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[ "To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that $ABCD$ can have any dimension. Give the rectangle dimensions of $AB = CD = 12$ and $BC = AD= 6$ , which is the easiest way to avoid fractions. Labelling the right midpoint as $M$ , and the bottom midpoint as $N$ , we know that $DN = NC = 6$ , and $BM = MC = 3$\n$[\\triangle ADN] = \\frac{1}{2}\\cdot 6\\cdot 6 = 18$\n$[\\triangle MNC] = \\frac{1}{2}\\cdot 3\\cdot 6 = 9$\n$[\\triangle ABM] = \\frac{1}{2}\\cdot 12\\cdot 3 = 18$\n$[\\triangle AMN] = [\\square ABCD] - [\\triangle ADN] - [\\triangle MNC] - [\\triangle ABM]$\n$[\\triangle AMN] = 72 - 18 - 9 - 18$\n$[\\triangle AMN] = 27$ , and the answer is $\\boxed{27}$", "The above answer is fast, but satisfying, and assumes that the area of $\\triangle AMN$ is independent of the dimensions of the rectangle. All in all, it's a very good answer though. However this is an alternative if you don't get the above answer. Label $AB = CD = l$ and $BC = DA = h$\nLabelling $m$ and $n$ as the right and lower midpoints respectively, and redoing all the work above, we get:\n$[\\triangle ABN] = \\frac{1}{2}\\cdot h\\cdot \\frac{l}{2} = \\frac{lh}{4}$\n$[\\triangle MNC] = \\frac{1}{2}\\cdot \\frac{l}{2}\\cdot \\frac{w}{2} = \\frac{lh}{8}$\n$[\\triangle ABM] = \\frac{1}{2}\\cdot l\\cdot \\frac{h}{2} = \\frac{lh}{4}$\n$[\\triangle AMN] = [\\square ABCD] - [\\triangle ADN] - [\\triangle MNC] - [\\triangle ABM]$\n$[\\triangle AMN] = lh - \\frac{lh}{4} - \\frac{lh}{8} - \\frac{lh}{4}$\n$[\\triangle AMN] = \\frac{3}{8}lh = \\frac{3}{8}\\cdot 72 = 27$ , and the answer is $\\boxed{27}$", "Let's assume, for simplicity, that the sides of the rectangle are $9$ and $8.$ The area of the 3 triangles would then be $8\\cdot\\frac{9}{2}\\cdot\\frac{1}{2} = 18,$ $4\\cdot\\frac{9}{2}\\cdot\\frac{1}{2} = 9,$ $4\\cdot 9\\cdot\\frac{1}{2} = 18.$ Adding these up, we get $45$ , and subtracting that from $72$ , we get $27$ , so the answer is $\\boxed{27}$" ]

[ "The area of a triangle is $\\frac12 bh.$ To maximize the base, let it be equal to the diameter of the semi circle, which is equal to $2r.$ To maximize the height, or altitude, choose the point directly in the middle of the arc connecting the endpoints of the diameter. It is equal to $r.$ Therefore the area is $\\frac12 \\cdot 2r \\cdot r = \\boxed{2}$" ]

[ "In order to attack this problem, we can use casework on the sign of $|x-y|$ and $|x+y|$\nCase 1: $|x-y|=x-y, |x+y|=x+y$\nSubstituting and simplifying, we have $x^2-6x+y^2=0$ , i.e. $(x-3)^2+y^2=3^2$ , which gives us a circle of radius $3$ centered at $(3,0)$\nCase 2: $|x-y|=y-x, |x+y|=x+y$\nSubstituting and simplifying again, we have $x^2+y^2-6y=0$ , i.e. $x^2+(y-3)^2=3^2$ . This gives us a circle of radius $3$ centered at $(0,3)$\nCase 3: $|x-y|=x-y, |x+y|=-x-y$\nDoing the same process as before, we have $x^2+y^2+6y=0$ , i.e. $x^2+(y+3)^2=3^2$ . This gives us a circle of radius $3$ centered at $(0,-3)$\nCase 4: $|x-y|=y-x, |x+y|=-x-y$\nOne last time: we have $x^2+y^2+6x=0$ , i.e. $(x+3)^2+y^2=3^2$ . This gives us a circle of radius $3$ centered at $(-3,0)$\nAfter combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like:\n Now, the area of the shaded region is just a square with side length $6$ with four semicircles of radius $3$ .\nThe area is $6\\cdot6+4\\cdot \\frac{9\\pi}{2} = 36+18\\pi$ . The answer is $36+18$ which is $\\boxed{54}$", "A somewhat faster variant of solution 1 is to use a bit of symmetry in order to show that the remaining three cases are identical to Case 1 in the above solution, up to rotations by $90^{\\circ}$ about the origin. This allows us to quickly sketch the region after solving Case 1.\nUpon simplifying Case 1, we obtain $(x-3)^2 + y^2 = 3^2$ which is a circle of radius 3 centered at $(3,0)$ . We remark that only the points on the semicircle where $x \\ge 3$ work here, since Case 1 assumes $x-y \\ge 0$ and $x+y \\ge 0$ . Next, we observe that an ordered pair is a solution to the given equation if and only if any of its $90^{\\circ}$ rotations about the origin is a solution. This follows as the value of $x^2+y^2-3(|x-y|+|x+y|)$ is invariant to $90^{\\circ}$ rotations, since $x^2+y^2$ simply represents the square of the distance to the origin (which is unchanged upon rotation), and $|x-y|+|x+y|$ is the sum of the distances to the lines $y=x$ and $y=-x$ , multiplied by $\\sqrt{2}$ (also unchanged upon $90^{\\circ}$ rotation).\nBy the above observation, we can quickly sketch the remainder of the region, and the area is $\\boxed{54}$ as above.", "Assume $y$ $0$ . We get that $x$ $6$ . That means that this figure must contain the points $(0,6), (6,0), (0, -6), (-6, 0)$ . Now, assume that $x$ $y$ . We get that $x$ $3 \\sqrt 3$ . We get the points $(3,3), (3,-3), (-3, 3), (-3, -3)$\nSince this contains $x^2 + y^2$ , assume that there are circles. Therefore, we can guess that there is a center square with area $6 \\cdot 6$ $36$ and $4$ semicircles with radius $3$ . We get $4$ semicircles with area $4.5 \\pi$ , and therefore the answer is $36+18$ $\\boxed{54}$" ]

[ "Let $[ABC]$ denote the area of figure $ABC$\nClearly, $[BEDC]=[ABCD]-[ABE]$ . Using basic area formulas,\nSince $AE+ED=BC=10$ and $ED=6$ $AE=4$ and the area of $\\triangle ABE$ is $4(4)=16$\nFinally, we have $[BEDC]=80-16=64\\rightarrow \\boxed{64}$" ]

[ "Lemma: If $x,y$ satisfy $px+qy=1$ , then the minimal value of $\\sqrt{x^2+y^2}$ is $\\frac{1}{\\sqrt{p^2+q^2}}$\nProof: Recall that the distance between the point $(x_0,y_0)$ and the line $px+qy+r = 0$ is given by $\\frac{|px_0+qy_0+r|}{\\sqrt{p^2+q^2}}$ . In particular, the distance between the origin and any point $(x,y)$ on the line $px+qy=1$ is at least $\\frac{1}{\\sqrt{p^2+q^2}}$\nLet the vertices of the right triangle be $(0,0),(5,0),(0,2\\sqrt{3}),$ and let $(a,0),(0,b)$ be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is $\\left(\\frac{a+b\\sqrt{3}}{2},\\frac{a\\sqrt{3}+b}{2}\\right)$ . This point must lie on the hypotenuse $\\frac{x}{5} + \\frac{y}{2\\sqrt{3}} = 1$ , i.e. $a,b$ must satisfy \\[\\frac{a+b\\sqrt{3}}{10}+\\frac{a\\sqrt{3}+b}{4\\sqrt{3}} = 1,\\] which can be simplified to \\[\\frac{7}{20}a + \\frac{11\\sqrt{3}}{60}b = 1.\\]\nBy the lemma, the minimal value of $\\sqrt{a^2+b^2}$ is \\[\\frac{1}{\\sqrt{\\left(\\frac{7}{20}\\right)^2 + \\left(\\frac{11\\sqrt{3}}{60}\\right)^2}} = \\frac{10\\sqrt{3}}{\\sqrt{67}},\\] so the minimal area of the equilateral triangle is \\[\\frac{\\sqrt{3}}{4} \\cdot \\left(\\frac{10\\sqrt{3}}{\\sqrt{67}}\\right)^2 = \\frac{\\sqrt{3}}{4} \\cdot \\frac{300}{67} = \\frac{75\\sqrt{3}}{67},\\] and hence the answer is $75+3+67=\\boxed{145}$", "Let $AB=2\\sqrt{3}, BC=5$ $D$ lies on $BC$ $F$ lies on $AB$ and $E$ lies on $AC$\nSet $D$ as the origin, $BD=a,BF=b$ $F$ can be expressed as $-a+bi$ in argand plane, the distance of $CD$ is $5-a$\nWe know that $(-a+bi)\\cdot\\cos(-\\frac{\\pi}{3})=(-\\frac{a+\\sqrt{3}b}{2}+\\frac{\\sqrt{3}a+b}{2}i)$ . We know that the slope of $AC$ is $-\\frac{2\\sqrt{3}}{5}$ , we have that $\\frac{5-a-(-\\frac{a+\\sqrt{3}b}{2})}{(\\sqrt{3}a+b)/2}=\\frac{5}{2\\sqrt{3}}$ , after computation, we have $11b+7\\sqrt{3}a=20\\sqrt{3}$\nNow the rest is easy with C-S inequality, $(a^2+b^2)(147+121)\\geq (7\\sqrt{3}a+11a)^2, a^2+b^2\\geq \\frac{300}{67}$ so the smallest area is $\\frac{\\sqrt{3}}{4}\\cdot \\frac{300}{67}=\\frac{75\\sqrt{3}}{67}$ , and the answer is $\\boxed{145}$", "Let $\\triangle ABC$ be the right triangle with sides $AB = x$ $AC = y$ , and $BC = z$ and right angle at $A$\nLet an equilateral triangle touch $AB$ $AC$ , and $BC$ at $D$ $E$ , and $F$ respectively, having side lengths of $c$\nNow, call $AD$ as $a$ and $AE$ as $b$ . Thus, $DB = x-a$ and $EC = y-b$\nBy Law of Sines on triangles $\\triangle DBF$ and $ECF$\n$BF = \\frac{z(a\\sqrt{3}+b)} {2y}$ and $CF = \\frac{z(a+b\\sqrt{3})} {2x}$\nSumming,\n$BF+CF = \\frac{z(a\\sqrt{3}+b)} {2y} + \\frac{z(a+b\\sqrt{3})} {2x} = BC = z$\nNow substituting $AB = x = 2\\sqrt{3}$ $AC = y = 5$ , and $BC = \\sqrt{37}$ and solving, $\\frac{7a}{20} + \\frac{11b\\sqrt{3}}{60} = 1$\nWe seek to minimize $[DEF] = c^2 \\frac{\\sqrt{3}}{4} = (a^2 + b^2) \\frac{\\sqrt{3}}{4}$\nThis is equivalent to minimizing $a^2+b^2$\nUsing the lemma from solution 1, we conclude that $\\sqrt{a^2+b^2} = \\frac{10\\sqrt{3}}{\\sqrt{67}}$\nThus, $[DEF] = \\frac{75\\sqrt{3}}{67}$ and our final answer is $\\boxed{145}$", "We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are $5$ and $2\\sqrt{3}i$ , respectively. Now let the vertex of the equilateral triangle on the real axis be $a$ and let the vertex of the equilateral triangle on the imaginary axis be $bi$ . Then, the third vertex of the equilateral triangle is given by: \\[(bi-a)e^{-\\frac{\\pi}{3}i}+a=(bi-a)(\\frac{1}{2}-\\frac{\\sqrt{3}}{2}i)+a=(\\frac{a}{2}+\\frac{b\\sqrt{3}}{2})+(\\frac{a\\sqrt{3}}{2}+\\frac{b}{2})i\\]\nFor this to be on the hypotenuse of the right triangle, we also have the following: \\[\\frac{\\frac{a\\sqrt{3}}{2}+\\frac{1}{2}}{\\frac{a}{2}+\\frac{b\\sqrt{3}}{2}-5}=-\\frac{2\\sqrt{3}}{5}\\iff 7\\sqrt{3}a+11b=20\\sqrt{3}\\]\nNote that the area of the equilateral triangle is given by $\\frac{\\sqrt{3}(a^2+b^2)}{4}$ , so we seek to minimize $a^2+b^2$ . This can be done by using the Cauchy Schwarz Inequality on the relation we derived above: \\[1200=(7\\sqrt{3}a+11b)^2\\leq ((7\\sqrt{3})^2+11^2)(a^2+b^2)\\implies a^2+b^2\\geq \\frac{1200}{268}\\]\nThus, the minimum we seek is simply $\\frac{\\sqrt{3}}{4}\\cdot\\frac{1200}{268}=\\frac{75\\sqrt{3}}{67}$ , so the desired answer is $\\boxed{145}$", "We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be $a$ and the point on the imaginary axis be $bi$ . Then, we see that $(a-bi)\\left(\\text{cis}\\frac{\\pi}{3}\\right)+bi=(a-bi)\\left(\\frac{1}{2}+i\\frac{\\sqrt{3}}{2}\\right)+bi=\\left(\\frac{1}{2}a+\\frac{\\sqrt{3}}{2}b\\right)+i\\left(\\frac{\\sqrt{3}}{2}a+\\frac{1}{2}b\\right).$ Now we switch back to Cartesian coordinates. The equation of the hypotenuse is $y=-\\frac{2\\sqrt{3}}{5}x+2\\sqrt{3}.$ This means that the point $\\left(\\frac{1}{2}a+\\frac{\\sqrt{3}}{2}b,\\frac{\\sqrt{3}}{2}a+\\frac{1}{2}b\\right)$ is on the line. Plugging the numbers in, we have $\\frac{\\sqrt{3}}{2}a+\\frac{1}{2}b=-\\frac{\\sqrt{3}}{5}a-\\frac{3}{5}b+2\\sqrt{3} \\implies 7\\sqrt{3}a+11b=20\\sqrt{3}.$ Now, we note that the side length of the equilateral triangle is $a^2+b^2$ so it suffices to minimize that. By Cauchy-Schwarz, we have $(a^2+b^2)(147+121)\\geq(7\\sqrt{3}a+11b)^2 \\implies (a^2+b^2)\\geq\\frac{300}{67}.$ Thus, the area of the smallest triangle is $\\frac{300}{67}\\cdot\\frac{\\sqrt{3}}{4}=\\frac{75\\sqrt{3}}{67}$ so our desired answer is $\\boxed{145}$", "Let $S$ be the triangle with side lengths $2\\sqrt{3},~5,$ and $\\sqrt{37}$\nWe will think about this problem backwards, by constructing a triangle as large as possible (We will call it $T$ , for convenience) which is similar to $S$ with vertices outside of a unit equilateral triangle $\\triangle ABC$ , such that each vertex of the equilateral triangle lies on a side of $T$ . After we find the side lengths of $T$ , we will use ratios to trace back towards the original problem.\nFirst of all, let $\\theta = 90^{\\circ}$ $\\alpha = \\arctan\\left(\\frac{2\\sqrt{3}}{5}\\right)$ , and $\\beta = \\arctan\\left(\\frac{5}{2\\sqrt{3}}\\right)$ (These three angles are simply the angles of triangle $S$ ; out of these three angles, $\\alpha$ is the smallest angle, and $\\theta$ is the largest angle). Then let us consider a point $P$ inside $\\triangle ABC$ such that $\\angle APB = 180^{\\circ} - \\theta$ $\\angle BPC = 180^{\\circ} - \\alpha$ , and $\\angle APC = 180^{\\circ} - \\beta$ . Construct the circumcircles $\\omega_{AB}, ~\\omega_{BC},$ and $\\omega_{AC}$ of triangles $APB, ~BPC,$ and $APC$ respectively.\nFrom here, we will prove the lemma that if we choose points $X$ $Y$ , and $Z$ on circumcircles $\\omega_{AB}, ~\\omega_{BC},$ and $\\omega_{AC}$ respectively such that $X$ $B$ , and $Y$ are collinear and $Y$ $C$ , and $Z$ are collinear, then $Z$ $A$ , and $X$ must be collinear. First of all, if we let $\\angle PAX = m$ , then $\\angle PBX = 180^{\\circ} - m$ (by the properties of cyclic quadrilaterals), $\\angle PBY = m$ (by adjacent angles), $\\angle PCY = 180^{\\circ} - m$ (by cyclic quadrilaterals), $\\angle PCZ = m$ (adjacent angles), and $\\angle PAZ = 180^{\\circ} - m$ (cyclic quadrilaterals). Since $\\angle PAX$ and $\\angle PAZ$ are supplementary, $Z$ $A$ , and $X$ are collinear as desired. Hence, $\\triangle XYZ$ has an inscribed equilateral triangle $ABC$\nIn addition, now we know that all triangles $XYZ$ (as described above) must be similar to triangle $S$ , as $\\angle AXB = \\theta$ and $\\angle BYC = \\alpha$ , so we have developed $AA$ similarity between the two triangles. Thus, $\\triangle XYZ$ is the triangle similar to $S$ which we were desiring. Our goal now is to maximize the length of $XY$ , in order to maximize the area of $XYZ$ , to achieve our original goal.\nNote that, all triangles $PYX$ are similar to each other if $Y$ $B$ , and $X$ are collinear. This is because $\\angle PYB$ is constant, and $\\angle PXB$ is also a constant value. Then we have $AA$ similarity between this set of triangles. To maximize $XY$ , we can instead maximize $PY$ , which is simply the diameter of $\\omega_{BC}$ . From there, we can determine that $\\angle PBY = 90^{\\circ}$ , and with similar logic, $PA$ $PB$ , and $PC$ are perpendicular to $ZX$ $XY$ , and $YZ$ respectively We have found our desired largest possible triangle $T$\nAll we have to do now is to calculate $YZ$ , and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within $S$ . First of all, we will prove that $\\angle ZPY = \\angle ACB + \\angle AXB$ . By the properties of cyclic quadrilaterals, $\\angle AXB = \\angle PAB + \\angle PBA$ , which means that $\\angle ACB + \\angle AXB = 180^{\\circ} - \\angle PAC - \\angle PBC$ . Now we will show that $\\angle ZPY = 180^{\\circ} - \\angle PAC - \\angle PBC$ . Note that, by cyclic quadrilaterals, $\\angle YZP = \\angle PAC$ and $\\angle ZYP = \\angle PBC$ . Hence, $\\angle ZPY = 180^{\\circ} - \\angle PAC - \\angle PBC$ (since $\\angle ZPY + \\angle YZP + \\angle ZYP = 180^{\\circ}$ ), proving the aforementioned claim. Then, since $\\angle ACB = 60^{\\circ}$ and $\\angle AXB = \\theta = 90^{\\circ}$ $\\angle ZPY = 150^{\\circ}$\nNow we calculate $PY$ and $PZ$ , which are simply the diameters of circumcircles $\\omega_{BC}$ and $\\omega_{AC}$ , respectively. By the extended law of sines, $PY = \\frac{BC}{\\sin{BPC}} = \\frac{\\sqrt{37}}{2\\sqrt{3}}$ and $PZ = \\frac{CA}{\\sin{CPA}} = \\frac{\\sqrt{37}}{5}$\nWe can now solve for $ZY$ with the law of cosines:\n\\[(ZY)^2 = \\frac{37}{25} + \\frac{37}{12} - \\left(\\frac{37}{5\\sqrt{3}}\\right)\\left(-\\frac{\\sqrt{3}}{2}\\right)\\]\n\\[(ZY)^2 = \\frac{37}{25} + \\frac{37}{12} + \\frac{37}{10}\\]\n\\[(ZY)^2 = \\frac{37 \\cdot 67}{300}\\]\n\\[ZY = \\sqrt{37} \\cdot \\frac{\\sqrt{67}}{10\\sqrt{3}}\\]\nNow we will apply this discovery towards our original triangle $S$ . Since the ratio between $ZY$ and the hypotenuse of $S$ is $\\frac{\\sqrt{67}}{10\\sqrt{3}}$ , the side length of the equilateral triangle inscribed within $S$ must be $\\frac{10\\sqrt{3}}{\\sqrt{67}}$ (as $S$ is simply as scaled version of $XYZ$ , and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within $S$ is $\\frac{75\\sqrt{3}}{67}$ , implying that the answer is $\\boxed{145}$", "Let the right triangle's lower-left point be at $O(0,0)$ . Notice the 2 other points will determine a unique equilateral triangle. Let 2 points be on the $x$ -axis ( $B$ ) and the $y$ -axis ( $A$ ) and label them $(b, 0)$ and $(0, a)$ respectively. The third point ( $C$ ) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of $a$ and $b$\n1. Find the slope of $AB$ and take the negative reciprocal of it to find the slope of the line containing $C$ . Notice the line contains the midpoint of $AB$ so we can then have an equation of the line.\n2. Let $AB=x.$ For $ABC$ to be an equilateral triangle, the altitude from $C$ to $AB$ must be $\\frac{x\\sqrt{3}}{2}.$\nWe then have two equations and two variables, so we can solve for $C$ 's coordinates.\nWe can find $C(\\frac{a+b\\sqrt{3}}{2}), (\\frac{b+a\\sqrt{3}}{2}).$ Also, note that $C$ must be on the hypotenuse of the triangle $\\frac{x}{5}+\\frac{y}{2\\sqrt{3}}=1.$ We can plug in $x$ and $y$ as the coordinates of $C$ , which simplifies to\n\\[11b+7\\sqrt{3}a=20\\sqrt{3}.\\]\nWe aim to minimize the side length of the triangle, which is $\\sqrt{a^2+b^2}.$ Applying the Cauchy inequality gives us\n\\[(a^2+b^2)(7\\sqrt{3}^2+11^2)\\geq (11b+7\\sqrt3a)^2 = 1200\\]\nFrom which we obtain $\\sqrt{a^2+b^2} \\geq \\sqrt{\\frac{300}{67}}.$ Thus, the area of the triangle = $\\frac{75\\sqrt{3}}{67}$ which leads to the answer $75+3+67=\\boxed{145}.$" ]

[ "Draw a square circumscribed around the circle. (Alternately, the circle is inscribed in the square.) If the circle has radius $4$ , it has diameter $8$ . Two of the diameters of the circle will run parallel to the sides of the square. Thus, the smallest square that contains it has side length $8$ , and area $8\\times8=64$ $\\boxed{64}$" ]

[ "Since the area of the whole figure is $100$ , each square has an area of $25$ and the side length is $5$\nThere are $10$ sides of this length, so the perimeter is $10(5)=50\\rightarrow \\boxed{50}$" ]

[ "Using the formula for the area of a trapezoid, we have $164=8(\\frac{BC+AD}{2})$ . Thus $BC+AD=41$ . Drop perpendiculars from $B$ to $AD$ and from $C$ to $AD$ and let them hit $AD$ at $E$ and $F$ respectively. Note that each of these perpendiculars has length $8$ . From the Pythagorean Theorem, $AE=6$ and $DF=15$ thus $AD=BC+21$ . Substituting back into our original equation we have $BC+BC+21=41$ thus $BC=\\boxed{10}$" ]

[ "The shaded region is a right trapezoid. Assume WLOG that $YZ=8$ . Then because the area of $\\triangle XYZ$ is equal to 8, the height of the triangle $XC=2$ . Because the line $AB$ is a midsegment, the top base of the trapezoid is $\\frac12 AB = \\frac14 YZ = 2$ . Also, $AB$ divides $XC$ in two, so the height of the trapezoid is $\\frac12 (2) = 1$ . The bottom base is $\\frac12 YZ = 4$ . The area of the shaded region is $\\frac12 (2+4)(1) = \\boxed{3}$", "Since $A$ and $B$ are the midpoints of $XY$ and $XZ$ , respectively, $AY=AX=BX=BZ$ . \nDraw segments $AC$ and $BC$ . \nDrawing an altitude in an isoceles triangle splits the triangle into 2 congruent triangles and we also know that $YC=CZ$ $AB$ is the line that connects the midpoints of two sides of a triangle together, which means that $AB$ is parallel to and half in length of $YZ$ . Then $AB=YC=CZ$ .\nSince $AB$ is parallel to $YZ$ , and $XY$ is the transversal, $\\angle XAB=\\angle AYC.$ Similarly, $\\angle XBA=\\angle BZC.$ Then, by SAS, $\\triangle XAB=\\triangle AYC=\\triangle BZC$ .\nSince corresponding parts of congruent triangles are congruent, $AC=BC=XA$ .\nSince ACY and BCZ are now isosceles triangles, $\\angle AYC=\\angle ACY$ and $\\angle BZC=\\angle BCZ$ Using the fact that $AB$ is parallel to $YZ$ $\\angle ACY=\\angle CAB$ and $\\angle BCZ=\\angle CBA$ . \nNow $\\triangle XAB=\\triangle AYC=\\triangle BZC=\\triangle ABC$ .\nDraw an altitude through each of them such that each triangle is split into two congruent right triangles. Now there are a total of 8 congruent small triangles, each with area 1. The shaded area has three of these triangles, so it has area 3.\nBasically the proof is to show $\\triangle XAB=\\triangle AYC=\\triangle BZC=\\triangle ABC$ . If you just look at the diagram you can easily see that the triangles are congruent and you would solve this a lot faster. Anyways, since those triangles are congruent, you can split each in half to find eight congruent triangles with area 1, and since the shaded region has three of these triangles, its area is $\\boxed{3}$", "We know the area of triangle $XYZ$ is $8$ square inches. The area of a triangle can also be represented as $\\frac{bh}{2}$ or in this problem $\\frac{XC\\cdot YZ}{2}$ . By solving, we have \\[\\frac{XC\\cdot YZ}{2} = 8,\\] \\[XC\\cdot YZ = 16.\\]\nWith SAS congruence, triangles $XCY$ and $XCZ$ are congruent. Hence, triangle $XCY = \\frac{8}{2} = 4$ . (Let's say point $D$ is the intersection between line segments $XC$ and $AB$ .) We can find the area of the trapezoid $ADCY$ by subtracting the area of triangle $XAD$ from $4$\nWe find the area of triangle $XAD$ by the $\\frac{bh}{2}$ formula- $\\frac{XD\\cdot AD}{2} = \\frac{\\frac{XC}{2}\\cdot AD}{2}$ $AD$ is $\\frac{1}{4}$ of $YZ$ from solution 1. The area of $XAD$ is \\[\\frac{\\frac{XC}{2}\\cdot \\frac{YZ}{4}}{2} = \\frac{16}{16} = 1\\]\nTherefore, the area of the shaded area- trapezoid $ADCY$ has area $4-1 = \\boxed{3}$", "$\\usepackage[dvipsnames]{xcolor} \\textcolor{BlueViolet}{\\text{Super fast after convincing yourself they are congruent}}$ \nWe draw the triangles as shown above. What seems like a half-square (or a corner of a square) indicates $90$ °. (This is not a random idea - the two original two triangles on the top inspires us to do so.)\nNote that $\\overline{AD}$ is perpendicular to $\\overline{YC}$ $\\overline{BE}$ is perpendicular to $\\overline{CZ}$\n$\\overline{AB}$ is perpendicular to $\\overline{YZ}$ , so because of corresponding angles theorem, $\\angle YAD=\\angle AXF$ . We are also told $\\overline{YA}=\\overline{AX}$ . Both $\\bigtriangleup \\text{XAF}$ and $\\bigtriangleup \\text{AYD}$ have a $90$ ° angle. Because of AAS Congruency (Angle-Angle-Side; $90$ °, $\\angle YAD$ , and side $\\overline{YA}$ ), $\\bigtriangleup \\text{XAF}$ is congruent to $\\bigtriangleup \\text{AYD}$ . Because of symmetry, the same goes for $\\bigtriangleup \\text{XFB}$ and $\\bigtriangleup \\text{BEZ}$\n$\\bigtriangleup \\text{YAD}$ is congruent to $\\bigtriangleup \\text{CAD}$ , which is congruent to $\\bigtriangleup \\text{ACF}$ because of symmetry. So there are $4$ congruent triangles on the left side. There is also $4$ congruent triangles on the right side because of symmetry.\nThere are $8$ congruent triangles in all. We are told the area of the triangle $\\bigtriangleup \\text{XYZ}$ is 8. $8\\div8=1$ , so each of the small congruent triangles is $1$ . The shaded area contains $3$ small triangles, so the area of the shaded section is $1\\cdot3=\\boxed{3}$" ]

[ "Since the arithmetic mean of the $50$ numbers is $38$ , their sum must be $50*38 = 1900$ . After $45$ and $55$ are discarded, the sum decreases by $45 + 55 = 100$ , so it must become $1900 - 100 = 1800$ . \nBut this means that the new mean of the remaining $50 - 2 = 48$ numbers must be $\\frac{1800}{48} = 37.5$ . Thusly, our answer is $\\boxed{37.5}$ , and we are done." ]

[ "Say that the four numbers are $a, b, c,$ $97$ . Then $\\frac{a+b+c+97}{4} = 85$ . What we are trying to find is $\\frac{a+b+c}{3}$ . Solving, \\[\\frac{a+b+c+97}{4} = 85\\] \\[a+b+c+97 = 340\\] \\[a+b+c = 243\\] \\[\\frac{a+b+c}{3} = \\boxed{81}\\]" ]

[ "We have $\\frac{1}{2}\\cdot \\left(\\frac{x + a}{x} + \\frac{x - a}{x}\\right) = \\frac{2}{2} = \\boxed{1}$" ]

[ "If the arithmetic mean of a set of $50$ numbers is $38$ , then the sum of the $50$ numbers equals $1900$ . Since $45$ and $55$ are being removed, subtract $100$ to get the sum of the remaining $48$ numbers, which is $1800$ . Therefore, the new mean is $37.5$ , which is answer choice $\\boxed{37.5}$" ]

[ "We wish to find $\\frac{9+99+\\cdots +999999999}{9}$ , or $\\frac{9(1+11+111+\\cdots +111111111)}{9}=123456789$ . This doesn't have the digit 0, so the answer is $\\boxed{0}$", "Notice that the final number is guaranteed to have the digits $\\{1, 3, 5, 7, 9\\}$ and that each of these digits can be paired with an even number adding up to 9. $\\boxed{0}$ can be taken out, with the other digits fulfilling divisibility by 9.", "The arithmetic mean is $\\frac{(10^1-1)+(10^2-1)+\\ldots+(10^9-1)}{9}=\\frac{1111111101}{9}=123456789$ . So select $\\boxed{0}$ .\n~hastapasta" ]

[ "We wish to find $\\frac{9+99+\\cdots +999999999}{9}$ , or $\\frac{9(1+11+111+\\cdots +111111111)}{9}=123456789$ . This doesn't have the digit 0, so the answer is $\\boxed{0}$", "Notice that the final number is guaranteed to have the digits $\\{1, 3, 5, 7, 9\\}$ and that each of these digits can be paired with an even number adding up to 9. $\\boxed{0}$ can be taken out, with the other digits fulfilling divisibility by 9.", "The arithmetic mean is $\\frac{(10^1-1)+(10^2-1)+\\ldots+(10^9-1)}{9}=\\frac{1111111101}{9}=123456789$ . So select $\\boxed{0}$ .\n~hastapasta" ]

[ "If the average of the numbers is $10$ , then their sum is $10\\times 10=100$\nTo maximize the largest number of the ten, we minimize the other nine. Since they must be distinct, positive whole numbers, we let them be $1,2,3,4,5,6,7,8,9$ . Their sum is $45$\nThe sum of nine of the numbers is $45$ , and the sum of all ten is $100$ so the last number must be $100-45=55\\rightarrow \\boxed{55}$" ]

[ "Since the average of the first $20$ numbers is $30$ , their sum is $20\\cdot30=600$\nSince the average of $30$ other numbers is $20$ , their sum is $30\\cdot20=600$\nSo the sum of all $50$ numbers is $600+600=1200$\nTherefore, the average of all $50$ numbers is $\\frac{1200}{50}=\\boxed{24}$" ]

[ "The sum of the ages of the fifth graders is $33 * 11$ , while the sum of the ages of the parents is $55 * 33$ . Therefore, the total sum of all their ages must be $2178$ , and given $33 + 55 = 88$ people in total, their average age is $\\frac{2178}{88} = \\frac{99}{4} = \\boxed{24.75}$" ]

[ "The sum of the ages of the fifth graders is $33 * 11$ , while the sum of the ages of the parents is $55 * 33$ . Therefore, the total sum of all their ages must be $2178$ , and given $33 + 55 = 88$ people in total, their average age is $\\frac{2178}{88} = \\frac{99}{4} = \\boxed{24.75}$" ]

[ "Let $x$ be the average of the remaining $4$ people.\nThe equation we get is $\\frac{4x + 18}{5} = 30$\nSimplify,\n$4x + 18 = 150$\n$4x = 132$\n$x = 33$\nTherefore, the answer is $\\boxed{33}$", "Since an $18$ year old left from a group of people averaging $30$ , The remaining people must total $30 - 18 = 12$ years older than $30$ . Therefore, the average is $\\dfrac{12}{4} = 3$ years over $30$ . Giving us $\\boxed{33}$", "The total ages would be $30*5=150$ . Then, if one $18$ year old leaves, we subtract $18$ from $150$ and get $132$ . Then, we divide $132$ by $4$ to get the new average, $\\boxed{33}$" ]

[ "The total of all their ages over the number of people is\n\\[\\frac{6 \\cdot 40 + 4\\cdot 25}{6+4} = \\frac{340}{10} = \\boxed{34}.\\]" ]

[ "First, find the total amount of the girl's ages and add it to the total amount of the boy's ages. It equals $(20)(15)+(15)(16)=540$ . The total amount of everyone's ages can be found from the average age, $17\\cdot40=680$ . Then you do $680-540=140$ to find the sum of the adult's ages. The average age of an adult is divided among the five of them, $140\\div5=\\boxed{28}$" ]

[ "The percent decrease is (the amount of decrease)/(original amount)\nthe amount of decrease is $41 - 7 = 34$\nso the percent decrease is $\\frac{34}{41}$ which is about $\\boxed{80}$" ]

[ "Let the $5$ numbers be $a, b, c, d$ , and $e$ . Thus $\\frac{a+b+c+d+e}{5}=54$ and $a+b+c+d+e=270$ . Since $\\frac{a+b}{2}=48$ $a+b=96$ . Substituting back into our original equation, we have $96+c+d+e=270$ and $c+d+e=174$ . Dividing by $3$ gives the average of $\\boxed{58}$" ]

[ "Let the total value, in cents, of the coins Paula has originally be $v$ , and the number of coins she has be $n$ . Then $\\frac{v}{n}=20\\Longrightarrow v=20n$ and $\\frac{v+25}{n+1}=21$ . Substituting yields: $20n+25=21(n+1),$ so $n=4$ $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if she had $3$ quarters and $1$ nickel in her purse. Thus, she has $\\boxed{0}$ dimes.", "If the new coin was worth $20$ cents, adding it would not change the mean. The additional $5$ cents raise the mean by $1$ , thus the new number of coins must be $5$ . Therefore there were $4$ coins worth a total of $4\\times20=80$ cents. As in the previous solution, we conclude that the only way to get $80$ cents using $4$ coins is $25+25+25+5$ . Thus, having three quarters, one nickel, and no dimes $\\boxed{0}.$" ]

[ "Let the total value, in cents, of the coins Paula has originally be $v$ , and the number of coins she has be $n$ . Then $\\frac{v}{n}=20\\Longrightarrow v=20n$ and $\\frac{v+25}{n+1}=21$ . Substituting yields: $20n+25=21(n+1),$ so $n=4$ $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if she had $3$ quarters and $1$ nickel in her purse. Thus, she has $\\boxed{0}$ dimes.", "If the new coin was worth $20$ cents, adding it would not change the mean. The additional $5$ cents raise the mean by $1$ , thus the new number of coins must be $5$ . Therefore there were $4$ coins worth a total of $4\\times20=80$ cents. As in the previous solution, we conclude that the only way to get $80$ cents using $4$ coins is $25+25+25+5$ . Thus, having three quarters, one nickel, and no dimes $\\boxed{0}.$" ]

[ "Let the $6$ boys have total weight $S_B$ and let the $4$ girls have total weight $S_G$ . We are given\n\\begin{align*} \\frac{S_B}{6} &= 150 \\\\ \\frac{S_G}{4} &= 120 \\end{align*}\nWe want the average of the $10$ children, which is \\[\\frac{S_B+S_G}{10}\\] From the first two equations , we can determine that $S_B=900$ and $S_G=480$ , so $S_B+S_G=1380$ . Therefore, the average we desire is \\[\\frac{1380}{10}=138 \\rightarrow \\boxed{138}\\]" ]

[ "The total frequency is $50+60+40+60+40=250$ , with the blue frequency of $60$ . Therefore, the precentage that preferred blue is $\\frac{60}{250}=\\boxed{24}$" ]

[ "The area of a triangle is shown by $\\frac{1}{2}bh$ . We set the base equal to $24$ , and the area equal to $60$ , and we get the height, or altitude, of the triangle to be $5$ . In this isosceles triangle, the height bisects the base, so by using the Pythagorean Theorem, $a^2+b^2=c^2$ , we can solve for one of the legs of the triangle (it will be the the hypotenuse, $c$ ). $a = 12$ $b = 5$ $c = 13$ .\nThe answer is $\\boxed{13}$" ]

[ "Numbers in binary work similar to their decimal counterparts, where the multiplier associated with each place is multiplied by two every single place to the left. For example, $1111_2$ $1111$ in base $2$ ) would equate to $1 * 2^3 + 1 * 2^2 + 1 * 2^1 + 1 * 2^0 = 8+4+2+1 = 15$\nUsing this same logic, $10011_2$ would be $1*2^4 + 1*2^1 + 1 * 2^0 = \\boxed{19}$" ]

[ "Recall that $9\\equiv-1\\pmod{5}.$ We expand $N$ by the definition of bases: \\begin{align*} N&=27006000052_9 \\\\ &= 2\\cdot9^{10} + 7\\cdot9^9 + 6\\cdot9^6 + 5\\cdot9 + 2 \\\\ &\\equiv 2\\cdot(-1)^{10} + 7\\cdot(-1)^9 + 6\\cdot(-1)^6 + 5\\cdot(-1) + 2 &&\\pmod{5} \\\\ &\\equiv 2-7+6-5+2 &&\\pmod{5} \\\\ &\\equiv -2 &&\\pmod{5} \\\\ &\\equiv \\boxed{3} ~Aidensharp ~Kante314 ~MRENTHUSIASM", "We need to first convert $N$ into a regular base- $10$ number: \\[N = 27006000052_9 = 2\\cdot9^{10} + 7\\cdot9^9 + 6\\cdot9^6 + 5\\cdot9 + 2.\\]\nNow, consider how the last digit of $9$ changes with changes of the power of $9:$ \\begin{align*} 9^0&=1, \\\\ 9^1&=9, \\\\ 9^2&=\\ldots 1, \\\\ 9^3&=\\ldots 9, \\\\ 9^4&=\\ldots 1, \\\\ & \\ \\vdots \\end{align*} Note that if $x$ is odd, then $9^x \\equiv 4\\pmod{5}.$ On the other hand, if $x$ is even, then $9^x \\equiv 1\\pmod{5}.$\nTherefore, we have \\begin{align*} N&\\equiv 2\\cdot(1) + 7\\cdot(4) + 6\\cdot(1) + 5\\cdot(4) + 2\\cdot(1) &&\\pmod{5} \\\\ &\\equiv 2+28+6+20+2 &&\\pmod{5} \\\\ &\\equiv 58 &&\\pmod{5} \\\\ &\\equiv \\boxed{3}$ may simplify the process further, as given by Solution 1." ]

[ "Recall that $9\\equiv-1\\pmod{5}.$ We expand $N$ by the definition of bases: \\begin{align*} N&=27006000052_9 \\\\ &= 2\\cdot9^{10} + 7\\cdot9^9 + 6\\cdot9^6 + 5\\cdot9 + 2 \\\\ &\\equiv 2\\cdot(-1)^{10} + 7\\cdot(-1)^9 + 6\\cdot(-1)^6 + 5\\cdot(-1) + 2 &&\\pmod{5} \\\\ &\\equiv 2-7+6-5+2 &&\\pmod{5} \\\\ &\\equiv -2 &&\\pmod{5} \\\\ &\\equiv \\boxed{3} ~Aidensharp ~Kante314 ~MRENTHUSIASM", "We need to first convert $N$ into a regular base- $10$ number: \\[N = 27006000052_9 = 2\\cdot9^{10} + 7\\cdot9^9 + 6\\cdot9^6 + 5\\cdot9 + 2.\\]\nNow, consider how the last digit of $9$ changes with changes of the power of $9:$ \\begin{align*} 9^0&=1, \\\\ 9^1&=9, \\\\ 9^2&=\\ldots 1, \\\\ 9^3&=\\ldots 9, \\\\ 9^4&=\\ldots 1, \\\\ & \\ \\vdots \\end{align*} Note that if $x$ is odd, then $9^x \\equiv 4\\pmod{5}.$ On the other hand, if $x$ is even, then $9^x \\equiv 1\\pmod{5}.$\nTherefore, we have \\begin{align*} N&\\equiv 2\\cdot(1) + 7\\cdot(4) + 6\\cdot(1) + 5\\cdot(4) + 2\\cdot(1) &&\\pmod{5} \\\\ &\\equiv 2+28+6+20+2 &&\\pmod{5} \\\\ &\\equiv 58 &&\\pmod{5} \\\\ &\\equiv \\boxed{3}$ may simplify the process further, as given by Solution 1." ]

[ "We can figure out $H = 0$ by noticing that $19!$ will end with $3$ zeroes, as there are three factors of $5$ in its prime factorization, so there would be 3 powers of 10 meaning it will end in 3 zeros. Next, we use the fact that $19!$ is a multiple of both $11$ and $9$ . Their divisibility rules (see Solution 2) tell us that $T + M \\equiv 3 \\;(\\bmod\\; 9)$ and that $T - M \\equiv 7 \\;(\\bmod\\; 11)$ . By guess and checking, we see that $T = 4, M = 8$ is a valid solution. Therefore the answer is $4 + 8 + 0 = \\boxed{12}$", "We know that $H = 0$ , because $19!$ ends in three zeroes (see Solution 1). Furthermore, we know that $9$ and $11$ are both factors of $19!$ . We can simply use the divisibility rules for $9$ and $11$ for this problem to find $T$ and $M$ . For $19!$ to be divisible by $9$ , the sum of digits must simply be divisible by $9$ . Summing the digits, we get that $T + M + 33$ must be divisible by $9$ . This leaves either $\\text{A}$ or $\\text{C}$ as our answer choice. Now we test for divisibility by $11$ . For a number to be divisible by $11$ , the alternating sum must be divisible by $11$ (for example, with the number $2728$ $2-7+2-8 = -11$ , so $2728$ is divisible by $11$ ). Applying the alternating sum test to this problem, we see that $T - M - 7$ must be divisible by 11. By inspection, we can see that this holds if $T=4$ and $M=8$ . The sum is $8 + 4 + 0 = \\boxed{12}$", "Multiplying it out, we get $19! = 121,645,100,408,832,000$ . Evidently, $T = 4$ $M = 8$ , and $H = 0$ . The sum is $8 + 4 + 0 = \\boxed{12}$", "7, 11, 13 are < 19 and 1001 = 7 * 11 * 13. Check the alternating sum of block 3: H00 - 832 + 40M - 100 + 6T5 - 121 and it is divisible by 1001. HTM + 5 - 53 = 0 (mod 1001) => HTM = 48.\nThe answer is $4 + 8 + 0 = \\boxed{12}$ . ~ AliciaWu" ]

[ "Since a card from B is placed on the bottom of the new stack, notice that cards from pile B will be marked as an even number in the new pile, while cards from pile A will be marked as odd in the new pile. Since 131 is odd and retains its original position in the stack, it must be in pile A. Also to retain its original position, exactly $131 - 1 = 130$ numbers must be in front of it. There are $\\frac{130}{2} = 65$ cards from each of piles A, B in front of card 131. This suggests that $n = 131 + 65 = 196$ ; the total number of cards is $196 \\cdot 2 = \\boxed{392}$", "If you index the final stack $1,2,\\dots,2n$ , you notice that pile A resides only in the odd indices and has maintained its original order aside from flipping over. The same has happened to pile B except replace odd with even. Thus, if 131 is still at index 131, an odd number, then 131 must be from pile A. The numbers in pile A are the consecutive integers $1,2,\\dots, n$ . This all leads us to the following equation. \\[131=2n-2(131)+1\\implies2n=\\boxed{392}\\]" ]

[ "The octahedron is just two congruent pyramids glued together by their base. The base of one pyramid is a rhombus with diagonals $4$ and $5$ , for an area $A=10$ . The height $h$ , of one pyramid, is $\\frac{3}{2}$ , so the volume of one pyramid is $\\frac{Ah}{3}=5$ . Thus, the octahedron has volume $2\\cdot5=\\boxed{10}$" ]

[ "The \"base\" of the octahedron is half the base of the rectangular prism because it is connected by the midpoints. Additionally, the volume of an octahedron is $\\dfrac{1}{3}$ of its respective prism. Thus, the octahedron's volume is $\\dfrac{1}{2} \\cdot \\dfrac{1}{3} = \\dfrac{1}{6}$ of the rectangular prism's volume, meaning that the answer is $3 \\cdot 4 \\cdot 5 \\cdot \\dfrac{1}{6} = \\boxed{10}$" ]

[ "\nLet $A$ be the center of the circle with radius $5$ , and $B$ be the center of the circle with radius $4$ . Let $\\overline{CD}$ be the common internal tangent of circle $A$ and circle $B$ . Extend $\\overline{BD}$ past $D$ to point $E$ such that $\\overline{BE}\\perp\\overline{AE}$ . Since $\\overline{AC}\\perp\\overline{CD}$ and $\\overline{BD}\\perp\\overline{CD}$ $ACDE$ is a rectangle. Therefore, $AC=DE$ and $CD=AE$\nSince the centers of the two circles are $41$ inches apart, $AB=41$ . Also, $BE=4+5=9$ . Using the Pythagorean Theorem on right triangle $ABE$ $CD=AE=\\sqrt{41^2-9^2}=\\sqrt{1600}=40$ . The length of the common internal tangent is $\\boxed{40}$" ]

[ "Let $s$ be the cost of one sandwich, $c$ be the cost of one cup of coffee, and $p$ be the price of one piece of pie. With the information,\n\\[3s+7c+p=3.15\\] \\[4s+10c+p=4.20\\]\nSubtract the first equation from the second to get\n\\[s+3c=1.05\\]\nThat means $s=1.05-3c$ . Substituting it back in the second equation results in.\n\\[4.20-12c+10c+p=4.20\\]\nSolving for $p$ yields $p=2c$ . With the substitutions, the cost of one sandwich, one cup of coffee, and one slice of pie is $(1.05-3c)+c+(2c) = \\boxed{1.05}$" ]

[ "Call $M$ and $N$ the feet of the altitudes from $O$ to $BC$ and $AB$ , respectively. Let $OB = r$ . Notice that $\\triangle{OMB} \\sim \\triangle{QOB}$ because both are right triangles, and $\\angle{OBQ} \\cong \\angle{OBM}$ . By $\\frac{MB}{BO}=\\frac{BO}{BQ}$ $MB = r\\left(\\frac{r}{4.5}\\right) = \\frac{r^2}{4.5}$ . However, since $O$ is the circumcenter of triangle $ABC$ $OM$ is a perpendicular bisector by the definition of a circumcenter. Hence, $\\frac{r^2}{4.5} = 2 \\implies r = 3$ . Since we know $BN=\\frac{5}{2}$ and $\\triangle BOP \\sim \\triangle BNO$ , we have $\\frac{BP}{3} = \\frac{3}{\\frac{5}{2}}$ . Thus, $BP = \\frac{18}{5}$ $m + n=\\boxed{023}$", "Minor arc $BC = 2A$ so $\\angle{BOC}=2A$ . Since $\\triangle{BOC}$ is isosceles ( $BO$ and $OC$ are radii), $\\angle{CBO}=(180-2A)/2=90-A$ $\\angle{CBO}=90-A$ , so $\\angle{BQO}=A$ . From this we get that $\\triangle{BPQ}\\sim \\triangle{BCA}$ . So $\\dfrac{BP}{BC}=\\dfrac{BQ}{BA}$ , plugging in the given values we get $\\dfrac{BP}{4}=\\dfrac{4.5}{5}$ , so $BP=\\dfrac{18}{5}$ , and $m+n=\\boxed{023}$", "Let $r=BO$ . Drawing perpendiculars, $BM=MC=2$ and $BN=NA=2.5$ . From there, \\[OM=\\sqrt{r^2-4}\\] Thus, \\[OQ=\\frac{\\sqrt{4r^2+9}}{2}\\] Using $\\triangle{BOQ}$ , we get $r=3$ . Now let's find $NP$ . After some calculations with $\\triangle{BON}$ $\\triangle{OPN}$ ${NP=11/10}$ . Therefore, \\[BP=\\frac{5}{2}+\\frac{11}{10}=18/5\\] $18+5=\\boxed{023}$", "Let $\\angle{BQO}=\\alpha$ . Extend $OB$ to touch the circumcircle at a point $K$ . Then, note that $\\angle{KAC}=\\angle{CBK}=\\angle{QBO}=90^\\circ-\\alpha$ . But since $BK$ is a diameter, $\\angle{KAB}=90^\\circ$ , implying $\\angle{CAB}=\\alpha$ . It follows that $APCQ$ is a cyclic quadrilateral.\nLet $BP=x$ . By Power of a Point, \\[5x=4\\cdot\\frac 9 2\\implies x=\\frac{18}{5}.\\] The answer is $18+5=\\boxed{023}$", "$\\textit{Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.}$\nDenote the circumradius of $ABC$ to be $R$ , the circumcircle of $ABC$ to be $O$ , and the shortest distance from $Q$ to circle $O$ to be $x$\nUsing Power of a Point on $Q$ relative to circle $O$ , we get that $x(x+2r) = 0.5 \\cdot 4.5 = \\frac{9}{4}$ . Using Pythagorean Theorem on triangle $QOB$ to get $(x + r)^2 + r^2 = \\frac{81}{4}$ . Subtracting the first equation from the second, we get that $2r^2 = 18$ and therefore $r = 3$ . Now, set $\\cos{ABC} = y$ . Using law of cosines on $ABC$ to find $AC$ in terms of $y$ and plugging that into the extended law of sines, we get $\\frac{\\sqrt{4^2 + 5^2 - 2 \\cdot 4 \\cdot 5 x}}{\\sqrt{1 - x^2}} = 2R = 6$ . Squaring both sides and cross multiplying, we get $36x^2 - 40x + 5 = 0$ . Now, we get $x = \\frac{10 \\pm \\sqrt{55}}{18}$ using quadratic formula. If you drew a decent diagram, $B$ is acute and therefore $x = \\frac{10 + \\sqrt{55}}{18}$ (You can also try plugging in both in the end and seeing which gives a rational solution). Note that $BP = 3\\frac{1}{\\sin{OPB}} = \\frac{3}{\\cos{\\angle ABC - \\angle QBO}}.$ Using the cosine addition formula and then plugging in what we know about $QBO$ , we get that $BP = \\frac{162}{2\\cos{B} + \\sqrt{5}\\sin{B}}$ . Now, the hard part is to find what $\\sin{B}$ is. We therefore want $\\frac{\\sqrt{324 - (10 + \\sqrt{55})^2}}{18} = \\frac{\\sqrt{169 - 20\\sqrt{55}}}{18}$ . For the numerator, by inspection $(a + b\\sqrt{55})^2$ will not work for integers $a$ and $b$ . The other case is if there is $(a\\sqrt{5} + b\\sqrt{11})^2$ . By inspection, $5\\sqrt{5} - 2\\sqrt{11}$ works. Therefore, plugging all this in yields the answer, $\\frac{18}{5} \\rightarrow \\boxed{23}$ . Solution by hyxue", " Reflect $A$ $P$ across $OB$ to points $A'$ and $P'$ , respectively with $A'$ on the circle and $P, O, P'$ collinear. Now, $\\angle A'CQ = 180^{\\circ} - \\angle A'CB = \\angle A'AB = \\angle P'PB$ by parallel lines. From here, $\\angle P'PB = \\angle PP'B = \\angle A'P'Q$ as $P, P', Q$ collinear. From here, $A'P'QC$ is cyclic, and by power of a point we obtain $\\frac{18}{5} \\implies \\boxed{023}$ .\n~awang11's sol" ]

[ "By the Binomial Theorem , each term of the expansion is $\\binom{8}{n}\\left(\\frac{x^2}{2}\\right)^{8-n}\\left(\\frac{-2}{x}\\right)^n$\nWe want the exponent of $x$ to be $7$ , so \\[2(8-n)-n=7\\] \\[16-3n=7\\] \\[n=3\\]\nIf $n=3$ , then the corresponding term is \\[\\binom{8}{3}\\left(\\frac{x^2}{2}\\right)^{5}\\left(\\frac{-2}{x}\\right)^3\\] \\[56 \\cdot \\frac{x^{10}}{32} \\cdot \\frac{-8}{x^3}\\] \\[-14x^7\\]\nThe answer is $\\boxed{14}$" ]

[ "Squaring, we find that $(9 + bi)^2 = 81 + 18bi - b^2$ . Cubing and ignoring the real parts of the result, we find that $(81 + 18bi - b^2)(9 + bi) = \\ldots + (9\\cdot 18 + 81)bi - b^3i$\nSetting these two equal, we get that $18bi = 243bi - b^3i$ , so $b(b^2 - 225) = 0$ and $b = -15, 0, 15$ . Since $b > 0$ , the solution is $\\boxed{015}$" ]

[ "Let the complex number $z$ equal $a+bi$ . Then the preceding equation can be expressed as \\[a+bi+\\sqrt{a^2+b^2} = 2+8i\\] Because $a$ and $b$ must both be real numbers, we immediately have that $bi = 8i$ , giving $b = 8$ . Plugging this in back to our equation gives us $a+\\sqrt{a^2+64} = 2$ . \nRearranging this into $2-a = \\sqrt{a^2+64}$ , we can square each side of the equation resulting in \\[4-4a+a^2 = a^2+64\\] Further simplification will yield $60 = -4a$ meaning that $-15 = a$ . Knowing both $a$ and $b$ , we can plug them in into $a^2+b^2$ . Our final answer is $\\boxed{289}$" ]

[ "Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{143} = z.$ We know that $z \\neq 0,$ because we are given the imaginary part of $z,$ so we can divide by $z$ to get $z^{142} = 1.$ So, $z$ must be a $142$ nd root of unity, and thus, by De Moivre's theorem, the imaginary part of $z$ will be of the form $\\sin{\\frac{2k\\pi}{142}} = \\sin{\\frac{k\\pi}{71}},$ where $k \\in \\{1, 2, \\ldots, 70\\}.$ Note that $71$ is prime and $k<71$ by the conditions of the problem, so the denominator in the argument of this value will always be $71.$ Thus, $n = \\boxed{071}.$" ]

[ "Multiplying the two equations together gives us \\[zw + 32i - \\frac{240}{zw} = -30 + 46i\\] and multiplying by $zw$ then gives us a quadratic in $zw$ \\[(zw)^2 + (30-14i)zw - 240 =0.\\] Using the quadratic formula, we find the two possible values of $zw$ to be $7i-15 \\pm \\sqrt{(15-7i)^2 + 240}$ $6+2i,$ $12i - 36.$ The smallest possible value of $\\vert zw\\vert^2$ is then obviously $6^2 + 2^2 = \\boxed{040}$" ]

[ "The triangle has base $6,$ which means its height satisfies \\[\\dfrac{6h}{2}=3h=12.\\] This means that $h=4,$ so the answer is $7+4=\\boxed{11}$", " Label point $D(3,7)$ as the point at which $CD\\perp DA$ . We now have $[\\triangle ABC] = [\\triangle BCD] - [\\triangle ACD]$ , where the brackets denote areas. On the right hand side, both of these triangles are right, so we can just compute the two sides of each triangle. The two side lengths of $\\triangle ACD$ are $y-7$ and $5-3=2$ . The two side lengths of $\\triangle BCD$ are $y-7$ and $11-3 = 8.$ Now,\n\\[[\\triangle ABC] = 12 = \\frac{1}{2}\\cdot (y-7)\\cdot 8 - \\frac{1}{2}\\cdot (y-7)\\cdot 2 = 3(y-7)\\]\nDividing by $3$ gives $y -7 = 4,$ so $y = \\boxed{11}.$", "By the Shoelace Theorem, $\\triangle ABC$ has area \\[\\frac{1}{2}|(y \\cdot 11 + 7 \\cdot 5 + 7 \\cdot 3) - (3 \\cdot 7 + 11 \\cdot 7 + 5 \\cdot y)| = \\frac{1}{2}|(11y + 56) - (98 + 5y)| = \\frac{1}{2}|6y - 42|\\] . From the problem, this is equal to $12$ . We now solve for y.\n$\\frac{1}{2}|6y - 42| = 12$\n$|6y-42| = 24$\n$6y - 42 = 24$ OR $6y - 42 = -24$\n$6y = 66$ OR $6y = 18$\n$y = 11$ OR $y = 3$\nHowever, since, as stated in the problem, $y > 7$ , our only valid solution is $\\boxed{11}$", "As in the figure, the triangle is determined by the vectors $\\begin{bmatrix}-2 \\\\ y-7\\end{bmatrix}$ and $\\begin{bmatrix}6\\\\0\\end{bmatrix}$ . Recall that the absolute value of the determinant of these vectors is the area of the parallelogram determined by those vectors; the triangle has half the area of that parallelogram. Then we must have that $\\frac{1}{2}|\\begin{vmatrix}-2 & y-7\\\\6 & 0\\end{vmatrix}| = 12 \\implies \\begin{vmatrix}-2 & y-7\\\\6 & 0\\end{vmatrix} = \\pm 24$ . Expanding the determinants, we find that $-6(y-7) = 24$ or $-6(y-7) = -24$ . Solving each equation individually, we find that $y = 3$ or $y = 11$ . However, the problem states that $y > 7$ , so the only valid solution is $\\boxed{11}$" ]

[ "For simplicity, we translate the points so that $A$ is on the origin and $D = (1,7)$ . Suppose $B$ has integer coordinates; then $\\overrightarrow{AB}$ is a vector with integer parameters (vector knowledge is not necessary for this solution). We construct the perpendicular from $A$ to $\\overline{CD}$ , and let $D' = (a,b)$ be the reflection of $D$ across that perpendicular. Then $ABCD'$ is a parallelogram , and $\\overrightarrow{AB} = \\overrightarrow{D'C}$ . Thus, for $C$ to have integer coordinates, it suffices to let $D'$ have integer coordinates.\nLet the slope of the perpendicular be $m$ . Then the midpoint of $\\overline{DD'}$ lies on the line $y=mx$ , so $\\frac{b+7}{2} = m \\cdot \\frac{a+1}{2}$ . Also, $AD = AD'$ implies that $a^2 + b^2 = 1^2 + 7^2 = 50$ . Combining these two equations yields\n\\[a^2 + \\left(7 - (a+1)m\\right)^2 = 50\\]\nSince $a$ is an integer, then $7-(a+1)m$ must be an integer. There are $12$ pairs of integers whose squares sum up to $50,$ namely $( \\pm 1, \\pm 7), (\\pm 7, \\pm 1), (\\pm 5, \\pm 5)$ . We exclude the cases $(\\pm 1, \\pm 7)$ because they lead to degenerate trapezoids (rectangle, line segment, vertical and horizontal sides). Thus we have\n\\[7 - 8m = \\pm 1, \\quad 7 + 6m = \\pm 1, \\quad 7 - 6m = \\pm 5, 7 + 4m = \\pm 5\\]\nThese yield $m = 1, \\frac 34, -1, -\\frac 43, 2, \\frac 13, -3, - \\frac 12$ . Therefore, the corresponding slopes of $\\overline{AB}$ are $-1, -\\frac 43, 1, \\frac 34, -\\frac 12, -3, \\frac 13$ , and $2$ . The sum of their absolute values is $\\frac{119}{12}$ . The answer is $m+n= \\boxed{131}$", "A very natural solution:\n. Shift $A$ to the origin. Suppose point $B$ was $(x, kx)$ . Note $k$ is the slope we're looking for. Note that point $C$ must be of the form: $(x \\pm 1, kx \\pm 7)$ or $(x \\pm 7, kx \\pm 1)$ or $(x \\pm 5, kx \\pm 5)$ . Note that we want the slope of the line connecting $D$ and $C$ so also be $k$ , since $AB$ and $CD$ are parallel.\nInstead of dealing with the 12 cases, we consider \npoint $C$ of the form $(x \\pm Y, kx \\pm Z)$ where \nwe plug in the necessary values for $Y$ and $Z$ after simplifying.\nSince the slopes of $AB$ and $CD$ must both be $k$ $\\frac{7 - kx \\pm Z}{1 - x \\pm Y} = k \\implies k = \\frac{7 \\pm Z}{1 \\pm Y}$ . Plugging in the possible values of $\\pm 7, \\pm 1, \\pm 5$ in their respective pairs and ruling out degenerate cases, we find the sum is $\\frac{119}{12} \\implies m + n = \\boxed{131}$ - whatRthose" ]

[ "The interquartile range is defined as $Q3 - Q1$ , which is $43 - 33 = 10$ $1.5$ times this value is $15$ , so all values more than $15$ below $Q1$ $33 - 15 = 18$ is an outlier. The only one that fits this is $6$ . All values more than $15$ above $Q3 = 43 + 15 = 58$ are also outliers, of which there are none so there is only $\\boxed{1}$ outlier in total." ]

[ "We have\n\\[\\dfrac{1}{20^{20}} = \\dfrac{1}{(10\\cdot2)^{20}}=\\dfrac{1}{10^{20}\\cdot2^{20}}\\]\nNow we do some estimation. Notice that $2^{20} = 1024^2$ , which means that $2^{20}$ is a little more than $1000^2=1,000,000$ . Multiplying it with $10^{20}$ , we get that the denominator is about $1\\underbrace{00\\dots0}_{26 \\text{ zeros}}$ . Notice that when we divide $1$ by an $n$ digit number as long as n is not a power of 10, there are $n-1$ zeros before the first nonzero digit. This means that when we divide $1$ by the $27$ digit integer $1\\underbrace{00\\dots0}_{26 \\text{ zeros}}$ , there are $\\boxed{26}$ zeros in the initial string after the decimal point. -PCChess", "First rewrite $\\frac{1}{20^{20}}$ as $\\frac{5^{20}}{10^{40}}$ . Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in ${5^{20}}$\n$\\log{5^{20}} = 20\\log{5}$ and memming $\\log{5}\\approx0.69$ (alternatively use the fact that $\\log{5} = 1 - \\log{2}$ ), $\\lfloor{20\\log{5}}\\rfloor+1=\\lfloor{20\\cdot0.69}\\rfloor+1=13+1=14$ digits.\nOur answer is $\\boxed{26}$", "Just as in Solution $2,$ we rewrite $\\dfrac{1}{20^{20}}$ as $\\dfrac{5^{20}}{10^{40}}.$ We then calculate $5^{20}$ entirely by hand, first doing $5^5 \\cdot 5^5,$ then multiplying that product by itself, resulting in $95,367,431,640,625.$ Because this is $14$ digits, after dividing this number by $10$ fourteen times, the decimal point is before the $9.$ Dividing the number again by $10$ twenty-six more times allows a string of $\\boxed{26}$ zeroes to be formed. -OreoChocolate", "We see that $\\frac{1}{20^{20}} = 9.5367432 \\cdot \\cdot \\cdot \\times 10^{-27}$ . We see that this has $27-1=26$ zeros after the decimal point before coming to $9$\nTherefore, the answer is $\\boxed{26}$", "\\begin{align*}|\\lceil \\log \\dfrac{1}{20^{20}} \\rceil| &= |\\lceil \\log 20^{-20} \\rceil| \\\\ &= |\\lceil -20 \\log(20) \\rceil| \\\\ &= |\\lceil -20(\\log 10 + \\log 2) \\rceil| \\\\ &= |\\lceil -20(1 + 0.301) \\rceil| \\\\ &= |\\lceil -26.02 \\rceil| \\\\ &= |-26| \\\\ &= \\boxed{26}" ]

[ "Angle-chasing using the small triangles:\nUse the line below and to the left of the $110^\\circ$ angle to find that the rightmost angle in the small lower-left triangle is $180 - 110 = 70^\\circ$\nThen use the small lower-left triangle to find that the remaining angle in that triangle is $180 - 70 - 40 = 70^\\circ$\nUse congruent vertical angles to find that the lower angle in the smallest triangle containing $A$ is also $70^\\circ$\nNext, use line segment $AB$ to find that the other angle in the smallest triangle containing $A$ is $180 - 100 = 80^\\circ$\nThe small triangle containing $A$ has a $70^\\circ$ angle and an $80^\\circ$ angle. The remaining angle must be $180 - 70 - 80 = \\boxed{30}$", "The third angle of the triangle containing the $100^\\circ$ angle and the $40^\\circ$ angle is $180^\\circ - 100^\\circ - 40^\\circ = 40^\\circ$ . It follows that $A$ is the third angle of the triangle consisting of the found $40^\\circ$ angle and the given $110^\\circ$ angle. Thus, $A$ is a $180^\\circ - 110^\\circ - 40^\\circ = 30^\\circ$ angle, and so the answer is $\\boxed{30}$" ]

[ "Another way to solve this problem would be to use exterior angles. Exterior angles of any polygon add up to $360^{\\circ}$ . Since there are $18$ exterior angles in an 18-gon, the average measure of an exterior angles is $\\frac{360}{18}=20^\\circ$ . We know from the problem that since the exterior angles must be in an arithmetic sequence, the median and average of them is $20$ . Since there are even number of exterior angles, the middle two must be $19^\\circ$ and $21^\\circ$ , and the difference between terms must be $2$ . Check to make sure the smallest exterior angle is greater than $0$ $19-2(8)=19-16=3^\\circ$ . It is, so the greatest exterior angle is $21+2(8)=21+16=37^\\circ$ and the smallest interior angle is $180-37=\\boxed{143}$", "The sum of the angles in a 18-gon is $(18-2) \\cdot 180^\\circ = 2880 ^\\circ.$ Because the angles are in an arithmetic sequence, we can also write the sum of the angles as $a+(a+d)+(a+2d)+\\dots+(a+17d)=18a+153d,$ where $a$ is the smallest angle and $d$ is the common difference. Since these two are equal, we know that $18a+153d = 2880 ^\\circ,$ or $2a+17d = 320^\\circ.$ The smallest value of $d$ that satisfies this is $d=2,$ so $a=143.$ Other values of $d$ and $a$ satisfy that equation, but if we tried any of them the last angle would be greater than $180,$ so the only value of $a$ that works is $a=\\boxed{143}$", "Each individual angle in a $18$ -gon is $\\frac {(18-2) \\cdot 180^\\circ}{18} = 160^\\circ$ . Since no angle in a convex polygon can be larger than $180^\\circ$ , the smallest angle possible is in the set $159, 161, 157, 163, 155, 165, 153, 167, 151, 169, 149, 171, 147, 173, 145, 175, 143, 177$\nOur smallest possible angle is $\\boxed{143}$" ]

[ "It becomes $(x^{8}+...)(x^{9}+...)$ with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or $\\boxed{17}$" ]

[ "We label our points using coordinates $0 \\le x,y \\le 3$ , with the bottom-left point being $(0,0)$ . By the Pythagorean Theorem , the distance between two points is $\\sqrt{d_x^2 + d_y^2}$ where $0 \\le d_x, d_y \\le 3$ ; these yield the possible distances (in decreasing order) \\[\\sqrt{18},\\ \\sqrt{13},\\ \\sqrt{10},\\ \\sqrt{9},\\ \\sqrt{8},\\ \\sqrt{5},\\ \\sqrt{4},\\ \\sqrt{2},\\ \\sqrt{1}\\] As these define $9$ lengths, so the maximum value of $m$ is $10$ . For now, we assume that $m = 10$ is achievable. Because it is difficult to immediately impose restrictions on a path with increasing distances, we consider the paths in shrinking fashion. Note that the shrinking paths and growing paths are equivalent, but there are restrictions upon the locations of the first edges of the former.\nThe $\\sqrt{18}$ length is only possible for one of the long diagonals, so our path must start with one of the $4$ corners of the grid. Without loss of generality (since the grid is rotationally symmetric), we let the vertex be $(0,0)$ and the endpoint be $(3,3)$\nThe $\\sqrt{13}$ length can now only go to $2$ points; due to reflectional symmetry about the main diagonal, we may WLOG let the next endpoint be $(1,0)$\nFrom $(1,0)$ , there are two possible ways to move $\\sqrt{10}$ away, either to $(0,3)$ or $(2,3)$ . However, from $(0,3)$ , there is no way to move $\\sqrt{9}$ away, so we discard it as a possibility.\nFrom $(2,3)$ , the lengths of $\\sqrt{8},\\ \\sqrt{5},\\ \\sqrt{4},\\ \\sqrt{2}$ fortunately are all determined, with the endpoint sequence being $(2,3)-(2,0)-(0,2)-(2,1)-(0,1)-(1,2)$\nFrom $(1,2)$ , there are $3$ possible lengths of $\\sqrt{1}$ (to either $(1,1),(2,2),(1,3)$ ). Thus, the number of paths is $r = 4 \\cdot 2 \\cdot 3 = 24$ , and the answer is $mr = 10 \\cdot 24 = \\boxed{240}$" ]

[ "The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is $\\boxed{4}$ . Similarly, the arrow would be pointing downward at 6:00. It would already have completed three 180 degree turns. Therefore, two 180 degree turns would be completed at 4:00.", "The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. $\\boxed{4}$", "The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so, therefore, the disk must travel its circumference before the arrow goes up. Its circumference is $20\\pi$ , so that is $20\\pi$ traveled on a $60\\pi$ arrow path. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of 4 o' clock. $\\boxed{4}$", "Suppose that the small disk also had a clock face on it, and that both disks were toothed wheels, free to rotate around their centers. The part of the picture where they engage would look like this: The small cog has half the radius, and therefore half the circumference. If the large cog turns $30^\\circ$ anticlockwise (i.e. 1 hour), the small cog turns $60^\\circ$ clockwise (i.e. 2 hours). However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above $30^\\circ$ clockwise to see what happens when the small cog rolls around it. It turns out that, when the point of tangency moves $30^\\circ$ clockwise (one hour), from our point of view the small disk rotates $90^\\circ$ clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of $360^\\circ$ (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is $\\boxed{4}$", "We can approach this problem with angle measures. As the circumference of the disk is $10\\pi,$ and the clock $20 \\pi,$ we have that 30 degrees, or the angular measure between hours, of the disk is only 15 degrees of the clock. This yields that every two hour ticks that the clock rotates, on the third one, the ticks will meet. However, the disk must rotate 360 degrees in order to come back to its original position, so the angular measure that the disk has covered relative to the clock is simply $12 \\cdot 15 \\cdot \\frac{2}{3},$ or $120^\\circ$ from the 12 starting point, so $\\boxed{4}$", "If the big clock were a flat plane, then the smaller clock could travel $\\dfrac{40\\pi}{20\\pi}=2$ full revolutions.\nBut we also need to account for rotation. If we mark a red dot on the bottom of the small clock/bottom of the arrow, and then drag it around the clock, the direction of the arrow would still change. After traveling around the big clock, the small clock would travel $1$ full rotation.\nConsidering these two movements, the small clock travels 3 full rotations around the big clock so the arrow is next pointing upwards at $\\dfrac{12 \\text{ o'clock}}{3}=\\boxed{4}$", "The center of rotation is in the center of the smaller circle, but extends to the center of the larger circle. That means the circumference of the circle in relation to the arrow is $60 \\pi$ . The other circle is $20 \\pi$ and so that is $\\frac{1}{3}$ . So $\\frac{12}{3} = 4$ which is $\\boxed{4}$", "Let the center of the disk be Planet X with orbit eccentricity $0$ and the center of the clock be the Sun. Note that the question would then be asking for the solar day of Planet X, rather than the sidereal day. Because planet X is rotating around its axis in the same direction as it is revolving around the Sun, the solar day can be calculated as $d_{solar}=\\frac{1}{\\frac{1}{y}+\\frac{1}{d_{siderial}}}$ , where $y$ is the length of the year of Planet X and $d_{siderial}$ is one sidereal day. It is easy to see that $y=12$ and $d_{siderial}=6$ , therefore the answer is $\\frac{1}{\\frac{1}{12}+\\frac{1}{6}}=\\frac{1}{\\frac{1}{4}}=\\boxed{4}$ ~sigmapie" ]

[ "In a $1$ -foot-by- $1$ -foot floor, there is $1$ white tile. In a $3$ -by- $3$ , there are $4$ . Continuing on, you can deduce the $n^{th}$ positive odd integer floor has $n^2$ white tiles. $15$ is the $8^{th}$ odd integer, so there are $\\boxed{64}$ white tiles." ]

[ "Call the squares' side lengths from smallest to largest $a_1,\\ldots,a_9$ , and let $l,w$ represent the dimensions of the rectangle.\nThe picture shows that \\begin{align*} a_1+a_2 &= a_3\\\\ a_1 + a_3 &= a_4\\\\ a_3 + a_4 &= a_5\\\\ a_4 + a_5 &= a_6\\\\ a_2 + a_3 + a_5 &= a_7\\\\ a_2 + a_7 &= a_8\\\\ a_1 + a_4 + a_6 &= a_9\\\\ a_6 + a_9 &= a_7 + a_8.\\end{align*}\nExpressing all terms 3 to 9 in terms of $a_1$ and $a_2$ and substituting their expanded forms into the previous equation will give the expression $5a_1 = 2a_2$\nWe can guess that $a_1 = 2$ . (If we started with $a_1$ odd, the resulting sides would not be integers and we would need to scale up by a factor of $2$ to make them integers; if we started with $a_1 > 2$ even, the resulting dimensions would not be relatively prime and we would need to scale down.) Then solving gives $a_9 = 36$ $a_6=25$ $a_8 = 33$ , which gives us $l=61,w=69$ . These numbers are relatively prime, as desired. The perimeter is $2(61)+2(69)=\\boxed{260}$", "We can just list the equations: \\begin{align*} s_3 &= s_1 + s_2 \\\\ s_4 &= s_3 + s_1 \\\\ s_5 &= s_4 + s_3 \\\\ s_6 &= s_5 + s_4 \\\\ s_7 &= s_5 + s_3 + s_2 \\\\ s_8 &= s_7 + s_2 \\\\ s_9 &= s_8 + s_2 - s_1 \\\\ s_9 + s_8 &= s_7 + s_6 + s_5 \\end{align*} We can then write each $s_i$ in terms of $s_1$ and $s_2$ as follows \\begin{align*} s_4 &= 2s_1 + s_2 \\\\ s_5 &= 3s_1 +2s_2 \\\\ s_6 &= 5s_1 + 3s_2 \\\\ s_7 &= 4s_1 + 4s_2 \\\\ s_8 &= 4s_1 + 5s_2 \\\\ s_9 &= 3s_1 + 6s_2 \\\\ \\end{align*} Since $s_9 + s_8 = s_7 + s_6 + s_5 \\implies (3s_1 + 6s_2) + (4s_1 + 5s_2) = (4s_1 + 4s_2) + (5s_1 + 3s_2) + (3s_1 + 2s_2),$ \\[2s_2 = 5s_1 \\implies \\frac{2}{5}s_2 = s_1.\\] Since the side lengths of the rectangle are relatively prime, we can see that $s_1 = 2$ and $s_2 = 5.$ Therefore, $2(2s_9 + s_6 + s_8) = 30s_1 + 40s_2 = \\boxed{260}.$ ~peelybonehead", "We set the side length of the smallest square to 1, and set the side length of square $a_4$ in the previous question to a. We do some \"side length chasing\" and get $4a - 4 = 2a + 5$ . Solving, we get $a = 4.5$ and the side lengths are $61$ and $69$ . Thus, the perimeter of the rectangle is $2(61 + 69) = \\boxed{260}.$" ]

[ "Clearly the arrow marks a value between $10.25$ and $10.5$ , so only $\\text{C}$ and $\\text{D}$ are possible.\nLooking, we see that the arrow is closer to $10.3$ , so $\\boxed{10.3}$" ]

[ "After 4 hours, we see that Bjorn biked 45 miles, and Alberto biked 60. Thus the answer is $60-45=15$ $\\boxed{15}$", "We see that each dot is $15$ units away from the nearest one above it. So the answer is $\\boxed{15}$" ]

[ "Let the radius of the circles be $r$ . The longer dimension of the rectangle can be written as $14r$ , and by the Pythagorean Theorem , we find that the shorter dimension is $2r\\left(\\sqrt{3}+1\\right)$\nTherefore, $\\frac{14r}{2r\\left(\\sqrt{3}+1\\right)}= \\frac{7}{\\sqrt{3} + 1} \\cdot \\left[\\frac{\\sqrt{3}-1}{\\sqrt{3}-1}\\right] = \\frac{1}{2}\\left(7\\sqrt{3} - 7\\right) = \\frac{1}{2}\\left(\\sqrt{p}-q\\right)$ . Thus we have $p=147$ and $q=7$ , so $p+q=\\boxed{154}$", "Since we only care about the ratio between the longer side and shorter side, we can set the longer side to $14$ . So, this means that each of the radii is $1$ . Now, we connect the radii of three circles such that they form an equilateral triangle with side length 4. Obviously, the height this triangle is $2\\sqrt{3}$ , and the shorter side if the triangle is therefore $2\\sqrt{3}+2$ and we use simplification similar to as showed above, and we reach the result $\\frac{1}{2} \\cdot (\\sqrt{147}-7)$ and the final answer is $147+7 = \\boxed{154}$" ]

[ "Let the two digits be $a$ and $b$ . Then, $5a + 5b = 10a + b - 10b - a = 9a - 9b$ , or $2a = 7b$ . This yields $a = 7$ and $b = 2$ because $a, b < 10$ . Then, $72 + 27 = \\boxed{99}.$", "We start like above. Let the two digits be $a$ and $b$ . Therefore, $5(a+b) = 10a+b-10b-a=9(a-b)$ . Since we are looking for $10a+b+10b+a=11(a+b)$ and we know that $a+b$ must be a multiple of $9$ , the only answer choice that works is $\\boxed{99}.$" ]

[ "Denote the $2$ roots of this quadratic as $r_1$ and $r_2$ . Note that $(r_1-r_2)^2=(r_1+r_2)^2-4r_1r_2$ . By Vieta's Formula's $r_1+r_2=7$ , and $r_1r_2=-9$ . Thus, $r_1-r_2=\\sqrt{49+4\\cdot 9}=\\boxed{85}$" ]

[ "The highest digit sum for three-digit numbers is $9+9+9=27$ . Therefore, the only possible digit combination is $9, 9, 8$ . Of course, of the three possible numbers, only $998$ works. Thus, the answer is $\\boxed{1}$" ]

[ "We see that since $QRS$ is divisible by $5$ $S$ must equal either $0$ or $5$ , but it cannot equal $0$ , so $S=5$ . We notice that since $PQR$ must be even, $R$ must be either $2$ or $4$ . However, when $R=2$ , we see that $T \\equiv 2 \\pmod{3}$ , which cannot happen because $2$ and $5$ are already used up; so $R=4$ . This gives $T \\equiv 3 \\pmod{4}$ , meaning $T=3$ . Now, we see that $Q$ could be either $1$ or $2$ , but $14$ is not divisible by $4$ , but $24$ is. This means that $Q=2$ and $P=\\boxed{1}$", "We know that out of $PQRST,$ $QRS$ is divisible by $5$ . Therefore $S$ is obviously 5 because $QRS$ is divisible by 5. So we now have $PQR5T$ as our number. Next, let's move on to the second piece of information that was given to us. $RST$ is divisible by 3. So, according to the divisibility by 3 rule, the sum of $RST$ has to be a multiple of 3. The only 2 big enough are 9 and 12 and since 5 is already given. The possible sums of $RT$ are 4 and 7. So, the possible values for $R$ are 1,3,4,3 and the possible values of $T$ are 3,1,3,4. So, using this we can move on to the fact that $PQR$ is divisible by 4. So, using that we know that $R$ has to be even so 4 is the only possible value for $R$ . Using that we also know that 3 is the only possible value for 3. So, we have $PQRST$ $PQ453$ so the possible values are 1 and 2 for $P$ and $Q$ . Using the divisibility rule of 4 we know that $QR$ has to be divisible by 4. So, either 14 or 24 are the possibilities, and 24 is divisible by 4. So the only value left for $P$ is 1. $P=\\boxed{1}$", "We know that $QRS$ is divisible by $5$ , so $S$ would be either $5$ or $0$ . However, $0$ is not a choice, so $S=5$ . Also, $PQR$ is divisible by $4$ , so this means that $QR$ is $12$ $32$ $24$ , or $52$ . If $R=2$ , then $T$ has to be $2$ or $5$ $RST$ is divisible by $3$ ), but both are taken. So, $R=4 \\Rightarrow QR=24$ $R+S+T$ must equal $9$ or $12$ , but because $4+5=9$ $R+S+T=12 \\Rightarrow T=3$ . This leaves $P=\\boxed{1}$", "We can simply try each of the answer choice, and we will see which one works. Trying $P=\\boxed{1}$" ]

[ "First, let us consider the case where $0$ is a base: This would result in the entire expression being $0.$ Contrastingly, if $0$ is an exponent, we will get a value greater than $0.$ $3^2\\times2^0=9$ is greater than $2^3\\times2^0=8$ and $2^2\\times3^0=4.$ Therefore, the answer is $\\boxed{9}.$", "The maximum possible value of using the digits $2,0,2,$ and $3$ : We can maximize our value by keeping the $3$ and $2$ together in one power (the biggest with the biggest and the smallest with the smallest). This shows $3^{2}\\times2^{0}=9\\times1=9.$ (We don't want $0^{2}$ because that is $0$ .) It is going to be $\\boxed{9}.$", "Trying all $12$ distinct orderings, we see that the only possible values are $0,4,8,$ and $9,$ the greatest of which is $\\boxed{9}.$" ]

[ "Since the number is even, the last digit must be $2$ or $4$ . To make the smallest possible number, the ten-thousands digit must be as small as possible, so the ten-thousands digit is $1$ . Simillarly, the thousands digit has second priority, so it must also be as small as possible once the ten-thousands digit is decided, so the thousands digit is $2$ . Similarly, the hundreds digit needs to be the next smallest number, so it is $3$ . However, for the tens digit, we can't use $4$ , since we already used $2$ and the number must be even, so the units digit must be $4$ and the tens digit is $9, \\boxed{9}$ (The number is $12394$ .)" ]

[ "A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form ${\\underline{(n+3)}}\\,{\\underline{(n+2)}}\\,{\\underline{( n+1)}}\\,{\\underline {(n)}}$ $= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n$ , for $n \\in \\lbrace0, 1, 2, 3, 4, 5, 6\\rbrace$\nNow, note that $3\\cdot 37 = 111$ so $30 \\cdot 37 = 1110$ , and $90 \\cdot 37 = 3330$ so $87 \\cdot 37 = 3219$ . So the remainders are all congruent to $n - 9 \\pmod{37}$ . However, these numbers are negative for our choices of $n$ , so in fact the remainders must equal $n + 28$\nAdding these numbers up, we get $(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\\cdot28 = \\boxed{217}$" ]

[ "Using the perimeter and area formulas, \\[2(x+y) = \\frac{2}{3}(a+b)\\] \\[x+y = \\frac{a+b}{3}\\] \\[xy = \\frac{ab}{3}\\] Dividing the second equation by the last equation results in \\[\\frac1y + \\frac1x = \\frac1b + \\frac1a\\] Since $x,y < a$ $\\tfrac1a < \\tfrac1x, \\tfrac1y$ . Since $a < b$ $\\tfrac1b < \\tfrac1a$ . That means \\[\\tfrac1x + \\tfrac1y > \\tfrac1a + \\tfrac1a > \\tfrac1a + \\tfrac1b\\] This is a contradiction, so there are $\\boxed{0}$ rectangles that satisfy the conditions." ]

[ "Given an arbitrary product and an arbitrary amount of terms to multiply to get that product, to maximize the sum, make all of the terms $1$ with the last one being the number. To minimize the sum, make all of the terms equal to each other. (This is a corollary that follows from the $AM-GM$ proof.) Since $2002$ is not a perfect cube, we have to make the terms as close to each other as possible. A good rule of thumb is the memorize the prime factorization of the AMC year that you are doing, which is $2002= 2 \\cdot 7 \\cdot 11 \\cdot 13$ . The three terms that are closest to each other that multiply to $2002$ are $11, 13,$ and $14$ , so our answer is $11+13+14=\\boxed{38}$" ]

[ "If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$ . If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$ . Thus, if the number of columns is $n$ , the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \\leq 14$ . In fact, when $n = 14$ we have $n(n + 7) = 14\\cdot 21 = 294 = 17^2 + 5$ , so this number works and no larger number can. Thus, the answer is $\\boxed{294}$", "Let there be $m$ members and $n$ members for the square and $c$ for the number of columns of the other formation. We have $n^2 +5 = c(c+7) \\implies n^2+5 = \\left(c+\\frac{7}{2}\\right)^2 -\\frac{49}{4} \\implies n^2 - \\left(c+\\frac{7}{2}\\right)^2 = -\\frac{69}{4} \\implies \\left(n-c-\\frac{7}{2}\\right)\\left(n + c +\\frac{7}{2}\\right) \\implies (2n-2c-7)(2n+2n+7) = -69.$\nTo maximize this we let $2n+2c+7 = 68$ and $2n-2c-7 = 1.$ Solving we find $n = 17$ so the desired number of members is $17^2 + 5 = \\boxed{294}.$", "Think of the process of moving people from the last column to new rows. Since there are less columns than rows, for each column removed, there are people discarded to the \"extra\" pile to be placed at the end. To maximize the number of \"extra\" people to fill in the last few rows. We remove 3 columns and add 4 rows. For the first new row, one more person will be discarded. For the second, 1 extra person are added since there is one more row now, and there are 2 less columns. Thus, there are 3 extra people discarded. Similarly, 5 extra people are discarded for the third column. Now there are $5+1+3+5=14$ people in the extra pile to put as the last row, so there are $14(14+7)=\\boxed{294}$ people.", "Note: Only do this if you have a LOT of time (and you've memorized all your perfect squares up to 1000).\nWe can see that the number of members in the band must be of the form $n(n + 7)$ for some positive integer $n$ . When $n = 28$ , this product is $980$ , and since AIME answers are nonnegative integers less than $1000$ , we don't have to check any higher $n$ . Also, we know that this product must be 5 more than a perfect square, so we can make a table as shown and bash: \\[\\begin{tabular}{|c|c|c|} n & n(n+7) & 5 more than a perfect square?\\\\ \\hline 1 & 8 & no\\\\ \\hline 2 & 18 & no\\\\ \\hline 3 & 30 & yes\\\\ \\hline 4 & 44 & no\\\\ \\hline 5 & 60 & no\\\\ \\hline 6 & 78 & no\\\\ \\hline 7 & 98 & no\\\\ \\hline 8 & 120 & no\\\\ \\hline 9 & 144 & no\\\\ \\hline 10 & 170 & no\\\\ \\hline 11 & 198 & no\\\\ \\hline 12 & 228 & no\\\\ \\hline 13 & 260 & no\\\\ \\hline 14 & 294 & yes\\\\ \\hline 15 & 330 & no\\\\ \\hline 16 & 368 & no\\\\ \\hline 17 & 408 & no\\\\ \\hline 18 & 450 & no\\\\ \\hline 19 & 494 & no\\\\ \\hline 20 & 540 & no\\\\ \\hline 21 & 588 & no\\\\ \\hline 22 & 638 & no\\\\ \\hline 23 & 690 & no\\\\ \\hline 24 & 744 & no\\\\ \\hline 25 & 800 & no\\\\ \\hline 26 & 858 & no\\\\ \\hline 27 & 918 & no\\\\ \\hline 28 & 980 & no\\\\ \\hline \\end{tabular}\\] Thus, we can see that our largest valid $n(n+7)$ is $\\boxed{294}$" ]

[ "There are $21$ pairs of consecutive exits. To find the maximum number of miles of one of these, the other $20$ must be equal to the minimum number yielding a total of $(5)(20)=100$ miles. The longest distance must be $118-100=\\boxed{18}$" ]

[ "We start with the same method as above. The domain of the arcsin function is $[-1, 1]$ , so $-1 \\le \\log_{m}(nx) \\le 1$\n\\[\\frac{1}{m} \\le nx \\le m\\] \\[\\frac{1}{mn} \\le x \\le \\frac{m}{n}\\] \\[\\frac{m}{n} - \\frac{1}{mn} = \\frac{1}{2013}\\] \\[n = 2013m - \\frac{2013}{m}\\]\nFor $n$ to be an integer, $m$ must divide $2013$ , and $m > 1$ . To minimize $n$ $m$ should be as small as possible because increasing $m$ will decrease $\\frac{2013}{m}$ , the amount you are subtracting, and increase $2013m$ , the amount you are adding; this also leads to a small $n$ which clearly minimizes $m+n$\nWe let $m$ equal $3$ , the smallest factor of $2013$ that isn't $1$ . Then we have $n = 2013*3 - \\frac{2013}{3} = 6039 - 671 = 5368$\n$m + n = 5371$ , so the answer is $\\boxed{371}$", "Note that we need $-1\\le f(x)\\le 1$ , and this eventually gets to $\\frac{m^2-1}{mn}=\\frac{1}{2013}$ . From there, break out the quadratic formula and note that \\[m= \\frac{n+\\sqrt{n^2+4026^2}}{2013\\times 2}.\\] Then we realize that the square root, call it $a$ , must be an integer. Then $(a-n)(a+n)=4026^2.$\nObserve carefully that $4026^2 = 2\\times 2\\times 3\\times 3\\times 11\\times 11\\times 61\\times 61$ ! It is not difficult to see that to minimize the sum, we want to minimize $n$ as much as possible. Seeing that $2a$ is even, we note that a $2$ belongs in each factor. Now, since we want to minimize $a$ to minimize $n$ , we want to distribute the factors so that their ratio is as small as possible (sum is thus minimum). The smallest allocation of $2, 61, 61$ and $2, 11, 3, 3, 11$ fails; the next best is $2, 61, 11, 3, 3$ and $2, 61, 11$ , in which $a=6710$ and $n=5368$ . That is our best solution, upon which we see that $m=3$ , thus $\\boxed{371}$" ]

[ "For all real numbers $a,b,$ and $c$ such that $b>0$ and $b\\neq1,$ note that:\nTherefore, we have \\begin{align*} \\log_{\\frac12}(\\log_4(\\log_{\\frac14}(\\log_{16}(\\log_{\\frac1{16}}x)))) \\text{ is defined} &\\implies \\log_4(\\log_{\\frac14}(\\log_{16}(\\log_{\\frac1{16}}x)))>0 \\\\ &\\implies \\log_{\\frac14}(\\log_{16}(\\log_{\\frac1{16}}x))>1 \\\\ &\\implies 0<\\log_{16}(\\log_{\\frac1{16}}x)<\\frac14 \\\\ &\\implies 1<\\log_{\\frac1{16}}x<2 \\\\ &\\implies \\frac{1}{256}<x<\\frac{1}{16}. \\end{align*} The domain of $f(x)$ is an interval of length $\\frac{1}{16}-\\frac{1}{256}=\\frac{15}{256},$ from which the answer is $15+256=\\boxed{271}.$", "For simplicity, let $a=\\log_{\\frac{1}{16}}{x},b=\\log_{16}a,c=\\log_{\\frac{1}{4}}b$ , and $d=\\log_4c$\nThe domain of $\\log_{\\frac{1}{2}}x$ is $x \\in (0, \\infty)$ , so $d \\in (0, \\infty)$ .\nThus, $\\log_4{c} \\in (0, \\infty) \\Rightarrow c \\in (1, \\infty)$ .\nSince $c=\\log_{\\frac{1}{4}}b$ we have $b \\in \\left(0, \\left(\\frac{1}{4}\\right)^1\\right)=\\left(0, \\frac{1}{4}\\right)$ .\nSince $b=\\log_{16}{a}$ , we have $a \\in (16^0,16^{1/4})=(1,2)$ .\nFinally, since $a=\\log_{\\frac{1}{16}}{x}$ $x \\in \\left(\\left(\\frac{1}{16}\\right)^2,\\left(\\frac{1}{16}\\right)^1\\right)=\\left(\\frac{1}{256},\\frac{1}{16}\\right)$\nThe length of the $x$ interval is $\\frac{1}{16}-\\frac{1}{256}=\\frac{15}{256}$ and the answer is $\\boxed{271}$", "The domain of $f(x)$ is the range of the inverse function $f^{-1}(x)=\\left(\\frac1{16}\\right)^{16^{\\left(\\frac14\\right)^{4^{\\left(\\frac12\\right)^x}}}}$ . Now $f^{-1}(x)$ can be seen to be strictly decreasing, since $\\left(\\frac12\\right)^x$ is decreasing, so $4^{\\left(\\frac12\\right)^x}$ is decreasing, so $\\left(\\frac14\\right)^{4^{\\left(\\frac12\\right)^x}}$ is increasing, so $16^{\\left(\\frac14\\right)^{4^{\\left(\\frac12\\right)^x}}}$ is increasing, therefore $\\left(\\frac1{16}\\right)^{16^{\\left(\\frac14\\right)^{4^{\\left(\\frac12\\right)^x}}}}$ is decreasing.\nTherefore, the range of $f^{-1}(x)$ is the open interval $\\left(\\lim_{x\\to\\infty}f^{-1}(x), \\lim_{x\\to-\\infty}f^{-1}(x)\\right)$ . We find: \\begin{align*} \\lim_{x\\to-\\infty}\\left(\\frac1{16}\\right)^{16^{\\left(\\frac14\\right)^{4^{\\left(\\frac12\\right)^x}}}}&= \\lim_{a\\to\\infty}\\left(\\frac1{16}\\right)^{16^{\\left(\\frac14\\right)^{4^a}}}\\\\ &= \\lim_{b\\to\\infty}\\left(\\frac1{16}\\right)^{16^{\\left(\\frac14\\right)^b}}\\\\ &= \\left(\\frac1{16}\\right)^{16^0}\\\\ &= \\frac{1}{16}. \\end{align*} Similarly, \\begin{align*} \\lim_{x\\to\\infty}\\left(\\frac1{16}\\right)^{16^{\\left(\\frac14\\right)^{4^{\\left(\\frac12\\right)^x}}}}&=\\left(\\frac1{16}\\right)^{16^{\\left(\\frac14\\right)^{4^0}}}\\\\ &= \\left(\\frac1{16}\\right)^{16^{\\frac14}}\\\\ &= \\left(\\frac1{16}\\right)^2\\\\ &= \\frac{1}{256}. \\end{align*} Hence the range of $f^{-1}(x)$ (which is then the domain of $f(x)$ ) is $\\left(\\frac{1}{256},\\frac{1}{16}\\right)$ and the answer is $\\boxed{271}$" ]

[ "Let $x$ be the number of students taking both a math and a foreign language class.\nBy P-I-E, we get $70 + 54 - x$ $93$\nSolving gives us $x = 31$\nBut we want the number of students taking only a math class,\nwhich is $70 - 31 = 39$\n$\\boxed{39}$", "We have $70 + 54 = 124$ people taking classes. However, we over-counted the number of people who take both classes. If we subtract the original amount of people who take classes we get that $31$ people took the two classes. To find the amount of people who took only math class, we subtract the people who didn't take only one math class, so we get $70 - 31 = \\boxed{39}$", "We are looking for students in math only, which is the complement (exactly the rest of the students) compared to those taking a language class. Since $54$ students take a language (with or without math), we subtract that from the total number of students. Since $93-54= 39,$ our answer is $\\boxed{39}.$ (It's not necessary to know that $70$ students take math.)\n~hailstone" ]

[ "\nWe know that the sum of all three areas is $93$ So, we have: \\[93 = 70-x+x+54-x\\] \\[93 = 70+54-x\\] \\[93 = 124 - x\\] \\[-31=-x\\] \\[x=31\\]\nWe are looking for the number of students in only math. This is $70-x$ . Substituting $x$ with $31$ , our answer is $\\boxed{39}$" ]

[ "Observe that all tables must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each table, there exists a valid table diagonally symmetrical across the diagonal extending from the top left to the bottom right.\n\\[\\begin{tabular}{|c|c|c|} \\hline 1&2&\\\\ \\hline 3&4&8\\\\ \\hline &&9\\\\ \\hline \\end{tabular} \\;\\;\\; \\begin{tabular}{|c|c|c|} \\hline 1&2&\\\\ \\hline 3&4&\\\\ \\hline &8&9\\\\ \\hline \\end{tabular}\\]\n3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. $2*6=12$\n\\[\\begin{tabular}{|c|c|c|} \\hline 1&2&3\\\\ \\hline 4&5&\\\\ \\hline &8&9\\\\ \\hline \\end{tabular} \\;\\;\\; \\begin{tabular}{|c|c|c|} \\hline 1&2&\\\\ \\hline 3&5&\\\\ \\hline &8&9\\\\ \\hline \\end{tabular} \\;\\;\\; \\begin{tabular}{|c|c|c|} \\hline 1&2&\\\\ \\hline 3&5&8\\\\ \\hline &&9\\\\ \\hline \\end{tabular} \\;\\;\\; \\begin{tabular}{|c|c|c|} \\hline 1&2&3\\\\ \\hline 4&5&8\\\\ \\hline &&9\\\\ \\hline \\end{tabular}\\]\nHere, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that $4<5$ , logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, $2*9=18$\nBy inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured. $2*6=12$\n\\[12+18+12=\\boxed{42}\\]", "This solution is trivial by the hook length theorem. The hooks look like this:\n$\\begin{tabular}{|c|c|c|} \\hline 5 & 4 & 3 \\\\ \\hline 4 & 3 & 2\\\\ \\hline 3 & 2 & 1\\\\ \\hline \\end{tabular}$\nSo, the answer is $\\frac{9!}{5 \\cdot 4 \\cdot 3 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 3 \\cdot 2 \\cdot 1}$ $\\boxed{42}$" ]

[ "Observe that all tables must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each table, there exists a valid table diagonally symmetrical across the diagonal extending from the top left to the bottom right.\n\\[\\begin{tabular}{|c|c|c|} \\hline 1&2&\\\\ \\hline 3&4&8\\\\ \\hline &&9\\\\ \\hline \\end{tabular} \\;\\;\\; \\begin{tabular}{|c|c|c|} \\hline 1&2&\\\\ \\hline 3&4&\\\\ \\hline &8&9\\\\ \\hline \\end{tabular}\\]\n3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. $2*6=12$\n\\[\\begin{tabular}{|c|c|c|} \\hline 1&2&3\\\\ \\hline 4&5&\\\\ \\hline &8&9\\\\ \\hline \\end{tabular} \\;\\;\\; \\begin{tabular}{|c|c|c|} \\hline 1&2&\\\\ \\hline 3&5&\\\\ \\hline &8&9\\\\ \\hline \\end{tabular} \\;\\;\\; \\begin{tabular}{|c|c|c|} \\hline 1&2&\\\\ \\hline 3&5&8\\\\ \\hline &&9\\\\ \\hline \\end{tabular} \\;\\;\\; \\begin{tabular}{|c|c|c|} \\hline 1&2&3\\\\ \\hline 4&5&8\\\\ \\hline &&9\\\\ \\hline \\end{tabular}\\]\nHere, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that $4<5$ , logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, $2*9=18$\nBy inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured. $2*6=12$\n\\[12+18+12=\\boxed{42}\\]", "This solution is trivial by the hook length theorem. The hooks look like this:\n$\\begin{tabular}{|c|c|c|} \\hline 5 & 4 & 3 \\\\ \\hline 4 & 3 & 2\\\\ \\hline 3 & 2 & 1\\\\ \\hline \\end{tabular}$\nSo, the answer is $\\frac{9!}{5 \\cdot 4 \\cdot 3 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 3 \\cdot 2 \\cdot 1}$ $\\boxed{42}$" ]

[ "Let $t = 1/x$ . After multiplying the equation by $t^{10}$ $1 + (13 - t)^{10} = 0\\Rightarrow (13 - t)^{10} = - 1$\nUsing DeMoivre, $13 - t = \\text{cis}\\left(\\frac {(2k + 1)\\pi}{10}\\right)$ where $k$ is an integer between $0$ and $9$\n$t = 13 - \\text{cis}\\left(\\frac {(2k + 1)\\pi}{10}\\right) \\Rightarrow \\bar{t} = 13 - \\text{cis}\\left(-\\frac {(2k + 1)\\pi}{10}\\right)$\nSince $\\text{cis}(\\theta) + \\text{cis}(-\\theta) = 2\\cos(\\theta)$ $t\\bar{t} = 170 - 26\\cos \\left(\\frac {(2k + 1)\\pi}{10}\\right)$ after expanding. Here $k$ ranges from 0 to 4 because two angles which sum to $2\\pi$ are involved in the product.\nThe expression to find is $\\sum t\\bar{t} = 850 - 26\\sum_{k = 0}^4 \\cos \\frac {(2k + 1)\\pi}{10}$\nBut $\\cos \\frac {\\pi}{10} + \\cos \\frac {9\\pi}{10} = \\cos \\frac {3\\pi}{10} + \\cos \\frac {7\\pi}{10} = \\cos \\frac {\\pi}{2} = 0$ so the sum is $\\boxed{850}$", "Divide both sides by $x^{10}$ to get \\[1 + \\left(13-\\dfrac{1}{x}\\right)^{10}=0\\]\nRearranging: \\[\\left(13-\\dfrac{1}{x}\\right)^{10} = -1\\]\nThus, $13-\\dfrac{1}{x} = \\omega$ where $\\omega = e^{i(\\pi n/5+\\pi/10)}$ where $n$ is an integer.\nWe see that $\\dfrac{1}{x}=13-\\omega$ . Thus, \\[\\dfrac{1}{x\\overline{x}}=(13\\, -\\, \\omega)(13\\, -\\, \\overline{\\omega})=169-13(\\omega\\, +\\, \\overline{\\omega})\\, +\\, \\omega\\overline{\\omega}=170\\, -\\, 13(\\omega\\, +\\, \\overline{\\omega})\\]\nSumming over all terms: \\[\\dfrac{1}{r_1\\overline{r_1}}+\\cdots + \\dfrac{1}{r_5\\overline{r_5}} = 5\\cdot 170 - 13(e^{i\\pi/10}+\\cdots +e^{i(9\\pi/5+\\pi/10)})\\]\nHowever, note that $e^{i\\pi/10}+\\cdots +e^{i(9\\pi/5+\\pi/10)}=0$ from drawing the numbers on the complex plane, our answer is just \\[5\\cdot 170=\\boxed{850}\\]" ]

[ "We may factor the equation as:\n\\begin{align*} 2000x^6+100x^5+10x^3+x-2&=0\\\\ 2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\\\ 2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\\\ 2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\\\ (20x^2+x-2)(100x^4+10x^2+1)&=0\\\\ \\end{align*}\nNow $100x^4+10x^2+1\\ge 1>0$ for real $x$ . Thus the real roots must be the roots of the equation $20x^2+x-2=0$ . By the quadratic formula the roots of this are:\n\\[x=\\frac{-1\\pm\\sqrt{1^2-4(-2)(20)}}{40} = \\frac{-1\\pm\\sqrt{1+160}}{40} = \\frac{-1\\pm\\sqrt{161}}{40}.\\]\nThus $r=\\frac{-1+\\sqrt{161}}{40}$ , and so the final answer is $-1+161+40 = \\boxed{200}$", "It would be really nice if the coefficients were symmetrical. What if we make the substitution, $x = -\\frac{i}{\\sqrt{10}}y$ . The the polynomial becomes\n$-2y^6 - (\\frac{i}{\\sqrt{10}})y^5 + (\\frac{i}{\\sqrt{10}})y^3 - (\\frac{i}{\\sqrt{10}})y - 2$\nIt's symmetric! Dividing by $y^3$ and rearranging, we get\n$-2(y^3 + \\frac{1}{y^3}) - (\\frac{i}{\\sqrt{10}})(y^2 + \\frac{1}{y^2}) + (\\frac{i}{\\sqrt{10}})$\nNow, if we let $z = y + \\frac{1}{y}$ , we can get the equations\n$z = y + \\frac{1}{y}$\n$z^2 - 2 = y^2 + \\frac{1}{y^2}$\nand\n$z^3 - 3z = y^3 + \\frac{1}{y^3}$\n(These come from squaring $z$ and subtracting $2$ , then multiplying that result by $z$ and subtracting $z$ ) \nPlugging this into our polynomial, expanding, and rearranging, we get\n$-2z^3 - (\\frac{i}{\\sqrt{10}})z^2 + 6z + (\\frac{3i}{\\sqrt{10}})$\nNow, we see that the two $i$ terms must cancel in order to get this polynomial equal to $0$ , so what squared equals 3? Plugging in $z = \\sqrt{3}$ into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying $z = -\\sqrt{3}$ , we see that it also works! Great, we use long division on the polynomial by\n$(z - \\sqrt{3})(z + \\sqrt{3}) = (z^2 - 3)$ and we get\n$2z -(\\frac{i}{\\sqrt{10}}) = 0$\nWe know that the other two solutions for z wouldn't result in real solutions for $x$ since we have to solve a quadratic with a negative discriminant, then multiply by $-(\\frac{i}{\\sqrt{10}})$ . We get that $z = (\\frac{i}{-2\\sqrt{10}})$ . Solving for $y$ (using $y + \\frac{1}{y} = z$ ) we get that $y = \\frac{-i \\pm \\sqrt{161}i}{4\\sqrt{10}}$ , and multiplying this by $-(\\frac{i}{\\sqrt{10}})$ (because $x = -(\\frac{i}{\\sqrt{10}})y$ ) we get that $x = \\frac{-1 \\pm \\sqrt{161}}{40}$ for a final answer of $-1 + 161 + 40 = \\boxed{200}$", "Observe that the given equation may be rearranged as $2000x^6-2+(100x^5+10x^3+x)=0$ .\nThe expression in parentheses is a geometric series with common factor $10x^2$ . Using the geometric sum formula, we rewrite as $2000x^6-2+\\frac{1000x^7-x}{10x^2-1}=0, 10x^2-1\\neq0$ .\nFactoring a bit, we get $2(1000x^6-1)+(1000x^6-1)\\frac{x}{10x^2-1}=0, 10x^2-1\\neq0 \\implies$ $(1000x^6-1)(2+\\frac{x}{10x^2-1})=0, 10x^2-1\\neq0$ .\nNote that setting $1000x^6-1=0$ gives $10x^2-1=0$ , which is clearly extraneous.\nHence, we set $2+\\frac{x}{10x^2-1}=0$ and use the quadratic formula to get the desired root $x=\\frac{-1+\\sqrt{161}}{40} \\implies -1+161+40=\\boxed{200}$" ]

[ "Let $y = 2^{111x}$ . Then our equation reads $\\frac{1}{4}y^3 + 4y = 2y^2 + 1$ or $y^3 - 8y^2 + 16y - 4 = 0$ . Thus, if this equation has roots $r_1, r_2$ and $r_3$ , by Vieta's formulas we have $r_1\\cdot r_2\\cdot r_3 = 4$ . Let the corresponding values of $x$ be $x_1, x_2$ and $x_3$ . Then the previous statement says that $2^{111\\cdot(x_1 + x_2 + x_3)} = 4$ so taking a logarithm of that gives $111(x_1 + x_2 + x_3) = 2$ and $x_1 + x_2 + x_3 = \\frac{2}{111}$ . Thus the answer is $111 + 2 = \\boxed{113}$" ]

[ "$x + \\sqrt{x-2} = 4$ Original Equation\n$\\sqrt{x-2} = 4 - x$ Subtract x from both sides\n$x-2 = 16 - 8x + x^2$ Square both sides\n$x^2 - 9x + 18 = 0$ Get all terms on one side\n$(x-6)(x-3) = 0$ Factor\n$x = \\{6, 3\\}$\nIf you put down A as your answer, it's wrong. You need to check for extraneous roots.\n$6 + \\sqrt{6 - 2} = 6 + \\sqrt{4} = 6 + 2 = 8 \\ne 4$\n$3 + \\sqrt{3-2} = 3 + \\sqrt{1} = 3 + 1 = 4 \\checkmark$\nThere is $\\boxed{1}$", "We can create symmetry in the equation: \\[x+\\sqrt{x-2} = 4\\] \\[x-2+\\sqrt{x-2} = 2.\\] Let $y = \\sqrt{x-2}$ , then we have \\[y^2+y-2 = 0\\] \\[(y+2)(y-1) = 0\\] The two roots are $\\sqrt{x-2} = -2, 1$\nNotice, that the first root is extraneous as the range for the square root function is always the non-negative numbers (remember, negative numbers in square roots give imaginary numbers - imaginary numbers in square roots don't give negative numbers); thus, the only real root for $x$ occurs for the second root; squaring both sides and solving for $x$ gives $x=3 \\Rightarrow \\boxed{1}$" ]

[ "According to Wolfram-Alpha, the answer is $\\boxed{118}$" ]

[ "We shall introduce another factor to make the equation easier to solve. If $r$ is a root of $z^6+z^3+1$ , then $0=(r^3-1)(r^6+r^3+1)=r^9-1$ . The polynomial $x^9-1$ has all of its roots with absolute value $1$ and argument of the form $40m^\\circ$ for integer $m$ (the ninth degree roots of unity ). Now we simply need to find the root within the desired range that satisfies our original equation $x^6 + x^3 + 1 = 0$\nThis reduces $\\theta$ to either $120^{\\circ}$ or $160^{\\circ}$ . But $\\theta$ can't be $120^{\\circ}$ because if $r=\\cos 120^\\circ +i\\sin 120^\\circ$ , then $r^6+r^3+1=3$ . (When we multiplied by $r^3 - 1$ at the beginning, we introduced some extraneous solutions, and the solution with $120^\\circ$ was one of them.) This leaves $\\boxed{160}$", "The substitution $y=z^3$ simplifies the equation to $y^2+y+1 = 0$ . Applying the quadratic formula gives roots $y=-\\frac{1}{2}\\pm \\frac{\\sqrt{3}i}{2}$ , which have arguments of $120$ and $240,$ respectively. \nWe can write them as $z^3 = \\cos 240^\\circ + i\\sin 240^\\circ$ and $z^3 = \\cos 120^\\circ + i\\sin 120^\\circ$ . \nSo we can use De Moivre's theorem (which I would suggest looking at if you never heard of it before) to find the fractional roots of the expressions above! \nFor $\\cos 240^\\circ + i\\sin 240$ we have $(\\cos 240^\\circ + i\\sin 240^\\circ)^{1/3}$ $\\Rightarrow$ $\\cos 80^\\circ + i\\sin 80^\\circ, \\cos 200^\\circ + i\\sin200^\\circ,$ and $\\cos 320^\\circ + i\\sin320^\\circ.$ Similarly for $(\\cos 120^\\circ + i\\sin 120^\\circ)^{1/3}$ , we have $\\cos 40^\\circ + i\\sin 40^\\circ, \\cos 160^\\circ + i\\sin 160^\\circ,$ and $\\cos 280^\\circ + i\\sin 280^\\circ.$ The only argument out of all these roots that fits the description is $\\theta = \\boxed{160}$" ]

[ "$2x + 7 = 3 \\Longrightarrow x = -2, \\quad -2b - 10 = -2 \\Longrightarrow -2b = 8 \\Longrightarrow b = \\boxed{4}$" ]

[ "$2x + 7 = 3 \\Longrightarrow x = -2, \\quad -2b - 10 = -2 \\Longrightarrow -2b = 8 \\Longrightarrow b = \\boxed{4}$" ]

[ "Solution by e_power_pi_times_i\n$4n = m$ , as stated in the question. In the line $L_1$ , draw a triangle with the coordinates $(0,0)$ $(1,0)$ , and $(1,m)$ . Then $m = \\tan(\\theta_1)$ . Similarly, $n = \\tan(\\theta_2)$ . Since $4n = m$ and $\\theta_1 = 2\\theta_2$ $\\tan(2\\theta_2) = 4\\tan(\\theta_2)$ . Using the angle addition formula for tangents, $\\dfrac{2\\tan(\\theta_2)}{1-\\tan^2(\\theta_2)} = 4\\tan(\\theta_2)$ . Solving, we have $\\tan(\\theta_2) = 0, \\dfrac{\\sqrt{2}}{2}$ . But line $L_1$ is not horizontal, so therefore $(m,n) = (2\\sqrt{2},\\dfrac{\\sqrt{2}}{2})$ . Looking at the answer choices, it seems the answer is $(2\\sqrt{2})(\\dfrac{\\sqrt{2}}{2}) = \\boxed{2}$" ]

[ "By the Multinomial Theorem , the summands can be written as\n\\[\\sum_{a+b+c=2006}{\\frac{2006!}{a!b!c!}x^ay^bz^c}\\]\nand\n\\[\\sum_{a+b+c=2006}{\\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},\\]\nrespectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:\n\\[{2006+2\\choose 2} = 2015028\\]\nterms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of $y$ and $z$ must be opposite. Now we find a pattern:\nif the exponent of $y$ is $1$ , the exponent of $z$ can be all even integers up to $2004$ , so there are $1003$ terms.\nif the exponent of $y$ is $3$ , the exponent of $z$ can go up to $2002$ , so there are $1002$ terms.\n$\\vdots$\nif the exponent of $y$ is $2005$ , then $z$ can only be 0, so there is $1$ term.\nIf we add them up, we get $\\frac{1003\\cdot1004}{2}$ terms. However, we can switch the exponents of $y$ and $z$ and these terms will still have a negative sign. So there are a total of $1003\\cdot1004$ negative terms.\nBy subtracting this number from 2015028, we obtain $\\boxed{1,008,016}$ or $1,008,016$ as our answer.", "Define $P$ such that $P=y+z$ . Then the expression in the problem becomes: $(x+P)^{2006}+(x-P)^{2006}$\nExpanding this using binomial theorem gives $(x^n+Px^{n-1}+...+P^{n-1}x+P^n)+(x^n-Px^{n-1}+...-P^{n-1}x+P^n)$ , where $n=2006$ (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves).\nSimplifying gives: $2(x^n+x^{n-2}P^2+...+x^2P^{n-2}+P^n)$ . Note that two terms that come out of different powers of $P$ cannot combine and simplify, as their exponent of $x$ will differ. Therefore, we simply add the number of terms produced from each addend. By the Binomial Theorem, $P^k=(y+z)^k$ will have $k+1$ terms, so the answer is $1+3+5+...+2007=1004^2=1,008,016 \\implies \\boxed{1,008,016}$", "Using stars and bars we know that $(x+y+z)^{2006}$ has ${2006+2\\choose 2}$ or $2015028$ different terms. Now we need to find out how many of these terms are canceled out by $(x-y-z)^{2006}$ . We know that for any term(let's say $x^{a}(-y)^{b}(-z)^{c}$ ) where $a+b+c=2006$ of the expansion of $(x-y-z)^{2006}$ is only going to cancel out with the corresponding term $x^{a}y^{b}z^{c}$ if only $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. Now let's do some casework to see how many terms fit this criteria:\nCase 1: $a$ is even\nNow we know that $a$ is even and $a+b+c=2006$ . Thus $b+c$ is also even or both $b$ and $c$ are odd or both $b$ and $c$ are even. This case clearly fails the above criteria and there are 0 possible solutions.\nCase 2: $a$ is odd\nNow we know that $a$ is odd and $a+b+c=2006$ . Thus $b+c$ is odd and $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. All terms that have $a$ being odd work.\nWe now need to figure out how many of the terms have $a$ as a odd number. We know that $a$ can be equal to any number between 0 and 2006. There are 1003 odd numbers between this range and 2007 total numbers. Thus $\\frac{1003}{2007}$ of the $2015028$ terms will cancel out and $\\frac{1004}{2007}$ of the terms will work. Thus there are $(\\frac{1004}{2007})(2015028)$ terms. This number comes out to be $1,008,016$ $\\implies \\boxed{1,008,016}$ (Author: David Camacho)", "Noticing how $y$ and $z$ are negative in the second part of the expression, let $x=a$ and $y+z = b$ . Then we get \\[(a+b)^{2006} + (a-b)^{2006}\\] We know that the terms that don't cancel out have even powers of $a$ and $b$ . Our expansion will be in the form of \\[2a^{2006} + x_1a^{2004}b^{2} + x_2a^{2002}b^{4} + \\cdots + 2b^{2006}\\] Note that $b^n = (x+y)^n$ has $n+1$ terms. Furthermore, the current expression is irreducible as each term has a different $x$ power. Thus, when we write $a$ and $b$ back to their original terms, we will have $1+3+5+ \\cdots + 2007 = 1004^2 = 1,008,016 \\implies \\boxed{1,008,016}$" ]

[ "Let us recall $\\text{PEMDAS}$ . We calculate the exponent first. $(-2)^{-2}=\\frac{1}{(-2)^2}=\\frac{1}{4}$ When we substitute, we get $81^{\\frac{1}{4}}=\\sqrt[4]{81}=\\boxed{3}$" ]

[ "We write the given expression as a single fraction: \\[\\frac{2021}{2020} - \\frac{2020}{2021} = \\frac{2021\\cdot2021-2020\\cdot2020}{2020\\cdot2021}\\] by cross multiplication. Then by factoring the numerator, we get \\[\\frac{2021\\cdot2021-2020\\cdot2020}{2020\\cdot2021}=\\frac{(2021-2020)(2021+2020)}{2020\\cdot2021}.\\] The question is asking for the numerator, so our answer is $2021+2020=4041,$ giving $\\boxed{4041}$", "Denote $a = 2020$ . Hence, \\begin{align*} \\frac{2021}{2020} - \\frac{2020}{2021} & = \\frac{a + 1}{a} - \\frac{a}{a + 1} \\\\ & = \\frac{\\left( a + 1 \\right)^2 - a^2}{a \\left( a + 1 \\right)} \\\\ & = \\frac{2 a + 1}{a \\left( a + 1 \\right)} . \\end{align*}\nWe observe that ${\\rm gcd} \\left( 2a + 1 , a \\right) = 1$ and ${\\rm gcd} \\left( 2a + 1 , a + 1 \\right) = 1$\nHence, ${\\rm gcd} \\left( 2a + 1 , a \\left( a + 1 \\right) \\right) = 1$\nTherefore, $p = 2 a + 1 = 4041$\nTherefore, the answer is $\\boxed{4041}$" ]

[ "$\\frac{2x^2-x}{(x+1)(x-2)}-\\frac{4+x}{(x+1)(x-2)} = \\frac{2x^2-2x-4}{(x+1)(x-2)}$\nThis can be factored as $\\frac{(2)(x^2-x-2)}{(x+1)(x-2)} \\implies \\frac{(2)(x+1)(x-2)}{(x+1)(x-2)}$ , which cancels out to $2 \\implies \\boxed{2}$" ]

[ "Using the FOIL method, we see that $(a+b)(a+c) = a^2 + ab + ac + bc.$ We want to solve \\[a + bc = a^2 + ab + ac + bc\\] \\[a = a^2 + ab + ac\\] \\[a((a + b + c) - 1) = 0\\] These expressions are $\\boxed{1.}$" ]

[ "It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of $7$ ones is the same as the number of ways to take away a certain number of ones from an assortment of $7$ $6$ s.\nSo, we can match up the values to find the sum with the same probability as $10$ . We can start by noticing that $7$ is the smallest possible roll and $42$ is the largest possible roll. The pairs with the same probability are as follows:\n$(7, 42), (8, 41), (9, 40), (10, 39), (11, 38)...$\nHowever, we need to find the number that matches up with $10$ . So, we can stop at $(10, 39)$ and deduce that the sum with equal probability as $10$ is $39$ . So, the correct answer is $\\boxed{39}$ , and we are done.", "Let's call the unknown value $x$ . By symmetry, we realize that the difference between 10 and the minimum value of the rolls is equal to the difference between the maximum and $x$ . So,\n$10 - 7 = 42- x$\n$x = 39$ and our answer is $\\boxed{39}$ By: Soccer_JAMS", "For the sums to have equal probability, the average sum of both sets of $7$ dies has to be $(6+1)\\cdot 7 = 49$ . Since having $10$ is similar to not having $10$ , you just subtract 10 from the expected total sum. $49 - 10 = 39$ so the answer is $\\boxed{39}$", "The expected value of the sums of the die rolls is $3.5\\cdot7=24.5$ , and since the probabilities should be distributed symmetrically on both sides of $24.5$ , the answer is $24.5+(24.5-10)=39$ , which is $\\boxed{39}$", "Another faster and easier way of doing this, without using almost any math at all, is realizing that the possible sums are ${7,8,9,10,...,39,40,41,42}$ . By symmetry, (and doing a few similar problems in the past), you can realize that the probability of obtaining $7$ is the same as the probability of obtaining $42$ $P(8)=P(41)$ and on and on and on. This means that $P(10)=P(39)$ , and thus the correct answer is $\\boxed{39}$" ]

[ "Let the $n$ th term of the series be $ar^{n-1}$ . Because \\[\\frac {8!}{7!} = \\frac {ar^7}{ar^4} = r^3 = 8,\\] it follows that $r = 2$ and the first term is $a = \\frac {7!}{r^4} = \\frac {7!}{16} = \\boxed{315}$" ]

[ "There are $7$ segments whose lengths are $2 \\sin \\frac{\\pi}{7}$ $7$ segments whose lengths are $2 \\sin \\frac{2 \\pi}{7}$ $7$ segments whose lengths are $2 \\sin \\frac{3\\pi}{7}$\nTherefore, the sum of the $4$ th powers of these lengths is \\begin{align*} 7 \\cdot 2^4 \\sin^4 \\frac{\\pi}{7} + 7 \\cdot 2^4 \\sin^4 \\frac{2 \\pi}{7} + 7 \\cdot 2^4 \\sin^4 \\frac{3 \\pi}{7} & = \\frac{7 \\cdot 2^4}{(2i)^4} \\left( e^{i \\frac{\\pi}{7}} - e^{i \\frac{\\pi}{7}} \\right)^4 + \\frac{7 \\cdot 2^4}{(2i)^4} \\left( e^{i \\frac{2 \\pi}{7}} - e^{i \\frac{2 \\pi}{7}} \\right)^4 + \\frac{7 \\cdot 2^4}{(2i)^4} \\left( e^{i \\frac{3 \\pi}{7}} - e^{i \\frac{4 \\pi}{7}} \\right)^4 \\\\ & = 7 \\left( e^{i \\frac{4 \\pi}{7}} - 4 e^{i \\frac{2 \\pi}{7}} + 6 - 4 e^{- i \\frac{2 \\pi}{7}} + e^{- i \\frac{4 \\pi}{7}} \\right) \\\\ & \\quad + 7 \\left( e^{i \\frac{8 \\pi}{7}} - 4 e^{i \\frac{4 \\pi}{7}} + 6 - 4 e^{- i \\frac{4 \\pi}{7}} + e^{- i \\frac{8 \\pi}{7}} \\right) \\\\ & \\quad + 7 \\left( e^{i \\frac{12 \\pi}{7}} - 4 e^{i \\frac{6 \\pi}{7}} + 6 - 4 e^{- i \\frac{6 \\pi}{7}} + e^{- i \\frac{12 \\pi}{7}} \\right) \\\\ & = 7 \\left( e^{i \\frac{4 \\pi}{7}} + e^{i \\frac{8 \\pi}{7}} + e^{i \\frac{12 \\pi}{7}} + e^{-i \\frac{4 \\pi}{7}} + e^{-i \\frac{8 \\pi}{7}} + e^{-i \\frac{12 \\pi}{7}} \\right) \\\\ & \\quad - 7 \\cdot 4 \\left( e^{i \\frac{2 \\pi}{7}} + e^{i \\frac{4 \\pi}{7}} + e^{i \\frac{6 \\pi}{7}} + e^{-i \\frac{2 \\pi}{7}} + e^{-i \\frac{4 \\pi}{7}} + e^{-i \\frac{6 \\pi}{7}} \\right) \\\\ & \\quad + 7 \\cdot 6 \\cdot 3 \\\\ & = 7 \\left( e^{i \\frac{4 \\pi}{7}} + e^{-i \\frac{6 \\pi}{7}} + e^{-i \\frac{2 \\pi}{7}} + e^{-i \\frac{4 \\pi}{7}} + e^{i \\frac{6 \\pi}{7}} + e^{i \\frac{2 \\pi}{7}} \\right) \\\\ & \\quad - 7 \\cdot 4 \\left( e^{i \\frac{2 \\pi}{7}} + e^{i \\frac{4 \\pi}{7}} + e^{i \\frac{6 \\pi}{7}} + e^{-i \\frac{2 \\pi}{7}} + e^{-i \\frac{4 \\pi}{7}} + e^{-i \\frac{6 \\pi}{7}} \\right) \\\\ & \\quad + 7 \\cdot 6 \\cdot 3 \\\\ & = -7 + 7 \\cdot 4 + 7 \\cdot 6 \\cdot 3 \\\\ & = \\boxed{147} ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)", "There are $7$ segments whose lengths are $2 \\sin \\frac{\\pi}{7}$ $7$ segments whose lengths are $2 \\sin \\frac{2 \\pi}{7}$ $7$ segments whose lengths are $2 \\sin \\frac{3\\pi}{7}$\nTherefore, the sum of the $4$ th powers of these lengths is \\begin{align*} & 7 \\cdot 2^4 \\sin^4 \\frac{\\pi}{7} + 7 \\cdot 2^4 \\sin^4 \\frac{2 \\pi}{7} + 7 \\cdot 2^4 \\sin^4 \\frac{3 \\pi}{7} \\\\ & = 7 \\cdot 2^4 \\left( \\frac{1 - \\cos \\frac{2 \\pi}{7}}{2} \\right)^2 + 7 \\cdot 2^4 \\left( \\frac{1 - \\cos \\frac{4 \\pi}{7}}{2} \\right)^2 + 7 \\cdot 2^4 \\left( \\frac{1 - \\cos \\frac{6 \\pi}{7}}{2} \\right)^2 \\\\ & = 7 \\cdot 2^2 \\left( 1 - 2 \\cos \\frac{2 \\pi}{7} + \\cos^2 \\frac{2 \\pi}{7} \\right) + 7 \\cdot 2^2 \\left( 1 - 2 \\cos \\frac{4 \\pi}{7} + \\cos^2 \\frac{4 \\pi}{7} \\right) + 7 \\cdot 2^2 \\left( 1 - 2 \\cos \\frac{6 \\pi}{7} + \\cos^2 \\frac{6 \\pi}{7} \\right) \\\\ & = 7 \\cdot 2^2 \\cdot 3 - 7 \\cdot 2^3 \\left( \\cos \\frac{2 \\pi}{7} + \\cos \\frac{4 \\pi}{7} + \\cos \\frac{6 \\pi}{7} \\right) + 7 \\cdot 2^2 \\left( \\cos^2 \\frac{2 \\pi}{7} + \\cos^2 \\frac{4 \\pi}{7} + \\cos^2 \\frac{6 \\pi}{7} \\right) \\\\ & = 7 \\cdot 2^2 \\cdot 3 - 7 \\cdot 2^3 \\left( \\cos \\frac{2 \\pi}{7} + \\cos \\frac{4 \\pi}{7} + \\cos \\frac{6 \\pi}{7} \\right) + 7 \\cdot 2^2 \\left( \\frac{1 + \\cos \\frac{4 \\pi}{7} }{2} + \\frac{1 + \\cos \\frac{8 \\pi}{7} }{2} + \\frac{1 + \\cos \\frac{12 \\pi}{7} }{2} \\right) \\\\ & = 7 \\cdot 2^2 \\cdot 3 - 7 \\cdot 2^3 \\left( \\cos \\frac{2 \\pi}{7} + \\cos \\frac{4 \\pi}{7} + \\cos \\frac{6 \\pi}{7} \\right) + 7 \\cdot 2 \\cdot 3 + 7 \\cdot 2 \\left( \\cos \\frac{4 \\pi}{7} + \\cos \\frac{8 \\pi}{7} + \\cos \\frac{12 \\pi}{7} \\right) \\\\ & = 7 \\cdot 2^2 \\cdot 3 - 7 \\cdot 2^3 \\left( \\cos \\frac{2 \\pi}{7} + \\cos \\frac{4 \\pi}{7} + \\cos \\frac{6 \\pi}{7} \\right) + 7 \\cdot 2 \\cdot 3 + 7 \\cdot 2 \\left( \\cos \\frac{4 \\pi}{7} + \\cos \\frac{6 \\pi}{7} + \\cos \\frac{2 \\pi}{7} \\right) \\\\ & = 7 \\cdot 2 \\cdot 3 \\left( 2 + 1 \\right) - 7 \\cdot 2 \\left( 4 - 1 \\right) \\left( \\cos \\frac{4 \\pi}{7} + \\cos \\frac{6 \\pi}{7} + \\cos \\frac{2 \\pi}{7} \\right) \\\\ & = 7 \\cdot 2 \\cdot 3 \\left( 2 + 1 \\right) - 7 \\cdot 2 \\left( 4 - 1 \\right) \\cdot \\left( - \\frac{1}{2} \\right) \\\\ & = \\boxed{147} . \\]", "As explained in Solutions 1 and 2, what we are trying to find is $7 \\cdot 2^4 \\sin^4 \\frac{\\pi}{7} + 7 \\cdot 2^4 \\sin^4 \\frac{2 \\pi}{7} + 7 \\cdot 2^4 \\sin^4 \\frac{3 \\pi}{7}$ . Using trig we get \\begin{align*} & \\sin^4 \\frac{\\pi}{7} + \\sin^4 \\frac{2 \\pi}{7} + \\sin^4 \\frac{3 \\pi}{7} \\\\ = & \\sin^2 \\frac{\\pi}{7} \\left(1 - \\cos^2 \\frac{\\pi}{7} \\right) + \\sin^2 \\frac{2\\pi}{7} \\left(1 - \\cos^2 \\frac{2\\pi}{7} \\right) + \\sin^2 \\frac{3\\pi}{7} \\left(1 - \\cos^2 \\frac{3\\pi}{7} \\right) \\\\ = & \\sin^2 \\frac{\\pi}{7} - \\left(\\frac{1}{2} \\sin \\frac{2\\pi}{7}\\right)^2 + \\sin^2 \\frac{2\\pi}{7} - \\left(\\frac{1}{2} \\sin \\frac{4\\pi}{7}\\right)^2 + \\sin^2 \\frac{3\\pi}{7} - \\left(\\frac{1}{2} \\sin \\frac{6\\pi}{7}\\right)^2\\\\ = & \\sin^2 \\frac{\\pi}{7} - \\frac{1}{4} \\sin^2 \\frac{2\\pi}{7} + \\sin^2 \\frac{2\\pi}{7} - \\frac{1}{4} \\sin^2 \\frac{4\\pi}{7} + \\sin^2 \\frac{3\\pi}{7} - \\frac{1}{4} \\sin^2 \\frac{6\\pi}{7} \\\\ = & \\frac{3}{4} \\left(\\sin^2 \\frac{\\pi}{7} + \\sin^2 \\frac{2\\pi}{7} + \\sin^2 \\frac{3\\pi}{7}\\right) \\\\ = & \\frac{3}{4} \\cdot \\frac{1}{2} \\left(1 - \\cos \\frac{2\\pi}{7} + 1 - \\cos \\frac{4\\pi}{7} + 1 - \\cos \\frac{6\\pi}{7} \\right)\\\\ = & \\frac{3}{4} \\cdot \\frac{1}{2} \\left(3 - \\left(-\\frac{1}{2}\\right)\\right) \\\\ = & \\frac{21}{16}. \\end{align*} Like in the second solution, we also use the fact that $\\cos \\frac{2\\pi}{7} + \\cos \\frac{4\\pi}{7} + \\cos \\frac{6\\pi}{7} = -\\frac{1}{2}$ , which admittedly might need some explanation.\nFor explanation see supplement\nNotice that \\begin{align*} \\cos \\frac{2\\pi}{7} + \\cos \\frac{4\\pi}{7} + \\cos \\frac{6\\pi}{7} & = \\frac{1}{2}\\left(e^\\frac{2i\\pi}{7}+ e^{-\\frac{2i\\pi}{7}} \\right) + \\frac{1}{2}\\left( e^\\frac{4i\\pi}{7}+ e^{-\\frac{4i\\pi}{7}} \\right) + \\frac{1}{2}\\left( e^\\frac{6i\\pi}{7}+ e^{-\\frac{6i\\pi}{7}}\\right) \\\\ & = \\frac{1}{2}\\left(e^\\frac{2i\\pi}{7}+ e^{-\\frac{2i\\pi}{7}}+ e^\\frac{4i\\pi}{7}+ e^{-\\frac{4i\\pi}{7}} +e^\\frac{6i\\pi}{7}+ e^{-\\frac{6i\\pi}{7}} + 1\\right) - \\frac{1}{2} \\end{align*} In the brackets we have the sum of the roots of the polynomial $x^7 - 1 = 0$ . These sum to $0$ by Vieta’s formulas , and the desired identity follows. See Roots of unity if you have not seen this technique.\nGoing back to the question: \\[7 \\cdot 2^4 \\sin^4 \\frac{\\pi}{7} + 7 \\cdot 2^4 \\sin^4 \\frac{2 \\pi}{7} + 7 \\cdot 2^4 \\sin^4 \\frac{3 \\pi}{7} = 7 \\cdot 2^4 \\left(\\sin^4 \\frac{\\pi}{7} + \\sin^4 \\frac{2 \\pi}{7} + \\sin^4 \\frac{3 \\pi}{7}\\right) = 7 \\cdot 2^4 \\cdot \\frac{21}{16} = \\boxed{147}.\\] ~obscene_kangaroo", "This solution follows the same steps as the trigonometry solutions (Solutions 2 and 3), except it gives an alternate way to prove the statement below true without complex numbers:\n\\[\\cos \\frac{2\\pi}{7} + \\cos \\frac{4\\pi}{7} + \\cos \\frac{6\\pi}{7} = -\\frac{1}{2}\\]\n\\begin{align*} S &= \\cos \\frac{2\\pi}{7} + \\cos \\frac{4\\pi}{7} + \\cos \\frac{6\\pi}{7}, \\\\ S^2 &= \\cos^2 \\frac{2\\pi}{7} + \\cos^2 \\frac{4\\pi}{7} + \\cos^2 \\frac{6\\pi}{7} + 2\\cos \\frac{2\\pi}{7}\\cos \\frac{4\\pi}{7} + 2\\cos \\frac{2\\pi}{7}\\cos \\frac{6\\pi}{7} + 2\\cos \\frac{4\\pi}{7}\\cos \\frac{6\\pi}{7} \\\\ &= \\left(\\frac{1+ \\cos \\frac{4\\pi}{7}}{2}\\right) + \\left(\\frac{1+ \\cos \\frac{8\\pi}{7}}{2}\\right) + \\left(\\frac{1+ \\cos \\frac{12\\pi}{7}}{2}\\right) + 2\\cos \\frac{2\\pi}{7}\\cos \\frac{4\\pi}{7} + 2\\cos \\frac{2\\pi}{7}\\cos \\frac{6\\pi}{7} + 2\\cos \\frac{4\\pi}{7}\\cos \\frac{6\\pi}{7} \\\\ &= \\frac{1}{2}(3 + S) + \\left(\\cos \\frac{6\\pi}{7} + \\cos \\frac{2\\pi}{7}\\right) + \\left(\\cos \\frac{8\\pi}{7} + \\cos \\frac{4\\pi}{7}\\right) + \\left(\\cos \\frac{10\\pi}{7} + \\cos \\frac{2\\pi}{7}\\right) \\\\ &= \\frac{1}{2}(3 + S) + 2\\cos \\frac{2\\pi}{7} + 2\\cos \\frac{4\\pi}{7} + 2\\cos \\frac{6\\pi}{7}\\\\ &= \\frac{1}{2}(3 + S) + 2S. \\\\ \\end{align*} We end up with \\[2S^2 - 5S - 3 = 0.\\] Using the quadratic formula, we find the solutions for $S$ to be $-\\frac{1}{2}$ and $3$ . Because $3$ is impossible, $S = -\\frac{1}{2}$ . \nWith this result, following similar to steps to Solutions 2 and 3 will get $\\boxed{147}$", "Let $x$ $y$ , and $z$ be the lengths of the chords with arcs $\\frac{2\\pi}{7}$ $\\frac{4\\pi}{7}$ and $\\frac{6\\pi}{7}$ respectively.\nThen by the law of cosines we get: \\begin{align*} x^2 &= 2\\left(1-\\cos\\frac{2\\pi}{7}\\right), \\\\ y^2 &= 2\\left(1-\\cos\\frac{4\\pi}{7}\\right), \\\\ z^2 &= 2\\left(1-\\cos\\frac{6\\pi}{7}\\right). \\end{align*} The answer is then just $7(x^4+y^4+z^4)$ (since there's $7$ of each diagonal/side), obtained by summing the squares of the above equations and then multiplying by $7$ \\begin{align*} & \\quad7\\cdot2^2\\left( \\left(1-\\cos\\frac{2\\pi}{7}\\right)^2 + \\left(1-\\cos\\frac{4\\pi}{7}\\right)^2 + \\left(1-\\cos\\frac{6\\pi}{7}\\right)^2 \\right) \\\\ &= 7\\cdot4\\left( \\left(1-2\\cos\\frac{2\\pi}{7}+\\cos^2\\frac{2\\pi}{7}\\right) + \\left(1-2\\cos\\frac{4\\pi}{7}+\\cos^2\\frac{4\\pi}{7}\\right) + \\left(1-2\\cos\\frac{6\\pi}{7}+\\cos^2\\frac{6\\pi}{7}\\right) \\right) \\\\ &= 7\\cdot4\\left(3-2\\left(\\cos\\frac{2\\pi}{7}+\\cos\\frac{4\\pi}{7}+\\cos\\frac{6\\pi}{7}\\right) + \\left(\\cos^2\\frac{2\\pi}{7}+\\cos^2\\frac{4\\pi}{7}+\\cos^2\\frac{6\\pi}{7}\\right) \\right) \\\\ &= 7\\cdot4\\left(3-2\\left(\\cos\\frac{2\\pi}{7}+\\cos\\frac{4\\pi}{7}+\\cos\\frac{6\\pi}{7}\\right) + \\frac12 \\left(1+\\cos\\frac{4\\pi}{7}+1+\\cos\\frac{8\\pi}{7}+1+\\cos\\frac{12\\pi}{7}\\right) \\right) \\\\ &= 7\\cdot4\\left(\\frac{9}{2}-2\\left(\\cos\\frac{2\\pi}{7}+\\cos\\frac{4\\pi}{7}+\\cos\\frac{6\\pi}{7}\\right) + \\frac12 \\left(\\cos\\frac{4\\pi}{7}+\\cos\\frac{6\\pi}{7}+\\cos\\frac{2\\pi}{7}\\right) \\right) \\\\ &= 7\\cdot4\\left(\\frac{9}{2}-\\frac{3}{2}\\left(\\cos\\frac{2\\pi}{7}+\\cos\\frac{4\\pi}{7}+\\cos\\frac{6\\pi}{7}\\right) \\right) \\\\ &= 7\\cdot4\\left(\\frac{9}{2}-\\frac{3}{2}\\left(-\\frac{1}{2}\\right)\\right) \\\\ &= \\boxed{147}", "Hope you had a ruler handy! This problem can be done with a ruler and basic estimation.\nFirst, measuring the radius of the circle obtains $2.9$ cm (when done on the paper version). Thus, any other measurement we get for the sides/diagonals should be divided by $2.9$\nMeasuring the sides of the circle gets $2.5$ cm. The shorter diagonals are $4.5$ cm, and the longest diagonals measure $5.6$ cm. Thus, we'd like to estimate \\[7\\left(\\frac{2.5}{2.9}\\right)^4 + 7\\left(\\frac{4.5}{2.9}\\right)^4 + 7\\left(\\frac{5.6}{2.9}\\right)^4.\\]\nWe know $\\left(\\frac{2.5}{2.9}\\right)^4$ is slightly less than $1.$ Let's approximate it as 1 for now. Thus, $7\\left(\\frac{2.5}{2.9}\\right)^4 \\approx 7.$\nNext, $\\left(\\frac{4.5}{2.9}\\right)^4$ is slightly more than $\\left(\\frac{4.5}{3}\\right)^4.$ We know $\\left(\\frac{4.5}{3}\\right)^4 = 1.5^4 = \\frac{81}{16},$ slightly more than $5,$ so we can approximate $\\left(\\frac{4.5}{2.9}\\right)^4$ as $5.5.$ Thus, $7\\left(\\frac{4.5}{2.9}\\right)^4 \\approx 38.5.$\nFinally, $\\left(\\frac{5.6}{2.9}\\right)^4$ is slightly less than $\\left(\\frac{5.6}{2.8}\\right)^4 = 2^4 = 16.$ We say it's around $15,$ so then $7\\left(\\frac{5.6}{2.9}\\right)^4 \\approx 105.$\nAdding what we have, we get $105 + 38.5 + 1 = 144.5$ as our estimate. We see $\\boxed{147}$ is very close to our estimate, so we have successfully finished the problem.", "Place the figure on the complex plane and let $w = e^{\\frac{2\\pi}{7}}$ (a $7$ th root of unity). The vertices of the $7$ -gon are $w^0,w^1,w^2,\\dots,w^6$ . We wish to find \\[\\sum_{i=0}^5\\sum_{j=i+1}^6\\lvert w^i-w^j\\rvert^4 = \\frac{1}{2}\\sum_{i=0}^6\\sum_{j=0}^6\\lvert w^i-w^j\\rvert^4.\\] The second expression is more convenient to work with. The factor of $\\tfrac{1}{2}$ is because it double-counts each edge (and includes when $i=j$ , but the length is $0$ so it doesn't matter).\nRecall the identity $|z|^2 = z\\overline{z}$ . Since $|w| = 1$ $\\overline{w} = w^{-1}$ , and \\begin{align*} \\lvert w^i-w^j\\rvert^4 &= ((w^i-w^j)(w^{-i}-w^{-j}))^2 \\\\ &= (2-w^{i-j}-w^{j-i})^2 \\\\ &= 4+w^{2i-2j}+w^{2j-2i}+2(-2w^{i-j}-2w^{j-i}+1) \\\\ &= 6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}. \\end{align*}\nBut notice that, for a fixed value of $i$ , over all values of $j$ from $0$ to $6$ , by properties of modular arithmetic with $w^7 = w^0$ (or expanding),\neach of the $w^\\bullet$ -terms takes on every value in $w^0,w^1,w^2,\\dots,w^6$ exactly once. Since $\\sum_{j=0}^6 w^j = 0$ \\begin{align*} \\frac{1}{2}\\sum_{i=0}^6\\sum_{j=0}^6\\lvert w^i-w^j\\rvert^4 &= \\frac{1}{2}\\sum_{i=0}^6\\sum_{j=0}^6(6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}) \\\\ &= \\frac{1}{2}\\sum_{i=0}^6\\left(\\sum_{j=0}^6 6-4\\sum_{j=0}^6 w^{i-j}-4\\sum_{j=0}^6 w^{j-i}+\\sum_{j=0}^6 w^{2i-2j}+\\sum_{j=0}^6 w^{2j-2i}\\right) \\\\ &= \\frac{1}{2}\\sum_{i=0}^6\\left(\\sum_{j=0}^6 6-4\\sum_{j=0}^6 w^j-4\\sum_{j=0}^6 w^j+\\sum_{j=0}^6 w^j+\\sum_{j=0}^6 w^j\\right) \\\\ &= \\frac{1}{2}\\sum_{i=0}^6\\sum_{j=0}^6 6 \\\\ &= \\frac{1}{2}\\cdot7\\cdot7\\cdot6 \\\\ &= \\boxed{147}", "This is how I solve this problem:\nIt's easy to solve for $3$ -gon, $4$ -gon, and $6$ -gon inscribed in a unit circle. (Okay, it's just the weird names for triangle, square, and hexagon)\nFor $3$ -gon, the sum is equal to $3$ times the $4$ th power of an edge. Thus, \\[S_3=3\\,\\cdot\\,\\left(\\sqrt{3}\\right)^4=27.\\]\nFor $4$ -gon, the sum is equal to $4$ times the $4$ th power of an edge, and $2$ times the $4$ th power of the diagonal. Thus, \\[S_4=4\\,\\cdot\\,\\left(\\sqrt{2}\\right)^4+2\\,\\cdot\\,\\left(2\\right)^4=48.\\]\nFor $6$ -gon, the sum is equal to $6$ times the $4$ th power of an edge, $6$ times the $4$ th power of the short diagonal, and $3$ times the $4$ th power of the long diagonal. Thus, \\[S_6=6\\,\\cdot\\,\\left(1\\right)^4+6\\,\\cdot\\,\\left(\\sqrt{3}\\right)^4+3\\,\\cdot\\,\\left(2\\right)^4=108.\\]\nThen, I quickly noticed that $27=3\\,\\cdot\\,3^2$ $48=3\\,\\cdot\\,4^2$ , and $108=3\\,\\cdot\\,6^2$ . So reasonably, it will work out this formula, $S_n=3n^2$ . (This step is purely out of guessing, maybe have a look at Solution 8 for more info...)\nBy inductive reasoning, we got $S_7=3\\,\\cdot\\,7^2=\\boxed{147}$" ]

[ "First, we put the figure in the coordinate plane with the center of the circle at the origin and a vertex on the positive x-axis. Thus, the coordinates of the vertices will be the terminal points of integer multiples of the angle $\\frac{2\\pi}{7},$ which are \\[\\left(\\cos\\dfrac{2\\pi n}{7}, \\sin\\dfrac{2\\pi n}{7}\\right)\\] for integer $n.$ Then, we notice there are three types of diagonals: the ones with chords of arcs $\\dfrac{2\\pi}{7}, \\dfrac{4\\pi}{7},$ and $\\dfrac{6\\pi}{7}.$ We notive there are here are $7$ of each type of diagonal. Then, we use the pythagorean theorem to find the distance from $\\left(\\cos\\frac{2\\pi n}{7}, \\sin\\frac{2\\pi n}{7}\\right)$ to $(1, 0)$ \\[\\left(\\sqrt{\\left(\\cos\\frac{2\\pi n}{7}-1\\right)^2+\\sin^2\\frac{2\\pi n}{7}}\\right)^4=\\left(\\sqrt{\\cos^2\\frac{2\\pi n}{7}+\\sin^2\\frac{2\\pi n}{7}-2\\cos\\frac{2\\pi n}{7}+1}\\right)^4\\] \\[=\\left(\\sqrt{2-2\\cos\\frac{2\\pi n}{7}}\\right)^4\\] \\[=4\\cos^2\\frac{2\\pi n}{7}-8\\cos\\frac{2\\pi n}{7}+4.\\] By the cosine double angle identity, $\\cos{2\\theta}=2\\cos^2\\theta-1.$ This means that $2\\cos^2\\theta=\\cos{2\\theta}+1.$ Substituting this in, \\[4\\cos^2\\frac{2\\pi n}{7}-8\\cos\\frac{2\\pi n}{7}+4=2\\left(\\cos\\frac{4\\pi n}{7}+1\\right)-8\\cos\\frac{2\\pi n}{7}+4=2\\cos\\frac{4\\pi n}{7}-8\\cos\\frac{2\\pi n}{7}+6.\\] Summing this up for $n=1,2,3,$ \\[2\\cos\\frac{4\\pi}{7}-8\\cos\\frac{2\\pi}{7}+6+2\\cos\\frac{8\\pi}{7}-8\\cos\\frac{4\\pi}{7}+6+2\\cos\\frac{12\\pi}{7}-8\\cos\\frac{6\\pi}{7}+6\\] \\[=2\\left(\\cos\\frac{4\\pi}{7}+\\cos\\frac{8\\pi}{7}+\\cos\\frac{12\\pi}{7}\\right)-8\\left(\\cos\\frac{2\\pi}{7}+\\cos\\frac{4\\pi}{7}+\\cos\\frac{6\\pi}{7}\\right)+18\\] \\[=2\\left(\\cos\\frac{4\\pi}{7}+\\cos\\frac{6\\pi}{7}+\\cos\\frac{2\\pi}{7}\\right)-8\\left(\\cos\\frac{2\\pi}{7}+\\cos\\frac{4\\pi}{7}+\\cos\\frac{6\\pi}{7}\\right)+18\\] \\[=2(-0.5)-8(-0.5)+18=21.\\] (These equalities are based on $\\cos\\theta=\\cos(2\\pi-\\theta)$ and $\\cos\\frac{2\\pi}{7}+\\cos\\frac{4\\pi}{7}+\\cos\\frac{6\\pi}{7}=-\\frac{1}{2}.$ ) (For explanation of this see supplement .)\nFinally, because there are $7$ of each type of diagonal, the answer is $7\\cdot 21=\\boxed{147}.$" ]

[ "Note that of the $12$ cities, $6$ of them ( $2$ on the top, $2$ on the bottom, and $1$ on each side) have $3$ edges coming into/out of them (i.e., in graph theory terms, they have degree $3$ ). Therefore, at least $1$ edge connecting to each of these cities cannot be used. Additionally, the same applies to the start and end points, since we don't want to return to them. Thus there are $6+2=8$ vertices that we know have $1$ unused edge, and we have $17-13=4$ unused edges to work with (since there are $17$ edges in total, and we must use exactly $13$ of them). It is not hard to find that there is only one configuration satisfying these conditions:\nNote: $\\text{X}$ s represent unused edges.\n\nObserve that at each of the $2$ cities marked with an $\\text{O}$ on a path, there are $2$ possibilities: we can either continue straight and cross back over the path later, or we can make a left turn, then turn right when we approach the junction again. This gives us a total of $2\\cdot 2 = \\boxed{4}$ valid paths.", "Let the bottom-left vertex be $(0,0)$ , and let each of the edges have length $1$ , so that all of the vertices are at lattice points.\nFirstly, notice that for any vertex $V$ on the graph (other than $A$ or $L$ ), we can visit it at most $M(V) = \\left\\lfloor \\frac{\\text{deg}(V)}{2} \\right\\rfloor$ times, where $\\text{deg}(V)$ is, as usual, the degree of $V$ (i.e. the number of edges coming into/out of $V$ ). This is because to visit any of these vertices, we would have to enter and exit it by two different edges, in order to avoid using any portion of a road more than once. (Those who know some graph theory will recognise this a well-known principle.) We will label each vertex with this number. Additionally, notice that if we visit $n$ vertices (not necessarily distinct, i.e. counting a vertex which is visited twice as two vertices) along our path, we must traverse $n-1$ edges (this is quite easy to prove). Thus, if we want to traverse $13$ edges in total, we have to visit $14$ vertices. We must visit $A$ and $L$ , leaving $12$ vertices from the rest of the graph to visit.\nIf we sum the maximum numbers of visits to each vertex, we find that this is exactly equal to the $12$ found above. This means that we have to visit each vertex $M(V)$ times, and must traverse $2\\cdot M(V)$ edges connected to each vertex. Specifically, we must traverse all $4$ of the edges connected to the two central vertices at $(1,1)$ and $(1,2)$ , as well as both edges connected to the $2$ corner vertices (excluding $A$ and $L$ ), and any $2$ edges connected to the other vertices along the outside edge of the rectangle.\nWith this information, we can now proceed by dividing into cases.\nCase 1 : We first move down from $A$\nIn this case, we see that we must immediately move right to $(1,1)$ in order to traverse the edge from $(0,1)$ to $(1,1)$ , since we can never revisit $(0,1)$ . However, by the same logic, we must immediately move to $(0,0)$ . This is a contradiction, so there are no possible paths in this case.\nCase 2 : We first move right from $A$\nSimilar to the last case, we see that we must immediately move to $(1,1)$ . By symmetry, we conclude that our last two moves must be $(2,1) \\rightarrow (2,0) \\rightarrow (3,0)$ . With this information, we have reduced the problem to traveling from $(1,1)$ to $(2,1)$ with the same constraints as before. We redraw the graph, removing the edges we have already used and updating our labels (note that $(1,1)$ and $(2,1)$ are still labeled with $2$ since we will pass through them twice, at the start and the end). Then, we remove the vertices with label $0$ and the edges we know we can never traverse, giving:\n Now, it is clear that we must complete a cycle in the lower left square, return to $(1,1)$ , go to $(2,1)$ , and complete a cycle in the top right square, returning to $(2,1)$ .\nThere are two ways to traverse each cycle (clockwise and anti-clockwise), giving us a total of $2\\cdot 2 = \\boxed{4}$ paths of length 13 from $A$ to $L$" ]

[ "Observe that only the two central vertices can be visited twice. Since the path is of length 13, we need to repeat a vertex. Using casework on each vertex, we can find there are two paths that go through each central vertex twice, for an answer of $\\boxed{4}.$", "Looking at the answer choices we see that it would probably be easy enough to count the different paths. After some experimenting, we find that we must cross one of the middle vertexes twice, making counting easier.\nWe finally find that there are $\\boxed{4}$ paths that satisfy the conditions. Luckily, $4$ is the largest answer choice so we know it must be correct." ]

[ "While imagining the folding, $\\overline{AB}$ goes on $\\overline{BC},$ $\\overline{AH}$ goes on $\\overline{CI},$ and $\\overline{EF}$ goes on $\\overline{FG}.$ So, $BJ=CI=8$ and $FG=BC=8.$ Also, $\\overline{HJ}$ becomes an edge parallel to $\\overline{FG},$ so that means $HJ=8.$\nSince $GH=14,$ then $JG=14-8=6.$ So, the area of $\\triangle BJG$ is $\\frac{8\\cdot6}{2}=24.$ If we let $\\triangle BJG$ be the base, then the height is $FG=8.$ So, the volume is $24\\cdot8=\\boxed{192}.$" ]

[ "Choose a section to start coloring. Assume, WLOG, that this section is color $1$ . We proceed coloring clockwise around the ring. Let $f(n,C)$ be the number of ways to color the first $n$ sections (proceeding clockwise) such that the last section has color $C$ . In general (except for when we complete the coloring), we see that \\[f(n,C_i)=\\sum_{j\\ne i} f(n-1,C_j),\\] i.e., $f(n,C_i)$ is equal to the number of colorings of $n-1$ sections that end in any color other than $C_i$ . Using this, we can compute the values of $f(n,C)$ in the following table.\n$\\begin{tabular}{c|c|c|c|c } \\multicolumn{1}{c}{}&\\multicolumn{4}{c}{\\(C\\)}\\\\ \\(n\\)&1 & 2 & 3& 4 \\\\ \\hline 1& 1 & 0 & 0 & 0\\\\ 2 & 0 & 1 & 1 & 1 \\\\ 3& 3 & 2 & 2 & 2 \\\\ 4 & 6 & 7 & 7 & 7 \\\\ 5 & 21 & 20 & 20 & 20\\\\ 6& 0& 61 & 61 & 61\\\\ \\end{tabular}$\nNote that $f(6, 1)=0$ because then $2$ adjacent sections are both color $1$ . We multiply this by $4$ to account for the fact that the initial section can be any color. Thus the desired answer is $(61+61+61) \\cdot 4 = \\boxed{732}$", "We use complementary counting. There are $4^6$ total colorings of the ring without restriction. To count the complement, we wish to count the number of colorings in which at least one set of adjacent sections are the same color. There are six possible sets of adjacent sections that could be the same color (think of these as borders). Call these $B_1,B_2,\\dots,B_6$ . Let $\\mathcal{A}_1, \\mathcal{A}_2,\\dots,\\mathcal{A}_6$ be the sets of colorings of the ring where the sections on both sides of $B_1,B_2,\\dots,B_6$ are the same color. We wish to determine $|\\mathcal{A}_1\\cup\\mathcal{A}_2\\cup\\cdots\\cup\\mathcal{A}_6|$ . Note that all of these cases are symmetric, and in general, $|\\mathcal{A}_i|=4^5$ . There are $6$ such sets $\\mathcal{A}_i$ . Also, $|\\mathcal{A}_i\\cup\\mathcal{A}_j|=4^4$ , because we can only change colors at borders, so if we have two borders along which we cannot change colors, then there are four borders along which we have a choice of color. There are $\\binom{6}{2}$ such pairs $\\mathcal{A}_i\\cup\\mathcal{A}_j$ . Similarly, $|\\mathcal{A}_i\\cup \\mathcal{A}_j\\cup \\mathcal{A}_k|=4^3$ , with $\\binom{6}{3}$ such triples, and we see that the pattern will continue for 4-tuples and 5-tuples. For 6-tuples, however, these cases occur when there are no changes of color along the borders, i.e., each section has the same color. Clearly, there are four such possibilities.\nTherefore, by PIE, \\[|\\mathcal{A}_1\\cup\\mathcal{A}_2\\cup\\cdots\\cup\\mathcal{A}_6|=\\binom{6}{1}\\cdot 4^5-\\binom{6}{2}\\cdot 4^4+\\binom{6}{3}\\cdot 4^3-\\binom{6}{4}\\cdot 4^2+\\binom{6}{5}\\cdot 4^1-4.\\] We wish to find the complement of this, or \\[4^6-\\left(\\binom{6}{1}\\cdot 4^5-\\binom{6}{2}\\cdot 4^4+\\binom{6}{3}\\cdot 4^3-\\binom{6}{4}\\cdot 4^2+\\binom{6}{5}\\cdot 4^1-4\\right).\\] By the Binomial Theorem, this is $(4-1)^6+3=\\boxed{732}$", "We use generating functions. Suppose that the colors are $0,1,2,3$ . Then as we proceed around a valid coloring of the ring in the clockwise direction, we know that between two adjacent sections with colors $s_i$ and $s_{i+1}$ , there exists a number $d_i\\in\\{1,2,3\\}$ such that $s_{i+1}\\equiv s_i+d_i\\pmod{4}$ . Therefore, we can represent each border between sections by the generating function $(x+x^2+x^3)$ , where $x,x^2,x^3$ correspond to increasing the color number by $1,2,3\\pmod4$ , respectively. Thus the generating function that represents going through all six borders is $A(x)=(x+x^2+x^3)^6$ , where the coefficient of $x^n$ represents the total number of colorings where the color numbers are increased by a total of $n$ as we proceed around the ring. But if we go through all six borders, we must return to the original section, which is already colored. Therefore, we wish to find the sum of the coefficients of $x^n$ in $A(x)$ with $n\\equiv 0\\pmod4$\nNow we note that if $P(x)=x^n$ , then \\[P(x)+P(ix)+P(-x)+P(-ix)=\\begin{cases}4x^n&\\text{if } n\\equiv0\\pmod{4}\\\\0&\\text{otherwise}.\\end{cases}\\] Therefore, the sum of the coefficients of $A(x)$ with powers congruent to $0\\pmod 4$ is \\[\\frac{A(1)+A(i)+A(-1)+A(-i)}{4}=\\frac{3^6+(-1)^6+(-1)^6+(-1)^6}{4}=\\frac{732}{4}.\\] We multiply this by $4$ to account for the initial choice of color, so our answer is $\\boxed{732}$", "Let $f(n)$ be the number of valid ways to color a ring with $n$ sections (which we call an $n$ -ring), so the answer is given by $f(6)$ . For $n=2$ , we compute $f(n)=4\\cdot3=12$ . For $n \\ge 3$ , we can count the number of valid colorings as follows: choose one of the sections arbitrarily, which we may color in $4$ ways. Moving clockwise around the ring, there are $3$ ways to color each of the $n-1$ other sections. Therefore, we have $4 \\cdot 3^{n-1}$ colorings of an $n$ -ring.\nHowever, note that the first and last sections could be the same color under this method. To count these invalid colorings, we see that by \"merging\" the first and last sections into one, we get a valid coloring of an $(n-1)$ -ring. That is, there are $f(n-1)$ colorings of an $n$ -ring in which the first and last sections have the same color. Thus, $f(n) = 4 \\cdot 3^{n-1} - f(n-1)$ for all $n \\ge 3$\nTo compute the requested value $f(6)$ , we repeatedly apply this formula: \\begin{align*} f(6)&=4\\cdot3^5-f(5)\\\\&=4\\cdot3^5-4\\cdot3^4+f(4)\\\\&=4\\cdot3^5-4\\cdot3^4+4\\cdot3^3-f(3)\\\\&=4\\cdot3^5-4\\cdot3^4+4\\cdot3^3-4\\cdot3^2+f(2)\\\\&=4(3^5-3^4+3^3-3^2+3)\\\\&=4\\cdot3\\cdot\\frac{3^5+1}{3+1}\\\\&=\\boxed{732} (Solution by MSTang.)", "Label the sections 1, 2, 3, 4, 5, 6 clockwise. We do casework on the colors of sections 1, 3, 5.\nCase 1: the colors of the three sections are the same.\nIn this case, each of sections 2, 4, 6 can be one of 3 colors, so this case yields $4 \\times 3^3 = 108$ ways.\nCase 2: two of sections 1, 3, 5 are the same color.\nNote that there are 3 ways for which two of the three sections have the same color, and $4 \\times 3 = 12$ ways to determine their colors. After this, the section between the two with the same color can be one of 3 colors and the other two can be one of 2 colors. This case yields $3 \\times (4 \\times 3) \\times (3 \\times 2 \\times 2) = 432$ ways.\nCase 3: all three sections of 1, 3, 5 are of different colors.\nClearly, there are $4 \\times 3 \\times 2 = 24$ choices for which three colors to use, and there are 2 ways to choose the colors of each of sections 2, 4, 6. Thus, this case gives $4 \\times 3 \\times 2 \\times 2^3 = 192$ ways.\nIn total, there are $108 + 432 + 192 = \\boxed{732}$ valid colorings.", "We will take a recursive approach to this problem. We can start by writing out the number of colorings for a circle with $1, 2,$ and $3$ compartments, which are $4, 12,$ and $24.$ Now we will try to find a recursive formula, $C(n)$ , for a circle with an arbitrary number of compartments $n.$ We will do this by focusing on the $n-1$ section in the circle. This section can either be the same color as the first compartment, or it can be a different color as the first compartment. We will focus on each case separately.\nCase 1:\nIf they are the same color, we can say there are $C(n-2)$ to fill the first $n-1$ compartments. The $nth$ compartment must be different from the first and second to last compartments, which are the same color. Hence this case adds $3*C(n-2)$ to our recursive formula.\nCase 2:\nIf they are different colors, we can say that there are $C(n-1)$ to fill the first $n-1$ compartments, and for the the $nth$ compartment, there are $2$ ways to color it because the $n-1$ and $1$ compartments are different colors. Hence this case adds $2*C(n-1).$\nSo our recursive formula, $C(n)$ , is $3*C(n-2) + 2*C(n-1).$ Using the initial values we calculated, we can evaluate this recursive formula up to $n=6.$ When $n=6,$ we get $\\boxed{732}$ valid colorings.", "WLOG, color the top left section $1$ and the top right section $2$ . Then the left section can be colored $2$ $3$ , or $4$ , and the right section can be colored $1$ $3$ , or $4$ . There are $3 \\cdot 3 = 9$ ways to color the left and right sections. We split this up into two cases.\nCase 1: The left and right sections are of the same color. There are $2$ ways this can happen: either they both are $3$ or they both are $4$ . We have $3$ colors to choose for the bottom left, and $2$ remaining colors to choose for the bottom right, for a total of $2 \\cdot 3 \\cdot 2$ cases.\nCase 2: The left and right sections are of different colors. There are $9 - 2 = 7$ ways this can happen. Assume the left is $3$ and the right is $4$ . Then the bottom left can be $1$ $2$ , or $4$ , and the bottom right can be $1$ $2$ , or $3$ . However the bottom sections cannot both be $1$ or both be $2$ , so there are $3 \\cdot 3 - 2 = 7$ ways to color the bottom sections, for a total for $7 \\cdot 7 = 49$ colorings.\nSince there were $4 \\cdot 3 = 12$ ways to color the top sections, the answer is $12 \\cdot (12 + 49) = \\boxed{732}$", "Label the four colors $1, 2, 3,$ and $4$ , respectively.\nNow let's imagine a circle with the four numbers $1, 2, 3,$ and $4$ written clockwise. We'll say that a bug is standing on number $a$ . It is easy to see that for the bug to move to a different number, it must walk $1, 2,$ or $3$ steps clockwise. (This is since adjacent numbers can't be the same, as stated in the problem). Note that the sixth number in the bug's walking sequence must not equal the first number. Thus, our total number of ways, $N$ , given the bug's starting number $k$ , is simply the number of ordered quintuplets of positive integers $(a_1, a_2, a_3, a_4, a_5)$ that satisfy $a_i \\in \\{1, 2, 3\\}$ for all $1 \\leq i \\leq 5$ and \\[\\sum_{i=1}^{5} a_i \\neq 8, 12,\\] since the bug cannot land on $k$ again on his fifth and last step.\nWe know that the number of ordered quintuplets of positive integers $(a_1, a_2, a_3, a_4, a_5)$ that satisfy $a_i \\in \\{1, 2, 3\\}$ without the other restriction is just $3^5$ , so we aim to find the number of quintuplets such that \\[a_1 + a_2 + a_3 + a_4 + a_5 = 8, 12.\\] Note that the number of ordered quintuplets satisfying $\\sum_{i=1}^{5} a_i = 8$ is the same as the number of them satisfying $\\sum_{i=1}^{5} a_i = 12$ due to symmetry. By stars and bars, there are $\\dbinom{7}{4} = 35$ ways to distribute the three \"extra\" units to the five variables $a_1, a_2, a_3, a_4, a_5$ (since $a_i \\geq 1$ ), but ways of distribution such that one variable is equal to $4$ are illegal, so the actual number of ways is $35 - 5 = 30$ . Since there are four possible values of $k$ (or the starting position for the bug), we obtain \\[N = 4 \\times (3^5 - 2 \\times 30) = \\boxed{732}.\\] -fidgetboss_4000", "Let's number the regions $1,2,\\dots 6$ . Suppose we color regions $1,2,3$ . Then, how many ways are there to color $4,5,6$\nNote: the numbers are numbered as shown:\n\n$\\textbf{Case 1:}$ The colors of $1,2,3$ are $BAB$ , in that order.\nThen the colors of $6,5,4$ can be $AXA$ $AXC$ $CXA$ $CXC$ , or $CXD$ in that order, where $X$ is any color not equal to its surroundings. Then there are $4$ choices for $A$ $3$ choices for $B$ (it cannot be $A$ ), $2$ choices for $C$ , and $1$ for $D$ , the last color. So, summing up, we have \\[4*3(1*3+2*2+2*2+2*3+2*1*2)=12*21=252\\] colorings.\n$\\textbf{Case 2:}$ The colors of $1,2,3$ are $BAC$ , in that order.\nAgain, we list out the possible arrangements of $6,5,4$ $AXA$ $AXB$ $CXA$ $AXD$ $DXA$ $CXB$ $CXD$ $DXB$ , or $DXD$ . (Easily simplified; listed here for clarity.) Then there are $4$ choices for $A$ as usual, $3$ choices for $B$ , and so on. Hence we have \\[4*3*2(1*3+1*2+1*2+1*2+1*2+1*2+1*2+1*2+1*3)=24*20=480\\] colorings in this case.\nAdding up, we have $252+480=\\boxed{732}$ as our answer.", "We quickly notice that this is just the cycle graph with 6 vertices. The chromatic polynomial for a cycle is $(k-1)^n+(-1)^n (k-1)$ where we use $k$ colors on a cycle of $n$ vertices. Plugging in $k=4$ and $n=6$ we arrive at $\\boxed{732}$ .\n~chrisdiamond10", "Let's label the regions as $1,2,3,4,5,6$ in that order. We start with region $1$ . There are no restrictions on the color of region $1$ so it can be any of the four colors. We know move on the region $2$ . It can be any color but color used for region $1$ , giving us $3$ choices. Section $3$ is where it gets a bit complicated; we will have to do casework based on whether the color of region $3$ is that of region $1$ or not.\nIf we have the color of region $3$ being different from that of region $1$ (in which we color do so in $2$ ways), then we have need for another casework at region $4$ . If the color of region $4$ is different from that of region $1$ (which can be achieved in $2$ ways), then we have yet another casework split.\nIf the color of region $5$ is different from that of region $1$ (which can be done in $2$ ways, then we would have a total of $2$ possible colorings for region $6$ (for it cannot be the color of regions $1$ nor $5$ ). Moving on to the case where the color of section $5$ is the same as that of section $1$ (which can be done in $1$ way), we will have $3$ ways (region $6$ cannot be that color of both region $1$ and $5$ ). Thus, if the color of region $4$ is different of that of region $1$ , then we have $2\\cdot 2 + 3 = 7$ ways.\nIf the color of region $4$ is the same as that of region $1$ (which can be done in one way), then the color of region $5$ have to be different from that of sector $1$ $3$ ways). That means there will be $2$ choices for the color of region $6$ . So if the color of region $4$ is the same as region $1$ , then we will have $3\\cdot 2 = 6$ . That means if the color of section $3$ is different from that of region $1$ , then there are $2\\cdot 7 + 6 = 20$\nNow, moving on to the case where section $3$ has the same color of region $1$ . That means section $4$ will have to be a different color of that of region $1$ $3$ ways). So, that means we have region $5$ to be either the same color or different color as region $1$ . If it is different (which can be done in $2$ ways), then there will be $2$ possibilities on the color of sector $6$ . If it is same (which can be done in $1$ way), then there are $3$ ways to color region $6$ . So, if section $3$ has the same color as section $1$ , then we have $3(2\\cdot 2 + 3) = 3(7) = 21$\nNow, in overall we will have $4\\cdot 3 (2(20)+21) = 12(61) = \\boxed{732}$", "Let the top-right segment be segment $1,$ and remaining segments are numbered $2,3,4,5,6$ in clockwise order.\nWe have $4$ choices for segment $1,$ and $3$ choices for segments $2,3,4,5.$ For segment $6,$ we wish to find the expected value of the number of choices for segment $6$ 's color, which depends on whether segment $5$ is red. If we let $P(n)$ denote the probability of segment $P$ being the same color as segment $1$ (for simplicity, denote segment $1$ 's color as red), we get the following recurrence relation: \\[P(n)=\\dfrac{1-P(n-1)}{3}.\\] This is because that you cannot have two reds in a row (hence the $1-P(n-1)$ ) and if segment $n-1$ is not red, there are three possible colors, one of which is red (hence the divide by $3$ ). Using the obvious fact that $P(1)=1$ by the definition of $P(n),$ we find that \\[P(5)=\\dfrac{7}{27}.\\] If segment $5$ is red, then there are three possible colors for segment $6.$ If it is not, there are $2$ possible choices for segment $6.$ This means the expected number of color choices for segment $6$ is \\[\\dfrac{7}{27}\\cdot3+\\dfrac{20}{27}\\cdot2,\\] and the total number of colorings of the ring is \\[4\\cdot3\\cdot3\\cdot3\\cdot3\\cdot\\left(\\dfrac{7}{27}\\cdot3+\\dfrac{20}{27}\\cdot2\\right)=\\boxed{732}.\\]" ]

[ "It is simple to notice that in each and every row, there is always one more black square than the white squares. Since there are $8$ rows, there are $8$ more black squares than the white squares. $8\\rightarrow \\boxed{8}$" ]

[ "\nDraw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$ , they each have area $2\\cdot\\frac{1}{2}\\cdot\\frac{4\\sqrt{3}}{4}=\\sqrt{3}$ . For triangle $ACD$ , we can see that $AC=AD=2\\sqrt{3}$ and $CD=2$ . Using Pythagorean Theorem, the altitude for this triangle is $\\sqrt{11}$ , so the area is $\\sqrt{11}$ . Adding each part up, we get $2\\sqrt{3}+\\sqrt{11}=\\sqrt{12}+\\sqrt{11} \\implies \\boxed{23}$" ]

[ "\nDraw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$ , they each have area $2\\cdot\\frac{1}{2}\\cdot\\frac{4\\sqrt{3}}{4}=\\sqrt{3}$ . For triangle $ACD$ , we can see that $AC=AD=2\\sqrt{3}$ and $CD=2$ . Using Pythagorean Theorem, the altitude for this triangle is $\\sqrt{11}$ , so the area is $\\sqrt{11}$ . Adding each part up, we get $2\\sqrt{3}+\\sqrt{11}=\\sqrt{12}+\\sqrt{11} \\implies \\boxed{23}$" ]