💡Okay, so I have this problem here: Among 2017 natural numbers, there is at least one two-digit number, and for any two of these numbers, at least one is a three-digit number. I need to determine the number of three-digit numbers among these 2017 numbers.Alright, let me try to break this down. First, natural numbers are just positive integers, right? So, we're dealing with numbers like 1, 2, 3, and so on. Two-digit numbers would be from 10 to 99, and three-digit numbers would be from 100 to 999. The problem says there's at least one two-digit number. So, in this collection of 2017 numbers, there's at least one number between 10 and 99. That's straightforward.Now, the second condition is a bit trickier: for any two numbers chosen from this collection, at least one of them is a three-digit number. Hmm, so if I pick any two numbers, one of them has to be a three-digit number. That means I can't have two two-digit numbers in the collection because if I did, then those two numbers would both be two-digit, violating the condition.Wait, so does that mean there can only be one two-digit number in the entire collection? Because if there were two or more, then I could pick those two, and neither would be a three-digit number, which isn't allowed. So, that makes sense. There can only be one two-digit number.So, if there's only one two-digit number, then the rest must be three-digit numbers. Since there are 2017 numbers in total, subtracting the one two-digit number, we get 2016 three-digit numbers.Let me just double-check that. If there's only one two-digit number, then any pair involving that two-digit number will have one two-digit and one three-digit number, which satisfies the condition because at least one is three-digit. Any pair of three-digit numbers will obviously satisfy the condition because both are three-digit. So, that works.Is there any other possibility? Could there be zero two-digit numbers? But the problem says there's at least one, so zero isn't allowed. So, one is the minimum, and as we saw, it can't be more than one because of the second condition.Therefore, the number of three-digit numbers must be 2016.**Final Answer**The number of three-digit numbers among these 2017 numbers is boxed{2016}.

💡Alright, so I have this geometry problem to prove that the median of a triangle is less than the half-sum of the sides enclosing it. Hmm, okay, let me try to wrap my head around this.First, let me recall what a median is in a triangle. A median is a line segment joining a vertex to the midpoint of the opposite side. So, in any triangle, there are three medians, each corresponding to one side.Now, the problem is asking me to show that this median is less than the average of the two sides that form the angle at the vertex from which the median is drawn. Wait, is that right? Or is it the half-sum of the sides enclosing the median? Hmm, maybe I need to clarify that.Let me consider a triangle ABC, with vertices A, B, and C. Let's say the median is drawn from vertex C to the midpoint of side AB. So, the median is CM, where M is the midpoint of AB. The sides enclosing this median would be sides AC and BC, right? So, the half-sum of these sides would be (AC + BC)/2.So, the problem is to prove that CM < (AC + BC)/2.Okay, that makes sense. So, I need to show that the length of the median from C to AB is less than the average of the lengths of sides AC and BC.How can I approach this? Maybe using some geometric theorems or properties. I remember there's something called the triangle inequality, which states that the sum of any two sides of a triangle is greater than the third side. Maybe that could be useful here.Alternatively, I recall there's a formula for the length of a median in terms of the sides of the triangle. Maybe I can use that formula and then compare it to the half-sum of the two sides.Let me try to recall the formula for the median. I think it's something like:CM² = (2AC² + 2BC² - AB²)/4Yes, that sounds familiar. So, if I can express CM in terms of AC and BC, maybe I can compare it to (AC + BC)/2.Let me write that down:CM = sqrt[(2AC² + 2BC² - AB²)/4]And I need to show that this is less than (AC + BC)/2.So, if I can show that sqrt[(2AC² + 2BC² - AB²)/4] < (AC + BC)/2, then I've proven the statement.To make it simpler, let's square both sides to eliminate the square root:(2AC² + 2BC² - AB²)/4 < (AC + BC)² / 4Multiplying both sides by 4:2AC² + 2BC² - AB² < (AC + BC)²Expanding the right side:(AC + BC)² = AC² + 2AC*BC + BC²So, substituting back:2AC² + 2BC² - AB² < AC² + 2AC*BC + BC²Simplifying the left side:2AC² + 2BC² - AB² = AC² + BC² + AC² + BC² - AB²But I'm not sure if that helps. Maybe I should subtract AC² + BC² from both sides:2AC² + 2BC² - AB² - AC² - BC² < 2AC*BCWhich simplifies to:AC² + BC² - AB² < 2AC*BCHmm, interesting. So, I need to show that AC² + BC² - AB² is less than 2AC*BC.Wait a minute, I remember that in the Law of Cosines, AB² = AC² + BC² - 2AC*BC*cos(theta), where theta is the angle at vertex C.So, substituting that into the inequality:AC² + BC² - (AC² + BC² - 2AC*BC*cos(theta)) < 2AC*BCSimplifying:AC² + BC² - AC² - BC² + 2AC*BC*cos(theta) < 2AC*BCWhich reduces to:2AC*BC*cos(theta) < 2AC*BCDividing both sides by 2AC*BC (assuming AC and BC are positive, which they are in a triangle):cos(theta) < 1Well, cos(theta) is always less than or equal to 1, and in a triangle, theta is between 0 and 180 degrees, so cos(theta) is strictly less than 1 because theta can't be 0 degrees (as that would collapse the triangle). Therefore, the inequality holds.So, going back, that means:AC² + BC² - AB² < 2AC*BCWhich implies:2AC² + 2BC² - AB² < AC² + 2AC*BC + BC²Which in turn implies:sqrt[(2AC² + 2BC² - AB²)/4] < (AC + BC)/2Therefore, CM < (AC + BC)/2So, that seems to work out. But let me double-check my steps to make sure I didn't make any mistakes.1. I started with the formula for the median: CM² = (2AC² + 2BC² - AB²)/42. I set up the inequality CM < (AC + BC)/23. Squared both sides to get rid of the square root: (2AC² + 2BC² - AB²)/4 < (AC + BC)² / 44. Multiplied both sides by 4: 2AC² + 2BC² - AB² < AC² + 2AC*BC + BC²5. Subtracted AC² + BC² from both sides: AC² + BC² - AB² < 2AC*BC6. Applied the Law of Cosines: AB² = AC² + BC² - 2AC*BC*cos(theta)7. Substituted into the inequality: AC² + BC² - (AC² + BC² - 2AC*BC*cos(theta)) < 2AC*BC8. Simplified to: 2AC*BC*cos(theta) < 2AC*BC9. Divided both sides by 2AC*BC: cos(theta) < 110. Since in a triangle, theta is between 0 and 180 degrees, cos(theta) is indeed less than 1.Therefore, all the steps check out, and the proof is valid.Alternatively, I could think about this problem using vector geometry or coordinate geometry, but I think the approach using the median formula and the Law of Cosines is straightforward and sufficient.Another way might be to consider reflecting the triangle over the median and using properties of parallelograms, but that might complicate things unnecessarily.I think the key insight here is recognizing the formula for the median and then using algebraic manipulation along with the Law of Cosines to establish the required inequality.So, in summary, by expressing the median in terms of the sides of the triangle and applying some algebraic and trigonometric identities, we can successfully prove that the median is indeed less than the half-sum of the sides enclosing it.

💡First, I need to determine the total number of people joining the country club, which includes John and his 3 family members, making a total of 4 people.Next, I'll calculate the one-time joining fee. Since the fee is 4000 per person, the total joining fee for 4 people is:4 people × 4000/person = 16,000.Then, I'll calculate the monthly cost. Each person has a monthly cost of 1000, so for 4 people, the monthly cost is:4 people × 1000/person = 4000/month.Over a year, which has 12 months, the total monthly cost is:4000/month × 12 months = 48,000.Adding the one-time joining fee to the total monthly cost gives the total cost for the first year:16,000 + 48,000 = 64,000.Since John pays half of this total cost, his share for the first year is:64,000 ÷ 2 = 32,000.

💡Okay, so I have this geometry problem here: On the circumcircle of triangle (ABC), find the point (M) such that the distance between its projections onto the lines (AC) and (BC) is maximized. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, let me visualize triangle (ABC) with its circumcircle. The circumcircle is the circle that passes through all three vertices of the triangle. So, any point (M) on this circle will be such that (M) is equidistant from the center of the circle. Now, I need to find a specific point (M) on this circle where the distance between its projections onto (AC) and (BC) is as large as possible.Projections onto lines... Okay, so if I have a point (M), its projection onto line (AC) would be the foot of the perpendicular from (M) to (AC). Similarly, its projection onto (BC) would be the foot of the perpendicular from (M) to (BC). Let me denote these projections as (P) and (Q) respectively. So, (P) is the foot on (AC) and (Q) is the foot on (BC).Now, the distance between (P) and (Q) is what I need to maximize. So, essentially, I need to find the point (M) on the circumcircle such that the distance (PQ) is as large as possible.Let me think about how to express (PQ) in terms of other quantities. Maybe using coordinates or trigonometric relationships?Wait, maybe I can use some properties of projections and circles. Since (M) is on the circumcircle, perhaps there's a relationship between (M), (C), and the projections (P) and (Q). Let me consider the quadrilateral formed by (M), (P), (Q), and (C). Since (P) and (Q) are projections, (MP) and (MQ) are both perpendicular to (AC) and (BC) respectively. So, quadrilateral (MPQC) has two right angles at (P) and (Q).Hmm, quadrilaterals with two right angles... I wonder if this makes it cyclic? Wait, if a quadrilateral has two opposite right angles, it might be cyclic. But in this case, the right angles are at (P) and (Q), which are adjacent if we consider the order (M), (P), (C), (Q). Hmm, not sure if that's cyclic.Alternatively, maybe I can use some trigonometric identities or the law of sines or cosines here. Let me consider triangle (MPQ). Wait, but (P) and (Q) are projections, so (MP) and (MQ) are perpendicular to (AC) and (BC). Maybe I can express (PQ) in terms of (MC) and the angle at (C).Let me denote (angle ACB) as (theta). Since (M) is on the circumcircle, the angle (angle AMC) is equal to (angle ABC) because they subtend the same arc. Wait, is that right? Actually, in a circle, angles subtended by the same arc are equal. So, (angle AMC = angle ABC). Hmm, but I need to relate this to the projections.Alternatively, maybe I can consider the coordinates. Let me place triangle (ABC) in a coordinate system to make things clearer. Let me set point (C) at the origin ((0,0)), point (A) at ((a,0)), and point (B) at ((0,b)). Then, the circumcircle can be determined based on these coordinates.But before I get too deep into coordinates, maybe there's a more geometric approach. Let me think about the projections. Since (P) is the projection of (M) onto (AC), the distance (MP) is the length of the perpendicular from (M) to (AC). Similarly, (MQ) is the length of the perpendicular from (M) to (BC).Wait, but I need the distance between (P) and (Q), not the lengths of the perpendiculars. Hmm, so perhaps I can express (PQ) in terms of these perpendiculars and the angle between (AC) and (BC).Let me denote the lengths (MP = h_1) and (MQ = h_2). Then, the distance (PQ) can be found using the Pythagorean theorem if I can find the angle between (MP) and (MQ). But since (MP) is perpendicular to (AC) and (MQ) is perpendicular to (BC), the angle between (MP) and (MQ) is equal to the angle between (AC) and (BC), which is (angle ACB). So, if I denote (angle ACB = theta), then the distance (PQ) can be expressed as:[PQ = sqrt{h_1^2 + h_2^2 - 2h_1h_2costheta}]Wait, is that right? Because if I have two vectors with magnitudes (h_1) and (h_2) and angle (theta) between them, the distance between their endpoints would be given by the law of cosines. So, yes, that seems correct.But I need to express (h_1) and (h_2) in terms of (M)'s position on the circumcircle. Hmm, maybe I can relate these heights to the sides of the triangle or the circumradius.Alternatively, perhaps I can use trigonometric identities involving the circumradius. Let me recall that in any triangle, the circumradius (R) is given by (R = frac{a}{2sin A}), where (a) is the length of the side opposite angle (A). But I'm not sure if that directly helps here.Wait, maybe I can consider the coordinates approach after all. Let me set up a coordinate system with (C) at the origin, (AC) along the x-axis, and (BC) along the y-axis. So, point (C) is ((0,0)), point (A) is ((c,0)), and point (B) is ((0,b)). Then, the circumcircle of triangle (ABC) can be found using the circumcircle formula.The general equation of a circle is (x^2 + y^2 + Dx + Ey + F = 0). Since points (A), (B), and (C) lie on this circle, we can plug in their coordinates to find (D), (E), and (F).Plugging in (C(0,0)): (0 + 0 + 0 + 0 + F = 0 Rightarrow F = 0).Plugging in (A(c,0)): (c^2 + 0 + Dc + 0 + 0 = 0 Rightarrow D = -c).Plugging in (B(0,b)): (0 + b^2 + 0 + Eb + 0 = 0 Rightarrow E = -b).So, the equation of the circumcircle is (x^2 + y^2 - cx - by = 0). Completing the squares:[x^2 - cx + y^2 - by = 0 left(x - frac{c}{2}right)^2 + left(y - frac{b}{2}right)^2 = left(frac{c}{2}right)^2 + left(frac{b}{2}right)^2]So, the center of the circumcircle is at (left(frac{c}{2}, frac{b}{2}right)) and the radius is (frac{sqrt{b^2 + c^2}}{2}).Now, any point (M) on the circumcircle can be parameterized as:[Mleft(frac{c}{2} + frac{sqrt{b^2 + c^2}}{2}costheta, frac{b}{2} + frac{sqrt{b^2 + c^2}}{2}sinthetaright)]where (theta) is the parameter varying from (0) to (2pi).Now, I need to find the projections (P) and (Q) of (M) onto (AC) and (BC) respectively.Since (AC) is along the x-axis, the projection (P) of (M) onto (AC) will have the same x-coordinate as (M) and y-coordinate 0. So,[Pleft(frac{c}{2} + frac{sqrt{b^2 + c^2}}{2}costheta, 0right)]Similarly, since (BC) is along the y-axis, the projection (Q) of (M) onto (BC) will have the same y-coordinate as (M) and x-coordinate 0. So,[Qleft(0, frac{b}{2} + frac{sqrt{b^2 + c^2}}{2}sinthetaright)]Now, the distance (PQ) can be found using the distance formula between points (P) and (Q):[PQ = sqrt{left(frac{c}{2} + frac{sqrt{b^2 + c^2}}{2}costheta - 0right)^2 + left(0 - left(frac{b}{2} + frac{sqrt{b^2 + c^2}}{2}sinthetaright)right)^2}]Simplifying this expression:[PQ = sqrt{left(frac{c}{2} + frac{sqrt{b^2 + c^2}}{2}costhetaright)^2 + left(-frac{b}{2} - frac{sqrt{b^2 + c^2}}{2}sinthetaright)^2}]Let me expand the squares:First term:[left(frac{c}{2} + frac{sqrt{b^2 + c^2}}{2}costhetaright)^2 = left(frac{c}{2}right)^2 + 2left(frac{c}{2}right)left(frac{sqrt{b^2 + c^2}}{2}costhetaright) + left(frac{sqrt{b^2 + c^2}}{2}costhetaright)^2][= frac{c^2}{4} + frac{csqrt{b^2 + c^2}}{2}costheta + frac{b^2 + c^2}{4}cos^2theta]Second term:[left(-frac{b}{2} - frac{sqrt{b^2 + c^2}}{2}sinthetaright)^2 = left(frac{b}{2}right)^2 + 2left(frac{b}{2}right)left(frac{sqrt{b^2 + c^2}}{2}sinthetaright) + left(frac{sqrt{b^2 + c^2}}{2}sinthetaright)^2][= frac{b^2}{4} + frac{bsqrt{b^2 + c^2}}{2}sintheta + frac{b^2 + c^2}{4}sin^2theta]Adding both terms together:[PQ^2 = left(frac{c^2}{4} + frac{csqrt{b^2 + c^2}}{2}costheta + frac{b^2 + c^2}{4}cos^2thetaright) + left(frac{b^2}{4} + frac{bsqrt{b^2 + c^2}}{2}sintheta + frac{b^2 + c^2}{4}sin^2thetaright)]Combine like terms:[PQ^2 = frac{c^2}{4} + frac{b^2}{4} + frac{csqrt{b^2 + c^2}}{2}costheta + frac{bsqrt{b^2 + c^2}}{2}sintheta + frac{b^2 + c^2}{4}(cos^2theta + sin^2theta)]Since (cos^2theta + sin^2theta = 1), this simplifies to:[PQ^2 = frac{c^2 + b^2}{4} + frac{sqrt{b^2 + c^2}}{2}(ccostheta + bsintheta) + frac{b^2 + c^2}{4}][= frac{c^2 + b^2}{4} + frac{c^2 + b^2}{4} + frac{sqrt{b^2 + c^2}}{2}(ccostheta + bsintheta)][= frac{c^2 + b^2}{2} + frac{sqrt{b^2 + c^2}}{2}(ccostheta + bsintheta)]So,[PQ = sqrt{frac{c^2 + b^2}{2} + frac{sqrt{b^2 + c^2}}{2}(ccostheta + bsintheta)}]Hmm, that looks a bit complicated. Maybe I can factor out (sqrt{b^2 + c^2}) from the second term:Let me denote (k = sqrt{b^2 + c^2}). Then,[PQ = sqrt{frac{k^2}{2} + frac{k}{2}(ccostheta + bsintheta)}][= sqrt{frac{k^2}{2} + frac{k}{2}(ccostheta + bsintheta)}]Let me factor out (frac{k}{2}):[PQ = sqrt{frac{k}{2}left(k + ccostheta + bsinthetaright)}]Hmm, not sure if that helps. Maybe I can write (ccostheta + bsintheta) as a single sinusoidal function. Let me recall that (Acostheta + Bsintheta = Rcos(theta - phi)), where (R = sqrt{A^2 + B^2}) and (tanphi = frac{B}{A}).So, in this case, (ccostheta + bsintheta = Rcos(theta - phi)), where (R = sqrt{c^2 + b^2} = k), and (tanphi = frac{b}{c}). So, (phi = arctanleft(frac{b}{c}right)).Therefore,[ccostheta + bsintheta = kcos(theta - phi)]Substituting back into the expression for (PQ):[PQ = sqrt{frac{k}{2}left(k + kcos(theta - phi)right)} = sqrt{frac{k}{2} cdot k(1 + cos(theta - phi))}][= sqrt{frac{k^2}{2}(1 + cos(theta - phi))}]Using the identity (1 + cosalpha = 2cos^2left(frac{alpha}{2}right)):[PQ = sqrt{frac{k^2}{2} cdot 2cos^2left(frac{theta - phi}{2}right)} = sqrt{k^2 cos^2left(frac{theta - phi}{2}right)} = k left|cosleft(frac{theta - phi}{2}right)right|]Since (k = sqrt{b^2 + c^2}) is positive, and the absolute value of cosine is just cosine of the angle within the appropriate range. So,[PQ = sqrt{b^2 + c^2} cosleft(frac{theta - phi}{2}right)]Now, to maximize (PQ), we need to maximize (cosleft(frac{theta - phi}{2}right)). The maximum value of cosine is 1, which occurs when the argument is 0. Therefore,[frac{theta - phi}{2} = 0 Rightarrow theta = phi]So, the maximum distance (PQ) is (sqrt{b^2 + c^2}), and this occurs when (theta = phi). But what does (theta = phi) mean in terms of the position of point (M)? Recall that (phi = arctanleft(frac{b}{c}right)), which is the angle that the line (BC) makes with the x-axis. So, when (theta = phi), point (M) is located such that the angle parameter in its parametrization equals (phi). Wait, let me think about the parametrization of point (M). Earlier, I wrote:[Mleft(frac{c}{2} + frac{sqrt{b^2 + c^2}}{2}costheta, frac{b}{2} + frac{sqrt{b^2 + c^2}}{2}sinthetaright)]So, when (theta = phi), the coordinates become:[x = frac{c}{2} + frac{sqrt{b^2 + c^2}}{2}cosphi y = frac{b}{2} + frac{sqrt{b^2 + c^2}}{2}sinphi]But since (phi = arctanleft(frac{b}{c}right)), we can express (cosphi) and (sinphi) in terms of (b) and (c):[cosphi = frac{c}{sqrt{b^2 + c^2}} sinphi = frac{b}{sqrt{b^2 + c^2}}]Substituting back into the coordinates of (M):[x = frac{c}{2} + frac{sqrt{b^2 + c^2}}{2} cdot frac{c}{sqrt{b^2 + c^2}} = frac{c}{2} + frac{c}{2} = c y = frac{b}{2} + frac{sqrt{b^2 + c^2}}{2} cdot frac{b}{sqrt{b^2 + c^2}} = frac{b}{2} + frac{b}{2} = b]So, point (M) is at ((c, b)). But wait, in our coordinate system, point (A) is at ((c, 0)) and point (B) is at ((0, b)). So, point (M) at ((c, b)) is actually the reflection of point (C) over the center of the circumcircle.Wait, the center of the circumcircle was at (left(frac{c}{2}, frac{b}{2}right)). So, reflecting (C(0,0)) over the center would give the point ((c, b)), which is exactly where (M) is located when (theta = phi). Therefore, point (M) is diametrically opposite to point (C) on the circumcircle.So, in other words, the point (M) that maximizes the distance between its projections onto (AC) and (BC) is the point such that (MC) is the diameter of the circumcircle.Let me verify this conclusion. If (MC) is the diameter, then (M) is the antipodal point of (C) on the circumcircle. Therefore, the angle (angle AMC) should be a right angle because any angle subtended by a diameter is a right angle. Wait, but in our case, we have projections, not necessarily right angles.Wait, no, actually, in our coordinate system, (M) is at ((c, b)), which is not necessarily forming a right angle with (A) and (B). Hmm, maybe my earlier conclusion is slightly off.Wait, but according to the parametrization, when (theta = phi), (M) is at ((c, b)), which is the reflection of (C) over the center. So, in that case, (MC) is indeed the diameter because the distance from (C(0,0)) to (M(c, b)) is (sqrt{c^2 + b^2}), which is equal to the diameter since the radius is (frac{sqrt{b^2 + c^2}}{2}). Therefore, the diameter is (sqrt{b^2 + c^2}), which matches.So, yes, (MC) is the diameter, and (M) is the antipodal point of (C). Therefore, this should be the point where the distance (PQ) is maximized.Let me think if there's another way to see this without coordinates. Since (P) and (Q) are projections, the quadrilateral (MPQC) has right angles at (P) and (Q). If (MC) is the diameter, then (M) is such that (angle MPC) and (angle MQC) are right angles. Wait, but in our case, (MP) and (MQ) are already perpendicular to (AC) and (BC), so they are right angles by definition.Hmm, maybe another approach is to consider that the distance (PQ) is related to the length (MC) and the angle between (AC) and (BC). Earlier, I had an expression (PQ = MC sin angle ACB). Wait, is that correct?Let me think. If (MC) is the diameter, then the angle (angle MAC) and (angle MBC) would be right angles because they subtend the diameter. But in our case, the projections are onto (AC) and (BC), not necessarily the sides (AB) or (BC).Wait, perhaps I can use vector projections. Let me denote vectors for (AC) and (BC). Let me denote vector (AC) as (vec{u}) and vector (BC) as (vec{v}). Then, the projections of (M) onto (AC) and (BC) can be expressed using dot products.But maybe that's complicating things. Alternatively, perhaps I can use complex numbers. Let me place the circumcircle on the complex plane with center at the origin for simplicity. Wait, but in our earlier coordinate system, the center was at (left(frac{c}{2}, frac{b}{2}right)). Maybe it's better to shift the coordinate system so that the center is at the origin.Let me try that. Let me translate the coordinate system so that the center of the circumcircle is at ((0,0)). Then, the coordinates of points (A), (B), and (C) would be shifted accordingly.Original coordinates:- (C(0,0))- (A(c,0))- (B(0,b))- Center at (left(frac{c}{2}, frac{b}{2}right))After translation by (left(-frac{c}{2}, -frac{b}{2}right)), the new coordinates are:- (Cleft(-frac{c}{2}, -frac{b}{2}right))- (Aleft(frac{c}{2}, -frac{b}{2}right))- (Bleft(-frac{c}{2}, frac{b}{2}right))- Center at ((0,0))Now, the circumradius is (frac{sqrt{b^2 + c^2}}{2}), so any point (M) on the circumcircle can be represented as (M(re^{itheta})), where (r = frac{sqrt{b^2 + c^2}}{2}).But this might not necessarily simplify things. Maybe I should stick with the previous coordinate system.Wait, going back, I had that (PQ = sqrt{b^2 + c^2} cosleft(frac{theta - phi}{2}right)), which is maximized when (cosleft(frac{theta - phi}{2}right) = 1), i.e., when (theta = phi). So, regardless of the coordinate system, this suggests that (M) is located such that (theta = phi), which in our coordinate system corresponds to the point ((c, b)), the antipodal point of (C).Therefore, the point (M) that maximizes the distance between its projections onto (AC) and (BC) is the point diametrically opposite to (C) on the circumcircle of triangle (ABC).Let me just check if this makes sense intuitively. If (M) is opposite to (C), then its projections onto (AC) and (BC) would be as far apart as possible because (M) is as far as possible from both (AC) and (BC). Hence, their projections would be maximally separated.Yes, that seems to make sense. So, I think I've arrived at the correct conclusion.**Final Answer**The point (M) that maximizes the distance between its projections onto (AC) and (BC) is the point diametrically opposite to (C) on the circumcircle of triangle (ABC). Therefore, the desired point (M) is such that (MC) is the diameter of the circumcircle.[boxed{M text{ is the point diametrically opposite to } C text{ on the circumcircle of } triangle ABC}]

💡Alright, so I have this problem here where I need to find the intersection of the domains of two functions. The functions are ( y = sqrt{4 - x^2} ) and ( y = ln(1 - x) ). The options given are A: (1,2), B: (1,2], C: (-2,1), and D: [-2,1). I need to figure out which one is correct.First, let me recall what the domain of a function is. The domain is all the possible input values (x-values) for which the function is defined. So, for each function, I need to determine the set of x-values that make the function valid.Starting with the first function, ( y = sqrt{4 - x^2} ). Since this is a square root function, the expression inside the square root must be greater than or equal to zero. That is, ( 4 - x^2 geq 0 ). Let me solve this inequality.So, ( 4 - x^2 geq 0 ) can be rewritten as ( x^2 leq 4 ). Taking square roots on both sides, we get ( |x| leq 2 ), which means ( -2 leq x leq 2 ). Therefore, the domain of this function, which I'll call set A, is all real numbers from -2 to 2, including both endpoints. So, A is the interval [-2, 2].Next, moving on to the second function, ( y = ln(1 - x) ). The natural logarithm function, ln, is only defined for positive arguments. That means the expression inside the logarithm, ( 1 - x ), must be greater than zero. So, I need to solve ( 1 - x > 0 ).Solving ( 1 - x > 0 ) gives ( -x > -1 ). When I multiply both sides by -1, I have to remember to reverse the inequality sign, so it becomes ( x < 1 ). Therefore, the domain of this function, set B, is all real numbers less than 1. So, B is the interval (-∞, 1).Now, I need to find the intersection of sets A and B, which is ( A cap B ). The intersection of two sets consists of all elements that are common to both sets.Set A is [-2, 2], and set B is (-∞, 1). So, graphically, if I imagine both intervals on the number line, A starts at -2 and goes up to 2, while B goes from negative infinity up to 1. The overlapping part of these two intervals would be from -2 up to 1.But I need to be careful about the endpoints. In set A, both -2 and 2 are included, but in set B, 1 is not included because it's an open interval. So, when taking the intersection, the overlapping interval will include -2 because it's included in A, but it will not include 1 because it's excluded in B.Therefore, the intersection ( A cap B ) is the interval from -2 to 1, including -2 but not including 1. In interval notation, that's [-2, 1).Looking back at the options provided:A: (1,2) – This is the interval from 1 to 2, which doesn't overlap with the intersection we found.B: (1,2] – Similarly, this is from 1 to 2, including 2, which is not the intersection.C: (-2,1) – This is from -2 to 1, but it's an open interval, meaning -2 is not included. However, in our intersection, -2 is included because it's part of set A.D: [-2,1) – This is from -2 to 1, including -2 but not including 1, which matches exactly what we found.So, the correct answer should be option D.Just to double-check, let me recap:- For ( sqrt{4 - x^2} ), the domain is [-2, 2].- For ( ln(1 - x) ), the domain is (-∞, 1).- The overlap between these two is from -2 up to, but not including, 1, which is [-2, 1).Yep, that seems right. I don't think I made any mistakes in solving the inequalities or interpreting the intervals. So, I'm confident that D is the correct choice.**Final Answer**boxed{D}

💡Okay, let's see. I have this problem: "What is the smallest integer k for which 64^k > some base^20, if k = 7? What is the base?" Hmm, that's a bit confusing at first glance. Let me try to break it down.So, the problem is asking for the smallest integer k such that 64 raised to the power of k is greater than some base raised to the 20th power. But wait, it also says "if k = 7." That seems contradictory because it's asking for the smallest k, but then it's giving k as 7. Maybe I'm misinterpreting it.Perhaps it's saying, "Given that k = 7, what is the smallest integer k for which 64^k > some base^20?" But that doesn't make much sense because k is already given as 7. Maybe the problem is asking for the base when k is 7? That would make more sense. So, if k is 7, what is the base such that 64^7 is greater than base^20?Alright, let's work with that interpretation. So, we have 64^7 > base^20. I need to find the base. To find the base, I can take both sides of the inequality to the power of 1/20 to isolate the base. That would give me base < (64^7)^(1/20).First, let's simplify 64^7. I know that 64 is 2^6, so 64^7 is (2^6)^7, which is 2^(6*7) = 2^42. So, 64^7 = 2^42.Now, plugging that back into the inequality: base < (2^42)^(1/20). Using the exponent rule (a^m)^n = a^(m*n), this becomes base < 2^(42/20). Simplifying 42/20, that's 2.1. So, base < 2^2.1.Now, 2^2 is 4, and 2^3 is 8. Since 2.1 is just a little more than 2, 2^2.1 should be just a bit more than 4. To get a better estimate, I can calculate 2^0.1. I remember that 2^0.1 is approximately 1.07177. So, 2^2.1 = 2^2 * 2^0.1 ≈ 4 * 1.07177 ≈ 4.287.Therefore, base < approximately 4.287. Since we're looking for the largest integer base that satisfies this inequality, the base must be 4.Wait, but the problem is asking for the smallest integer k for which 64^k > base^20, given that k = 7. But we've found that when k = 7, the base is 4. So, is 7 the smallest integer k for which 64^k > 4^20? Let's check with k = 6 to see if 64^6 is still greater than 4^20.Calculating 64^6: 64 is 2^6, so 64^6 = (2^6)^6 = 2^36. And 4^20 is (2^2)^20 = 2^40. So, 2^36 vs. 2^40. Clearly, 2^36 is less than 2^40. Therefore, 64^6 < 4^20. So, k = 7 is indeed the smallest integer k for which 64^k > 4^20.So, to summarize, when k = 7, the base is 4, and 7 is the smallest integer k for which 64^k exceeds 4^20.