💡Alright, so I have this geometry problem about a tetrahedron (ABCD). Let me try to visualize it first. A tetrahedron has four triangular faces, and in this case, the dihedral angle between faces (ABC) and (BCD) is (60^circ). That means if I were to look at the edge where these two faces meet, the angle between them is 60 degrees.Vertex (A) is projected onto face (BCD), and the projection is point (H), which is the orthocenter of triangle (BCD). Okay, so the orthocenter is the point where the three altitudes of a triangle intersect. That gives me some information about the position of (H) relative to triangle (BCD).Point (G) is the centroid of triangle (ABC). The centroid is the intersection point of the medians of the triangle, and it divides each median into a ratio of 2:1. Also, it's given that (AH = 4) and (AB = AC). So triangle (ABC) is isosceles with (AB = AC), and the length from (A) to its projection (H) is 4 units.I need to find the distance (GH). Hmm, so I have two points: (G), the centroid of (ABC), and (H), the orthocenter of (BCD). I need to figure out how far apart they are.Let me start by recalling some properties of centroids and orthocenters. The centroid (G) is located at the average of the coordinates of the three vertices of triangle (ABC). The orthocenter (H) is a bit more complicated, but since it's given as the projection of (A) onto (BCD), maybe I can use some coordinate geometry here.Maybe I should assign coordinates to the points to make this more concrete. Let me place point (B) at the origin ((0, 0, 0)), point (C) along the x-axis at ((c, 0, 0)), and point (D) somewhere in the plane (BCD). Since (H) is the orthocenter of (BCD), it must lie somewhere inside the triangle if it's acute, on the triangle if it's a right triangle, or outside if it's obtuse. But since (H) is the projection of (A), maybe it's inside.Wait, actually, since (H) is the projection of (A) onto (BCD), the line (AH) is perpendicular to the plane (BCD). So (AH) is the altitude from (A) to the plane (BCD), and its length is 4.Given that the dihedral angle between faces (ABC) and (BCD) is (60^circ), this angle is formed along the edge (BC). So the angle between the two planes is 60 degrees. That might relate to the angle between the normals of the two planes or something like that.Since (AB = AC), triangle (ABC) is isosceles with (AB = AC). So point (A) is equidistant from (B) and (C). That might help in setting up coordinates.Let me try to set up a coordinate system. Let me place point (B) at ((0, 0, 0)), point (C) at ((c, 0, 0)), and point (D) at ((d, e, 0)). Then, the plane (BCD) lies in the z=0 plane. Point (A) is somewhere above this plane, and its projection (H) is in the plane (BCD).Since (AH) is perpendicular to the plane (BCD), the coordinates of (A) would be ((h_x, h_y, 4)), where ((h_x, h_y, 0)) is point (H). But (H) is the orthocenter of triangle (BCD), so I need to find the coordinates of (H) based on triangle (BCD).Wait, maybe I can choose coordinates such that (H) is at the origin? No, because (H) is the projection of (A), and (A) is at ((h_x, h_y, 4)). Maybe it's better to set (H) at some point in the plane, say ((h_x, h_y, 0)), and then (A) is directly above it at ((h_x, h_y, 4)).But I also know that (AB = AC). So the distance from (A) to (B) is equal to the distance from (A) to (C). Let's write that out:[AB = AC implies sqrt{(h_x - 0)^2 + (h_y - 0)^2 + (4 - 0)^2} = sqrt{(h_x - c)^2 + (h_y - 0)^2 + (4 - 0)^2}]Simplifying both sides:[sqrt{h_x^2 + h_y^2 + 16} = sqrt{(h_x - c)^2 + h_y^2 + 16}]Squaring both sides:[h_x^2 + h_y^2 + 16 = (h_x - c)^2 + h_y^2 + 16]Simplify:[h_x^2 = h_x^2 - 2 c h_x + c^2]Subtract (h_x^2) from both sides:[0 = -2 c h_x + c^2]So:[-2 c h_x + c^2 = 0 implies c(-2 h_x + c) = 0]Assuming (c neq 0) (since (B) and (C) are distinct points), we have:[-2 h_x + c = 0 implies h_x = frac{c}{2}]So the x-coordinate of (H) is (c/2). That makes sense because in an isosceles triangle (ABC), the projection of (A) onto (BC) should be the midpoint of (BC). Wait, but (H) is the orthocenter of (BCD), not necessarily the midpoint of (BC). Hmm, that might not hold.Wait, no, in this case, since (AB = AC), the projection of (A) onto (BCD) is such that it lies along the median from (A) to (BC). So in the plane (BCD), the projection (H) lies along the line perpendicular to (BC) at its midpoint. But (H) is also the orthocenter of (BCD). So maybe triangle (BCD) is such that its orthocenter lies at the midpoint of (BC). That would mean triangle (BCD) is an isosceles triangle with (BD = CD), making the orthocenter lie at the midpoint.Alternatively, maybe triangle (BCD) is equilateral? If (BCD) is equilateral, then the orthocenter coincides with the centroid and the circumcenter, which would be at the midpoint of (BC). But I don't know if (BCD) is equilateral.Wait, but in the problem, it's not specified whether triangle (BCD) is equilateral or not. So maybe I need to make some assumptions or find relations based on the given dihedral angle.The dihedral angle between faces (ABC) and (BCD) is (60^circ). The dihedral angle is the angle between the two planes along their line of intersection, which is edge (BC). So the angle between the normals of the two planes is (60^circ).To find the dihedral angle, I can use the normals of the two planes. The normal to plane (ABC) can be found using the cross product of vectors (AB) and (AC), and the normal to plane (BCD) can be found using the cross product of vectors (BC) and (BD).But since I don't have coordinates yet, maybe I can relate the dihedral angle to the angle between the normals. The dihedral angle is equal to the angle between the normals if the normals are pointing towards each other, or 180 minus that angle. So depending on the orientation, it's either (60^circ) or (120^circ). But since it's given as (60^circ), I can assume it's the smaller angle.Alternatively, maybe I can use the relationship between the height (AH = 4) and the dihedral angle to find some lengths.Since (AH) is perpendicular to the plane (BCD), and the dihedral angle between (ABC) and (BCD) is (60^circ), perhaps the length (AF) (where (F) is the foot of the perpendicular from (A) to (BC)) can be related to (AH) through the dihedral angle.Wait, actually, if I consider the dihedral angle, it's the angle between the two planes, so the angle between the normals is (60^circ). The height (AH) is along the normal to plane (BCD), and the height from (A) to (BC) in plane (ABC) would form an angle of (60^circ) with (AH).So if I let (F) be the foot of the perpendicular from (A) to (BC) in plane (ABC), then triangle (AFH) is a right triangle with angle at (F) being (60^circ). So (AF) is the adjacent side, (AH) is the opposite side, and the hypotenuse is (FH).Wait, actually, no. Since (AH) is perpendicular to the plane (BCD), and (AF) is in plane (ABC), the angle between (AF) and (AH) is the dihedral angle, which is (60^circ). So in triangle (AFH), angle at (A) is (60^circ), with (AH = 4), and (AF) being the adjacent side.So using trigonometry, (AF = frac{AH}{sin 60^circ} = frac{4}{sqrt{3}/2} = frac{8}{sqrt{3}}).Wait, is that correct? Let me think. If angle at (A) is (60^circ), then (AF) is adjacent, (AH) is opposite. So:[tan 60^circ = frac{AH}{AF} implies sqrt{3} = frac{4}{AF} implies AF = frac{4}{sqrt{3}} = frac{4 sqrt{3}}{3}]Wait, now I'm confused. Which one is it?Hold on, maybe I need to draw a diagram mentally. The dihedral angle is between the two planes, so the angle between the normals is (60^circ). The height (AH) is along the normal to (BCD), and the height (AF) is along the normal to (BC) in plane (ABC). So the angle between (AH) and (AF) is (60^circ).So in triangle (AFH), we have:- (AH = 4)- angle at (A) is (60^circ)- (AF) is the length we need to findWait, no, actually, (AF) is in plane (ABC), and (AH) is perpendicular to plane (BCD). So the angle between (AF) and (AH) is (60^circ). So in triangle (AFH), which is a right triangle? Wait, no, because (AF) is not necessarily perpendicular to (FH).Wait, maybe I need to consider the relationship between the heights and the dihedral angle.Alternatively, perhaps I can use vectors. Let me assign coordinates to the points to make this more precise.Let me place point (B) at ((0, 0, 0)), point (C) at ((c, 0, 0)), and point (D) at ((d, e, 0)). Then, the plane (BCD) is the xy-plane. Point (A) is at some point ((x, y, 4)), since its projection (H) is at ((x, y, 0)).Given that (AB = AC), the distance from (A) to (B) equals the distance from (A) to (C):[sqrt{(x - 0)^2 + (y - 0)^2 + (4 - 0)^2} = sqrt{(x - c)^2 + (y - 0)^2 + (4 - 0)^2}]Simplifying:[x^2 + y^2 + 16 = (x - c)^2 + y^2 + 16]Which gives:[x^2 = x^2 - 2 c x + c^2 implies 0 = -2 c x + c^2 implies x = frac{c}{2}]So the x-coordinate of (H) is (c/2). Therefore, point (H) is at ((c/2, y, 0)).Now, since (H) is the orthocenter of triangle (BCD), it must satisfy the condition that the altitudes from each vertex intersect at (H). So, in triangle (BCD), the altitude from (B) must pass through (H), the altitude from (C) must pass through (H), and the altitude from (D) must pass through (H).Given that (B) is at ((0, 0, 0)), (C) is at ((c, 0, 0)), and (D) is at ((d, e, 0)), let's find the equations of the altitudes.First, the altitude from (B) to side (CD). The side (CD) goes from ((c, 0, 0)) to ((d, e, 0)). The slope of (CD) is (frac{e - 0}{d - c} = frac{e}{d - c}). Therefore, the slope of the altitude from (B) is the negative reciprocal, which is (frac{c - d}{e}).Since this altitude passes through (B(0,0)), its equation is (y = frac{c - d}{e} x).Similarly, the altitude from (C) to side (BD). The side (BD) goes from ((0, 0, 0)) to ((d, e, 0)). The slope of (BD) is (frac{e - 0}{d - 0} = frac{e}{d}). Therefore, the slope of the altitude from (C) is the negative reciprocal, which is (-frac{d}{e}).Since this altitude passes through (C(c, 0)), its equation is (y = -frac{d}{e}(x - c)).Now, the orthocenter (H) is the intersection of these two altitudes. So let's solve the two equations:1. (y = frac{c - d}{e} x)2. (y = -frac{d}{e}(x - c))Set them equal:[frac{c - d}{e} x = -frac{d}{e}(x - c)]Multiply both sides by (e):[(c - d) x = -d(x - c)]Expand the right side:[(c - d) x = -d x + d c]Bring all terms to the left:[(c - d) x + d x - d c = 0]Simplify:[c x - d x + d x - d c = 0 implies c x - d c = 0 implies c(x - d) = 0]Assuming (c neq 0), we have (x = d). Plugging back into the first equation:[y = frac{c - d}{e} cdot d = frac{d(c - d)}{e}]So the orthocenter (H) is at ((d, frac{d(c - d)}{e}, 0)). But earlier, we found that (H) is at ((c/2, y, 0)). Therefore, we have:[d = frac{c}{2}]So (d = c/2). Plugging this back into the expression for (y):[y = frac{(c/2)(c - c/2)}{e} = frac{(c/2)(c/2)}{e} = frac{c^2}{4 e}]Therefore, point (H) is at ((c/2, c^2/(4 e), 0)).Now, since (H) is the orthocenter, the third altitude from (D) must also pass through (H). The altitude from (D) to side (BC). Side (BC) is along the x-axis from ((0,0,0)) to ((c,0,0)). The slope of (BC) is 0, so the altitude from (D) is vertical. Therefore, the altitude from (D) is the vertical line (x = d = c/2). Since (H) is on this altitude, its x-coordinate is indeed (c/2), which matches our earlier result.So now, we have coordinates for all points:- (B(0, 0, 0))- (C(c, 0, 0))- (D(c/2, e, 0)) (Wait, no. Earlier, (D) was at ((d, e, 0)), and (d = c/2). So (D(c/2, e, 0)).)- (H(c/2, c^2/(4 e), 0))- (A(c/2, c^2/(4 e), 4))Wait, hold on. Earlier, I thought (A) was at ((x, y, 4)), which is ((c/2, y, 4)). But from the orthocenter condition, (H) is at ((c/2, c^2/(4 e), 0)). So (A) is directly above (H), so its coordinates are ((c/2, c^2/(4 e), 4)).Now, let's find the centroid (G) of triangle (ABC). The centroid is the average of the coordinates of (A), (B), and (C).So:[G_x = frac{c/2 + 0 + c}{3} = frac{3 c/2}{3} = frac{c}{2}][G_y = frac{c^2/(4 e) + 0 + 0}{3} = frac{c^2}{12 e}][G_z = frac{4 + 0 + 0}{3} = frac{4}{3}]So (G) is at ((c/2, c^2/(12 e), 4/3)).Now, point (H) is at ((c/2, c^2/(4 e), 0)).To find (GH), we can compute the distance between (G) and (H).The distance formula in 3D is:[GH = sqrt{(x_G - x_H)^2 + (y_G - y_H)^2 + (z_G - z_H)^2}]Plugging in the coordinates:[GH = sqrt{left(frac{c}{2} - frac{c}{2}right)^2 + left(frac{c^2}{12 e} - frac{c^2}{4 e}right)^2 + left(frac{4}{3} - 0right)^2}]Simplify each component:- (x)-component: (0)- (y)-component: (frac{c^2}{12 e} - frac{c^2}{4 e} = frac{c^2}{12 e} - frac{3 c^2}{12 e} = -frac{2 c^2}{12 e} = -frac{c^2}{6 e})- (z)-component: (frac{4}{3})So:[GH = sqrt{0 + left(-frac{c^2}{6 e}right)^2 + left(frac{4}{3}right)^2} = sqrt{frac{c^4}{36 e^2} + frac{16}{9}}]Hmm, this seems a bit complicated. Maybe I need to find a relationship between (c) and (e) using the dihedral angle.The dihedral angle between faces (ABC) and (BCD) is (60^circ). The dihedral angle can be found using the normals of the two planes.First, let's find the normal to plane (ABC). Points (A(c/2, c^2/(4 e), 4)), (B(0,0,0)), and (C(c, 0, 0)).Vectors in plane (ABC):- (AB = B - A = (-c/2, -c^2/(4 e), -4))- (AC = C - A = (c/2, -c^2/(4 e), -4))The normal vector (N_{ABC}) is the cross product of (AB) and (AC):[N_{ABC} = AB times AC]Calculating the cross product:Let me denote (AB = (-c/2, -c^2/(4 e), -4)) and (AC = (c/2, -c^2/(4 e), -4)).Compute the determinant:[N_{ABC} = begin{vmatrix}mathbf{i} & mathbf{j} & mathbf{k} - c/2 & - c^2/(4 e) & -4 c/2 & - c^2/(4 e) & -4 end{vmatrix}]Calculating each component:- (i)-component: ((- c^2/(4 e))(-4) - (-4)(- c^2/(4 e)) = (c^2/(e)) - (c^2/(e)) = 0)- (j)-component: (- [ (-c/2)(-4) - (-4)(c/2) ] = - [ 2 c - (-2 c) ] = - [4 c] = -4 c)- (k)-component: ((- c/2)(- c^2/(4 e)) - (- c^2/(4 e))(c/2) = (c^3)/(8 e) - (- c^3)/(8 e) = (c^3)/(8 e) + (c^3)/(8 e) = (c^3)/(4 e))So (N_{ABC} = (0, -4 c, c^3/(4 e))).Similarly, the normal vector to plane (BCD) is the cross product of vectors (BC) and (BD).Vectors in plane (BCD):- (BC = C - B = (c, 0, 0))- (BD = D - B = (c/2, e, 0))Compute the cross product (N_{BCD} = BC times BD):[N_{BCD} = begin{vmatrix}mathbf{i} & mathbf{j} & mathbf{k} c & 0 & 0 c/2 & e & 0 end{vmatrix}]Calculating each component:- (i)-component: (0 cdot 0 - 0 cdot e = 0)- (j)-component: (- [ c cdot 0 - 0 cdot (c/2) ] = - [0 - 0] = 0)- (k)-component: (c cdot e - 0 cdot (c/2) = c e)So (N_{BCD} = (0, 0, c e)).Now, the dihedral angle between the two planes is the angle between their normals. The dihedral angle is given as (60^circ), so:[cos 60^circ = frac{N_{ABC} cdot N_{BCD}}{||N_{ABC}|| cdot ||N_{BCD}||}]Compute the dot product:[N_{ABC} cdot N_{BCD} = (0)(0) + (-4 c)(0) + (c^3/(4 e))(c e) = 0 + 0 + (c^4)/4 = c^4 / 4]Compute the magnitudes:[||N_{ABC}|| = sqrt{0^2 + (-4 c)^2 + (c^3/(4 e))^2} = sqrt{16 c^2 + c^6/(16 e^2)}][||N_{BCD}|| = sqrt{0^2 + 0^2 + (c e)^2} = c e]So:[cos 60^circ = frac{c^4 / 4}{sqrt{16 c^2 + c^6/(16 e^2)} cdot c e}]Simplify:[frac{1}{2} = frac{c^4 / 4}{c e sqrt{16 c^2 + c^6/(16 e^2)}}]Multiply both sides by the denominator:[frac{1}{2} cdot c e sqrt{16 c^2 + c^6/(16 e^2)} = c^4 / 4]Multiply both sides by 2:[c e sqrt{16 c^2 + c^6/(16 e^2)} = c^4 / 2]Divide both sides by (c e) (assuming (c neq 0), (e neq 0)):[sqrt{16 c^2 + c^6/(16 e^2)} = c^3 / (2 e)]Square both sides:[16 c^2 + c^6/(16 e^2) = c^6 / (4 e^2)]Multiply both sides by (16 e^2) to eliminate denominators:[16 e^2 cdot 16 c^2 + c^6 = 4 c^6]Simplify:[256 c^2 e^2 + c^6 = 4 c^6]Bring all terms to one side:[256 c^2 e^2 + c^6 - 4 c^6 = 0 implies 256 c^2 e^2 - 3 c^6 = 0]Factor out (c^2):[c^2 (256 e^2 - 3 c^4) = 0]Since (c neq 0), we have:[256 e^2 - 3 c^4 = 0 implies 256 e^2 = 3 c^4 implies e^2 = frac{3 c^4}{256} implies e = frac{c^2 sqrt{3}}{16}]So (e = frac{c^2 sqrt{3}}{16}).Now, recall that point (H) is at ((c/2, c^2/(4 e), 0)). Plugging in (e):[y_H = frac{c^2}{4 e} = frac{c^2}{4 cdot (c^2 sqrt{3}/16)} = frac{c^2}{(c^2 sqrt{3}/4)} = frac{4}{sqrt{3}} = frac{4 sqrt{3}}{3}]So (H) is at ((c/2, 4 sqrt{3}/3, 0)).Similarly, point (G) is at ((c/2, c^2/(12 e), 4/3)). Plugging in (e):[y_G = frac{c^2}{12 e} = frac{c^2}{12 cdot (c^2 sqrt{3}/16)} = frac{c^2}{(3 c^2 sqrt{3}/4)} = frac{4}{3 sqrt{3}} = frac{4 sqrt{3}}{9}]So (G) is at ((c/2, 4 sqrt{3}/9, 4/3)).Now, let's compute (GH):[GH = sqrt{(c/2 - c/2)^2 + (4 sqrt{3}/9 - 4 sqrt{3}/3)^2 + (4/3 - 0)^2}]Simplify each component:- (x)-component: (0)- (y)-component: (4 sqrt{3}/9 - 4 sqrt{3}/3 = 4 sqrt{3}/9 - 12 sqrt{3}/9 = -8 sqrt{3}/9)- (z)-component: (4/3)So:[GH = sqrt{0 + left(-frac{8 sqrt{3}}{9}right)^2 + left(frac{4}{3}right)^2} = sqrt{frac{64 cdot 3}{81} + frac{16}{9}} = sqrt{frac{192}{81} + frac{144}{81}} = sqrt{frac{336}{81}} = sqrt{frac{112}{27}} = frac{sqrt{112}}{sqrt{27}} = frac{4 sqrt{7}}{3 sqrt{3}} = frac{4 sqrt{21}}{9}]So, after all that, the distance (GH) is (frac{4 sqrt{21}}{9}).**Final Answer**boxed{dfrac{4sqrt{21}}{9}}

💡Okay, so I have this math problem here about a new operation called "※". It's defined for any non-zero rational numbers m and n as m※n = 3m - n. The problem has two parts: first, I need to find the value of 2※10, and second, I have to determine whether this operation satisfies the distributive law, which is a※(b + c) = a※b + a※c. Alright, let's start with the first part. I need to compute 2※10. According to the definition, m※n is equal to 3 times m minus n. So, substituting m with 2 and n with 10, I get:2※10 = 3*2 - 10Let me calculate that step by step. 3 multiplied by 2 is 6, and then subtracting 10 from that gives me:6 - 10 = -4So, 2※10 equals -4. That seems straightforward. Now, moving on to the second part. I need to check if the operation "※" satisfies the distributive law. The distributive law states that for any numbers a, b, and c, the following should hold:a※(b + c) = a※b + a※cTo verify this, I should compute both sides of the equation separately and see if they are equal.First, let's compute the left side: a※(b + c). Using the definition of the operation, this would be:a※(b + c) = 3a - (b + c)Simplifying that, it becomes:3a - b - cNow, let's compute the right side: a※b + a※c. Using the definition again, each operation is:a※b = 3a - ba※c = 3a - cAdding these two results together:(3a - b) + (3a - c) = 3a - b + 3a - cCombining like terms:3a + 3a = 6aAnd the constants:-b - cSo, altogether, the right side simplifies to:6a - b - cNow, comparing both sides:Left side: 3a - b - cRight side: 6a - b - cHmm, these are not equal unless 3a equals 6a, which would only be true if a is zero. But the problem specifies that m and n are non-zero rational numbers, so a cannot be zero. Therefore, the two sides are not equal in general.Wait, let me double-check my calculations to make sure I didn't make a mistake. For the left side, a※(b + c) = 3a - (b + c) = 3a - b - c. That seems correct.For the right side, a※b + a※c = (3a - b) + (3a - c) = 6a - b - c. Yeah, that also looks right.So, unless a is zero, which it can't be, the two sides are different. Therefore, the operation "※" does not satisfy the distributive law.I think that's it. The first part is straightforward, and the second part shows that the distributive law doesn't hold because the left and right sides aren't equal unless a is zero, which isn't allowed here.

💡Alright, so I've got this problem here about sets A and B, and I need to find their union. Let me see... Okay, set A is defined as all x such that 0 < x < 2. That means A is the interval from 0 to 2, not including the endpoints. Got that.Now, set B is a bit trickier. It's defined as all x where x squared minus 1 is less than 0. Hmm, so x² - 1 < 0. I think I can solve this inequality to figure out what B is. Let me rewrite that inequality: x² < 1. So, x squared is less than 1. That means x has to be between -1 and 1 because if you square any number between -1 and 1, you get something less than 1. So, B is the interval from -1 to 1, not including the endpoints.Okay, so now I have A as (0, 2) and B as (-1, 1). I need to find the union of A and B, which is all the numbers that are in either A or B or both. Let me visualize this on a number line.Starting from the left, B goes from -1 to 1, and A goes from 0 to 2. So, if I combine these two intervals, the leftmost point is -1, and the rightmost point is 2. But wait, does that mean the union is from -1 to 2? Let me check.From -1 to 0, only B covers that part. From 0 to 1, both A and B overlap. From 1 to 2, only A covers that part. So, putting it all together, the union should indeed cover from -1 to 2. That makes sense because there are no gaps between -1 and 2 when combining these two intervals.Looking at the answer choices, option B is (-1, 2), which matches what I just figured out. So, I think that's the correct answer.But just to be thorough, let me make sure I didn't make any mistakes. Did I solve the inequality for B correctly? Yes, x² - 1 < 0 leads to x² < 1, which gives -1 < x < 1. That seems right.And for the union, since A starts at 0 and goes to 2, and B starts at -1 and goes to 1, combining them should cover from -1 to 2 without any breaks. Yeah, that checks out.I don't think I missed anything here. The union of A and B is definitely (-1, 2). So, option B is the one to go with.

💡First, identify the number of wrappers and bottle caps Danny currently has.Danny has 67 wrappers and 35 bottle caps.To determine how many more wrappers Danny has compared to bottle caps, subtract the number of bottle caps from the number of wrappers.67 wrappers - 35 bottle caps = 32Therefore, Danny has 32 more wrappers than bottle caps.

💡Okay, so I have this problem here where I need to find the minimum value of r such that the circle defined by the equation (x - cosθ)² + (y - sinθ)² = r² intersects with the line 2x - y - 10 = 0. θ is given to be in the interval [0, 2π), which means it covers all possible angles around the unit circle. First, I should recall that the general equation of a circle is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. In this case, the center of the circle is (cosθ, sinθ). So, the circle is centered at a point on the unit circle because cos²θ + sin²θ = 1, right? That means the center is always somewhere on the circumference of the unit circle.Now, the line given is 2x - y - 10 = 0. I need to find the minimum radius r such that this circle intersects with this line. I remember that for a circle and a line to intersect, the distance from the center of the circle to the line must be less than or equal to the radius. So, if I can find the distance from the center (cosθ, sinθ) to the line 2x - y - 10 = 0, then the radius r must be at least that distance. Therefore, the minimum value of r would be the minimum distance from the center to the line.The formula for the distance from a point (h, k) to the line Ax + By + C = 0 is |Ah + Bk + C| / sqrt(A² + B²). In this case, A is 2, B is -1, and C is -10. So, plugging in the center coordinates, the distance d would be |2*cosθ - 1*sinθ - 10| / sqrt(2² + (-1)²) = |2cosθ - sinθ - 10| / sqrt(5).So, d = |2cosθ - sinθ - 10| / sqrt(5). Since r must be greater than or equal to this distance for the circle and line to intersect, the minimum value of r is the minimum value of d. Therefore, I need to find the minimum value of |2cosθ - sinθ - 10| / sqrt(5) as θ varies from 0 to 2π.To find the minimum of this expression, I can first consider the expression inside the absolute value: 2cosθ - sinθ - 10. Let me denote this as E = 2cosθ - sinθ - 10. So, E = (2cosθ - sinθ) - 10. The term (2cosθ - sinθ) is a linear combination of sine and cosine, which can be expressed in the form Rcos(θ + φ), where R is the amplitude and φ is the phase shift.I remember that for an expression like a*cosθ + b*sinθ, the amplitude R is sqrt(a² + b²). So, in this case, a is 2 and b is -1. Therefore, R = sqrt(2² + (-1)²) = sqrt(4 + 1) = sqrt(5). So, 2cosθ - sinθ can be written as sqrt(5)cos(θ + φ), where φ is some angle.Therefore, E = sqrt(5)cos(θ + φ) - 10. The maximum and minimum values of E will depend on the maximum and minimum values of cos(θ + φ). Since cos(θ + φ) ranges between -1 and 1, the maximum value of E is sqrt(5)*1 - 10 = sqrt(5) - 10, and the minimum value is sqrt(5)*(-1) - 10 = -sqrt(5) - 10.But since we have the absolute value of E divided by sqrt(5), let's consider |E| / sqrt(5) = |sqrt(5)cos(θ + φ) - 10| / sqrt(5). Let's denote this as |sqrt(5)cos(θ + φ) - 10| / sqrt(5). To find the minimum of this expression, I can think about how the numerator behaves. The numerator is |sqrt(5)cos(θ + φ) - 10|. Since sqrt(5) is approximately 2.236, the term sqrt(5)cos(θ + φ) varies between -sqrt(5) and sqrt(5). So, the numerator is |something between -sqrt(5) and sqrt(5) minus 10|. Let me compute the numerator more carefully. The expression inside the absolute value is sqrt(5)cos(θ + φ) - 10. Since sqrt(5) is about 2.236, the maximum value of sqrt(5)cos(θ + φ) is 2.236, and the minimum is -2.236. Therefore, sqrt(5)cos(θ + φ) - 10 varies between -2.236 - 10 = -12.236 and 2.236 - 10 = -7.764. So, the expression inside the absolute value is always negative because the maximum is -7.764, which is still negative. Therefore, |sqrt(5)cos(θ + φ) - 10| = 10 - sqrt(5)cos(θ + φ).Therefore, the distance d becomes (10 - sqrt(5)cos(θ + φ)) / sqrt(5). So, d = (10 / sqrt(5)) - (sqrt(5)cos(θ + φ))/sqrt(5) = (10 / sqrt(5)) - cos(θ + φ).Simplifying, 10 / sqrt(5) is equal to 2*sqrt(5). Because 10 divided by sqrt(5) is (10*sqrt(5))/5 = 2*sqrt(5). So, d = 2*sqrt(5) - cos(θ + φ).Now, since cos(θ + φ) ranges between -1 and 1, the term -cos(θ + φ) ranges between -1 and 1 as well. Therefore, d ranges between 2*sqrt(5) - 1 and 2*sqrt(5) + 1. But wait, we're looking for the minimum value of d, which would correspond to the minimum value of r. So, the minimum value of d is 2*sqrt(5) - 1, because when cos(θ + φ) is 1, we subtract 1, making d as small as possible. Therefore, the minimum value of r is 2*sqrt(5) - 1. Let me verify this. If I take θ such that cos(θ + φ) = 1, then the distance d would be 2*sqrt(5) - 1. Since r must be at least d, the smallest possible r is 2*sqrt(5) - 1. To double-check, let's compute 2*sqrt(5). sqrt(5) is approximately 2.236, so 2*sqrt(5) is approximately 4.472. Subtracting 1 gives approximately 3.472. So, r must be at least about 3.472. Is there a way to confirm this? Maybe by plugging in specific θ values. Let's say θ is such that cosθ = 2/sqrt(5) and sinθ = -1/sqrt(5). Wait, is that possible? Because cos²θ + sin²θ should equal 1. Let's check: (2/sqrt(5))² + (-1/sqrt(5))² = 4/5 + 1/5 = 5/5 = 1. Yes, that works. So, if θ is such that cosθ = 2/sqrt(5) and sinθ = -1/sqrt(5), then the center of the circle is (2/sqrt(5), -1/sqrt(5)). Now, let's compute the distance from this center to the line 2x - y - 10 = 0. Using the distance formula: |2*(2/sqrt(5)) - (-1/sqrt(5)) - 10| / sqrt(4 + 1) = |4/sqrt(5) + 1/sqrt(5) - 10| / sqrt(5) = |5/sqrt(5) - 10| / sqrt(5) = |sqrt(5) - 10| / sqrt(5). Wait, that's |sqrt(5) - 10| / sqrt(5). Since sqrt(5) is about 2.236, sqrt(5) - 10 is negative, so the absolute value is 10 - sqrt(5). Therefore, the distance is (10 - sqrt(5)) / sqrt(5) = 10/sqrt(5) - sqrt(5)/sqrt(5) = 2*sqrt(5) - 1, which matches our earlier result. So, when θ is such that cosθ = 2/sqrt(5) and sinθ = -1/sqrt(5), the distance from the center to the line is exactly 2*sqrt(5) - 1, which is the minimum possible distance. Therefore, the minimum value of r is indeed 2*sqrt(5) - 1.I think that makes sense. I started by finding the distance from the center to the line, expressed it in terms of θ, transformed the trigonometric expression into a single cosine function to find its extrema, and then determined the minimum distance, which gives the minimum radius. I also double-checked with specific θ values to confirm the result. So, I'm pretty confident that the minimum value of r is 2*sqrt(5) - 1.

💡Okay, so I have this problem: I need to prove that in any convex polyhedron, there is either a triangular face or a vertex where three edges meet. Hmm, interesting. I remember that convex polyhedrons are three-dimensional shapes where all their faces are convex polygons, and every interior angle is less than 180 degrees. Also, they don't have any indentations or holes.First, let me think about what a convex polyhedron looks like. The most common examples are cubes, tetrahedrons, octahedrons, etc. In a cube, all faces are squares, and each vertex has three edges meeting there. So, in a cube, we have vertices where three edges meet. In a tetrahedron, all faces are triangles, so it's the other case. So, in these examples, the statement holds true.But I need to prove this for any convex polyhedron, not just these examples. Maybe I can use some properties of polyhedrons or some theorems related to them. I recall Euler's formula, which relates the number of vertices (V), edges (E), and faces (F) of a convex polyhedron: V - E + F = 2. Maybe that can help.Let me consider the contrapositive of the statement. If there are no triangular faces and no vertices where three edges meet, then the polyhedron cannot exist. So, assuming that all faces have at least four sides and every vertex has at least four edges, I should reach a contradiction.Okay, so let's assume that all faces are quadrilaterals or higher (i.e., each face has at least four edges). Also, assume that every vertex has at least four edges meeting there. Now, let's see what this implies about the number of edges, vertices, and faces.Each face has at least four edges, but each edge is shared by two faces. So, the total number of edges E is at least (4F)/2 = 2F. Similarly, each vertex has at least four edges, but each edge connects two vertices. So, the total number of edges E is also at least (4V)/2 = 2V.So, from these two, we have E ≥ 2F and E ≥ 2V. Now, using Euler's formula: V - E + F = 2. If I substitute E from the inequalities, I get V - 2F + F = 2, which simplifies to V - F = 2. Similarly, V - 2V + F = 2, which simplifies to -V + F = 2. Wait, that doesn't make sense because V - F = 2 and -V + F = 2 can't both be true unless V = F, which would imply 0 = 2, which is impossible.Hmm, so this contradiction suggests that our initial assumption is wrong. Therefore, it can't be the case that all faces are quadrilaterals or higher and all vertices have at least four edges. So, there must be at least one triangular face or at least one vertex where three edges meet.Wait, let me double-check my reasoning. I assumed E ≥ 2F and E ≥ 2V, then substituted into Euler's formula. Let me write it step by step:From E ≥ 2F and E ≥ 2V, we have:From E ≥ 2F: F ≤ E/2From E ≥ 2V: V ≤ E/2Now, substitute these into Euler's formula:V - E + F = 2Since V ≤ E/2 and F ≤ E/2, then:E/2 - E + E/2 ≤ 2Simplify:(E/2 + E/2) - E ≤ 2E - E ≤ 20 ≤ 2Which is true, but it doesn't give a contradiction. Hmm, maybe I did something wrong.Wait, maybe I should consider that E ≥ 2F and E ≥ 2V, so combining these, we have:From E ≥ 2F and E ≥ 2V, we get F ≤ E/2 and V ≤ E/2.Substitute into Euler's formula:V - E + F = 2But V ≤ E/2 and F ≤ E/2, so:E/2 - E + E/2 ≤ 2Which simplifies to 0 ≤ 2, which is always true, not a contradiction. So, my previous approach didn't lead to a contradiction. Maybe I need a different approach.Perhaps I should consider the average number of edges per face and the average number of edges per vertex. For a convex polyhedron, the average number of edges per face is 2E/F, and the average number of edges per vertex is 2E/V.If all faces have at least four edges, then 2E/F ≥ 4, which implies E ≥ 2F.Similarly, if all vertices have at least four edges, then 2E/V ≥ 4, which implies E ≥ 2V.Now, using Euler's formula: V - E + F = 2.From E ≥ 2F and E ≥ 2V, we have F ≤ E/2 and V ≤ E/2.Substitute into Euler's formula:V - E + F ≤ E/2 - E + E/2 = 0But Euler's formula says V - E + F = 2, so 0 ≥ 2, which is a contradiction.Ah, there we go. So, by assuming that all faces have at least four edges and all vertices have at least four edges, we end up with a contradiction because Euler's formula would require V - E + F = 2, but our assumptions lead to V - E + F ≤ 0, which is impossible. Therefore, our assumption must be wrong, meaning that there must be at least one face with three edges (a triangular face) or at least one vertex with three edges.So, that seems to be the proof. I think I got it now. By assuming the opposite and using Euler's formula, I reached a contradiction, which means the original statement must be true.Just to make sure, let me think of another example. Take a square pyramid. It has a square base and four triangular faces. So, it has triangular faces, so the statement holds. If I try to make a polyhedron without any triangular faces and without any vertices where only three edges meet, like a cube, but in a cube, all vertices have three edges, so it still satisfies the condition. Wait, but in a cube, all vertices have three edges, so it's the other case. So, in any case, either there's a triangular face or a vertex with three edges.Another example: a dodecahedron, which has all pentagonal faces. Each vertex has three edges meeting there, so again, it satisfies the condition.Okay, I think this makes sense now. The key was to use Euler's formula and the inequalities derived from the assumptions about the number of edges per face and per vertex. By showing that these assumptions lead to a contradiction with Euler's formula, we prove that at least one of the assumptions must be false, meaning there must be a triangular face or a vertex with three edges.