Grok2

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question:A sphere is inscribed in a cube, and the cube has a surface area of 24 square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?

answer:To solve this problem, I need to relate the surface area of the cube to the radius of the sphere and then the radius of the sphere to the edge length of the inner cube. Let's start with the outer cube. It has a surface area of 24 square meters, which means that each face has an area of 24/6 = 4 square meters. The edge length of the outer cube is then the square root of 4, which is 2 meters. The sphere is inscribed in the cube, so its diameter is equal to the edge length of the cube. That means the radius of the sphere is half of that, which is 1 meter. Now, the inner cube is inscribed in the sphere, so its longest diagonal is equal to the diameter of the sphere. The longest diagonal of a cube is the square root of 3 times the edge length, so I can set up an equation: sqrt(3) * x = 2, where x is the edge length of the inner cube. Solving for x, I get x = 2 / sqrt(3), which I can rationalize by multiplying the numerator and denominator by sqrt(3). That gives me x = 2 * sqrt(3) / 3. The surface area of the inner cube is 6 times the area of one face, which is x squared. So, the surface area is 6 * (2 * sqrt(3) / 3)^2. Simplifying, I get 6 * 4 * 3 / 9, which is 8 square meters.

question:The truncated right circular cone has a large base radius 8 cm and a small base radius of 4 cm. The height of the truncated cone is 6 cm. How many text{cm}^3 are in the volume of this solid? [asy] import olympiad; size(150); defaultpen(linewidth(0.8)); dotfactor=4; draw(ellipse((0,0),4,1)); draw(ellipse((0,3),2,1/2)); draw((3.97,.1)(1.97,3.1)^^(3.97,.1)(1.97,3.1)); [/asy]

answer:[asy] import olympiad; size(150); defaultpen(linewidth(0.8)); dotfactor=4; draw(ellipse((0,0),4,1)); draw(ellipse((0,3),2,1/2),gray(.7)); // draw((-3.97,.1)--(-1.97,3.1)^^(3.97,.1)--(1.97,3.1)); draw((-3.97,.1)--(0,6.07)--(3.97,.1)); draw((4,0)--(0,0)--(0,6.07),linewidth(0.8)); draw((2,3)--(0,3),linewidth(0.8)); label("4",(2,3)--(0,3),S); label("8",(4,0)--(0,0),S); label("6",(0,0)--(0,3),W); label("x",(0,2)--(0,6.07),W); [/asy] We "complete" the truncated cone by adding a smaller, similar cone atop the cut, forming a large cone. We don't know the height of the small cone, so call it x. Since the small and large cone are similar, we have x/4=(x+6)/8; solving yields x=6. Hence the small cone has radius 4, height 6, and volume (1/3)pi(4^2)(6)=32pi and the large cone has radius 8, height 12, and volume (1/3)pi(8^2)(12)=256pi. The frustum's volume is the difference of these two volumes, or 256pi-32pi=224pi cubic cm.

question:[asy] unitsize(27); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A,B,C,D,E,F,X,Y,Z; A=(3,3); B=(0,0); C=(6,0); D=(4,0); E=(4,2); F=(1,1); draw(ABCcycle); draw(AD); draw(BE); draw(CF); X=intersectionpoint(AD,CF); Y=intersectionpoint(BE,AD); Z=intersectionpoint(BE,CF); label(A,A,N); label(B,B,SW); label(C,C,SE); label(D,D,S); label(E,E,NE); label(F,F,NW); label(N_1,X,NE); label(N_2,Y,WNW); label(N_3,Z,S); [/asy] In the figure, overline{CD}, overline{AE} and overline{BF} are onethird of their respective sides. It follows that overline{AN_2}: overline{N_2N_1}: overline{N_1D} = 3: 3: 1, and similarly for lines BE and CF. Then the area of triangle N_1N_2N_3 is: text{(A) } frac {1}{10} triangle ABC qquad text{(B) } frac {1}{9} triangle ABC qquad text{(C) } frac{1}{7}triangle ABCqquad text{(D) } frac{1}{6}triangle ABCqquad text{(E) } text{none of these}

answer:Let [ABC]=K. Then [ADC] = frac{1}{3}K, and hence [N_1DC] = frac{1}{7} [ADC] = frac{1}{21}K. Similarly, [N_2EA]=[N_3FB] = frac{1}{21}K. Then [N_2N_1CE] = [ADC] - [N_1DC]-[N_2EA] = frac{5}{21}K, and same for the other quadrilaterals. Then [N_1N_2N_3] is just [ABC] minus all the other regions we just computed. That is,[[N_1N_2N_3] = K - 3left(frac{1}{21}Kright) - 3left(frac{5}{21}right)K = K - frac{6}{7}K = frac{1}{7}triangle ABC.]

question:The second hand on the clock pictured below is 6 cm long. How far in centimeters does the tip of this second hand travel during a period of 30 minutes? Express your answer in terms of pi. [asy] draw(Circle((0,0),20)); label(12,(0,20),S); label(9,(20,0),E); label(6,(0,20),N); label(3,(20,0),W); dot((0,0)); draw((0,0)(12,0)); draw((0,0)(8,10)); draw((0,0)(11,14),linewidth(1)); label(6cm,(5.5,7),SE); [/asy]

answer:In 30 minutes, the tip of the second hand travels 30 times around the circumference of a circle of radius 6cm. Since the circumference is 2pi cdot6 = 12pi, the tip of the second hand travels 12pi cdot 30 = 360pi centimeters.