answer：Okay, so I have this geometry problem here. Let me read it again to make sure I understand it properly."A median ( CM ) is drawn from the right angle ( C ) of the triangle ( ABC ). A circle inscribed in the triangle ( ACM ) touches the side ( CM ) at its midpoint. Find the angles of triangle ( ABC )."Alright, so we have a right triangle ( ABC ) with the right angle at ( C ). A median ( CM ) is drawn from ( C ) to the midpoint ( M ) of side ( AB ). Then, there's an inscribed circle in triangle ( ACM ), and this circle touches ( CM ) at its midpoint. We need to find the angles of triangle ( ABC ).First, let me visualize this. Triangle ( ABC ) is right-angled at ( C ), so sides ( AC ) and ( BC ) are the legs, and ( AB ) is the hypotenuse. The median ( CM ) connects ( C ) to the midpoint ( M ) of ( AB ). So, ( M ) is the midpoint, meaning ( AM = MB ).Now, in triangle ( ACM ), there's an inscribed circle (incircle) that touches all three sides. The point where this incircle touches ( CM ) is at its midpoint. That's interesting because it tells us something specific about the lengths related to the incircle.Let me recall some properties of incircles. The incircle of a triangle touches each side at a point, and the distances from the vertices to these points are related to the triangle's semiperimeter. Specifically, if a circle is inscribed in a triangle, the lengths from each vertex to the point where the incircle touches the opposite side are equal for the two sides meeting at that vertex.In other words, if the incircle touches side ( AC ) at point ( D ), side ( CM ) at point ( E ), and side ( AM ) at point ( F ), then ( AD = AF ), ( CD = CE ), and ( AE = AF ). Wait, maybe I should write this more carefully.Actually, for any triangle, the lengths from each vertex to the point where the incircle touches the opposite side are equal. So, in triangle ( ACM ), the incircle touches ( AC ) at some point, say ( D ), ( CM ) at its midpoint ( E ), and ( AM ) at some point ( F ). Then, the lengths satisfy:- ( AD = AF )- ( CD = CE )- ( AE = AF )Wait, maybe I need to correct that. Let me think again.In triangle ( ACM ), the incircle touches:- Side ( AC ) at point ( D )- Side ( CM ) at point ( E )- Side ( AM ) at point ( F )Then, the lengths from the vertices to the points of tangency are equal. So:- From ( A ): ( AF = AD )- From ( C ): ( CD = CE )- From ( M ): ( ME = MF )Since the incircle touches ( CM ) at its midpoint ( E ), that means ( CE = EM ). Therefore, ( CE = EM ), which implies that ( E ) is the midpoint of ( CM ).So, ( CE = EM ). From the property of the incircle, we also have ( CD = CE ). Therefore, ( CD = CE = EM ).Similarly, from ( A ), ( AF = AD ). Let me denote the lengths:Let ( AF = AD = x ).Then, ( CD = CE = y ).Since ( AC = AD + DC = x + y ).Similarly, ( CM = CE + EM = y + y = 2y ).And ( AM = AF + FM = x + FM ).But from the property of the incircle, ( FM = ME = y ). Therefore, ( AM = x + y ).Wait, so ( AM = x + y ), and ( AC = x + y ). Therefore, ( AM = AC ).Hmm, so ( AM = AC ). That's interesting.But ( AM ) is a side of triangle ( ACM ), and ( AC ) is another side. So, if ( AM = AC ), then triangle ( ACM ) is isoceles with ( AC = AM ).Wait, but ( ACM ) is a triangle with sides ( AC ), ( CM ), and ( AM ). If ( AC = AM ), then it's isoceles with base ( CM ).But earlier, we also found that ( CM = 2y ), and ( AC = x + y ). Since ( AM = AC ), then ( AM = x + y ), but ( AM = AF + FM = x + y ), which is consistent.Wait, maybe I should write the semiperimeter of triangle ( ACM ).The semiperimeter ( s ) of triangle ( ACM ) is given by:( s = frac{AC + CM + AM}{2} )But since ( AM = AC ), let's denote ( AC = AM = b ), and ( CM = 2y ).So, ( s = frac{b + 2y + b}{2} = frac{2b + 2y}{2} = b + y ).Now, the lengths from the vertices to the points of tangency are:- From ( A ): ( AF = AD = s - CM = (b + y) - 2y = b - y )- From ( C ): ( CD = CE = s - AM = (b + y) - b = y )- From ( M ): ( ME = MF = s - AC = (b + y) - b = y )Wait, so from ( A ), ( AF = AD = b - y ). But earlier, I denoted ( AF = AD = x ), so ( x = b - y ).But also, ( AC = AD + DC = x + y = (b - y) + y = b ), which is consistent.Similarly, ( CM = CE + EM = y + y = 2y ).And ( AM = AF + FM = (b - y) + y = b ), which is consistent with ( AM = AC = b ).So, everything checks out so far.Now, since triangle ( ACM ) has sides ( AC = AM = b ) and ( CM = 2y ), and it's isoceles with ( AC = AM ).But wait, if ( ACM ) is isoceles with ( AC = AM ), then the base angles at ( C ) and ( M ) are equal.But in triangle ( ACM ), the angles at ( C ) and ( M ) are equal.But in triangle ( ABC ), which is right-angled at ( C ), ( CM ) is the median to the hypotenuse ( AB ), so ( CM = frac{1}{2} AB ).Wait, that's a property of right-angled triangles: the median to the hypotenuse is equal to half the hypotenuse.So, ( CM = frac{1}{2} AB ).But in triangle ( ACM ), ( CM = 2y ), so ( AB = 2 CM = 4y ).But ( AB ) is the hypotenuse of triangle ( ABC ), so ( AB = sqrt{AC^2 + BC^2} ).But we don't know ( BC ) yet.Wait, maybe I can express ( AC ) in terms of ( y ). Since ( AC = b ), and ( b = AC = AM ), and ( AM ) is a side of triangle ( ACM ).But perhaps I can find the lengths in terms of ( y ).Wait, let me think differently. Since triangle ( ACM ) is isoceles with ( AC = AM = b ), and ( CM = 2y ), then by the Law of Cosines in triangle ( ACM ):( CM^2 = AC^2 + AM^2 - 2 cdot AC cdot AM cdot cos angle CAM )But since ( AC = AM = b ), this simplifies to:( (2y)^2 = b^2 + b^2 - 2b^2 cos angle CAM )So,( 4y^2 = 2b^2 - 2b^2 cos angle CAM )Divide both sides by 2:( 2y^2 = b^2 - b^2 cos angle CAM )So,( 2y^2 = b^2 (1 - cos angle CAM) )Hmm, not sure if that helps directly. Maybe I should consider the inradius.Wait, the inradius ( r ) of a triangle is given by ( r = frac{A}{s} ), where ( A ) is the area and ( s ) is the semiperimeter.In triangle ( ACM ), the inradius is the distance from the incenter to each side. Since the incircle touches ( CM ) at its midpoint, which is ( E ), and ( CE = y ), then the inradius ( r = CE = y ).Wait, is that correct? Wait, the inradius is the distance from the incenter to the side, which is the length of the perpendicular from the incenter to the side. But in this case, since the incircle touches ( CM ) at its midpoint, which is ( E ), and ( CE = y ), does that mean the inradius is ( y )?Wait, actually, the inradius is the distance from the incenter to the side, which is the same as the length of the tangent from the vertex to the point of tangency. But in this case, since the incircle touches ( CM ) at its midpoint, which is ( E ), and ( CE = y ), then the inradius ( r = CE = y ).Wait, but in general, the inradius is not necessarily equal to the length from the vertex to the point of tangency unless the triangle is equilateral or something. Hmm, maybe I need to think again.Wait, no, the inradius is the distance from the incenter to the side, which is the same as the length of the perpendicular from the incenter to the side. The length from the vertex to the point of tangency is not necessarily equal to the inradius unless the triangle is equilateral.So, perhaps I was wrong earlier. Let me correct that.In triangle ( ACM ), the inradius ( r ) is the distance from the incenter to each side, which is the same for all sides. The point where the incircle touches ( CM ) is at its midpoint ( E ), so the distance from the incenter to ( CM ) is ( r ), which is equal to the length of the perpendicular from the incenter to ( CM ).But since ( E ) is the midpoint of ( CM ), and the incircle touches ( CM ) at ( E ), the inradius ( r ) is equal to the distance from the incenter to ( CM ), which is the same as the length of the perpendicular from the incenter to ( CM ).But since ( E ) is the midpoint, and the incircle touches ( CM ) at ( E ), the inradius ( r ) is equal to the distance from the incenter to ( CM ), which is the same as the length of the perpendicular from the incenter to ( CM ).Wait, maybe I can use coordinates to model this problem. Let me try that.Let me place triangle ( ABC ) on a coordinate system with point ( C ) at the origin ( (0, 0) ), point ( A ) on the x-axis at ( (a, 0) ), and point ( B ) on the y-axis at ( (0, b) ). Then, the hypotenuse ( AB ) goes from ( (a, 0) ) to ( (0, b) ), and its midpoint ( M ) is at ( left( frac{a}{2}, frac{b}{2} right) ).So, the median ( CM ) goes from ( (0, 0) ) to ( left( frac{a}{2}, frac{b}{2} right) ).Now, triangle ( ACM ) has vertices at ( A (a, 0) ), ( C (0, 0) ), and ( M left( frac{a}{2}, frac{b}{2} right) ).We need to find the inradius of triangle ( ACM ) and the point where it touches ( CM ). According to the problem, this point is the midpoint of ( CM ), which is ( E left( frac{a}{4}, frac{b}{4} right) ).Wait, because ( CM ) goes from ( (0, 0) ) to ( left( frac{a}{2}, frac{b}{2} right) ), so its midpoint is indeed ( left( frac{a}{4}, frac{b}{4} right) ).So, the incircle of triangle ( ACM ) touches ( CM ) at ( E left( frac{a}{4}, frac{b}{4} right) ).Now, let's find the inradius and the coordinates of the incenter of triangle ( ACM ).First, let's compute the lengths of the sides of triangle ( ACM ).- Side ( AC ): from ( (a, 0) ) to ( (0, 0) ), length is ( a ).- Side ( CM ): from ( (0, 0) ) to ( left( frac{a}{2}, frac{b}{2} right) ), length is ( sqrt{left( frac{a}{2} right)^2 + left( frac{b}{2} right)^2} = frac{1}{2} sqrt{a^2 + b^2} ).- Side ( AM ): from ( (a, 0) ) to ( left( frac{a}{2}, frac{b}{2} right) ), length is ( sqrt{left( frac{a}{2} right)^2 + left( frac{b}{2} right)^2} = frac{1}{2} sqrt{a^2 + b^2} ).Wait, so sides ( CM ) and ( AM ) are equal in length, each being ( frac{1}{2} sqrt{a^2 + b^2} ). So, triangle ( ACM ) is isoceles with ( AC = a ) and the other two sides equal.Wait, but earlier, I thought ( AC = AM ), but actually, ( AC = a ), and ( AM = frac{1}{2} sqrt{a^2 + b^2} ). So, unless ( a = frac{1}{2} sqrt{a^2 + b^2} ), which would imply ( 2a = sqrt{a^2 + b^2} ), squaring both sides: ( 4a^2 = a^2 + b^2 ), so ( 3a^2 = b^2 ), so ( b = a sqrt{3} ).Wait, that's interesting. So, if ( AC = AM ), then ( b = a sqrt{3} ). But is that necessarily the case here?Wait, no, because in triangle ( ACM ), sides ( CM ) and ( AM ) are equal, making it isoceles with base ( AC ). So, the base is ( AC = a ), and the equal sides are ( CM = AM = frac{1}{2} sqrt{a^2 + b^2} ).But in our problem, the incircle touches ( CM ) at its midpoint ( E ). So, perhaps we can use this condition to find a relationship between ( a ) and ( b ).Let me recall that in a triangle, the distance from the incenter to a side is equal to the inradius ( r ). The inradius can also be expressed as ( r = frac{A}{s} ), where ( A ) is the area of the triangle and ( s ) is the semiperimeter.So, let's compute the semiperimeter ( s ) of triangle ( ACM ):( s = frac{AC + CM + AM}{2} = frac{a + frac{1}{2} sqrt{a^2 + b^2} + frac{1}{2} sqrt{a^2 + b^2}}{2} = frac{a + sqrt{a^2 + b^2}}{2} ).Now, the area ( A ) of triangle ( ACM ) can be found using coordinates. The coordinates of the vertices are ( A (a, 0) ), ( C (0, 0) ), and ( M left( frac{a}{2}, frac{b}{2} right) ).Using the shoelace formula:( A = frac{1}{2} | (a cdot 0 - 0 cdot frac{b}{2}) + (0 cdot frac{b}{2} - frac{a}{2} cdot 0) + (frac{a}{2} cdot 0 - a cdot frac{b}{2}) | )Simplify:( A = frac{1}{2} | 0 + 0 + (- frac{ab}{2}) | = frac{1}{2} cdot frac{ab}{2} = frac{ab}{4} ).So, the area is ( frac{ab}{4} ).Therefore, the inradius ( r = frac{A}{s} = frac{frac{ab}{4}}{frac{a + sqrt{a^2 + b^2}}{2}} = frac{ab}{4} cdot frac{2}{a + sqrt{a^2 + b^2}} = frac{ab}{2(a + sqrt{a^2 + b^2})} ).Now, the inradius ( r ) is also equal to the distance from the incenter to side ( CM ). Since the incircle touches ( CM ) at its midpoint ( E left( frac{a}{4}, frac{b}{4} right) ), the inradius ( r ) is the distance from the incenter to ( E ).Wait, actually, the inradius is the distance from the incenter to any side, which is the same as the length of the perpendicular from the incenter to that side. Since the incircle touches ( CM ) at ( E ), the inradius ( r ) is the distance from the incenter to ( CM ), which is the same as the length of the perpendicular from the incenter to ( CM ).But since ( E ) is the point of tangency, the incenter lies along the angle bisector of angle ( C ) in triangle ( ACM ). Wait, no, in triangle ( ACM ), the incenter is located at the intersection of the angle bisectors.Alternatively, perhaps I can find the coordinates of the incenter.The incenter of a triangle can be found as the weighted average of the vertices' coordinates, weighted by the lengths of the sides opposite those vertices.In triangle ( ACM ), the incenter ( I ) has coordinates:( I_x = frac{AC cdot x_M + CM cdot x_A + AM cdot x_C}{AC + CM + AM} )( I_y = frac{AC cdot y_M + CM cdot y_A + AM cdot y_C}{AC + CM + AM} )Wait, no, the formula is:The incenter coordinates are given by:( left( frac{a x_A + b x_B + c x_C}{a + b + c}, frac{a y_A + b y_B + c y_C}{a + b + c} right) )where ( a, b, c ) are the lengths of the sides opposite to vertices ( A, B, C ) respectively.Wait, in triangle ( ACM ), the sides are:- Opposite to ( A ): ( CM = frac{1}{2} sqrt{a^2 + b^2} )- Opposite to ( C ): ( AM = frac{1}{2} sqrt{a^2 + b^2} )- Opposite to ( M ): ( AC = a )So, the incenter coordinates would be:( I_x = frac{CM cdot x_A + AM cdot x_C + AC cdot x_M}{CM + AM + AC} )( I_y = frac{CM cdot y_A + AM cdot y_C + AC cdot y_M}{CM + AM + AC} )Plugging in the values:( I_x = frac{left( frac{1}{2} sqrt{a^2 + b^2} right) cdot a + left( frac{1}{2} sqrt{a^2 + b^2} right) cdot 0 + a cdot frac{a}{2}}{frac{1}{2} sqrt{a^2 + b^2} + frac{1}{2} sqrt{a^2 + b^2} + a} )Simplify numerator:( frac{1}{2} sqrt{a^2 + b^2} cdot a + 0 + a cdot frac{a}{2} = frac{a}{2} sqrt{a^2 + b^2} + frac{a^2}{2} )Denominator:( frac{1}{2} sqrt{a^2 + b^2} + frac{1}{2} sqrt{a^2 + b^2} + a = sqrt{a^2 + b^2} + a )So,( I_x = frac{frac{a}{2} sqrt{a^2 + b^2} + frac{a^2}{2}}{sqrt{a^2 + b^2} + a} )Similarly,( I_y = frac{left( frac{1}{2} sqrt{a^2 + b^2} right) cdot 0 + left( frac{1}{2} sqrt{a^2 + b^2} right) cdot 0 + a cdot frac{b}{2}}{sqrt{a^2 + b^2} + a} = frac{frac{ab}{2}}{sqrt{a^2 + b^2} + a} )So, the incenter ( I ) has coordinates:( I_x = frac{frac{a}{2} (sqrt{a^2 + b^2} + a)}{sqrt{a^2 + b^2} + a} = frac{a}{2} )( I_y = frac{frac{ab}{2}}{sqrt{a^2 + b^2} + a} )Simplify ( I_x ):( I_x = frac{a}{2} )That's interesting, the x-coordinate of the incenter is ( frac{a}{2} ).Now, the inradius ( r ) is the distance from the incenter ( I ) to side ( CM ). Since ( CM ) is the line from ( (0, 0) ) to ( left( frac{a}{2}, frac{b}{2} right) ), its equation can be found.The slope of ( CM ) is ( frac{frac{b}{2} - 0}{frac{a}{2} - 0} = frac{b}{a} ). So, the equation of ( CM ) is ( y = frac{b}{a} x ).The distance from point ( I left( frac{a}{2}, frac{ab}{2(sqrt{a^2 + b^2} + a)} right) ) to the line ( y = frac{b}{a} x ) is given by:( r = frac{| frac{b}{a} cdot frac{a}{2} - frac{ab}{2(sqrt{a^2 + b^2} + a)} |}{sqrt{left( frac{b}{a} right)^2 + (-1)^2}} )Simplify numerator:( | frac{b}{2} - frac{ab}{2(sqrt{a^2 + b^2} + a)} | )Factor out ( frac{b}{2} ):( frac{b}{2} left| 1 - frac{a}{sqrt{a^2 + b^2} + a} right| )Simplify the expression inside the absolute value:( 1 - frac{a}{sqrt{a^2 + b^2} + a} = frac{(sqrt{a^2 + b^2} + a) - a}{sqrt{a^2 + b^2} + a} = frac{sqrt{a^2 + b^2}}{sqrt{a^2 + b^2} + a} )So, numerator becomes:( frac{b}{2} cdot frac{sqrt{a^2 + b^2}}{sqrt{a^2 + b^2} + a} )Denominator of the distance formula:( sqrt{left( frac{b^2}{a^2} right) + 1} = sqrt{frac{b^2 + a^2}{a^2}}} = frac{sqrt{a^2 + b^2}}{a} )So, the distance ( r ) is:( r = frac{frac{b}{2} cdot frac{sqrt{a^2 + b^2}}{sqrt{a^2 + b^2} + a}}{frac{sqrt{a^2 + b^2}}{a}} = frac{b}{2} cdot frac{sqrt{a^2 + b^2}}{sqrt{a^2 + b^2} + a} cdot frac{a}{sqrt{a^2 + b^2}} )Simplify:( r = frac{ab}{2(sqrt{a^2 + b^2} + a)} )Which matches our earlier expression for ( r ).But we also know that the incircle touches ( CM ) at its midpoint ( E left( frac{a}{4}, frac{b}{4} right) ). So, the distance from the incenter ( I ) to ( E ) should be equal to the inradius ( r ).Wait, no, the inradius is the distance from the incenter to the side, which is the same as the distance from the incenter to the point of tangency ( E ). So, the distance between ( I ) and ( E ) should be equal to ( r ).Let me compute the distance between ( I left( frac{a}{2}, frac{ab}{2(sqrt{a^2 + b^2} + a)} right) ) and ( E left( frac{a}{4}, frac{b}{4} right) ):Distance squared:( left( frac{a}{2} - frac{a}{4} right)^2 + left( frac{ab}{2(sqrt{a^2 + b^2} + a)} - frac{b}{4} right)^2 )Simplify:( left( frac{a}{4} right)^2 + left( frac{ab}{2(sqrt{a^2 + b^2} + a)} - frac{b}{4} right)^2 )Compute each term:First term: ( left( frac{a}{4} right)^2 = frac{a^2}{16} )Second term:Let me find a common denominator:( frac{ab}{2(sqrt{a^2 + b^2} + a)} - frac{b}{4} = frac{2ab - b(sqrt{a^2 + b^2} + a)}{4(sqrt{a^2 + b^2} + a)} )Simplify numerator:( 2ab - bsqrt{a^2 + b^2} - ab = ab - bsqrt{a^2 + b^2} )So,( frac{ab - bsqrt{a^2 + b^2}}{4(sqrt{a^2 + b^2} + a)} = frac{b(a - sqrt{a^2 + b^2})}{4(sqrt{a^2 + b^2} + a)} )Factor out ( -1 ) from numerator:( frac{-b(sqrt{a^2 + b^2} - a)}{4(sqrt{a^2 + b^2} + a)} )Notice that ( (sqrt{a^2 + b^2} - a)(sqrt{a^2 + b^2} + a) = a^2 + b^2 - a^2 = b^2 )So,( frac{-b(sqrt{a^2 + b^2} - a)}{4(sqrt{a^2 + b^2} + a)} = frac{-b cdot frac{b^2}{sqrt{a^2 + b^2} + a}}{4(sqrt{a^2 + b^2} + a)} = frac{-b^3}{4(sqrt{a^2 + b^2} + a)^2} )Wait, that seems complicated. Maybe I made a mistake in simplifying.Wait, let's go back:We have:( frac{ab}{2(sqrt{a^2 + b^2} + a)} - frac{b}{4} = frac{2ab - b(sqrt{a^2 + b^2} + a)}{4(sqrt{a^2 + b^2} + a)} )Simplify numerator:( 2ab - bsqrt{a^2 + b^2} - ab = ab - bsqrt{a^2 + b^2} )So,( frac{ab - bsqrt{a^2 + b^2}}{4(sqrt{a^2 + b^2} + a)} = frac{b(a - sqrt{a^2 + b^2})}{4(sqrt{a^2 + b^2} + a)} )Factor out ( -1 ):( frac{-b(sqrt{a^2 + b^2} - a)}{4(sqrt{a^2 + b^2} + a)} )Now, multiply numerator and denominator by ( (sqrt{a^2 + b^2} - a) ):( frac{-b(sqrt{a^2 + b^2} - a)^2}{4(sqrt{a^2 + b^2} + a)(sqrt{a^2 + b^2} - a)} )Denominator becomes:( 4(sqrt{a^2 + b^2}^2 - a^2) = 4(b^2) )So,( frac{-b(sqrt{a^2 + b^2} - a)^2}{4b^2} = frac{-(sqrt{a^2 + b^2} - a)^2}{4b} )Therefore, the second term squared is:( left( frac{-(sqrt{a^2 + b^2} - a)^2}{4b} right)^2 = frac{(sqrt{a^2 + b^2} - a)^4}{16b^2} )So, the distance squared between ( I ) and ( E ) is:( frac{a^2}{16} + frac{(sqrt{a^2 + b^2} - a)^4}{16b^2} )But this distance should be equal to ( r^2 ), which is:( r^2 = left( frac{ab}{2(sqrt{a^2 + b^2} + a)} right)^2 = frac{a^2b^2}{4(sqrt{a^2 + b^2} + a)^2} )So, we have:( frac{a^2}{16} + frac{(sqrt{a^2 + b^2} - a)^4}{16b^2} = frac{a^2b^2}{4(sqrt{a^2 + b^2} + a)^2} )This equation seems quite complicated. Maybe there's a simpler approach.Wait, earlier, I found that in triangle ( ACM ), ( AC = AM ) if ( b = a sqrt{3} ). Maybe that's the case here because the incircle touches ( CM ) at its midpoint, which might impose that triangle ( ACM ) is equilateral or something similar.Wait, if ( ACM ) is equilateral, then all sides are equal, so ( AC = CM = AM ). But ( AC = a ), and ( CM = frac{1}{2} sqrt{a^2 + b^2} ). So, if ( AC = CM ), then:( a = frac{1}{2} sqrt{a^2 + b^2} )Multiply both sides by 2:( 2a = sqrt{a^2 + b^2} )Square both sides:( 4a^2 = a^2 + b^2 )So,( 3a^2 = b^2 )Thus,( b = a sqrt{3} )So, if ( b = a sqrt{3} ), then triangle ( ABC ) has legs ( AC = a ) and ( BC = a sqrt{3} ), making it a 30-60-90 triangle.Indeed, in a 30-60-90 triangle, the sides are in the ratio ( 1 : sqrt{3} : 2 ), so the angles are ( 30^circ, 60^circ, 90^circ ).But does this satisfy the condition that the incircle of ( ACM ) touches ( CM ) at its midpoint?Let me check.If ( b = a sqrt{3} ), then ( CM = frac{1}{2} sqrt{a^2 + b^2} = frac{1}{2} sqrt{a^2 + 3a^2} = frac{1}{2} sqrt{4a^2} = frac{1}{2} cdot 2a = a ).So, ( CM = a ), and ( AC = a ), so triangle ( ACM ) is isoceles with ( AC = CM = a ), and ( AM = a ) as well, making it equilateral.Wait, if ( ACM ) is equilateral, then all its sides are equal, so ( AC = CM = AM = a ). Therefore, the incircle would touch each side at its midpoint, which is consistent with the problem statement.Therefore, the condition is satisfied when ( b = a sqrt{3} ), making triangle ( ABC ) a 30-60-90 triangle.Thus, the angles of triangle ( ABC ) are ( 30^circ, 60^circ, ) and ( 90^circ ).**Final Answer**The angles of triangle ( ABC ) are (boxed{30^circ}), (boxed{60^circ}), and (boxed{90^circ}).

answer：Okay, so I have this problem about a geometric sequence. Let me try to figure it out step by step. First, the problem says that in the geometric sequence {a_n}, where each term is positive, the expression a₂a₄ + 2a₃a₅ + a₄a₆ equals 25. I need to find the value of a₃ + a₅, and the options are A:5, B:10, C:15, D:20.Alright, since it's a geometric sequence, I remember that each term is multiplied by a common ratio r. So, the nth term can be written as a_n = a₁ * r^(n-1). That means:a₂ = a₁ * ra₃ = a₁ * r²a₄ = a₁ * r³a₅ = a₁ * r⁴a₆ = a₁ * r⁵So, let me write down each of these terms:a₂ = a₁ * ra₄ = a₁ * r³a₃ = a₁ * r²a₅ = a₁ * r⁴a₆ = a₁ * r⁵Now, the expression given is a₂a₄ + 2a₃a₅ + a₄a₆. Let me substitute each term with the expressions in terms of a₁ and r.First, a₂a₄ = (a₁ * r) * (a₁ * r³) = a₁² * r⁴Then, 2a₃a₅ = 2 * (a₁ * r²) * (a₁ * r⁴) = 2 * a₁² * r⁶Next, a₄a₆ = (a₁ * r³) * (a₁ * r⁵) = a₁² * r⁸So, putting it all together, the expression becomes:a₁² * r⁴ + 2 * a₁² * r⁶ + a₁² * r⁸ = 25Hmm, that looks like a quadratic in terms of a₁² * r⁴. Let me see. Let me factor out a₁² * r⁴:a₁² * r⁴ (1 + 2r² + r⁴) = 25Wait, 1 + 2r² + r⁴ is actually (r² + 1)². Let me check:(r² + 1)² = r⁴ + 2r² + 1. Yes, that's correct.So, substituting that in, we have:a₁² * r⁴ * (r² + 1)² = 25Hmm, that's a bit complicated. Maybe I can express this differently. Let me think about the terms a₃ and a₅.a₃ = a₁ * r²a₅ = a₁ * r⁴So, a₃ + a₅ = a₁ * r² + a₁ * r⁴ = a₁ r² (1 + r²)Wait, that's interesting. Let me denote S = a₃ + a₅ = a₁ r² (1 + r²)Looking back at the expression we had earlier:a₁² * r⁴ * (1 + r²)² = 25Notice that (a₁ r² (1 + r²))² = (a₃ + a₅)² = S²So, S² = 25Since all terms are positive, S must be positive. Therefore, S = 5.So, a₃ + a₅ = 5, which is option A.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Expressed each term in terms of a₁ and r: correct.2. Calculated a₂a₄, 2a₃a₅, a₄a₆: correct.3. Factored out a₁² r⁴: correct.4. Recognized 1 + 2r² + r⁴ as (r² + 1)²: correct.5. Noted that a₃ + a₅ = a₁ r² (1 + r²): correct.6. Therefore, (a₃ + a₅)² = 25, so a₃ + a₅ = 5: correct.Yes, everything seems to check out. I think the answer is A:5.**Final Answer**boxed{A}

answer：Alright, so I've got these two inequalities to solve, and I need to figure out the basis for transforming them. Let's take them one at a time.Starting with the first one: If ( x - 1 > 2 ), then ( x > 3 ). Okay, so I need to understand what transformation is happening here to get from ( x - 1 > 2 ) to ( x > 3 ). Hmm, it looks like they're trying to isolate ( x ) in the inequality. To do that, I guess I need to get rid of the (-1) that's being subtracted from ( x ). So, if I have ( x - 1 > 2 ), and I want to get ( x ) by itself, I should probably add 1 to both sides of the inequality. That makes sense because addition is the inverse operation of subtraction. Let me try that:( x - 1 + 1 > 2 + 1 )Simplifying both sides:( x > 3 )Okay, that works out. So, the transformation here is adding 1 to both sides of the inequality to isolate ( x ). That seems straightforward.Now, moving on to the second inequality: If ( -4x > 8 ), then ( x < -2 ). This one looks a bit trickier because there's a negative coefficient in front of ( x ). I remember that when you multiply or divide both sides of an inequality by a negative number, the direction of the inequality sign flips. Let me see if that applies here.Starting with ( -4x > 8 ), I want to solve for ( x ), so I need to get rid of the (-4) that's being multiplied by ( x ). To do that, I should divide both sides by (-4). But wait, since I'm dividing by a negative number, I need to reverse the inequality sign. Let's do that:( frac{-4x}{-4} < frac{8}{-4} )Simplifying both sides:( x < -2 )Alright, that makes sense. The transformation here is dividing both sides by (-4) and flipping the inequality sign because we're dividing by a negative number.So, to summarize:1. For the first inequality, ( x - 1 > 2 ), the transformation is adding 1 to both sides to isolate ( x ), resulting in ( x > 3 ).2. For the second inequality, ( -4x > 8 ), the transformation is dividing both sides by (-4) and reversing the inequality sign, resulting in ( x < -2 ).I think that covers the basis for the transformations in both inequalities. It's important to remember that when dealing with inequalities, adding or subtracting the same number from both sides doesn't change the direction of the inequality, but multiplying or dividing by a negative number does flip the inequality sign. That was a bit confusing at first, but now it makes sense.I wonder if there are any exceptions or special cases where these rules don't apply. Maybe if the coefficient is zero or something? But in this problem, the coefficients are non-zero, so these transformations should work fine. It's also interesting how similar these operations are to solving equations, except for that crucial step when dealing with negative coefficients in inequalities.Overall, understanding these transformations helps in solving inequalities more systematically. It's like having a set of rules to follow, which makes the process less error-prone. I should practice more problems like this to get even more comfortable with the concepts.

answer：Okay, so I have this problem where I need to draw a regular hexagon, then make a square from each edge of the hexagon, and then form equilateral triangles by drawing an edge between every pair of neighboring squares. Then, I have to figure out the ratio between the number of triangles and the number of squares if this figure is continued symmetrically off to infinity. Hmm, that sounds a bit complex, but let me try to break it down step by step.First, I need to visualize a regular hexagon. A regular hexagon has six equal sides and six equal angles. Now, from each edge of this hexagon, I have to make a square. So, if I imagine each side of the hexagon, I'll attach a square to it. Since the hexagon has six sides, I'll end up with six squares attached to it.Next, the problem says to form equilateral triangles by drawing an edge between every pair of neighboring squares. So, each square is attached to the hexagon, and each square has a neighboring square. If I connect these neighboring squares with an edge, that should form an equilateral triangle. Since there are six squares, each connected to its neighbor, I should get six equilateral triangles.Now, the figure is continued symmetrically off to infinity. That means this pattern of adding squares and triangles continues indefinitely. I need to figure out the ratio of triangles to squares in this infinite figure.Let me think about how this pattern grows. Starting with the hexagon, I add six squares and six triangles. Now, if I continue this process, each new layer will involve adding more squares and triangles. But how does the number of squares and triangles grow with each layer?Maybe I can think of it as layers around the hexagon. The first layer has six squares and six triangles. If I go to the next layer, each square from the first layer will have a new square attached to it, and similarly, new triangles will be formed between these new squares. But wait, each square is attached to the hexagon, so adding a new square to each existing square might not be straightforward.Alternatively, perhaps each side of the hexagon leads to a chain of squares and triangles extending outward. So, for each side, there's a square, and then another square attached to that square, and so on, infinitely. Similarly, between each pair of neighboring squares, there's a triangle.If that's the case, then for each side of the hexagon, there's an infinite number of squares and triangles extending outward. But since the hexagon has six sides, the total number of squares and triangles would be six times the number in each chain.But wait, if it's continued symmetrically off to infinity, the number of squares and triangles would both be infinite. However, the ratio between them might still be finite because both are countably infinite.Let me try to think of the ratio in terms of the number of squares and triangles per unit length or per unit area, but since it's infinite, that might not be the right approach.Perhaps I can consider the pattern in terms of repeating units. Each square is connected to two triangles, one on each side. So, for each square, there are two triangles associated with it. But that might not be accurate because each triangle is shared between two squares.Wait, no. Each triangle is formed between two neighboring squares. So, each triangle is associated with two squares. Therefore, the number of triangles is equal to the number of squares because each triangle corresponds to a pair of neighboring squares.But in the initial step, we have six squares and six triangles, so the ratio is 1:1. If this pattern continues, each new layer adds the same number of squares and triangles, maintaining the ratio of 1:1.Let me test this with the next layer. If I add another layer of squares and triangles, starting from the first six squares, each square will have a new square attached to it, and between each pair of new squares, there will be a new triangle. So, for each existing square, we add a new square and a new triangle. Since there are six squares, we add six new squares and six new triangles. So, the total becomes 12 squares and 12 triangles, maintaining the ratio of 1:1.Continuing this process, each layer adds six squares and six triangles, keeping the ratio consistent. Therefore, as the figure extends to infinity, the ratio between the number of triangles and the number of squares remains 1:1.I think that makes sense. Each square is associated with one triangle, and each triangle is shared between two squares, but overall, the count of squares and triangles grows at the same rate, resulting in a ratio of 1:1.

answer：Okay, so I need to find the smallest positive integer that ends with a 9 and is divisible by 13. Hmm, let's break this down. First, the number has to end with 9, so it's like 9, 19, 29, 39, and so on. I guess I can list these numbers and check which one is divisible by 13. But that might take a while if I have to go through each one. Maybe there's a smarter way to do this.Let me think about divisibility by 13. I remember that 13 times 1 is 13, 13 times 2 is 26, 13 times 3 is 39, and so on. Wait, 39 ends with a 9? No, 39 ends with a 9? Wait, 39 is 3 times 13, right? So 39 ends with a 9? No, 39 ends with a 9? Wait, 39 is 3 times 13, which is 39, and 39 does end with a 9. So is 39 the answer? But wait, 39 is smaller than 129, which was the previous answer. Did I make a mistake earlier?Wait, let me double-check. 39 divided by 13 is 3, which is an integer, so 39 is divisible by 13. And it ends with a 9. So why did the previous answer say 129? Maybe I did something wrong before. Let me go back.In the previous thought process, I started checking from 9, then 19, 29, 39, and so on, up to 129. But if 39 is divisible by 13 and ends with a 9, then 39 should be the correct answer. So why did I go all the way to 129? Maybe I made a mistake in my initial checks.Let me verify 39. 39 divided by 13 is 3, which is correct. So 39 is indeed divisible by 13 and ends with a 9. Therefore, 39 is the smallest such number. So the previous answer of 129 must be incorrect. I must have made a mistake in my initial reasoning.Wait, maybe I confused the problem with another one. Let me make sure. The problem is to find the smallest positive integer that ends with 9 and is divisible by 13. So 39 fits both criteria. It's smaller than 129, so it should be the correct answer.But then why did I think 129 was the answer before? Maybe I was thinking of a different problem where the number had to be two digits or something. But no, the problem just says the smallest positive integer, so 39 is valid.Let me try another approach to confirm. If I represent the number as 10k + 9, where k is a non-negative integer, then I need 10k + 9 to be divisible by 13. So, 10k + 9 ≡ 0 mod 13. That means 10k ≡ -9 mod 13. Since -9 mod 13 is the same as 4 mod 13 (because 13 - 9 = 4), so 10k ≡ 4 mod 13.Now, I need to solve for k in this congruence. 10k ≡ 4 mod 13. To solve for k, I can multiply both sides by the modular inverse of 10 mod 13. The inverse of 10 mod 13 is a number m such that 10m ≡ 1 mod 13. Let's find m.Trying m=4: 10*4=40, 40 mod 13 is 1 (since 13*3=39, 40-39=1). So m=4 is the inverse. Therefore, k ≡ 4*4 ≡ 16 ≡ 3 mod 13. So k ≡ 3 mod 13. That means the smallest k is 3. Therefore, the number is 10*3 + 9 = 39.So, this confirms that 39 is indeed the smallest positive integer ending with 9 and divisible by 13. I must have made a mistake in my initial approach by not checking 39 properly. It's important to verify each step carefully to avoid errors.

answer：Okay, so I'm trying to figure out this bacteria growth problem. Let me read it again to make sure I understand what's being asked."A certain bacteria colony grows at a specific rate every day for 22 days, at which point it reaches the limit of its habitat and can no longer grow. If two bacteria colonies start growing simultaneously, they will take 21 days to reach the habitat's limit. What is the growth rate of the bacteria colony?"Alright, so there's one colony that takes 22 days to reach the habitat's limit, and if you have two colonies starting at the same time, they reach the limit in 21 days. I need to find the daily growth rate.First, I think I need to model the growth of the bacteria colony. Since it's growing at a specific rate every day, it sounds like exponential growth. So, maybe I can use the formula for exponential growth:[ N(t) = N_0 times r^t ]Where:- ( N(t) ) is the population at time t,- ( N_0 ) is the initial population,- ( r ) is the growth rate per day,- ( t ) is the time in days.Okay, so for one colony, it takes 22 days to reach the habitat's limit. Let's call that limit ( K ). So, at day 22:[ K = N_0 times r^{22} ]Now, if there are two colonies starting at the same time, their combined growth would be:[ 2 times N_0 times r^t ]And this combined growth reaches the same limit ( K ) in 21 days. So:[ K = 2 times N_0 times r^{21} ]Now, I have two equations:1. ( K = N_0 times r^{22} )2. ( K = 2 times N_0 times r^{21} )Since both equal ( K ), I can set them equal to each other:[ N_0 times r^{22} = 2 times N_0 times r^{21} ]Hmm, I can cancel out ( N_0 ) from both sides because it's the same initial population:[ r^{22} = 2 times r^{21} ]Now, I can divide both sides by ( r^{21} ) to simplify:[ r = 2 ]Wait, so the growth rate ( r ) is 2? That means the bacteria colony doubles every day. Let me check if that makes sense.If one colony doubles every day, starting from ( N_0 ), after 22 days it would be ( N_0 times 2^{22} ), which is the limit ( K ).If two colonies start at the same time, each doubling every day, their combined population would be ( 2 times N_0 times 2^{21} ). Simplifying that:[ 2 times N_0 times 2^{21} = N_0 times 2^{22} ]Which is the same as the limit ( K ). So, yes, it takes 21 days for two colonies to reach ( K ) because their combined growth effectively adds an extra day of growth.That seems to check out. So, the growth rate ( r ) is indeed 2 per day.I think that's the answer. The bacteria colony grows at a rate of doubling every day.