $S(a,1)=S(b,1)$ , then you do the same for the second letters of the strings. But you have to multiply the $4/9$ chance that $S(a,2)$ comes before $S(b,2)$ as there is a $4/9$ chance we will get to this step.\nSimilarly, if $S(a,2)=S(b,2)$ , then there is a $(4/9)^3$ chance that we will get to comparing the third letters and that $S(a)$ comes before $S(b)$\nSo we have $p=(4/9)+(4/9)^2+(4/9)^3=4/9+16/81+64/729=532/729$ . Therefore, the answer is $\\boxed{532}$", "Consider $n$ letter strings instead. If the first letters all get transmitted correctly, then the $a$ string will be first. Otherwise, the only way is for both of the first letters to be the same, and then we consider the next $n-1$ letter string following the first letter. This easily leads to a recursion: $p_n=\\frac23\\cdot\\frac23+2\\cdot\\frac23\\cdot\\frac13p_{n-1}=\\frac49+\\frac49p_{n-1}$ . Clearly, $p_0=0\\implies p_1=\\frac49\\implies p_2=\\frac{52}{81}\\implies p_3=\\frac{532}{729}$ . Therefore, the answer is $\\boxed{532}$", "The probability that $S_a$ will take the form $a$ _ _ and that $S_b$ will take the form $b$ _ _ is $\\frac{2}{3}\\cdot\\frac{2}{3} = \\frac{4}{9}$ . Then, the probability that both $S_a$ and $S_b$ will share the same first digit is $2\\cdot\\frac{2}{3}\\cdot\\frac{1}{3} = \\frac{4}{9}$ . Now if the first digits of either sequence are the same, then we must now consider these same probabilities for the second letter of each sequence. The probability that when the first two letters of both sequences are the same, that the second letter of $S_a$ is $a$ and that the second letter of $S_b$ is $b$ is $\\frac{2}{3}\\cdot\\frac{2}{3}=\\frac{4}{9}$ . Similarly, the probability that when the first two letters of both sequences are the same, that the second set of letters in both sets of sequences are the same is $2\\cdot\\frac{2}{3}\\cdot\\frac{1}{3} = \\frac{4}{9}$ . Now, if the last case is true then the probability that $S_a$ precedes $S_b$ is $\\frac{2}{3}\\cdot\\frac{2}{3} = \\frac{4}{9}$ . Therefore the total probability would be: $\\frac{4}{9} + \\frac{4}{9}\\left(\\frac{4}{9} + \\frac{4}{9}\\left(\\frac{4}{9}\\right)\\right) = \\frac{4}{9}+\\frac{4}{9}\\left(\\frac{52}{81}\\right) = \\frac{4}{9} + \\frac{208}{729} = \\frac{532}{729}$ . Therefore the answer is $\\boxed{532}$" ]

[ "Call the given grid \"Grid A\". Consider Grid B, where the vertices of Grid B fall in the centers of the squares of Grid A; thus, Grid B has dimensions $(n-1) \\times (n-1)$ . There is a one-to-one correspondence between the edges of Grid B and the number of adjacent pairs of unit squares in Grid A. The number of edges in Grid B is $2n(n-1)$ , and the number of ways to pick two squares out of Grid A is $\\dbinom{n^2}{2}$ . So, the probability that the two chosen squares are adjacent is $\\frac{2n(n-1)}{\\binom{n^2}{2}} = \\frac{2n(n-1)}{\\frac{n^2(n^2-1)}{2}} = \\frac{4}{n(n+1)}$ . We wish to find the smallest positive integer $n$ such that $\\frac{4}{n(n+1)} < \\frac{1}{2015}$ , and by inspection the first such $n$ is $\\boxed{090}$", "Consider a $3 \\times 3$ grid, where there are $4$ corner squares, $4$ edge squares, and $1$ center square. A $4 \\times 4$ grid has $4$ corner squares, $8$ edge squares, and $4$ center squares. By examining simple cases, we can conclude that for a grid that is $n \\times n$ , there are always $4$ corners squares, $4(n-2)$ edge squares, and $n^2-4n+4$ center squares.\nEach corner square is adjacent to $2$ other squares, edge squares to $3$ other squares, and center squares to $4$ other squares. In the problem, the second square can be any square that is not the first, which means there are $n^2-1$ to choose from. With this information, we can conclude that the probability that second unit square is adjacent to the first is $\\frac{2}{n^2-1}(\\frac{4}{n^2}) +\\frac{3}{n^2-1}(\\frac{4(n-2)}{n^2}) +\\frac{4}{n^2-1}(\\frac{n^2-4n+4}{n^2})$\nSimplifying, we get $\\frac{4}{n(n+1)}$ which we can set to be less than $\\frac{1}{2015}$ . By inspection, we find that the first such $n$ is $\\boxed{090}$", "There are 3 cases in this problem. Number one, the center squares. Two, the edges, and three, the corners. The probability of getting one center square and an adjacent square is $\\frac{(n-2)(n-2)}{n^2}$ multiplied by $\\frac{4}{n^2 -1}$ . Add that to the probability of an edge and an adjacent square( $\\frac{4n-8}{n^2}$ multiplied by $\\frac{3}{n^2-1}$ ) and the probability of a corner and an adjacent square( $\\frac{4}{n^2}$ multiplied by $\\frac{2}{n^2-1}$ ) to get $\\frac{4n^2-4n}{n^4-n^2}$ . Simplify to get $\\frac{4}{n^2+n}$ . With some experimentation, we realize that the smallest value of n is $\\boxed{090}$" ]

[ "We can use the formula for the diagonal of the rectangle, or $d=\\sqrt{a^2+b^2+c^2}$ The problem gives us $a=1, b=8,$ and $c=9.$ Solving gives us $9=\\sqrt{1^2 + 8^2 + c^2} \\implies c^2=9^2-8^2-1^2 \\implies c^2=16 \\implies c=\\boxed{4}.$" ]

[ "This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.\nThe first statement can be written as $p-2=3(c-2)$ . The second statement can be written as $p-4=4(c-4)$\nTo solve the system of equations:\n$p=3c-4$\n$p=4c-12$\n$3c-4=4c-12$\n$c=8$\n$p=20.$\nLet $x$ be the number of years until Pete is twice as old as his cousin.\n$20+x=2(8+x)$\n$20+x=16+2x$\n$x=4$\nThe answer is $\\boxed{4}$" ]

[ "This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.\nThe first statement can be written as $p-2=3(c-2)$ . The second statement can be written as $p-4=4(c-4)$\nTo solve the system of equations:\n$p=3c-4$\n$p=4c-12$\n$3c-4=4c-12$\n$c=8$\n$p=20.$\nLet $x$ be the number of years until Pete is twice as old as his cousin.\n$20+x=2(8+x)$\n$20+x=16+2x$\n$x=4$\nThe answer is $\\boxed{4}$" ]

[ "Working backwards, if $3/4$ of the chairs are taken and $6$ are empty, then there are three times as many taken chairs as empty chairs, or $3 \\cdot 6 = 18$ . If $x$ is the number of people in the room and $2/3$ are seated, then $\\frac23 x = 18$ and $x = \\boxed{27}$" ]

[ "There are $3$ possibilities for the meat and $4$ possibilites for the dessert, for a total of $4\\times3=12$ possibilities for the meat and the dessert. There are $4$ possibilities for the first vegetable and $3$ possibilities for the second, but order doesn't matter, so we overcounted by a factor of $2$ . For example, we counted 'baked beans and corn' and 'corn and baked beans' as $2$ different possibilities, so the total possibilites for the two vegetables is $\\frac{4\\times3}{2}=6$ , and the total number of possibilites is $12\\times6=\\boxed{72}.$" ]

[ "Let $x$ be the number of marbles Tyrone gave to Eric. Then, $97-x = 2\\cdot(11+x)$ . Solving for $x$ yields $75=3x$ and $x = 25$ . The answer is $\\boxed{25}$", "Since the number of balls Tyrone and Eric have a specific ratio when Tyrone gives some of his balls to Eric, we can divide the total amount of balls by their respective ratios. Altogether, they have $97+11$ balls, or $108$ balls, and it does not change when Tyrone gives some of his to Eric, since none are lost in the process. After Tyrone gives some of his balls away, the ratio between the balls is $2:1$ , and added together is $108$ . The two numbers add up to $3$ and we divide $108$ by $3$ , and the outcome is $36$ . This is the number of balls Eric has, and so doubling that results in the number of balls Tyrone has, which is $72$ $97-72$ is $25$ , and $36-11$ is $25$ , thus proving our statement true, and the answer is $\\boxed{25}$" ]

[ "Drop an altitude from $P$ to $AB$ and let its base be $x$ . Note that if we repeat this for $Q$ and $R$ , all four right triangles (including $\\triangle{RSC}$ ) will have the same trig ratios. By proportion, the hypotenuse $AP$ is $\\frac{x}{100}(120) = \\frac65 x$ , so $\\cos\\theta = \\frac{x}{(\\frac65x)} = \\frac56 \\Rightarrow \\theta = \\boxed{56}$" ]

[ "Let $h$ $l$ , and $w$ represent the height of the table and the length and width of the wood blocks, respectively, in inches. From Figure 1, we have $l+h-w=32$ , and from Figure 2, $w+h-l=28$ . Adding the equations gives $2h=60 \\implies h=30$ , which is $\\boxed{30}$" ]

[ "You need $2$ dimes, $1$ nickel, and $4$ pennies for the first $25$ cents. From $26$ cents to $50$ cents, you only need to add $1$ quarter. From $51$ cents to $75$ cents, you also only need to add $1$ quarter. The same for $76$ cents to $99$ cents. Notice that instead of $100$ , it is $99$ . We are left with $3$ quarters, $1$ nickel, $2$ dimes, and $4$ pennies. Thus, the correct answer is $3+2+1+4=\\boxed{10}$" ]

[ "There is 1 way to get from C to N. There is only one way to get from D to N, which is DCN.\nSince A can only go to C or D, which each only have 1 way to get to N each, there are $1+1=2$ ways to get from A to N.\nSince B can only go to A, C or N, and A only has 2 ways to get to N, C only has 1 way and to get from B to N is only 1 way, there are $2+1+1=4$ ways to get from B to N.\nM can only go to either B or A, A has 2 ways and B has 4 ways, so M has $4+2=6$ ways to get to N.\n6 is $\\boxed{6}$" ]

[ "Since a multiple-digit prime number is not divisible by either 2 or 5, it must end with 1, 3, 7, or 9 in the units place. The remaining digits given must therefore appear in the tens place. Hence our answer is $20 + 40 + 50 + 60 + 1 + 3 + 7 + 9 = 190\\Rightarrow\\boxed{190}$" ]

[ "There are $4!\\cdot 4$ \"words\" beginning with each of the first four letters alphabetically. From there, there are $3!\\cdot 3$ with $U$ as the first letter and each of the first three letters alphabetically. After that, the next \"word\" is $USAMO$ , hence our answer is $4\\cdot 4!+3\\cdot 3!+1=\\boxed{115}$", "Let $A = 1$ $M = 2$ $O = 3$ $S = 4$ , and $U = 5$ . Then counting backwards, $54321, 54312, 54231, 54213, 54132, 54123$ , so the answer is $\\boxed{115}$" ]

[ "Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. Now assume that the first equation is above the second, since order doesn't matter. Then $C>A$ and $B>D$ . Therefore the number of ways to choose the four integers is $\\tbinom{6}{2}\\tbinom{4}{2}=90$ , and the answer is $\\boxed{90}$", "Setting $y = Ax^2+B = Cx^2+D$ , we find that $Ax^2-Cx^2 = x^2(A-C) = D-B$ , so $x^2 = \\frac {D-B}{A-C} \\ge 0$ by the trivial inequality. This implies that $D-B$ and $A-C$ must both be positive or negative. If two distinct values are chosen for $(A, C)$ and $(B, D)$ respectively, there are $2$ ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). We must divide by $2$ at the end, however, since the $2$ curves aren't considered distinct. Calculating, we get \\[\\frac {1}{2} \\cdot \\binom {6}{2} \\binom {4}{2} \\cdot 2 = \\boxed{90}.\\] ~ike.chen", "Like in Solution 2 , we find $\\frac {D-B}{A-C} \\ge 0$ . Notice that, since $D \\ne B$ , this expression can never equal $0$ , and since $A \\ne C$ , there won't be a divide-by- $0$ . This means that every choice results in either a positive or a negative value.\nFor every choice of $(A,B,C,D)$ that results in a positive value, we can flip $B$ and $D$ to obtain a corresponding negative value. This is a bijection (we could flip $B$ and $D$ again to obtain the original choice (injectivity) and we could flip $B$ and $D$ from any negative choice to obtain the corresponding positive choice (surjectivity)), so half of the choices are positive (where the curves intersect) and half are negative (where they don't).\nThis means that of the $\\frac{6\\cdot5\\cdot4\\cdot3}{2} = 180$ total choices (dividing by $2$ because the order of the curves does not matter), half of them, or $\\frac{180}{2} = \\boxed{90}$ , lead to intersecting curves." ]

[ "Note that Viswam walks at a constant speed of $60$ blocks per hour as he takes $1$ minute to walk each block. After walking $5$ blocks, he has taken $5$ minutes, and he has $5$ minutes remaining, to walk $7$ blocks. Therefore, he must walk at a speed of $7 \\cdot 60 \\div 5 = 84$ blocks per hour to get to school on time, from the time he starts his detour. Since he normally walks $\\frac{1}{2}$ mile, which is equal to $10$ blocks, $1$ mile is equal to $20$ blocks. Therefore, he must walk at $84 \\div 20 = 4.2$ mph from the time he starts his detour to get to school on time, so the answer is $\\boxed{4.2}$", "Viswam walks $10$ blocks, or half a mile, in $10$ minutes. Therefore, he walks at a rate of $3$ mph. From the time he takes his detour, he must travel $7$ blocks instead of $5$ . Our final equation is $\\frac{7}{5} \\times 3 = \\frac{21}{5} = \\boxed{4.2}$", "We can cheese this problem.\nNotice that Viswam will need to walk $7$ blocks during the second half as opposed to his normal $5$ blocks. Since rate is equal to distance over time, this implies that the final answer will likely be a multiple of $7$ , since you will need to convert $7$ blocks to miles. The only answer choice that is a multiple of $7$ is $\\boxed{4.2}$", "To travel $\\frac{1}{2}$ of a mile in total, each block must be $\\frac{1}{20}$ of a mile long. Since Viswam takes $1$ minute to walk along each block, it would take him $10$ minutes normally.\nViswam has already travelled for $5$ mins by the time he encounters the detour, so he must travel $7$ block lengths in the remaining $5$ minutes. The distance he has to travel is $7\\cdot \\frac{1}{20} = \\frac{7}{20}$ of a mile. Therefore,\n\\[\\frac{7}{20} \\mathrm{\\ miles} = r\\cdot 5\\mathrm{\\ mins}.\\] \\[\\frac{7\\mathrm{\\ miles}}{100\\mathrm{\\ mins}} = r.\\]\nAs $60$ mins equals $1$ hour, we set up the following proportion:\n\\[r = \\frac{7\\mathrm{\\ miles}}{100\\mathrm{\\ mins}} = \\frac{m\\mathrm{\\ miles}}{60\\mathrm{\\ mins}}.\\]\nCross multiplying and cancelling units yields\n\\[m = \\frac{7\\cdot 60}{100},\\]\nor $\\boxed{4.2}$", "When calculating the speed of his normal route, we get:\n$\\frac{0.5 \\mathrm{miles}}{10 \\mathrm{minutes}}= \\frac{3 \\mathrm{miles}}{60 \\mathrm{minutes}}$\nKeep in mind that his route is $10$ blocks long. Since we count $7$ blocks when the detour starts, than this would mean that he has $\\frac{7}{10}$ miles left to walk, considering the normal amount of blocks that he usually walks. Multiplying by $6$ to get the rate, we get, $\\frac{42}{10}=4.2$\nTherefore, the answer is $\\boxed{4.2}$", "We can start by splitting the walk into two sections: The 5-block section before the detour and the 7-block section during and after the detour.\nSince Viswam is walking at his normal rate for the first 5, it took him 5 minutes to walk that much. Since it must take him a total of 10 min in order for him to reach on time, he has 5 min for the second section of his walk.\nWe can figure out that 1 block equals 0.05 mi by dividing 10 by 0.5. Now, we can convert 7 blocks per 5 min into 4.2 miles per 1 hour. So, we arrive at $\\boxed{4.2}$" ]

[ "There are $4\\frac{1}{2}$ hours from 7:30 a.m. to noon, and $4$ hours from noon to 4 p.m., meaning Walter was away from home for a total of $8\\frac{1}{2}$ hours, or $8.5\\times 60 = 510$ minutes.\nTallying up the times he was not on the bus, he has $6\\times 50 = 300$ minutes in classes, $30$ minutes at lunch, and $2\\times 60 = 120$ minutes of \"additional time\". That is a total of $300 + 30 + 120 = 450$ minutes that Walter was away from home, but not on the bus.\nTherefore, Walter spent $510 - 450 = 60$ minutes on the bus, and the answer is $\\boxed{60}$" ]

[ "Walter has $1 + 5 + 10 + 25 = 41$ cents in his pocket. There are $100$ cents in a dollar. Therefore, Walter has $\\frac{41}{100}\\cdot 100\\% = 41\\%$ of a dollar in his pocket, and the right answer is $\\boxed{41}$" ]

[ "We need to find the least common multiple of the four numbers given.\n$\\textrm{LCM}(3, 4, 6, 7) = \\textrm{LCM}(3, 2^2, 2 \\cdot 3, 7) = 2^2 \\cdot 3 \\cdot 7 = 84$\nSo the answer is $\\boxed{84}$" ]

[ "Since $-5>-2015$ , the product must end with a $5$\nThe multiplicands are the odd negative integers from $-1$ to $-2013$ . There are $\\frac{|-2013+1|}2+1=1006+1$ of these numbers. Since $(-1)^{1007}=-1$ , the product is negative.\nTherefore, the answer must be $\\boxed{5.}$" ]

[ "Since $-1$ raised to an odd integer is $-1$ and $-1$ raised to an even integer exponent is $1$\n$(-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = \\boxed{0}.$" ]

[ "We can synchronously multiply ${2^3}$ to the expresions both above and below the fraction bar. Thus, \\[\\frac{2^3+2^3}{2^{-3}+2^{-3}}\\\\=\\frac{2^6+2^6}{1+1}\\\\={2^6}.\\] Hence, the fraction equals to $\\boxed{64}$", "We have \\[\\frac{2^3+2^3}{2^{-3}+2^{-3}} = \\frac{8 + 8}{\\frac{1}{8} + \\frac{1}{8}} = \\frac{16}{\\frac{1}{4}} = 16 \\cdot 4 = 64,\\] so our answer is $\\boxed{64}$" ]