💡Okay, so I have this problem where I need to find the minimum value of the expression (frac{a}{3b} + frac{b}{6c} + frac{c}{9a}), where (a), (b), and (c) are positive real numbers. Hmm, I remember that for expressions like this, the AM-GM inequality might be useful. Let me try to recall how that works.The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is always greater than or equal to the geometric mean. In other words, for any non-negative numbers (x_1, x_2, ..., x_n), we have:[frac{x_1 + x_2 + ... + x_n}{n} geq sqrt[n]{x_1 x_2 ... x_n}]and equality holds when all the (x_i) are equal.In this problem, I have three terms: (frac{a}{3b}), (frac{b}{6c}), and (frac{c}{9a}). Maybe I can apply the AM-GM inequality to these three terms. Let me write that down.So, applying AM-GM:[frac{frac{a}{3b} + frac{b}{6c} + frac{c}{9a}}{3} geq sqrt[3]{frac{a}{3b} cdot frac{b}{6c} cdot frac{c}{9a}}]Multiplying both sides by 3 to get rid of the denominator on the left:[frac{a}{3b} + frac{b}{6c} + frac{c}{9a} geq 3 sqrt[3]{frac{a}{3b} cdot frac{b}{6c} cdot frac{c}{9a}}]Now, I need to simplify the product inside the cube root. Let me compute that step by step.First, multiply the numerators: (a cdot b cdot c = abc).Then, multiply the denominators: (3b cdot 6c cdot 9a). Let's compute that:- (3 times 6 = 18)- (18 times 9 = 162)- Then, the variables: (b cdot c cdot a = abc)So, the denominator is (162abc). Therefore, the product inside the cube root is:[frac{abc}{162abc} = frac{1}{162}]So, substituting back into the inequality:[frac{a}{3b} + frac{b}{6c} + frac{c}{9a} geq 3 sqrt[3]{frac{1}{162}}]Now, I need to compute (sqrt[3]{frac{1}{162}}). Let me see, 162 can be factored as (81 times 2), and 81 is (3^4). So, 162 is (3^4 times 2). Therefore, (sqrt[3]{162}) is (sqrt[3]{3^4 times 2}).Wait, maybe it's easier to write 162 as (2 times 81) and then as (2 times 3^4), so:[sqrt[3]{162} = sqrt[3]{2 times 3^4} = sqrt[3]{2} times sqrt[3]{3^4} = sqrt[3]{2} times 3^{4/3} = 3 times sqrt[3]{3 times 2} = 3 times sqrt[3]{6}]Wait, let me check that again. Actually, (sqrt[3]{3^4}) is (3^{4/3}), which is (3^{1 + 1/3}) or (3 times 3^{1/3}). So, putting it all together:[sqrt[3]{162} = sqrt[3]{2 times 3^4} = sqrt[3]{2} times sqrt[3]{3^4} = sqrt[3]{2} times 3^{4/3} = 3^{1} times 3^{1/3} times sqrt[3]{2} = 3 times sqrt[3]{3 times 2} = 3 times sqrt[3]{6}]Wait, that seems a bit convoluted. Maybe it's simpler to just leave it as (sqrt[3]{162}) for now. So, the cube root of 162 is approximately... well, I don't need the numerical value, just the exact form.So, going back to the inequality:[frac{a}{3b} + frac{b}{6c} + frac{c}{9a} geq 3 times frac{1}{sqrt[3]{162}} = frac{3}{sqrt[3]{162}}]So, the minimum value is (frac{3}{sqrt[3]{162}}). But wait, can I simplify this further? Let me see.Since 162 is (81 times 2) and 81 is (3^4), so:[sqrt[3]{162} = sqrt[3]{3^4 times 2} = 3^{4/3} times 2^{1/3} = 3 times 3^{1/3} times 2^{1/3} = 3 times sqrt[3]{3 times 2} = 3 times sqrt[3]{6}]So, substituting back:[frac{3}{sqrt[3]{162}} = frac{3}{3 times sqrt[3]{6}} = frac{1}{sqrt[3]{6}}]Wait, that simplifies nicely! So, the minimum value is (frac{1}{sqrt[3]{6}}). But let me double-check that.Starting from (sqrt[3]{162}), which is (sqrt[3]{81 times 2}). Since 81 is (3^4), so:[sqrt[3]{3^4 times 2} = 3^{4/3} times 2^{1/3} = 3^{1 + 1/3} times 2^{1/3} = 3 times 3^{1/3} times 2^{1/3} = 3 times (3 times 2)^{1/3} = 3 times sqrt[3]{6}]Yes, so (sqrt[3]{162} = 3 times sqrt[3]{6}). Therefore:[frac{3}{sqrt[3]{162}} = frac{3}{3 times sqrt[3]{6}} = frac{1}{sqrt[3]{6}}]So, the minimum value is (frac{1}{sqrt[3]{6}}). That seems simpler. But let me confirm if I did everything correctly.Wait, when I applied AM-GM, I assumed that the equality condition holds, which requires that all three terms are equal. So, for equality, we must have:[frac{a}{3b} = frac{b}{6c} = frac{c}{9a}]Let me set each of these equal to some constant (k). So:[frac{a}{3b} = k implies a = 3b k][frac{b}{6c} = k implies b = 6c k][frac{c}{9a} = k implies c = 9a k]Now, let's substitute these into each other to find a relationship between (a), (b), and (c).From the first equation, (a = 3b k). From the second, (b = 6c k). Substitute (b) into the first equation:[a = 3 times (6c k) times k = 18 c k^2]From the third equation, (c = 9a k). Substitute (a) from above:[c = 9 times (18 c k^2) times k = 162 c k^3]So, (c = 162 c k^3). Since (c) is positive, we can divide both sides by (c):[1 = 162 k^3 implies k^3 = frac{1}{162} implies k = frac{1}{sqrt[3]{162}}]Which is consistent with what we found earlier. So, the equality holds when (k = frac{1}{sqrt[3]{162}}), which means the minimum value is indeed (frac{3}{sqrt[3]{162}} = frac{1}{sqrt[3]{6}}).But wait, let me see if I can express (frac{1}{sqrt[3]{6}}) in a different form. Since (sqrt[3]{6} = 6^{1/3}), so (frac{1}{6^{1/3}} = 6^{-1/3}). Alternatively, it can be written as (sqrt[3]{frac{1}{6}}).Alternatively, rationalizing the denominator, but I don't think it's necessary here. So, the minimum value is (frac{1}{sqrt[3]{6}}).But just to be thorough, let me assign specific values to (a), (b), and (c) that satisfy the equality condition and check if the expression indeed equals (frac{1}{sqrt[3]{6}}).From earlier, we have:[a = 3b k, quad b = 6c k, quad c = 9a k]Let me choose (k = frac{1}{sqrt[3]{162}}). Let's pick a value for (c) and compute (a) and (b).Let me set (c = 1) for simplicity. Then:[c = 1]From (c = 9a k):[1 = 9a times frac{1}{sqrt[3]{162}} implies a = frac{sqrt[3]{162}}{9}]From (b = 6c k):[b = 6 times 1 times frac{1}{sqrt[3]{162}} = frac{6}{sqrt[3]{162}}]Now, let's compute each term of the original expression:[frac{a}{3b} = frac{frac{sqrt[3]{162}}{9}}{3 times frac{6}{sqrt[3]{162}}} = frac{sqrt[3]{162}}{9} times frac{sqrt[3]{162}}{18} = frac{(sqrt[3]{162})^2}{162}]Similarly, (frac{b}{6c} = frac{frac{6}{sqrt[3]{162}}}{6 times 1} = frac{1}{sqrt[3]{162}})And (frac{c}{9a} = frac{1}{9 times frac{sqrt[3]{162}}{9}} = frac{1}{sqrt[3]{162}})Wait, so two of the terms are (frac{1}{sqrt[3]{162}}) and one term is (frac{(sqrt[3]{162})^2}{162}). Let me compute (frac{(sqrt[3]{162})^2}{162}).Since (sqrt[3]{162} = 162^{1/3}), so ((sqrt[3]{162})^2 = 162^{2/3}). Therefore:[frac{162^{2/3}}{162} = 162^{2/3 - 1} = 162^{-1/3} = frac{1}{162^{1/3}} = frac{1}{sqrt[3]{162}}]So, all three terms are equal to (frac{1}{sqrt[3]{162}}), which confirms that the sum is indeed (3 times frac{1}{sqrt[3]{162}} = frac{3}{sqrt[3]{162}} = frac{1}{sqrt[3]{6}}).Therefore, the minimum value of the expression is (frac{1}{sqrt[3]{6}}).But just to make sure, let me compute the numerical value of (frac{1}{sqrt[3]{6}}) to see what it approximately equals. The cube root of 6 is approximately 1.817, so (frac{1}{1.817} approx 0.55). So, the minimum value is roughly 0.55.Alternatively, if I compute (frac{3}{sqrt[3]{162}}), since (sqrt[3]{162} approx 5.45), so (frac{3}{5.45} approx 0.55), which matches. So, both forms give the same approximate value.Therefore, I'm confident that the minimum value is (frac{1}{sqrt[3]{6}}), which can also be written as (frac{3}{sqrt[3]{162}}), but the former is simpler.So, in conclusion, the minimum possible value of the given expression is (frac{1}{sqrt[3]{6}}).

💡Alright, so I have this problem about a rectangular parallelepiped, which is like a 3D shape with six faces, all of which are rectangles. The specific one given is ABCDA₁B₁C₁D₁. The edges AB, AD, and AA₁ are given as 1 cm, 2 cm, and 1 cm respectively. I need to find the minimum area of triangle PA₁C, where point P lies on the line AB₁.First, I should probably visualize this shape. Let me try to sketch it mentally. So, ABCD is the base rectangle, with AB = 1 cm and AD = 2 cm. Then, the vertical edges like AA₁ are 1 cm, so the height of the parallelepiped is 1 cm. So, it's kind of a short box.Now, the points are labeled as follows: A, B, C, D are the base vertices, and A₁, B₁, C₁, D₁ are the top vertices directly above A, B, C, D respectively. So, AB is 1 cm, AD is 2 cm, and AA₁ is 1 cm.I need to find the minimum area of triangle PA₁C, where P is on AB₁. So, point P moves along the line segment from A to B₁. I have to figure out where along this line the area of triangle PA₁C is minimized.Okay, let's think about how to approach this. Maybe using coordinates would help. Let me assign coordinates to each vertex.Let's place point A at the origin (0, 0, 0). Then, since AB is 1 cm, point B would be at (1, 0, 0). AD is 2 cm, so point D is at (0, 2, 0). Since AA₁ is 1 cm, point A₁ is at (0, 0, 1). Then, point B₁, which is above B, would be at (1, 0, 1). Point C is diagonally opposite to A on the base, so its coordinates would be (1, 2, 0). Similarly, point C₁ would be at (1, 2, 1), but I don't think I need that for this problem.So, to recap, the coordinates are:- A: (0, 0, 0)- B: (1, 0, 0)- D: (0, 2, 0)- A₁: (0, 0, 1)- B₁: (1, 0, 1)- C: (1, 2, 0)Now, point P lies on AB₁. Let me parametrize point P. Since AB₁ is a line segment from A(0,0,0) to B₁(1,0,1), I can write the parametric equations for AB₁.Let me use a parameter t, where t ranges from 0 to 1. So, when t=0, P is at A, and when t=1, P is at B₁. So, the coordinates of P can be written as:P = A + t*(B₁ - A) = (0,0,0) + t*(1,0,1) = (t, 0, t)So, P(t) = (t, 0, t), where t ∈ [0,1].Now, I need to find the area of triangle PA₁C as a function of t and then find its minimum.To find the area of triangle PA₁C, I can use the formula involving vectors. The area is half the magnitude of the cross product of vectors PA₁ and PC.Wait, but actually, triangle PA₁C can be considered as a triangle with vertices at P(t), A₁, and C. So, vectors from A₁ to P and from A₁ to C can be used.Alternatively, I can use the formula for the area of a triangle given by three points in space. The area is (1/2) times the magnitude of the cross product of two sides.Let me denote the vectors:Vector PA₁ = A₁ - P = (0 - t, 0 - 0, 1 - t) = (-t, 0, 1 - t)Vector PC = C - P = (1 - t, 2 - 0, 0 - t) = (1 - t, 2, -t)Then, the cross product PA₁ × PC will give a vector perpendicular to both, and its magnitude will be equal to the area of the parallelogram formed by PA₁ and PC. So, half of that will be the area of the triangle.So, let's compute the cross product:PA₁ × PC = |i   j   k|              -t   0   1 - t              1 - t  2   -tCalculating the determinant:i * (0*(-t) - 2*(1 - t)) - j * (-t*(-t) - (1 - t)*(1 - t)) + k * (-t*2 - 0*(1 - t))Let me compute each component step by step.First, the i component:0*(-t) = 02*(1 - t) = 2 - 2tSo, 0 - (2 - 2t) = -2 + 2tSo, the i component is (-2 + 2t)iNext, the j component:(-t)*(-t) = t²(1 - t)*(1 - t) = (1 - t)² = 1 - 2t + t²So, t² - (1 - 2t + t²) = t² - 1 + 2t - t² = -1 + 2tBut since it's subtracted, it becomes -(-1 + 2t) = 1 - 2tSo, the j component is (1 - 2t)jWait, hold on. Let me double-check:The formula for the j component is:- [ (-t)*(-t) - (1 - t)*(1 - t) ]Which is - [ t² - (1 - 2t + t²) ] = - [ t² -1 + 2t - t² ] = - [ -1 + 2t ] = 1 - 2tYes, that's correct.Now, the k component:(-t)*2 = -2t0*(1 - t) = 0So, -2t - 0 = -2tSo, the k component is (-2t)kPutting it all together:PA₁ × PC = (-2 + 2t)i + (1 - 2t)j - 2t kNow, let's compute the magnitude of this cross product:|PA₁ × PC| = sqrt[ (-2 + 2t)^2 + (1 - 2t)^2 + (-2t)^2 ]Let me compute each term:(-2 + 2t)^2 = (2t - 2)^2 = 4t² - 8t + 4(1 - 2t)^2 = 1 - 4t + 4t²(-2t)^2 = 4t²Adding them up:4t² -8t +4 +1 -4t +4t² +4t²Combine like terms:4t² +4t² +4t² = 12t²-8t -4t = -12t4 +1 =5So, total is 12t² -12t +5Therefore, |PA₁ × PC| = sqrt(12t² -12t +5)Thus, the area of triangle PA₁C is (1/2)*sqrt(12t² -12t +5)So, we need to minimize this expression with respect to t in [0,1].Let me denote f(t) = (1/2)*sqrt(12t² -12t +5)To find the minimum, we can instead minimize the square of the function, since the square root is a monotonic function.So, let me define g(t) = (12t² -12t +5)We need to find the t that minimizes g(t).This is a quadratic function in t, which opens upwards (since the coefficient of t² is positive). Therefore, its minimum occurs at the vertex.The vertex of a quadratic function at² + bt + c is at t = -b/(2a)Here, a =12, b= -12So, t = -(-12)/(2*12) = 12/24 = 0.5So, the minimum occurs at t=0.5Therefore, the minimum value of g(t) is:g(0.5) =12*(0.5)^2 -12*(0.5) +5 =12*(0.25) -6 +5=3 -6 +5=2So, g(t) has a minimum value of 2 at t=0.5Therefore, the minimum area is (1/2)*sqrt(2) = sqrt(2)/2So, the minimum area of triangle PA₁C is sqrt(2)/2 cm².Wait, let me verify my steps to make sure I didn't make any mistakes.First, assigning coordinates: seems correct.Parametrizing P(t) as (t,0,t): correct.Vectors PA₁ and PC: correct.Cross product calculation:i component: 0*(-t) - 2*(1 - t) = 0 - 2 + 2t = -2 + 2tj component: - [ (-t)*(-t) - (1 - t)*(1 - t) ] = - [ t² - (1 - 2t + t²) ] = - [ -1 + 2t ] =1 - 2tk component: (-t)*2 -0*(1 - t)= -2tSo, cross product is (-2 + 2t, 1 - 2t, -2t). Correct.Magnitude squared: (-2 + 2t)^2 + (1 - 2t)^2 + (-2t)^2Which is 4t² -8t +4 +1 -4t +4t² +4t²Total: 12t² -12t +5. Correct.Then, since it's a quadratic, minimum at t=0.5. Plugging in, get g(0.5)=2. So, area is (1/2)*sqrt(2)=sqrt(2)/2.Yes, that seems correct.Alternatively, another approach is to realize that the minimal area occurs when the height from P to the line A₁C is minimized. Since the area is (1/2)*base*height, and base is fixed as |A₁C|, minimizing the height will minimize the area.So, perhaps another way is to compute the distance from P(t) to the line A₁C and find its minimum.But since I already have the cross product method, which gives the same result, I think my answer is correct.Therefore, the minimum area is sqrt(2)/2 cm².**Final Answer**The minimum area of triangle ( mathrm{PA}_{1}mathrm{C} ) is boxed{dfrac{sqrt{2}}{2}} cm².

💡Alright, so I've got this problem about orthogonal tetrahedrons. It's part (a) through (e), and I need to prove each statement. Let me start with part (a).**Part (a):** If two pairs of opposite edges of a tetrahedron are mutually perpendicular, then the third pair of edges is also mutually perpendicular, i.e., such a tetrahedron is orthogonal.Hmm, okay. So, a tetrahedron has four triangular faces and six edges. Opposite edges are those that don't intersect and aren't adjacent. So, in a tetrahedron ABCD, the opposite edges would be AB and CD, AC and BD, AD and BC.If two pairs of these opposite edges are perpendicular, then I need to show the third pair is also perpendicular.Let me visualize this. Maybe I can assign coordinates to the vertices to make it easier. Let's place vertex A at the origin (0,0,0). Let me assume edge AB is along the x-axis, so B is at (b,0,0). Then, since AB is perpendicular to CD, CD must lie in a plane perpendicular to AB. Similarly, if AC is perpendicular to BD, then BD must lie in a plane perpendicular to AC.Wait, maybe using vectors would help. Let me denote vectors for the edges. Let vector AB be **u**, AC be **v**, and AD be **w**. Then, the opposite edges would be CD, BD, and BC.But CD is from C to D, which is vector **w** - **v**. Similarly, BD is **w** - **u**, and BC is **v** - **u**.Given that AB is perpendicular to CD, so **u** · (**w** - **v**) = 0.Similarly, AC is perpendicular to BD, so **v** · (**w** - **u**) = 0.I need to show that AD is perpendicular to BC, which would mean **w** · (**v** - **u**) = 0.Let me write down the given conditions:1. **u** · (**w** - **v**) = 0 ⇒ **u** · **w** - **u** · **v** = 0 ⇒ **u** · **w** = **u** · **v**2. **v** · (**w** - **u**) = 0 ⇒ **v** · **w** - **v** · **u** = 0 ⇒ **v** · **w** = **v** · **u**From both equations, **u** · **w** = **v** · **u** and **v** · **w** = **v** · **u**, so **u** · **w** = **v** · **w**.Therefore, **w** · (**u** - **v**) = 0.Which means **w** is perpendicular to (**u** - **v**), which is vector BC.So, **w** · (**v** - **u**) = 0, which is exactly what we needed to show. Thus, AD is perpendicular to BC.Okay, that seems to work. So, if two pairs of opposite edges are perpendicular, the third pair must also be perpendicular. So, the tetrahedron is orthogonal.**Part (b):** The four altitudes of an orthogonal tetrahedron intersect at one point.Alright, altitudes in a tetrahedron are the perpendicular lines from each vertex to the opposite face.In an orthogonal tetrahedron, since all opposite edges are perpendicular, maybe this symmetry ensures that the altitudes intersect at a single point.I recall that in a tetrahedron, the altitudes intersect at a point called the orthocenter, similar to the orthocenter in a triangle.But in a general tetrahedron, the altitudes don't necessarily intersect at a single point. However, in an orthogonal tetrahedron, due to the perpendicularity of edges, perhaps this forces the altitudes to intersect.Let me think about coordinates again. Suppose I have an orthogonal tetrahedron with vertices at (0,0,0), (a,0,0), (0,b,0), and (0,0,c). This is a standard orthogonal tetrahedron where edges along the axes are perpendicular.In this case, the altitudes from each vertex would be along the coordinate axes, and they all intersect at the origin. So, in this case, the altitudes intersect at the origin.But is this always the case? Maybe in any orthogonal tetrahedron, regardless of the coordinate system, the altitudes intersect at one point.Alternatively, maybe I can use vector methods or properties of orthocenters in tetrahedrons.Wait, in a tetrahedron, if all opposite edges are perpendicular, then the tetrahedron is called orthogonal, and it's known that the altitudes intersect at a single point. So, this is a known property.But to prove it, maybe I can use the fact that in such a tetrahedron, the orthocenter exists and is unique.Alternatively, using coordinates, suppose the tetrahedron has vertices at A, B, C, D, with AB, AC, AD mutually perpendicular. Then, the altitudes from each vertex would indeed intersect at the orthocenter, which is the point where all three coordinate axes meet.But in the general case, where the edges are not necessarily along the coordinate axes, but still mutually perpendicular, the same logic applies.So, I think the key is that in an orthogonal tetrahedron, the altitudes must intersect at one point because of the inherent symmetry and perpendicularity.**Part (c):** If three altitudes of a tetrahedron intersect at one point, then this tetrahedron is orthogonal.Hmm, so the converse of part (b). If three altitudes intersect at one point, then the tetrahedron must be orthogonal.In a general tetrahedron, if three altitudes intersect at one point, does that imply that all opposite edges are perpendicular?I think yes, because the intersection of three altitudes imposes a certain symmetry on the tetrahedron, forcing the edges to be perpendicular.Let me think in terms of coordinates again. Suppose three altitudes intersect at a point. Then, maybe the tetrahedron can be embedded in a coordinate system where the edges are along the axes, making it orthogonal.Alternatively, using vector methods, if three altitudes intersect, then the corresponding vectors must satisfy certain orthogonality conditions.Wait, in a tetrahedron, the altitudes are related to the faces. If three altitudes intersect, then the faces must be arranged in such a way that their normals are orthogonal.Wait, maybe not directly, but perhaps the intersection of altitudes imposes that the edges are orthogonal.Alternatively, perhaps I can use the fact that if three altitudes intersect, then the tetrahedron is orthocentric, and in an orthocentric tetrahedron, if it's also isohedral or something, it becomes orthogonal.Wait, maybe I need to recall that in a tetrahedron, if three altitudes intersect, then the fourth altitude also intersects at the same point, making it an orthocentric tetrahedron.But being orthocentric doesn't necessarily mean it's orthogonal. So, maybe I need more.Wait, but in part (c), it's given that three altitudes intersect at one point, and we need to show the tetrahedron is orthogonal, meaning all opposite edges are perpendicular.So, perhaps I can use the fact that if three altitudes intersect, then certain edges must be perpendicular.Let me try to think in terms of coordinates. Suppose the tetrahedron has vertices A, B, C, D, and three altitudes intersect at point O.Then, O is the orthocenter. So, OA is perpendicular to the face BCD, OB is perpendicular to the face ACD, and OC is perpendicular to the face ABD.If I can show that the edges AB, AC, AD are mutually perpendicular, then the tetrahedron is orthogonal.Wait, but in general, the faces being perpendicular to the lines from O doesn't necessarily mean the edges are perpendicular.Hmm, maybe I need a different approach.Alternatively, perhaps I can use the fact that in a tetrahedron, if three altitudes intersect, then the tetrahedron is orthocentric, and in an orthocentric tetrahedron, if it's also isohedral, it's orthogonal.Wait, maybe I need to recall some properties.Alternatively, perhaps I can use the fact that if three altitudes intersect, then the tetrahedron satisfies certain vector conditions that imply the edges are perpendicular.Let me denote vectors for the edges again. Let **u**, **v**, **w** be vectors from vertex A to B, C, D respectively.If the altitudes from B, C, D intersect at a point O, then O is the orthocenter.So, the altitude from B is perpendicular to the face ACD, which is the plane containing A, C, D.Similarly, the altitude from C is perpendicular to the face ABD, and the altitude from D is perpendicular to the face ABC.If these three altitudes intersect at O, then O must satisfy certain conditions.Wait, maybe I can express O in terms of coordinates.Suppose A is at (0,0,0), B is at (a,0,0), C is at (0,b,0), D is at (0,0,c). Then, the tetrahedron is orthogonal, and the orthocenter is at (0,0,0).But in this case, the altitudes are along the coordinate axes, and they intersect at the origin.But if I have a general tetrahedron where three altitudes intersect at a point, does that mean it's orthogonal?Wait, maybe I can consider the reciprocal. If a tetrahedron is orthogonal, then its altitudes intersect at one point, as shown in part (b). Now, if three altitudes intersect, does that imply orthogonality?I think yes, because the intersection of three altitudes imposes that the edges must be perpendicular.Alternatively, maybe I can use the fact that in a tetrahedron, if three altitudes intersect, then the fourth altitude also intersects at the same point, making it orthocentric, and in such a tetrahedron, if it's also isohedral, it's orthogonal.Wait, but I'm not sure about that.Alternatively, maybe I can use the fact that in a tetrahedron, the altitudes intersect if and only if the tetrahedron is orthocentric, and in an orthocentric tetrahedron, if it's also isohedral, it's orthogonal.But I'm not sure if being orthocentric implies orthogonality.Wait, maybe I need to think differently. If three altitudes intersect, then the tetrahedron must satisfy certain conditions on its edges.Let me consider the vectors again. Suppose the tetrahedron has vertices A, B, C, D, and the altitudes from B, C, D intersect at O.Then, O is the orthocenter. So, the vector from A to O must be perpendicular to the face BCD, which is the plane containing B, C, D.Similarly, the vector from B to O must be perpendicular to the face ACD, and the vector from C to O must be perpendicular to the face ABD.If I can express these conditions in terms of vectors, maybe I can derive that the edges are perpendicular.Let me denote vectors:Let **a** = vector OA, **b** = vector OB, **c** = vector OC, **d** = vector OD.Since O is the orthocenter, **a** is perpendicular to the plane BCD, so **a** is perpendicular to vectors **b** - **d** and **c** - **d**.Similarly, **b** is perpendicular to the plane ACD, so **b** is perpendicular to vectors **a** - **d** and **c** - **d**.And **c** is perpendicular to the plane ABD, so **c** is perpendicular to vectors **a** - **d** and **b** - **d**.Wait, this seems complicated. Maybe I can consider the scalar products.Since **a** is perpendicular to plane BCD, then **a** · (**b** - **d**) = 0 and **a** · (**c** - **d**) = 0.Similarly, **b** · (**a** - **d**) = 0 and **b** · (**c** - **d**) = 0.And **c** · (**a** - **d**) = 0 and **c** · (**b** - **d**) = 0.This gives us a system of equations.Let me write them out:1. **a** · (**b** - **d**) = 0 ⇒ **a** · **b** - **a** · **d** = 02. **a** · (**c** - **d**) = 0 ⇒ **a** · **c** - **a** · **d** = 03. **b** · (**a** - **d**) = 0 ⇒ **a** · **b** - **b** · **d** = 04. **b** · (**c** - **d**) = 0 ⇒ **b** · **c** - **b** · **d** = 05. **c** · (**a** - **d**) = 0 ⇒ **a** · **c** - **c** · **d** = 06. **c** · (**b** - **d**) = 0 ⇒ **b** · **c** - **c** · **d** = 0From equations 1 and 3:From 1: **a** · **b** = **a** · **d**From 3: **a** · **b** = **b** · **d**Thus, **a** · **d** = **b** · **d**Similarly, from equations 2 and 5:From 2: **a** · **c** = **a** · **d**From 5: **a** · **c** = **c** · **d**Thus, **a** · **d** = **c** · **d**From equations 4 and 6:From 4: **b** · **c** = **b** · **d**From 6: **b** · **c** = **c** · **d**Thus, **b** · **d** = **c** · **d**So, from all these, we have:**a** · **d** = **b** · **d** = **c** · **d**Let me denote this common value as k.So, **a** · **d** = **b** · **d** = **c** · **d** = kNow, let's look at equations 1 and 2:From 1: **a** · **b** = kFrom 2: **a** · **c** = kSimilarly, from 4: **b** · **c** = kSo, **a** · **b** = **a** · **c** = **b** · **c** = kNow, let's consider the vectors **a**, **b**, **c**.If **a**, **b**, **c** are vectors from the orthocenter O to the vertices A, B, C, D.Wait, but in our setup, **a**, **b**, **c**, **d** are vectors from O to A, B, C, D respectively.So, the position vectors of A, B, C, D with respect to O are **a**, **b**, **c**, **d**.Given that, the edges of the tetrahedron are vectors like **b** - **a**, **c** - **a**, etc.But we need to show that the opposite edges are perpendicular.Opposite edges are AB and CD, AC and BD, AD and BC.So, vector AB is **b** - **a**, vector CD is **d** - **c**.We need to show that (**b** - **a**) · (**d** - **c**) = 0.Similarly for the other pairs.Let me compute (**b** - **a**) · (**d** - **c**):= **b** · **d** - **b** · **c** - **a** · **d** + **a** · **c**From earlier, we have:**b** · **d** = k**b** · **c** = k**a** · **d** = k**a** · **c** = kSo, substituting:= k - k - k + k = 0Thus, (**b** - **a**) · (**d** - **c**) = 0So, AB is perpendicular to CD.Similarly, let's check AC and BD.Vector AC is **c** - **a**, vector BD is **d** - **b**.Compute (**c** - **a**) · (**d** - **b**):= **c** · **d** - **c** · **b** - **a** · **d** + **a** · **b**Again, substituting:**c** · **d** = k**c** · **b** = k**a** · **d** = k**a** · **b** = kSo, = k - k - k + k = 0Thus, AC is perpendicular to BD.Finally, check AD and BC.Vector AD is **d** - **a**, vector BC is **c** - **b**.Compute (**d** - **a**) · (**c** - **b**):= **d** · **c** - **d** · **b** - **a** · **c** + **a** · **b**Substituting:**d** · **c** = k**d** · **b** = k**a** · **c** = k**a** · **b** = kSo, = k - k - k + k = 0Thus, AD is perpendicular to BC.Therefore, all opposite edges are perpendicular, so the tetrahedron is orthogonal.**Part (d):** If the foot of one of the altitudes of the tetrahedron is the orthocenter (the intersection point of the altitudes) of the corresponding face, then the tetrahedron is orthogonal, and vice versa.Alright, so if the foot of an altitude is the orthocenter of the corresponding face, then the tetrahedron is orthogonal, and conversely, if the tetrahedron is orthogonal, then the foot of each altitude is the orthocenter of the corresponding face.Let me first understand what this means.In a tetrahedron, each face is a triangle. The orthocenter of a face is the point where the three altitudes of that face intersect.If the foot of an altitude from a vertex to the opposite face is the orthocenter of that face, then that imposes a certain condition on the tetrahedron.I need to show that this condition implies that the tetrahedron is orthogonal, and conversely, if the tetrahedron is orthogonal, then this condition holds.Let me start with the forward direction: if the foot of one altitude is the orthocenter of the corresponding face, then the tetrahedron is orthogonal.Suppose in tetrahedron ABCD, the foot of the altitude from D to face ABC is the orthocenter of face ABC.Let me denote the foot as H. So, H is the orthocenter of triangle ABC, and DH is perpendicular to the plane ABC.In triangle ABC, the orthocenter H is the intersection of the altitudes of ABC.So, in triangle ABC, the altitudes from A, B, C intersect at H.But H is also the foot of the altitude from D.So, in the tetrahedron, DH is perpendicular to ABC, and H is the orthocenter of ABC.I need to show that this implies that the tetrahedron is orthogonal, i.e., all opposite edges are perpendicular.Let me consider the implications.Since H is the orthocenter of ABC, the altitudes from A, B, C in triangle ABC meet at H.Also, DH is perpendicular to ABC.So, in the tetrahedron, the line DH is perpendicular to the plane ABC, and H is the orthocenter of ABC.I need to relate this to the perpendicularity of opposite edges.Let me consider edges AB and CD.Since H is the orthocenter of ABC, the altitude from C to AB passes through H.Similarly, the altitude from B to AC passes through H.But since DH is perpendicular to ABC, and H is in ABC, then DH is the altitude from D to ABC.Now, let me consider the edges.In triangle ABC, since H is the orthocenter, we have:- AH ⊥ BC- BH ⊥ AC- CH ⊥ ABNow, since DH is perpendicular to ABC, and H is the orthocenter, perhaps we can relate the edges of the tetrahedron.Let me consider edge AB and edge CD.Edge AB is in ABC, and edge CD connects C to D.Since DH is perpendicular to ABC, and H is the orthocenter, perhaps CD is related to the altitude from C.Wait, in triangle ABC, the altitude from C is CH, which is perpendicular to AB.But in the tetrahedron, CD is not necessarily the same as CH, unless D lies along the line perpendicular to ABC at H.Wait, but if DH is perpendicular to ABC, then D lies along the line through H perpendicular to ABC.So, if I consider the coordinates again, let me place H at the origin for simplicity.Let me assign coordinates such that H is at (0,0,0), and the plane ABC is the xy-plane.Then, D is at (0,0,h) for some h.In triangle ABC, since H is the orthocenter, the altitudes from A, B, C meet at H.So, in the xy-plane, the coordinates of A, B, C must satisfy that the altitudes intersect at H.Let me denote A as (a,0,0), B as (0,b,0), and C as (0,0,0). Wait, but H is the orthocenter, so if C is at (0,0,0), then the altitude from C is the line x=0, y=0, which is just the z-axis, but H is at (0,0,0), so that doesn't make sense.Wait, maybe I need to choose coordinates differently.Let me consider triangle ABC with orthocenter at H=(0,0,0).Let me place A at (a,0,0), B at (0,b,0), and C at (0,0,c), but then the orthocenter would not necessarily be at (0,0,0).Wait, maybe it's better to use a different coordinate system.Alternatively, let me consider that in triangle ABC, the orthocenter H is at (0,0,0), and the plane ABC is the xy-plane.Let me denote A as (a,0,0), B as (0,b,0), and C as (0,0,0). Wait, but then the orthocenter of ABC would not be at (0,0,0) unless ABC is a right triangle.Wait, if ABC is a right triangle at C, then the orthocenter is at C.So, if I set C at (0,0,0), and A at (a,0,0), B at (0,b,0), then ABC is a right triangle at C, and the orthocenter is at C.So, in this case, H=C=(0,0,0).Then, D is at (0,0,h), since DH is perpendicular to ABC.So, D is at (0,0,h).Now, let's see the edges:AB is from (a,0,0) to (0,b,0).CD is from (0,0,0) to (0,0,h).So, vector AB is (-a, b, 0), and vector CD is (0,0,h).Their dot product is (-a)(0) + (b)(0) + (0)(h) = 0.So, AB is perpendicular to CD.Similarly, AC is from (a,0,0) to (0,0,0), which is (-a,0,0).BD is from (0,b,0) to (0,0,h), which is (0,-b,h).Their dot product is (-a)(0) + (0)(-b) + (0)(h) = 0.So, AC is perpendicular to BD.Similarly, AD is from (a,0,0) to (0,0,h), which is (-a,0,h).BC is from (0,b,0) to (0,0,0), which is (0,-b,0).Their dot product is (-a)(0) + (0)(-b) + (h)(0) = 0.So, AD is perpendicular to BC.Thus, in this case, the tetrahedron is orthogonal.But this is a specific case where ABC is a right triangle at C, and D is along the z-axis.But in general, if the foot of the altitude is the orthocenter of the face, does this imply that the tetrahedron is orthogonal?Wait, in the general case, if H is the orthocenter of ABC, and DH is perpendicular to ABC, then the coordinates would be similar to the above, leading to the edges being perpendicular.Thus, the tetrahedron must be orthogonal.Conversely, if the tetrahedron is orthogonal, then the foot of each altitude is the orthocenter of the corresponding face.In an orthogonal tetrahedron, as we saw earlier, the altitudes intersect at one point, which is the orthocenter.So, the foot of each altitude from a vertex to the opposite face is the orthocenter of that face.Thus, the statement holds in both directions.**Part (e):** In an orthogonal tetrahedron, the common perpendiculars of the three pairs of opposite edges pass through the intersection point of the altitudes.Alright, so in an orthogonal tetrahedron, the common perpendiculars of each pair of opposite edges pass through the orthocenter, which is the intersection point of the altitudes.I need to show that for each pair of opposite edges, their common perpendicular line passes through the orthocenter.Let me consider the common perpendicular of AB and CD.In an orthogonal tetrahedron, AB is perpendicular to CD, so their common perpendicular is the line connecting them, but since they are skew lines, their common perpendicular is the line that is perpendicular to both and connects them.But in an orthogonal tetrahedron, since AB and CD are perpendicular, their common perpendicular would pass through the orthocenter.Wait, maybe I need to think in terms of coordinates again.Let me consider the orthogonal tetrahedron with vertices at A=(a,0,0), B=(0,b,0), C=(0,0,c), D=(0,0,0).Wait, no, in this case, D is at (0,0,0), and the orthocenter is at (0,0,0).Wait, but in this case, the common perpendiculars of AB and CD would be the line through D and perpendicular to AB.But AB is from (a,0,0) to (0,b,0), so vector AB is (-a, b, 0).The common perpendicular to AB and CD would be the line through D=(0,0,0) in the direction perpendicular to both AB and CD.But CD is from C=(0,0,c) to D=(0,0,0), so vector CD is (0,0,-c).The cross product of AB and CD would give a vector perpendicular to both.Compute AB × CD:|i   j   k||-a   b   0||0   0  -c|= i*(b*(-c) - 0*0) - j*(-a*(-c) - 0*0) + k*(-a*0 - b*0)= i*(-bc) - j*(ac) + k*0= (-bc, -ac, 0)So, the common perpendicular direction is (-bc, -ac, 0).But in this case, the common perpendicular line would pass through D=(0,0,0) and have direction (-bc, -ac, 0).But the orthocenter is at (0,0,0), so the common perpendicular passes through the orthocenter.Similarly, for the other pairs of opposite edges.Thus, in this coordinate system, the common perpendiculars pass through the orthocenter.But in general, for any orthogonal tetrahedron, the common perpendiculars of opposite edges pass through the orthocenter.Alternatively, since in an orthogonal tetrahedron, the altitudes intersect at the orthocenter, and the common perpendiculars are related to the altitudes, they must pass through the orthocenter.Thus, the common perpendiculars of the three pairs of opposite edges pass through the intersection point of the altitudes.So, all parts seem to hold.

💡To determine how often Marcy takes a sip, we can follow these steps:1. **Convert the bottle size to milliliters:** [ 2 text{ liters} = 2000 text{ milliliters} ]2. **Calculate the number of sips needed to finish the bottle:** [ frac{2000 text{ ml}}{40 text{ ml/sip}} = 50 text{ sips} ]3. **Determine the time between each sip:** [ frac{250 text{ minutes}}{50 text{ sips}} = 5 text{ minutes per sip} ]

💡First, determine the number of adult tickets sold by subtracting the number of child tickets from the total number of attendees.Next, calculate the total revenue generated from adult tickets by multiplying the number of adult tickets by the cost of an adult ticket.Then, subtract the revenue from adult tickets from the total revenue to find the revenue from child tickets.Finally, divide the revenue from child tickets by the number of child tickets to find the cost of one child ticket.

💡First, I need to determine the area of each individual square. Since the entire structure is made up of five identical squares and the total area is 125 cm², I can find the area of one square by dividing the total area by the number of squares.Next, with the area of one square known, I can calculate the side length of each square by taking the square root of the area.Once I have the side length, I can analyze the cross-shaped figure. The cross consists of a central square with four additional squares attached to each side. This arrangement affects the overall perimeter because some sides of the squares are internal and not part of the outer perimeter.I will count the number of outer edges that contribute to the perimeter. Each square has four sides, but when squares are connected, some sides overlap and are not part of the perimeter. By carefully counting the outer edges, I can determine the total perimeter of the cross-shaped figure.Finally, I will multiply the number of outer edges by the side length of one square to find the total perimeter.