💡First, I recognize that there are 5 classes with a decreasing number of students in each subsequent class. The largest class has 32 students, and the total number of students in the school is 140.I'll denote the number of students in each class as follows:- 1st class: 32 students- 2nd class: 32 - x students- 3rd class: 32 - 2x students- 4th class: 32 - 3x students- 5th class: 32 - 4x studentsNext, I'll set up an equation to represent the total number of students:32 + (32 - x) + (32 - 2x) + (32 - 3x) + (32 - 4x) = 140Combining like terms:160 - 10x = 140Now, I'll solve for x:160 - 140 = 10x20 = 10xx = 2Therefore, each class has 2 fewer students than the previous class.

💡Alright, so I have this geometry problem here about two circles that are externally tangent at a point A. There's a common tangent t that touches the first circle γ at point T and the second circle γ' at point T'. Then, M is defined as the midpoint of the segment TT'. I need to show that MT equals MA and also equals MT'. Okay, let me try to visualize this. There are two circles touching at point A, and they have a common tangent t that touches each circle at T and T'. So, t is a line that just touches each circle at one point each. Since the circles are externally tangent, they touch at A, and their centers must lie along the line connecting A. Let me recall some properties of tangent lines and circles. A tangent to a circle is perpendicular to the radius at the point of tangency. So, if I draw radii from the centers of γ and γ' to T and T' respectively, those radii should be perpendicular to the tangent line t. Hmm, maybe I can use similar triangles or something related to midpoints here. Since M is the midpoint of TT', perhaps there's some symmetry involved. Also, since t is a common tangent, the lengths from T to A and T' to A might be related in some way.Wait, maybe I should draw the common tangent at point A. Since the circles are tangent at A, there's a common tangent there as well. Let me call this tangent t_A. Now, t_A is another tangent line that touches both circles at A. I remember that for two tangent circles, the common external tangent and the common tangent at the point of contact have some interesting properties. Maybe I can use the fact that the lengths from the midpoint M to T and T' are equal because M is the midpoint, but I need to relate this to MA.Let me think about the power of a point. The power of point M with respect to both circles should be equal because M lies on the radical axis of the two circles. But wait, the radical axis is the line perpendicular to the line joining the centers, and since the circles are tangent, their radical axis is the common tangent at A. So, M lies on the radical axis, which is t_A. Therefore, the power of M with respect to both circles is equal, which means MT^2 = MA^2 and MT'^2 = MA^2. Hence, MT = MA and MT' = MA. So, MT = MA = MT'. Wait, did I just jump to the conclusion? Let me make sure. The power of a point M with respect to a circle is equal to the square of the length of the tangent from M to the circle. So, if M lies on the radical axis, then its power with respect to both circles is equal. Since t is a common tangent, and M is the midpoint of TT', which are points of tangency, then the power of M with respect to both circles should indeed be equal. Therefore, MT^2 = MA^2 and MT'^2 = MA^2, leading to MT = MA = MT'. I think that makes sense. Maybe I should also consider drawing the lines and seeing if triangles involved are congruent or something. But the power of a point seems to give the result directly. Alternatively, maybe using coordinate geometry could help. Let me try setting up coordinates. Let me place point A at the origin (0,0). Let the centers of the circles be along the x-axis, say center O at (h,0) for circle γ and center O' at (k,0) for circle γ'. Since the circles are externally tangent at A, the distance between the centers should be equal to the sum of their radii. Let me denote the radius of γ as r and the radius of γ' as r'. Then, the distance between O and O' is h - k = r + r'. Since they are tangent at A, which is the origin, the point A lies on both circles, so the distance from O to A is r, and from O' to A is r'. Therefore, h = r and k = -r', so the distance between O and O' is r + r', which matches.Now, the common tangent t touches γ at T and γ' at T'. Let me find the coordinates of T and T'. Since t is a common tangent, it should be symmetric with respect to the line joining the centers, which is the x-axis in my setup. So, the tangent t should be above or below the x-axis. Let's assume it's above for simplicity.The slope of the tangent line t can be found using the formula for the slope of a tangent to two circles. Alternatively, since the tangent is common, the triangles formed by the centers and the points of tangency should be similar.Let me denote the slope of t as m. Then, the equation of t can be written as y = m x + c. Since it's tangent to both circles, the distance from the center O to the line t should be equal to the radius r, and similarly for O' and r'.The distance from a point (x0, y0) to the line a x + b y + c = 0 is |a x0 + b y0 + c| / sqrt(a^2 + b^2). In our case, the line t is y - m x - c = 0, so a = -m, b = 1, c' = -c.So, the distance from O(h,0) to t is | -m h + 0 - c | / sqrt(m^2 + 1) = | -m h - c | / sqrt(m^2 + 1) = r.Similarly, the distance from O'(k,0) to t is | -m k - c | / sqrt(m^2 + 1) = r'.So, we have two equations:| -m h - c | = r sqrt(m^2 + 1)| -m k - c | = r' sqrt(m^2 + 1)Since the tangent is above the x-axis, and assuming h > 0 and k < 0, the constants might have specific signs. Let me assume that -m h - c = -r sqrt(m^2 + 1) and -m k - c = r' sqrt(m^2 + 1). Subtracting these two equations:(-m h - c) - (-m k - c) = -r sqrt(m^2 + 1) - r' sqrt(m^2 + 1)Simplifying:-m h + m k = - (r + r') sqrt(m^2 + 1)Factor out m:m (k - h) = - (r + r') sqrt(m^2 + 1)But k - h = - (h - k) = - (r + r'), since h = r and k = -r'.So, m (- (r + r')) = - (r + r') sqrt(m^2 + 1)Divide both sides by - (r + r'):m = sqrt(m^2 + 1)Wait, that can't be right because sqrt(m^2 + 1) is always greater than |m|. So, maybe I made a mistake in the signs.Let me reconsider the distances. Since the tangent is above the x-axis, the y-intercept c should be positive. The distance from O(h,0) to t is | -m h - c | / sqrt(m^2 + 1) = r. Since both -m h - c and r are positive, we can drop the absolute value:- m h - c = - r sqrt(m^2 + 1)Similarly, for O'(k,0):- m k - c = - r' sqrt(m^2 + 1)So, we have:- m h - c = - r sqrt(m^2 + 1) ...(1)- m k - c = - r' sqrt(m^2 + 1) ...(2)Subtract equation (2) from equation (1):(- m h - c) - (- m k - c) = - r sqrt(m^2 + 1) - (- r' sqrt(m^2 + 1))Simplify:- m h + m k = - r sqrt(m^2 + 1) + r' sqrt(m^2 + 1)Factor out m and sqrt(m^2 + 1):m (k - h) = (r' - r) sqrt(m^2 + 1)But k - h = - (h - k) = - (r + r'), since h = r and k = -r'So:m (- (r + r')) = (r' - r) sqrt(m^2 + 1)Multiply both sides by -1:m (r + r') = (r - r') sqrt(m^2 + 1)Now, let's square both sides to eliminate the square root:m^2 (r + r')^2 = (r - r')^2 (m^2 + 1)Expand both sides:m^2 (r^2 + 2 r r' + r'^2) = (r^2 - 2 r r' + r'^2)(m^2 + 1)Let me distribute on both sides:Left side: m^2 r^2 + 2 m^2 r r' + m^2 r'^2Right side: (r^2 - 2 r r' + r'^2) m^2 + (r^2 - 2 r r' + r'^2)Now, subtract left side from right side:0 = (r^2 - 2 r r' + r'^2) m^2 + (r^2 - 2 r r' + r'^2) - m^2 r^2 - 2 m^2 r r' - m^2 r'^2Simplify term by term:For m^2 terms:(r^2 - 2 r r' + r'^2 - r^2 - 2 r r' - r'^2) m^2 = (-4 r r') m^2For constant terms:(r^2 - 2 r r' + r'^2)So, overall:0 = -4 r r' m^2 + (r^2 - 2 r r' + r'^2)Bring all terms to one side:4 r r' m^2 = r^2 - 2 r r' + r'^2Factor the right side:4 r r' m^2 = (r - r')^2So, m^2 = (r - r')^2 / (4 r r')Take square root:m = (r - r') / (2 sqrt(r r'))Since we assumed the tangent is above the x-axis, and if r > r', then m is positive; otherwise, negative. But let's keep it as m = (r - r') / (2 sqrt(r r')) for now.Now, let's find c from equation (1):- m h - c = - r sqrt(m^2 + 1)We know h = r, so:- m r - c = - r sqrt(m^2 + 1)Solve for c:c = - m r + r sqrt(m^2 + 1)Similarly, from equation (2):- m k - c = - r' sqrt(m^2 + 1)But k = -r', so:- m (-r') - c = - r' sqrt(m^2 + 1)Simplify:m r' - c = - r' sqrt(m^2 + 1)Solve for c:c = m r' + r' sqrt(m^2 + 1)Now, we have two expressions for c:c = - m r + r sqrt(m^2 + 1)c = m r' + r' sqrt(m^2 + 1)Set them equal:- m r + r sqrt(m^2 + 1) = m r' + r' sqrt(m^2 + 1)Bring all terms to one side:- m r - m r' + r sqrt(m^2 + 1) - r' sqrt(m^2 + 1) = 0Factor:- m (r + r') + (r - r') sqrt(m^2 + 1) = 0But from earlier, we have:m (r + r') = (r - r') sqrt(m^2 + 1)So, substituting:- (r - r') sqrt(m^2 + 1) + (r - r') sqrt(m^2 + 1) = 0Which simplifies to 0 = 0, so it's consistent.Okay, so now we have expressions for m and c. Let's try to find the coordinates of T and T'.The tangent line t has equation y = m x + c.For circle γ with center (h,0) = (r,0) and radius r, the point T lies on both the circle and the tangent line.So, the coordinates of T satisfy:(x - r)^2 + y^2 = r^2and y = m x + c.Substitute y into the circle equation:(x - r)^2 + (m x + c)^2 = r^2Expand:x^2 - 2 r x + r^2 + m^2 x^2 + 2 m c x + c^2 = r^2Combine like terms:(1 + m^2) x^2 + (-2 r + 2 m c) x + (r^2 + c^2 - r^2) = 0Simplify:(1 + m^2) x^2 + (-2 r + 2 m c) x + c^2 = 0Since t is tangent to the circle, this quadratic equation has exactly one solution, so the discriminant must be zero.Discriminant D:[ -2 r + 2 m c ]^2 - 4 (1 + m^2) c^2 = 0Expand:4 r^2 - 8 r m c + 4 m^2 c^2 - 4 c^2 - 4 m^2 c^2 = 0Simplify:4 r^2 - 8 r m c - 4 c^2 = 0Divide by 4:r^2 - 2 r m c - c^2 = 0So, r^2 - 2 r m c - c^2 = 0We can solve for x:From the quadratic equation, since D=0, x = [2 r - 2 m c] / [2 (1 + m^2)] = (r - m c) / (1 + m^2)So, x-coordinate of T is (r - m c)/(1 + m^2)Similarly, y-coordinate is y = m x + c = m (r - m c)/(1 + m^2) + c = [m r - m^2 c + c (1 + m^2)] / (1 + m^2) = [m r + c] / (1 + m^2)So, T = [(r - m c)/(1 + m^2), (m r + c)/(1 + m^2)]Similarly, for circle γ' with center (k,0) = (-r',0) and radius r', the point T' lies on both the circle and the tangent line.So, the coordinates of T' satisfy:(x + r')^2 + y^2 = r'^2and y = m x + cSubstitute y:(x + r')^2 + (m x + c)^2 = r'^2Expand:x^2 + 2 r' x + r'^2 + m^2 x^2 + 2 m c x + c^2 = r'^2Combine like terms:(1 + m^2) x^2 + (2 r' + 2 m c) x + (r'^2 + c^2 - r'^2) = 0Simplify:(1 + m^2) x^2 + (2 r' + 2 m c) x + c^2 = 0Again, since t is tangent, discriminant D=0:[2 r' + 2 m c]^2 - 4 (1 + m^2) c^2 = 0Expand:4 r'^2 + 8 r' m c + 4 m^2 c^2 - 4 c^2 - 4 m^2 c^2 = 0Simplify:4 r'^2 + 8 r' m c - 4 c^2 = 0Divide by 4:r'^2 + 2 r' m c - c^2 = 0So, r'^2 + 2 r' m c - c^2 = 0Similarly, solving for x:x = [ -2 r' - 2 m c ] / [2 (1 + m^2)] = (- r' - m c)/(1 + m^2)So, x-coordinate of T' is (- r' - m c)/(1 + m^2)y-coordinate is y = m x + c = m (- r' - m c)/(1 + m^2) + c = [ - m r' - m^2 c + c (1 + m^2) ] / (1 + m^2) = [ - m r' + c ] / (1 + m^2)So, T' = [ (- r' - m c)/(1 + m^2), (- m r' + c)/(1 + m^2) ]Now, we have coordinates for T and T'. Let's find the midpoint M of TT'.Midpoint M has coordinates:x_M = [ (r - m c)/(1 + m^2) + (- r' - m c)/(1 + m^2) ] / 2 = [ (r - m c - r' - m c) ] / [2 (1 + m^2)] = (r - r' - 2 m c) / [2 (1 + m^2)]y_M = [ (m r + c)/(1 + m^2) + (- m r' + c)/(1 + m^2) ] / 2 = [ m r + c - m r' + c ] / [2 (1 + m^2)] = (m (r - r') + 2 c) / [2 (1 + m^2)]So, M = [ (r - r' - 2 m c)/(2 (1 + m^2)), (m (r - r') + 2 c)/(2 (1 + m^2)) ]Now, we need to find MA, MT, and MT' and show they are equal.First, let's find point A, which is the point of tangency of the two circles. Since the circles are externally tangent at A, and in our coordinate system, A is at the origin (0,0).So, MA is the distance from M to A(0,0):MA^2 = x_M^2 + y_M^2Similarly, MT is the distance from M to T:MT^2 = (x_M - x_T)^2 + (y_M - y_T)^2Similarly for MT'.This seems quite involved, but maybe we can find a relationship between MA, MT, and MT' using the coordinates.Alternatively, perhaps there's a geometric property or theorem that can simplify this.Wait, earlier I thought about the power of a point. Since M is the midpoint of TT', and TT' is the common tangent, then M lies on the radical axis of the two circles. The radical axis is the set of points with equal power with respect to both circles, which in this case is the common tangent at A. Therefore, the power of M with respect to both circles is equal, meaning MT^2 = MA^2 and MT'^2 = MA^2, hence MT = MA = MT'.That seems to be a more straightforward approach without getting into coordinate calculations.But just to be thorough, let me see if I can confirm this with the coordinates.We have expressions for x_M, y_M, x_T, y_T, x_T', y_T'.Let me compute MA^2:MA^2 = x_M^2 + y_M^2Similarly, MT^2 = (x_M - x_T)^2 + (y_M - y_T)^2Let me compute MT^2:(x_M - x_T)^2 = [ (r - r' - 2 m c)/(2 (1 + m^2)) - (r - m c)/(1 + m^2) ]^2= [ (r - r' - 2 m c - 2 r + 2 m c) / (2 (1 + m^2)) ]^2= [ (- r - r') / (2 (1 + m^2)) ]^2= (r + r')^2 / [4 (1 + m^2)^2]Similarly, (y_M - y_T)^2 = [ (m (r - r') + 2 c)/(2 (1 + m^2)) - (m r + c)/(1 + m^2) ]^2= [ (m (r - r') + 2 c - 2 m r - 2 c) / (2 (1 + m^2)) ]^2= [ (- m r' - m r) / (2 (1 + m^2)) ]^2= (m (r + r'))^2 / [4 (1 + m^2)^2]So, MT^2 = (r + r')^2 / [4 (1 + m^2)^2] + (m^2 (r + r')^2) / [4 (1 + m^2)^2] = (r + r')^2 (1 + m^2) / [4 (1 + m^2)^2] = (r + r')^2 / [4 (1 + m^2)]Similarly, MA^2 = x_M^2 + y_M^2x_M = (r - r' - 2 m c)/(2 (1 + m^2))y_M = (m (r - r') + 2 c)/(2 (1 + m^2))So,x_M^2 = (r - r' - 2 m c)^2 / [4 (1 + m^2)^2]y_M^2 = (m (r - r') + 2 c)^2 / [4 (1 + m^2)^2]So, MA^2 = [ (r - r' - 2 m c)^2 + (m (r - r') + 2 c)^2 ] / [4 (1 + m^2)^2]Let me expand the numerator:(r - r' - 2 m c)^2 = (r - r')^2 - 4 m c (r - r') + 4 m^2 c^2(m (r - r') + 2 c)^2 = m^2 (r - r')^2 + 4 m c (r - r') + 4 c^2Adding them together:(r - r')^2 - 4 m c (r - r') + 4 m^2 c^2 + m^2 (r - r')^2 + 4 m c (r - r') + 4 c^2Simplify:(r - r')^2 + m^2 (r - r')^2 + 4 m^2 c^2 + 4 c^2= (1 + m^2)(r - r')^2 + 4 c^2 (1 + m^2)= (1 + m^2)[(r - r')^2 + 4 c^2]So, MA^2 = (1 + m^2)[(r - r')^2 + 4 c^2] / [4 (1 + m^2)^2] = [(r - r')^2 + 4 c^2] / [4 (1 + m^2)]Now, recall from earlier, we have:From equation (1):- m r - c = - r sqrt(m^2 + 1)So, c = - m r + r sqrt(m^2 + 1)Similarly, from equation (2):c = m r' + r' sqrt(m^2 + 1)So, c = - m r + r sqrt(m^2 + 1) = m r' + r' sqrt(m^2 + 1)Let me compute 4 c^2:4 c^2 = 4 [ - m r + r sqrt(m^2 + 1) ]^2= 4 [ m^2 r^2 - 2 m r^2 sqrt(m^2 + 1) + r^2 (m^2 + 1) ]= 4 [ m^2 r^2 - 2 m r^2 sqrt(m^2 + 1) + m^2 r^2 + r^2 ]= 4 [ 2 m^2 r^2 - 2 m r^2 sqrt(m^2 + 1) + r^2 ]Similarly, (r - r')^2 = r^2 - 2 r r' + r'^2So, MA^2 = [ r^2 - 2 r r' + r'^2 + 4 c^2 ] / [4 (1 + m^2)]Substituting 4 c^2:= [ r^2 - 2 r r' + r'^2 + 4 [ 2 m^2 r^2 - 2 m r^2 sqrt(m^2 + 1) + r^2 ] ] / [4 (1 + m^2)]= [ r^2 - 2 r r' + r'^2 + 8 m^2 r^2 - 8 m r^2 sqrt(m^2 + 1) + 4 r^2 ] / [4 (1 + m^2)]Combine like terms:= [ (1 + 4) r^2 - 2 r r' + r'^2 + 8 m^2 r^2 - 8 m r^2 sqrt(m^2 + 1) ] / [4 (1 + m^2)]= [ 5 r^2 - 2 r r' + r'^2 + 8 m^2 r^2 - 8 m r^2 sqrt(m^2 + 1) ] / [4 (1 + m^2)]This seems complicated. Maybe I should find a relationship between c and m.Recall that c = - m r + r sqrt(m^2 + 1)Let me compute c^2:c^2 = [ - m r + r sqrt(m^2 + 1) ]^2 = m^2 r^2 - 2 m r^2 sqrt(m^2 + 1) + r^2 (m^2 + 1) = 2 m^2 r^2 - 2 m r^2 sqrt(m^2 + 1) + r^2Similarly, from earlier, we have:From the power of point M, MT^2 = (r + r')^2 / [4 (1 + m^2)]And MA^2 = [ (r - r')^2 + 4 c^2 ] / [4 (1 + m^2)]Let me compute (r - r')^2 + 4 c^2:= r^2 - 2 r r' + r'^2 + 4 [2 m^2 r^2 - 2 m r^2 sqrt(m^2 + 1) + r^2 ]= r^2 - 2 r r' + r'^2 + 8 m^2 r^2 - 8 m r^2 sqrt(m^2 + 1) + 4 r^2= 5 r^2 - 2 r r' + r'^2 + 8 m^2 r^2 - 8 m r^2 sqrt(m^2 + 1)This is the same as the numerator in MA^2.So, MA^2 = [5 r^2 - 2 r r' + r'^2 + 8 m^2 r^2 - 8 m r^2 sqrt(m^2 + 1)] / [4 (1 + m^2)]Similarly, MT^2 = (r + r')^2 / [4 (1 + m^2)] = (r^2 + 2 r r' + r'^2) / [4 (1 + m^2)]I need to show that MA^2 = MT^2.So, set MA^2 = MT^2:[5 r^2 - 2 r r' + r'^2 + 8 m^2 r^2 - 8 m r^2 sqrt(m^2 + 1)] / [4 (1 + m^2)] = (r^2 + 2 r r' + r'^2) / [4 (1 + m^2)]Multiply both sides by 4 (1 + m^2):5 r^2 - 2 r r' + r'^2 + 8 m^2 r^2 - 8 m r^2 sqrt(m^2 + 1) = r^2 + 2 r r' + r'^2Subtract r^2 + 2 r r' + r'^2 from both sides:4 r^2 - 4 r r' + 8 m^2 r^2 - 8 m r^2 sqrt(m^2 + 1) = 0Factor out 4 r^2:4 r^2 [1 - (r')/r + 2 m^2 - 2 m sqrt(m^2 + 1)] = 0Since r ≠ 0, we have:1 - (r')/r + 2 m^2 - 2 m sqrt(m^2 + 1) = 0Let me recall that m = (r - r') / (2 sqrt(r r'))Let me compute m sqrt(m^2 + 1):m sqrt(m^2 + 1) = [ (r - r') / (2 sqrt(r r')) ] sqrt( [ (r - r')^2 / (4 r r') ] + 1 )= [ (r - r') / (2 sqrt(r r')) ] sqrt( (r^2 - 2 r r' + r'^2 + 4 r r') / (4 r r') )= [ (r - r') / (2 sqrt(r r')) ] sqrt( (r^2 + 2 r r' + r'^2) / (4 r r') )= [ (r - r') / (2 sqrt(r r')) ] * (r + r') / (2 sqrt(r r'))= (r - r')(r + r') / (4 r r')= (r^2 - r'^2) / (4 r r')So, m sqrt(m^2 + 1) = (r^2 - r'^2)/(4 r r')Now, let's compute 2 m sqrt(m^2 + 1):2 * (r^2 - r'^2)/(4 r r') = (r^2 - r'^2)/(2 r r')Similarly, 2 m^2:2 * [ (r - r')^2 / (4 r r') ] = (r - r')^2 / (2 r r')So, plugging back into the equation:1 - (r')/r + (r - r')^2 / (2 r r') - (r^2 - r'^2)/(2 r r') = 0Simplify term by term:1 - (r')/r + [ (r^2 - 2 r r' + r'^2) - (r^2 - r'^2) ] / (2 r r')= 1 - (r')/r + [ r^2 - 2 r r' + r'^2 - r^2 + r'^2 ] / (2 r r')= 1 - (r')/r + [ -2 r r' + 2 r'^2 ] / (2 r r')= 1 - (r')/r + [ -2 r r' + 2 r'^2 ] / (2 r r')Simplify the fraction:= 1 - (r')/r + [ -2 r r' ] / (2 r r') + [ 2 r'^2 ] / (2 r r')= 1 - (r')/r - 1 + (r')/r= 0So, the equation holds, meaning MA^2 = MT^2, hence MA = MT.Similarly, by symmetry, MA = MT'.Therefore, MT = MA = MT'.So, after going through the coordinate approach and verifying the algebra, it confirms that MT = MA = MT'.Another way to see this is by using the power of a point. Since M is the midpoint of TT', and TT' is the common tangent, M lies on the radical axis of the two circles, which is the common tangent at A. Therefore, the power of M with respect to both circles is equal, meaning MT^2 = MA^2 and MT'^2 = MA^2, hence MT = MA = MT'.This problem combines properties of tangent lines, midpoints, and the radical axis to establish the required equalities.

💡Okay, so I have this problem here: there's a unit square, which means each side is 1 unit long. Inside this square, there's a circle inscribed, meaning it touches all four sides of the square. Then, a diagonal of the square is drawn, and I need to find the total length of the segments of this diagonal that aren't inside the circle. Hmm, okay, let me try to visualize this.First, I should probably draw a unit square. Since it's a unit square, each side is 1 unit long. The circle inscribed in it would have a diameter equal to the side length of the square, right? So, the diameter of the circle is 1, which means the radius is 0.5 units. That makes sense because the circle touches the midpoint of each side of the square.Now, the diagonal of the square. In a square, the diagonal can be calculated using the Pythagorean theorem. Since each side is 1, the diagonal length should be sqrt(1^2 + 1^2) = sqrt(2). So, the diagonal is sqrt(2) units long. Got that.The problem is asking for the total length of the segments of this diagonal that are not inside the circle. So, essentially, the diagonal passes through the circle, and we need to find the parts of the diagonal that lie outside the circle and add them up.To do this, I think I need to find where the diagonal intersects the circle. Once I have those intersection points, I can calculate the lengths of the diagonal segments outside the circle and sum them.Let me set up a coordinate system to make this easier. Let's place the square with its bottom-left corner at (0,0) and top-right corner at (1,1). The center of the square, which is also the center of the inscribed circle, would then be at (0.5, 0.5). The circle has a radius of 0.5, so its equation is (x - 0.5)^2 + (y - 0.5)^2 = (0.5)^2.The diagonal of the square can be represented by the line from (0,0) to (1,1). The equation of this diagonal is y = x.Now, to find the points where the diagonal intersects the circle, I can substitute y = x into the circle's equation:(x - 0.5)^2 + (x - 0.5)^2 = 0.25Simplifying this:2*(x - 0.5)^2 = 0.25Divide both sides by 2:(x - 0.5)^2 = 0.125Take the square root of both sides:x - 0.5 = ±sqrt(0.125)So, x = 0.5 ± sqrt(0.125)Similarly, since y = x, the y-coordinates will be the same.Simplify sqrt(0.125). Since 0.125 is 1/8, sqrt(1/8) is (sqrt(2))/4, which is approximately 0.3535.So, the intersection points are at:x = 0.5 + (sqrt(2))/4 ≈ 0.5 + 0.3535 ≈ 0.8535andx = 0.5 - (sqrt(2))/4 ≈ 0.5 - 0.3535 ≈ 0.1465Therefore, the diagonal intersects the circle at approximately (0.8535, 0.8535) and (0.1465, 0.1465).Now, the diagonal has a total length of sqrt(2) ≈ 1.4142 units. The part of the diagonal inside the circle is the segment between these two intersection points. To find the length of this segment, I can calculate the distance between (0.8535, 0.8535) and (0.1465, 0.1465).Using the distance formula:Distance = sqrt[(0.8535 - 0.1465)^2 + (0.8535 - 0.1465)^2]Simplify:= sqrt[(0.707)^2 + (0.707)^2]= sqrt[0.5 + 0.5]= sqrt[1]= 1Wait, that's interesting. The length of the diagonal inside the circle is exactly 1 unit. Since the total length of the diagonal is sqrt(2), the parts outside the circle must be sqrt(2) - 1.But let me verify this because sometimes approximations can be misleading. Let's do it algebraically without approximating.We have the intersection points at x = 0.5 ± (sqrt(2))/4. So, the distance between these two points along the diagonal is the difference in their x-coordinates times sqrt(2) because the diagonal has a slope of 1, so the distance between two points (x1, y1) and (x2, y2) on the line y = x is |x1 - x2|*sqrt(2).The difference in x-coordinates is [0.5 + (sqrt(2))/4] - [0.5 - (sqrt(2))/4] = (sqrt(2))/2.Therefore, the distance between the two intersection points is (sqrt(2)/2)*sqrt(2) = (2)/2 = 1. So, yes, the length inside the circle is exactly 1.Therefore, the total length of the diagonal not inside the circle is sqrt(2) - 1.Wait, but the problem says "the total length of the segments of the diagonal not contained within the circle." Since the diagonal is split into three parts: from (0,0) to the first intersection point, then from there to the second intersection point (which is inside the circle), and then from the second intersection point to (1,1). So, the two segments outside the circle are each from (0,0) to (0.1465, 0.1465) and from (0.8535, 0.8535) to (1,1). Each of these segments has the same length because of symmetry.So, the length from (0,0) to (0.1465, 0.1465) is the distance between these two points. Using the distance formula:sqrt[(0.1465 - 0)^2 + (0.1465 - 0)^2] = sqrt[2*(0.1465)^2] = 0.1465*sqrt(2)Similarly, the length from (0.8535, 0.8535) to (1,1) is also 0.1465*sqrt(2). So, adding these two gives 2*0.1465*sqrt(2) ≈ 2*(0.1465)*(1.4142) ≈ 0.4142.But wait, sqrt(2) is approximately 1.4142, so sqrt(2) - 1 ≈ 0.4142, which matches the approximate value I just calculated. So, both methods give the same result.Therefore, the exact value is sqrt(2) - 1, which is approximately 0.4142.So, putting it all together, the total length of the segments of the diagonal not contained within the circle is sqrt(2) - 1.

💡First, let's define the variables:- Let ( b ) represent the number of boys.- Let ( g ) represent the number of girls.According to the problem, ( frac{3}{7} ) of the number of boys is equal to ( frac{6}{11} ) of the number of girls. This can be written as:[frac{3}{7}b = frac{6}{11}g]To find the ratio of boys to girls, we can rearrange the equation:[frac{b}{g} = frac{6}{11} div frac{3}{7} = frac{6}{11} times frac{7}{3} = frac{42}{33} = frac{14}{11}]So, the simplest integer ratio of boys to girls is ( 14:11 ).Next, we are given that there are 22 girls in the class. Using the ratio ( 14:11 ), we can set up a proportion to find the number of boys:[frac{14}{11} = frac{b}{22}]Solving for ( b ):[b = frac{14}{11} times 22 = 28]Finally, to find the total number of students, we add the number of boys and girls:[28 text{ boys} + 22 text{ girls} = 50 text{ students}]

💡Okay, so I have this problem here: In the polar coordinate system, I need to find the distance from the point (1,0) to the line given by the equation ρ(cosθ + sinθ) = 2. Hmm, polar coordinates can sometimes be tricky, especially when dealing with lines and distances. Let me try to break this down step by step.First, I remember that in polar coordinates, any point is represented as (ρ, θ), where ρ is the distance from the origin (or pole) and θ is the angle measured from the positive x-axis. The point (1,0) is straightforward—it's 1 unit away from the origin along the angle of 0 radians, which is just the positive x-axis. So, that point is essentially (1,0) in Cartesian coordinates as well.Now, the line is given by the equation ρ(cosθ + sinθ) = 2. I need to figure out what this line looks like in Cartesian coordinates because I'm more familiar with calculating distances in that system. I recall that in polar coordinates, ρcosθ is equal to the x-coordinate, and ρsinθ is equal to the y-coordinate. So, if I rewrite the equation:ρcosθ + ρsinθ = 2That translates to:x + y = 2Oh, okay, so the line is simply x + y = 2 in Cartesian coordinates. That makes it easier because now I can use the standard formula for the distance from a point to a line in Cartesian coordinates.The formula for the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is:Distance = |Ax₀ + By₀ + C| / sqrt(A² + B²)So, first, I need to write the equation of the line in the standard form Ax + By + C = 0. The given line is x + y = 2. Let me rearrange that:x + y - 2 = 0So, A = 1, B = 1, and C = -2.Now, the point from which I need to find the distance is (1,0). So, plugging into the distance formula:Distance = |1*1 + 1*0 - 2| / sqrt(1² + 1²)Distance = |1 + 0 - 2| / sqrt(1 + 1)Distance = |-1| / sqrt(2)Distance = 1 / sqrt(2)Hmm, 1 over square root of 2. I can rationalize the denominator if needed, so that would be sqrt(2)/2. But I think both forms are acceptable, though sometimes rationalized denominators are preferred.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the equation ρ(cosθ + sinθ) = 2, converting to Cartesian gives x + y = 2. The point is (1,0). Using the distance formula, yes, A=1, B=1, C=-2. Plugging in (1,0):|1*1 + 1*0 - 2| = |1 - 2| = |-1| = 1Denominator is sqrt(1 + 1) = sqrt(2). So, 1/sqrt(2) is correct. Rationalizing gives sqrt(2)/2.But wait, the question is in the context of polar coordinates. Is there a different way to approach this problem directly in polar coordinates without converting to Cartesian? Maybe I should try that method as well to confirm.In polar coordinates, the distance from a point (ρ₁, θ₁) to a line given by ρcos(θ - α) = p is |ρ₁cos(θ₁ - α) - p|. But I'm not sure if that's the exact formula. Let me recall.I think the general equation of a line in polar coordinates is ρ = e / (cos(θ - φ)), where e is the distance from the origin to the line, and φ is the angle that the perpendicular from the origin to the line makes with the positive x-axis. So, if I have a line in polar form, it can be expressed as ρcos(θ - φ) = e.Comparing that to the given equation ρ(cosθ + sinθ) = 2. Let me see if I can write cosθ + sinθ in terms of cos(θ - φ).I remember that cosθ + sinθ can be written as sqrt(2)cos(θ - 45°), because cosθ + sinθ = sqrt(2)cos(θ - 45°). Let me verify that.Using the cosine of difference identity: cos(θ - φ) = cosθ cosφ + sinθ sinφ. If I set φ = 45°, then cosφ = sinφ = sqrt(2)/2. So,sqrt(2)cos(θ - 45°) = sqrt(2)[cosθ cos45° + sinθ sin45°] = sqrt(2)[cosθ*(sqrt(2)/2) + sinθ*(sqrt(2)/2)] = sqrt(2)*(sqrt(2)/2)(cosθ + sinθ) = (2/2)(cosθ + sinθ) = cosθ + sinθ.Yes, that works. So, cosθ + sinθ = sqrt(2)cos(θ - 45°). Therefore, the given equation ρ(cosθ + sinθ) = 2 can be rewritten as ρ*sqrt(2)cos(θ - 45°) = 2, which simplifies to ρcos(θ - 45°) = 2/sqrt(2) = sqrt(2).So, now the equation is in the standard polar form ρcos(θ - φ) = e, where φ = 45° and e = sqrt(2). This tells me that the line is at a distance of sqrt(2) from the origin, and the perpendicular from the origin to the line makes an angle of 45° with the positive x-axis.Now, to find the distance from the point (1,0) to this line. In polar coordinates, the formula for the distance from a point (ρ₁, θ₁) to the line ρcos(θ - φ) = e is |ρ₁cos(θ₁ - φ) - e|.So, plugging in the values: ρ₁ = 1, θ₁ = 0°, φ = 45°, e = sqrt(2).Distance = |1*cos(0° - 45°) - sqrt(2)| = |cos(-45°) - sqrt(2)|Since cosine is even, cos(-45°) = cos(45°) = sqrt(2)/2.So, Distance = |sqrt(2)/2 - sqrt(2)| = |sqrt(2)/2 - 2*sqrt(2)/2| = |-sqrt(2)/2| = sqrt(2)/2.Wait, that's the same result as before! So, whether I convert to Cartesian coordinates and use the distance formula or use the polar coordinate formula directly, I get the same answer: sqrt(2)/2.But let me think again. The point (1,0) is in polar coordinates, which is (1,0) in Cartesian as well. The line x + y = 2 is in Cartesian. The distance from (1,0) to this line is 1/sqrt(2) or sqrt(2)/2. So, both methods confirm that.But wait, in my initial calculation using Cartesian coordinates, I got 1/sqrt(2), which is equal to sqrt(2)/2. So, both are correct, just different forms.But the question is in the context of polar coordinates. So, maybe the answer is expected in a specific form. Hmm, but both forms are acceptable. However, sometimes in exams, they prefer rationalized denominators, so sqrt(2)/2 might be the preferred form.Alternatively, maybe I can think of another way. Let me visualize this. The line x + y = 2 is a straight line in the Cartesian plane, and the point (1,0) is on the x-axis. The distance from (1,0) to the line should be the shortest distance, which is perpendicular.If I draw a perpendicular from (1,0) to the line x + y = 2, the length of that perpendicular is the distance we're looking for. Using the formula, we found it to be sqrt(2)/2.Alternatively, I can parametrize the line and find the minimum distance, but that might be more complicated. The formula method is straightforward.Wait, another thought: Since the line is x + y = 2, its slope is -1, so the perpendicular slope is 1. So, the equation of the line perpendicular to x + y = 2 passing through (1,0) is y = x - 1.Now, to find the intersection point of y = x - 1 and x + y = 2. Let's solve these two equations:From y = x - 1, substitute into x + y = 2:x + (x - 1) = 22x - 1 = 22x = 3x = 3/2Then y = (3/2) - 1 = 1/2So, the intersection point is (3/2, 1/2). Now, the distance from (1,0) to (3/2, 1/2) is sqrt[(3/2 - 1)^2 + (1/2 - 0)^2] = sqrt[(1/2)^2 + (1/2)^2] = sqrt[1/4 + 1/4] = sqrt[1/2] = sqrt(2)/2.Again, same result. So, three different methods all lead to the same answer: sqrt(2)/2.But wait, the initial problem was presented in polar coordinates. So, is there a way to do this entirely in polar coordinates without converting to Cartesian? Let me try.In polar coordinates, the distance from a point (ρ₁, θ₁) to a line given by ρ = e / cos(θ - φ) is |ρ₁ cos(θ₁ - φ) - e|. Wait, is that correct?Wait, no, actually, the standard form of a line in polar coordinates is ρ = e / cos(θ - φ), where e is the distance from the origin to the line, and φ is the angle of the perpendicular from the origin to the line.Given that, the distance from a point (ρ₁, θ₁) to this line is |ρ₁ cos(θ₁ - φ) - e|.In our case, the line is given by ρ(cosθ + sinθ) = 2, which we converted to ρcos(θ - 45°) = sqrt(2). So, comparing to the standard form ρ = e / cos(θ - φ), we have e = sqrt(2) and φ = 45°.So, using the distance formula in polar coordinates:Distance = |ρ₁ cos(θ₁ - φ) - e|Plugging in the values:ρ₁ = 1, θ₁ = 0°, φ = 45°, e = sqrt(2)Distance = |1 * cos(0° - 45°) - sqrt(2)| = |cos(-45°) - sqrt(2)| = |sqrt(2)/2 - sqrt(2)| = | - sqrt(2)/2 | = sqrt(2)/2.Same result again. So, regardless of the method—converting to Cartesian, using the distance formula in Cartesian, finding the intersection point, or using the polar coordinate distance formula—the distance is sqrt(2)/2.Therefore, I'm confident that the distance from the point (1,0) to the line ρ(cosθ + sinθ) = 2 is sqrt(2)/2.But just to make sure, let me think if there's any alternative perspective or potential mistake I might have made.One thing to note is that in polar coordinates, the equation of a line can sometimes be tricky because it's not as straightforward as in Cartesian coordinates. However, by converting it to Cartesian, I was able to use familiar formulas and verify the result multiple times.Another point is ensuring that when I converted cosθ + sinθ to sqrt(2)cos(θ - 45°), I did it correctly. I used the identity that a*cosθ + b*sinθ = sqrt(a² + b²)cos(θ - φ), where tanφ = b/a. In this case, a = 1 and b = 1, so tanφ = 1, which means φ = 45°, and sqrt(a² + b²) = sqrt(2). So, that conversion was correct.Also, when using the distance formula in Cartesian coordinates, I made sure to rearrange the line equation to standard form Ax + By + C = 0, which I did correctly: x + y - 2 = 0.Plugging in the point (1,0) into the formula, I correctly calculated the numerator as |1 - 2| = 1 and the denominator as sqrt(1 + 1) = sqrt(2), leading to 1/sqrt(2) or sqrt(2)/2.When solving for the intersection point, I correctly found the equations of the perpendicular line and solved the system of equations to find the intersection at (3/2, 1/2), then calculated the distance between (1,0) and (3/2, 1/2) as sqrt(2)/2.In the polar coordinate method, I correctly identified the standard form of the line and applied the distance formula, ensuring that the angle subtraction was handled properly, leading again to sqrt(2)/2.So, after multiple methods and checks, I'm certain that the distance is sqrt(2)/2.**Final Answer**The distance from the point (1,0) to the line is boxed{dfrac{sqrt{2}}{2}}.

💡Okay, so I need to find all the roots of the polynomial ( x^3 - 6x^2 + 11x - 6 ). Hmm, I remember that finding roots of a polynomial means finding the values of ( x ) that make the polynomial equal to zero. Since this is a cubic polynomial, I know there should be three roots, and they can be real or complex numbers. But I think in this case, since the coefficients are all real numbers, the complex roots would come in conjugate pairs if there are any. But let's see.First, I recall something called the Rational Root Theorem. It says that any possible rational root of the polynomial, when written in its simplest form ( frac{p}{q} ), ( p ) is a factor of the constant term, and ( q ) is a factor of the leading coefficient. In this polynomial, the constant term is -6, and the leading coefficient is 1. So, the possible rational roots are just the factors of -6, which are ( pm1, pm2, pm3, pm6 ).Alright, so I can test these possible roots by plugging them into the polynomial and see if the result is zero. Let's start with ( x = 1 ):( 1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0 ). Oh, that works! So, ( x = 1 ) is a root.Now, let's try ( x = 2 ):( 2^3 - 6(2)^2 + 11(2) - 6 = 8 - 24 + 22 - 6 = 0 ). Nice, ( x = 2 ) is also a root.Next, ( x = 3 ):( 3^3 - 6(3)^2 + 11(3) - 6 = 27 - 54 + 33 - 6 = 0 ). Perfect, ( x = 3 ) is another root.Wait a second, I already found three roots: 1, 2, and 3. But since it's a cubic polynomial, there should be exactly three roots, counting multiplicities. So, does that mean all the roots are distinct? It seems like it because each of these values made the polynomial zero, and they are all different.But just to be thorough, maybe I should check if any of these roots have multiplicities greater than one. To do that, I can perform polynomial division or use synthetic division to factor the polynomial and see if any of these roots repeat.Let's try synthetic division with ( x = 1 ):1 | 1 -6 11 -6 | 1 -5 6 ---------------- 1 -5 6 0So, after dividing by ( x - 1 ), the polynomial becomes ( x^2 - 5x + 6 ). Now, let's factor this quadratic:( x^2 - 5x + 6 = (x - 2)(x - 3) ). So, the polynomial factors completely as ( (x - 1)(x - 2)(x - 3) ). Therefore, all roots are distinct, and there are no repeated roots.Just to make sure, let's try plugging in ( x = 6 ):( 6^3 - 6(6)^2 + 11(6) - 6 = 216 - 216 + 66 - 6 = 60 ). That's not zero, so ( x = 6 ) is not a root.Similarly, ( x = -1 ):( (-1)^3 - 6(-1)^2 + 11(-1) - 6 = -1 - 6 - 11 - 6 = -24 ). Not zero.( x = -2 ):( (-2)^3 - 6(-2)^2 + 11(-2) - 6 = -8 - 24 - 22 - 6 = -60 ). Not zero.( x = -3 ):( (-3)^3 - 6(-3)^2 + 11(-3) - 6 = -27 - 54 - 33 - 6 = -120 ). Not zero.So, none of the negative possible rational roots are actual roots. Therefore, the only roots are 1, 2, and 3, each with multiplicity one.I think that's it. I've checked all possible rational roots, factored the polynomial, and confirmed that all roots are distinct. So, the roots are 1, 2, and 3.