answer：Alright, so I have this geometry problem here: there are three equal circles that all intersect at a common point H. The other points where they intersect form an acute-angled triangle ABC. I need to prove that H is the orthocenter of triangle ABC. Hmm, okay, let's break this down step by step.First, let me visualize what's going on. There are three circles, all the same size, intersecting at point H. So, H is a common point for all three circles. Then, each pair of circles intersects at another point, forming the vertices of triangle ABC. So, A, B, and C are the other intersection points of the circles. Since the triangle is acute-angled, all its angles are less than 90 degrees. That might be important later.Now, I need to recall what an orthocenter is. The orthocenter of a triangle is the point where the three altitudes intersect. An altitude is a perpendicular line from a vertex to the opposite side (or its extension). So, to show that H is the orthocenter, I need to show that H lies on all three altitudes of triangle ABC.Since all three circles are equal, they have the same radius. Let me denote the centers of the circles as O1, O2, and O3. Since all circles are equal, the distances from each center to H are equal, meaning O1H = O2H = O3H.Wait, but I don't know the positions of the centers. Maybe I can find another way. Let me think about the properties of the points where the circles intersect.Each pair of circles intersects at H and another point. So, for example, circle 1 and circle 2 intersect at H and A, circle 1 and circle 3 intersect at H and B, and circle 2 and circle 3 intersect at H and C. So, points A, B, and C lie on the circumferences of two circles each.Since the circles are equal, the distances from H to A, B, and C must be equal because they are radii of the same circles. Wait, is that true? Let me think. If all circles are equal, then the distance from each center to H is equal, but the distances from H to A, B, and C depend on the positions of A, B, and C relative to H.Hmm, maybe that's not the right approach. Let me consider the angles instead. Since all circles are equal, the arcs subtended by the chords HA, HB, and HC should be equal. Therefore, the angles subtended by these chords at the centers of the circles should be equal.But again, I don't know the positions of the centers. Maybe I can use the fact that the angles at H are related to the angles of triangle ABC. Let me try that.Consider angle BAH. Since A lies on the intersection of two circles, the angle BAH is equal to the angle subtended by the chord BH in one of the circles. Similarly, angle BCH is equal to the angle subtended by the chord AH in another circle. Since the circles are equal, these angles should be equal.Wait, maybe I should formalize this. Let me denote the centers of the circles as O1, O2, and O3. Then, since all circles are equal, O1A = O1B = O1C = radius, and similarly for O2 and O3.But I'm getting confused. Maybe I should use coordinate geometry. Let me place point H at the origin for simplicity. Then, the centers of the circles can be placed at some points equidistant from H. But without loss of generality, I can assume specific coordinates to make calculations easier.Alternatively, maybe I can use properties of cyclic quadrilaterals. Since points A, B, and C lie on the circles, certain quadrilaterals formed by these points and H are cyclic. For example, quadrilateral HAO1O2 is cyclic because both HA and O1O2 are radii of the same circle.Wait, maybe that's overcomplicating things. Let me think about the orthocenter properties. If H is the orthocenter, then the altitudes from A, B, and C must pass through H. So, I need to show that the lines from A, B, and C perpendicular to the opposite sides pass through H.Let me consider one altitude first. Let's take the altitude from A to BC. I need to show that H lies on this altitude. To do this, I can show that AH is perpendicular to BC.Since A lies on two circles, the circles centered at O1 and O2, perhaps I can relate the angles at H to the sides of the triangle. Maybe the angles formed at H by the lines HA, HB, and HC relate to the angles of triangle ABC.Wait, another thought: in an acute-angled triangle, the orthocenter lies inside the triangle. Since ABC is acute, H must be inside ABC. That makes sense because all three circles intersect at H, which is inside the triangle.Let me try to use the property that in a circle, the angle subtended by a chord at the center is twice the angle subtended at the circumference. So, if I can relate the angles at H to the angles at the centers, maybe I can find some relationships.Suppose I consider the circle passing through A and B with center O1. Then, the angle at H, angle AHB, is equal to half the measure of the central angle AOB. But I don't know where O1 is, so maybe that's not helpful.Wait, but all three circles are equal, so the central angles for chords HA, HB, and HC should be equal. Therefore, the angles at H, angle AHB, angle BHC, and angle AHC, are all equal? Hmm, not necessarily, because the chords might not be equal.Wait, no, since all circles are equal, the lengths HA, HB, and HC are equal because they are radii of the same circles. So, HA = HB = HC. Therefore, triangle HAB, HBC, and HAC are all equilateral? Wait, no, not necessarily equilateral, but at least isosceles.Wait, if HA = HB = HC, then triangle HAB is isosceles with HA = HB, triangle HBC is isosceles with HB = HC, and triangle HAC is isosceles with HA = HC. So, that gives us some angle equalities.For example, in triangle HAB, angles at A and B are equal. Similarly, in triangle HBC, angles at B and C are equal, and in triangle HAC, angles at A and C are equal.But how does that relate to triangle ABC? Maybe I can relate the angles at H to the angles of triangle ABC.Let me denote some angles. Let’s say angle BAH is equal to angle BCH because they subtend the same chord BH in their respective circles. Similarly, angle ABH is equal to angle ACH, and angle CBH is equal to angle CAH.So, if I denote angle BAH as α, angle ABH as β, and angle CBH as γ, then angle BCH is also α, angle ACH is β, and angle CAH is γ.Now, looking at point H, the sum of the angles around H should be 360 degrees. But since H is inside triangle ABC, the sum of the angles at H related to the triangle should be 180 degrees.Wait, actually, in triangle ABC, the sum of angles at H would be angle AHB + angle BHC + angle AHC = 360 degrees. But since ABC is a triangle, the angles at H are related to the angles of ABC.Wait, maybe I should consider the triangle angles. Let me think about triangle ABC. The sum of its internal angles is 180 degrees. Also, the angles at H are related to the angles of ABC.Given that angle BAH = angle BCH = α, angle ABH = angle ACH = β, and angle CBH = angle CAH = γ, then in triangle ABC, the angles at A, B, and C can be expressed in terms of α, β, and γ.Specifically, angle at A is angle BAC = angle BAH + angle CAH = α + γ.Similarly, angle at B is angle ABC = angle ABH + angle CBH = β + γ.And angle at C is angle BCA = angle BCH + angle ACH = α + β.So, the angles of triangle ABC are α + γ, β + γ, and α + β.Since triangle ABC is acute, all these angles are less than 90 degrees.Now, the sum of the angles in triangle ABC is:(α + γ) + (β + γ) + (α + β) = 2α + 2β + 2γ = 180 degrees.Therefore, α + β + γ = 90 degrees.That's interesting. So, the sum of α, β, and γ is 90 degrees.Now, going back to point H, let's consider the angles around H. The sum of the angles at H is 360 degrees, but since H is inside triangle ABC, the angles related to the triangle would be angle AHB, angle BHC, and angle AHC.But from the earlier expressions, angle AHB is equal to 180 degrees minus angle BAC, which is 180 - (α + γ).Similarly, angle BHC is 180 - (β + γ), and angle AHC is 180 - (α + β).Adding these up:(180 - (α + γ)) + (180 - (β + γ)) + (180 - (α + β)) = 540 - 2(α + β + γ).But we know that α + β + γ = 90 degrees, so:540 - 2*90 = 540 - 180 = 360 degrees, which checks out.But how does this help me? Maybe I can relate these angles to the altitudes.Let me consider the altitude from A to BC. If I can show that AH is perpendicular to BC, then H lies on the altitude from A.To show that AH is perpendicular to BC, I need to show that angle AHB is 90 degrees plus angle BAC or something like that. Wait, no, if AH is perpendicular to BC, then angle AHB would be 90 degrees.Wait, no, that's not necessarily true. If AH is an altitude, then angle between AH and BC is 90 degrees, but angle AHB is not necessarily 90 degrees.Wait, maybe I should use the fact that in triangle ABC, the orthocenter H satisfies the property that the angles at H relate to the angles of the triangle.Alternatively, maybe I can use coordinate geometry. Let me assign coordinates to the points.Let me place point H at the origin (0,0). Let me assume that the three circles have centers at points O1, O2, and O3, each at a distance r from H, since all circles are equal. So, O1, O2, O3 lie on a circle of radius r centered at H.Let me choose specific coordinates for simplicity. Let me place O1 at (r,0), O2 at (r cos 120°, r sin 120°), and O3 at (r cos 240°, r sin 240°). So, the centers are equally spaced around H.Now, the circles centered at O1, O2, and O3, each with radius r, intersect at H and at points A, B, and C.Let me find the coordinates of points A, B, and C.First, find the intersection of circles O1 and O2. The equations are:Circle O1: (x - r)^2 + y^2 = r^2Circle O2: (x - r cos 120°)^2 + (y - r sin 120°)^2 = r^2Solving these equations simultaneously.First, expand both equations.For circle O1:x^2 - 2rx + r^2 + y^2 = r^2Simplify:x^2 - 2rx + y^2 = 0For circle O2:x^2 - 2r cos 120° x + r^2 cos^2 120° + y^2 - 2r sin 120° y + r^2 sin^2 120° = r^2Simplify:x^2 - 2r cos 120° x + y^2 - 2r sin 120° y + r^2 (cos^2 120° + sin^2 120°) = r^2Since cos^2 θ + sin^2 θ = 1, this simplifies to:x^2 - 2r cos 120° x + y^2 - 2r sin 120° y + r^2 = r^2Which further simplifies to:x^2 - 2r cos 120° x + y^2 - 2r sin 120° y = 0Now, subtract the equation of circle O1 from the equation of circle O2:(x^2 - 2r cos 120° x + y^2 - 2r sin 120° y) - (x^2 - 2r x + y^2) = 0 - 0Simplify:-2r cos 120° x + (-2r sin 120° y) - (-2r x) = 0Which is:-2r cos 120° x - 2r sin 120° y + 2r x = 0Factor out 2r:2r [ -cos 120° x - sin 120° y + x ] = 0Divide both sides by 2r (since r ≠ 0):(-cos 120° x - sin 120° y + x) = 0Simplify:x (1 - cos 120°) - sin 120° y = 0Compute cos 120° and sin 120°:cos 120° = -1/2sin 120° = √3/2So,x (1 - (-1/2)) - (√3/2) y = 0Simplify:x (3/2) - (√3/2) y = 0Multiply both sides by 2:3x - √3 y = 0So, the line of intersection of circles O1 and O2 is 3x - √3 y = 0.Now, we can find the intersection points by solving this line with one of the circle equations, say circle O1:x^2 - 2r x + y^2 = 0From the line equation, y = (3x)/√3 = √3 xSubstitute y = √3 x into the circle equation:x^2 - 2r x + (√3 x)^2 = 0Simplify:x^2 - 2r x + 3x^2 = 0Combine like terms:4x^2 - 2r x = 0Factor:2x (2x - r) = 0So, x = 0 or x = r/2If x = 0, then y = √3 * 0 = 0, which is point H (0,0).If x = r/2, then y = √3 * (r/2) = (r√3)/2So, the other intersection point is (r/2, (r√3)/2). Let's call this point A.Similarly, we can find points B and C by intersecting circles O1 and O3, and O2 and O3.But for brevity, let's assume symmetry. Since the centers are placed at 120° apart, the triangle ABC formed by the other intersection points should be equilateral. Wait, is that true?Wait, no, because the triangle ABC is given as acute-angled, but not necessarily equilateral. However, in this specific coordinate setup, with centers equally spaced, the triangle ABC might turn out to be equilateral.But in the problem, it's just an acute-angled triangle, so maybe my assumption of equally spaced centers is too restrictive. Maybe the centers are not necessarily equally spaced, but just that the circles are equal.Hmm, perhaps I should not assume the centers are equally spaced. Maybe that's why the triangle ABC is just acute-angled, not necessarily equilateral.So, perhaps my coordinate approach is making the problem more specific than it needs to be. Maybe I should think more generally.Let me go back to the angle relationships. Earlier, I had that α + β + γ = 90 degrees.Also, the angles of triangle ABC are α + γ, β + γ, and α + β.Since the sum of these is 180 degrees, and each is less than 90 degrees because the triangle is acute.Now, to show that H is the orthocenter, I need to show that the altitudes pass through H.Let me consider the altitude from A to BC. If I can show that AH is perpendicular to BC, then H lies on this altitude.Similarly, I need to show that BH is perpendicular to AC and CH is perpendicular to AB.So, let's try to show that AH is perpendicular to BC.To do this, I can show that the product of their slopes is -1, but since I don't have coordinates, maybe I can use vector methods or geometric properties.Alternatively, maybe I can use the fact that in triangle ABC, the orthocenter H satisfies certain angle relationships.Wait, another idea: in triangle ABC, the orthocenter H has the property that the angles at H are equal to 180 degrees minus the angles of the triangle.Wait, no, more precisely, the angles at H are equal to 180 degrees minus the angles at the opposite vertices.Wait, let me recall: in the orthocenter configuration, the angles at H are equal to 180 degrees minus the angles of the triangle.So, angle BHC = 180° - angle BAC.Similarly, angle AHB = 180° - angle ACB, and angle AHC = 180° - angle ABC.Wait, if that's the case, then in our earlier notation, angle BHC = 180° - (α + γ).But from earlier, angle BHC = 180° - (β + γ). Wait, that seems contradictory.Wait, no, let me clarify.In the orthocenter configuration, angle BHC is equal to 180° minus angle BAC.Given that angle BAC = α + γ, then angle BHC = 180° - (α + γ).But from earlier, we had angle BHC = 180° - (β + γ).Therefore, 180° - (α + γ) = 180° - (β + γ), which implies α + γ = β + γ, so α = β.Similarly, angle AHB = 180° - angle ACB = 180° - (α + β).But from earlier, angle AHB = 180° - (α + γ).Therefore, 180° - (α + β) = 180° - (α + γ), which implies β = γ.Similarly, angle AHC = 180° - angle ABC = 180° - (β + γ).From earlier, angle AHC = 180° - (α + β).Therefore, 180° - (β + γ) = 180° - (α + β), which implies γ = α.So, from these, we get α = β = γ.But earlier, we had α + β + γ = 90°, so 3α = 90°, which implies α = 30°, and similarly β = γ = 30°.Wait, so all angles α, β, γ are 30 degrees.Therefore, the angles of triangle ABC are:Angle BAC = α + γ = 30° + 30° = 60°Angle ABC = β + γ = 30° + 30° = 60°Angle BCA = α + β = 30° + 30° = 60°So, triangle ABC is equilateral.Wait, but the problem states that ABC is an acute-angled triangle, which is true because all angles are 60°, which are acute. But does this mean that ABC must be equilateral? Or is this a special case?Wait, in my earlier steps, I assumed that the centers are equally spaced, which led to ABC being equilateral. But in the general case, the problem doesn't specify that the centers are equally spaced, only that the circles are equal.So, maybe my coordinate approach was too restrictive, leading to ABC being equilateral. But the problem states ABC is just acute-angled, not necessarily equilateral.Therefore, perhaps my earlier conclusion that α = β = γ is only valid in the equilateral case, but in the general case, α, β, γ can be different as long as their sum is 90°.Wait, but how does that affect the orthocenter?Wait, let me think differently. Maybe I can use the fact that in triangle ABC, the orthocenter H satisfies the property that the reflections of H over the sides lie on the circumcircle of ABC.But I'm not sure if that helps directly.Alternatively, maybe I can use trigonometric identities or Ceva's theorem.Wait, Ceva's theorem states that for concurrent cevians, the product of certain ratios equals 1. But since H is the orthocenter, the cevians are the altitudes.But I don't know the ratios, so maybe that's not helpful.Wait, another idea: in triangle ABC, the orthocenter H has the property that the angles between the sides and the lines from the vertices to H are related to the angles of the triangle.Specifically, if I consider the altitude from A to BC, then angle between AH and BC is 90°, and angle BAH is equal to 90° - angle ABC.Wait, let me recall: in a triangle, the angle between a side and the altitude is equal to 90° minus the opposite angle.So, angle between AH and BC is 90°, and angle BAH is equal to 90° - angle ABC.Similarly, angle ABH is equal to 90° - angle ACB.Wait, if that's the case, then in our earlier notation, angle BAH = α = 90° - angle ABC.But angle ABC = β + γ.So, α = 90° - (β + γ).But we also have α + β + γ = 90°, so substituting:α = 90° - (β + γ) = αWhich is consistent.Wait, so angle BAH = 90° - angle ABC.Similarly, angle ABH = 90° - angle ACB.And angle CBH = 90° - angle BAC.But from earlier, we have:angle BAH = α = 90° - (β + γ)angle ABH = β = 90° - (α + γ)angle CBH = γ = 90° - (α + β)But since α + β + γ = 90°, we can substitute:From α = 90° - (β + γ), and since β + γ = 90° - α, then α = 90° - (90° - α) = α, which is consistent.Similarly for β and γ.So, this shows that the angles at H are related to the angles of triangle ABC in a way consistent with H being the orthocenter.Therefore, combining all these observations, it seems that H must be the orthocenter of triangle ABC.But to make this rigorous, maybe I should use the property that in triangle ABC, the orthocenter is the intersection of the altitudes, and show that H lies on all three altitudes.Given that H is the common intersection point of the three circles, and the angles at H relate to the angles of ABC in a way that satisfies the orthocenter properties, it follows that H is indeed the orthocenter.So, after considering the angle relationships and the properties of the orthocenter, I can conclude that H is the orthocenter of triangle ABC.

answer：First, I need to evaluate the innermost function application, which is ( N(O(2)) ).Calculating ( O(2) ):[O(2) = 2^2 = 4]Then, applying ( N ) to the result:[N(4) = 2sqrt{4} = 2 times 2 = 4]Next, apply ( P ) to this result:[P(4) = 4 + 1 = 5]Now, evaluate ( N(O(5)) ):[O(5) = 5^2 = 25][N(25) = 2sqrt{25} = 2 times 5 = 10]Apply ( P ) again:[P(10) = 10 + 1 = 11]Finally, calculate ( N(O(11)) ):[O(11) = 11^2 = 121][N(121) = 2sqrt{121} = 2 times 11 = 22]Thus, the final result of the sequence of operations ( N(O(P(N(O(P(N(O(2)))))))) ) is (boxed{22}).

answer：Okay, so I have this problem about a geometric sequence. The sum of the first 4 terms is 2, and the sum of the first 8 terms is 6. I need to find the sum of the first 12 terms. Hmm, let me think about how to approach this.First, I remember that in a geometric sequence, each term is multiplied by a common ratio, r. The sum of the first n terms of a geometric sequence can be calculated using the formula:[ S_n = a frac{1 - r^n}{1 - r} ]where ( a ) is the first term and ( r ) is the common ratio. Given that ( S_4 = 2 ) and ( S_8 = 6 ), I can set up two equations:1. ( S_4 = a frac{1 - r^4}{1 - r} = 2 )2. ( S_8 = a frac{1 - r^8}{1 - r} = 6 )Hmm, maybe I can divide the second equation by the first to eliminate ( a ). Let me try that:[ frac{S_8}{S_4} = frac{a frac{1 - r^8}{1 - r}}{a frac{1 - r^4}{1 - r}} = frac{1 - r^8}{1 - r^4} = frac{6}{2} = 3 ]Simplifying the left side:[ frac{1 - r^8}{1 - r^4} = 1 + r^4 ]So,[ 1 + r^4 = 3 ][ r^4 = 2 ][ r = sqrt[4]{2} ]Wait, but ( r ) could be positive or negative. Since the sum is positive, I think ( r ) is positive. So, ( r = sqrt[4]{2} ).Now, let's find ( a ) using ( S_4 = 2 ):[ 2 = a frac{1 - (sqrt[4]{2})^4}{1 - sqrt[4]{2}} ][ 2 = a frac{1 - 2}{1 - sqrt[4]{2}} ][ 2 = a frac{-1}{1 - sqrt[4]{2}} ][ a = 2 times frac{1 - sqrt[4]{2}}{-1} ][ a = -2(1 - sqrt[4]{2}) ][ a = -2 + 2sqrt[4]{2} ]Hmm, that seems a bit complicated. Maybe there's a simpler way. I remember that in a geometric sequence, the sums of terms also form a geometric sequence. So, ( S_4 ), ( S_8 - S_4 ), and ( S_{12} - S_8 ) should form a geometric sequence.Given that ( S_4 = 2 ) and ( S_8 = 6 ), then ( S_8 - S_4 = 6 - 2 = 4 ). So, the terms of this new geometric sequence are 2, 4, and the next term should be ( 4 times r ), where ( r ) is the common ratio.Wait, let me check the common ratio. Since it's a geometric sequence, the ratio between consecutive terms should be constant. So,[ r = frac{4}{2} = 2 ]Therefore, the next term after 4 should be ( 4 times 2 = 8 ). So, ( S_{12} - S_8 = 8 ). Therefore,[ S_{12} = S_8 + 8 = 6 + 8 = 14 ]Wait, that seems straightforward. So, the sum of the first 12 terms is 14. Let me verify this with the initial approach.Earlier, I found ( r = sqrt[4]{2} ) and ( a = -2 + 2sqrt[4]{2} ). Let me compute ( S_{12} ):[ S_{12} = a frac{1 - r^{12}}{1 - r} ][ = (-2 + 2sqrt[4]{2}) frac{1 - (sqrt[4]{2})^{12}}{1 - sqrt[4]{2}} ][ = (-2 + 2sqrt[4]{2}) frac{1 - 2^{3}}{1 - sqrt[4]{2}} ][ = (-2 + 2sqrt[4]{2}) frac{1 - 8}{1 - sqrt[4]{2}} ][ = (-2 + 2sqrt[4]{2}) frac{-7}{1 - sqrt[4]{2}} ][ = (-2 + 2sqrt[4]{2}) times frac{-7}{1 - sqrt[4]{2}} ][ = (2 - 2sqrt[4]{2}) times frac{7}{1 - sqrt[4]{2}} ][ = 2(1 - sqrt[4]{2}) times frac{7}{1 - sqrt[4]{2}} ][ = 2 times 7 = 14 ]Okay, that checks out. So, both methods lead to ( S_{12} = 14 ). I think that's the correct answer.

answer：To find the equation of the tangent line to the parabola ( y = x^2 ) that is parallel to the line ( 2x - y + 4 = 0 ), I start by determining the slope of the given line.The given line can be rewritten in slope-intercept form ( y = mx + b ) as ( y = 2x + 4 ), which shows that the slope ( m ) is 2.Next, I find the derivative of the parabola ( y = x^2 ) to determine the slope of the tangent line at any point ( x ). The derivative is ( y' = 2x ).Setting the derivative equal to the slope of the given line, ( 2x = 2 ), I solve for ( x ) and find ( x = 1 ).Substituting ( x = 1 ) back into the equation of the parabola, ( y = (1)^2 = 1 ), I determine the point of tangency is ( (1, 1) ).Using the point-slope form of a line, ( y - y_1 = m(x - x_1) ), with the point ( (1, 1) ) and slope 2, I write the equation of the tangent line as ( y - 1 = 2(x - 1) ). Simplifying this, I get ( y = 2x - 1 ).Finally, rearranging the equation to standard form, the equation of the tangent line is ( 2x - y - 1 = 0 ).

answer：First, I recognize that for both ( P - Q ) and ( P + Q ) to be prime numbers, they must be odd. This implies that one of ( P ) or ( Q ) must be even, and the other must be odd. Since ( Q ) is a prime number, the only even prime is 2. Therefore, ( Q = 2 ).Next, I need to find a prime number ( P ) such that both ( P - 2 ) and ( P + 2 ) are also prime. The smallest set of three consecutive odd numbers that are all primes are 3, 5, and 7. This means ( P = 5 ) and ( Q = 2 ).Now, I calculate the sum of these four primes:[P + Q + (P - Q) + (P + Q) = 5 + 2 + (5 - 2) + (5 + 2) = 5 + 2 + 3 + 7 = 17]Since 17 is a prime number, the sum of these four primes is prime.

answer：First, calculate the number of baby grasshoppers. Since a dozen is 12, 2 dozen baby grasshoppers would be 2 multiplied by 12, which equals 24 baby grasshoppers.Next, determine the number of grasshoppers in the 3 groups in the bushes. Each group had 9 grasshoppers, so 3 groups would have 3 multiplied by 9, which equals 27 grasshoppers.Now, add up all the grasshoppers Jemma found before the increase:7 (on the African daisy plant) + 24 (baby grasshoppers) + 27 (in the bushes) = 58 grasshoppers.The total number of grasshoppers increased by 20% after adding those nearby. To find the new total, calculate 20% of 58 and then add it to the original total.20% of 58 is 0.20 multiplied by 58, which equals 11.6. Since we can't have a fraction of a grasshopper, we'll round this to the nearest whole number, which is 12.Finally, add this increase to the original total:58 (original total) + 12 (increase) = 70 grasshoppers.So, there were 70 grasshoppers in the new total.