answer:I. megoldás: 1. **Visualize the Problem:** Let's assume the triangle is already constructed and label the vertices as shown in the reference figure. 2. **Extend Side BC and Construct Equal Sides:** Extend the side BC = a beyond point B to A_1 such that BA_1 = BA = c. Similarly, extend BC beyond point C to A_2 such that CA_2 = CA = b. 3. **Angles in Isosceles Triangles:** In the isosceles triangles BAA_1 and CAA_2, the angles at the base are frac{beta}{2} and frac{gamma}{2} respectively, because in any triangle, the sum of the two interior angles is equal to the exterior angle opposite to those two angles. 4. **Construct Triangle AA_1A_2:** In the triangle AA_1A_2, side A_1A_2 is equal to the perimeter k of the original triangle, and the angle at A is: [ A = alpha + frac{beta}{2} + frac{gamma}{2} = frac{2alpha + beta + gamma}{2} = frac{180^circ + alpha}{2} = 90^circ + frac{alpha}{2}. ] 5. **Location of Point A:** Point A is on a circular arc above A_1A_2 = k such that the section A_1A_2 is seen under an angle of 90^circ + frac{alpha}{2} from A. Moreover, it is on a line parallel to A_1A_2 at a distance m_a from it. 6. **Construction Steps:** - Draw the segment A_1A_2 = k. - Construct the two mentioned geometric loci: the circular arc and the parallel line at distance m_a. - Their intersection point A (and possibly A^*) is a vertex of the sought triangle. - Bisect segments AA_1 and AA_2 perpendicularly to obtain the points B and C on A_1A_2. 7. **Multiplicity of Solutions:** Only one solution exists as A^* is the mirror image of A concerning the perpendicular bisector of A_1A_2. 8. **Condition for Constructibility:** [ m_a leq MN = ON - OM = OA_1 - OM = frac{A_1M}{cos frac{alpha}{2}} - A_1M tan frac{alpha}{2} = frac{k}{2 cos frac{alpha}{2}} - frac{k sin frac{alpha}{2}}{2 cos frac{alpha}{2}} = frac{k}{2} cdot frac{1 - sin frac{alpha}{2}}{cos frac{alpha}{2}}. ] If the equality holds, we obtain an isosceles triangle. II. megoldás: 1. **Understanding Excircle Tangent Points:** It is known that the points where the excircle touches the extensions of the sides of a triangle are at a distance of half the perimeter from the tangents' intersection. 2. **Constructing the Circle:** Between the given angle alpha, construct a circle that is tangent to the angle's legs at points E_1 and E_2 which are frac{k}{2} distance away from vertex A. 3. **Intersection of Circles:** - Draw a circle around point A with radius m_a. - The intersection between the constructed circle and this circle, marked as A_1 and E, will determine the locations of points B and C on the angle legs. 4. **Verification:** Check that the triangle formed satisfies the given conditions: [ text{Perimeter } = AB + BE + EC + CA = AB + BE_1 + E_2C + CA = AE_1 + AE_2 = frac{k}{2} + frac{k}{2} = k, ] and that the height from vertex A to side BC is indeed m_a. 5. **Condition for Constructibility:** Ensure that the two circles do not intersect at more than one point: [ m_a leq AO - OE_1 = frac{k}{2 cos frac{alpha}{2}} - frac{k sin frac{alpha}{2}}{2 cos frac{alpha}{2}}, ] Likewise to the first solution. (boxed{text{Final Answer}})

answer:We are given the equations: [ a + b + c + d = m^2 ] [ a^2 + b^2 + c^2 + d^2 = 1989 ] and the condition that the largest of (a, b, c, d) is (n^2). We need to determine the values of (m) and (n). 1. **Check the Parity of (a, b, c, d):** Since all numbers are perfect squares and their sum ( m^2 ) is odd, ( m ) must be an odd integer. 2. **Estimate Upper Bound for (m^2):** [ m^2 = a + b + c + d leq 4 cdot sqrt{frac{a^2 + b^2 + c^2 + d^2}{4}} = 2 sqrt{1989} ] [ 2 sqrt{1989} leq 2 cdot 45 = 90 quad Rightarrow quad m^2 < 90 ] Possible odd perfect squares less than 90 are: [m^2 in {1, 9, 25, 49, 81}] 3. **Lower Bound for (m^2):** From the given, we have: [ (a + b + c + d)^2 > a^2 + b^2 + c^2 + d^2 = 1989 quad Rightarrow quad m^4 > 1989 quad Rightarrow quad m^2 > 45 ] Thus possible values are: [m^2 in {49, 81}] 4. **Case ( m^2 = 49 ):** [ a + b + c + d = 49 ] The largest number among (a, b, c, d) is (d = n^2 Rightarrow d = 25 text{ or } 36). - For (d = 25), [ a + b + c = 49 - 25 = 24, quad a^2 + b^2 + c^2 = 1989 - 25^2 = 1364 ] [ (a + b + c)^2 = 24^2 = 576 < 1364 quad text{(Contradiction)} ] - For (d = 36), [ a + b + c = 49 - 36 = 13, quad a^2 + b^2 + c^2 = 1989 - 36^2 = 693 ] [ (a + b + c)^2 = 13^2 = 169 < 693 quad text{(Contradiction)} ] Therefore, ( m^2 neq 49 ). 5. **Case ( m^2 = 81 ):** [ a + b + c + d = 81 ] The largest number among (a, b, c, d) is (d = n^2 Rightarrow d = 25 text{ or } 36). - For (d = 25 rightarrow n = 5), [ a + b + c = 81 - 25 = 56, quad a^2 + b^2 + c^2 = 1989 - 25^2 = 1364 ] [ 4 mid 1364 quad Rightarrow a, b, c text{ are even numbers} quad a = 2a_1, b = 2b_1, c = 2c_1 ] [ a_1 + b_1 + c_1 = 28, quad a_1^2 + b_1^2 + c_1^2 = 341 ] But ( a_1, b_1, c_1 ) both being even and odd gives contradictions. - For (d = 36 rightarrow n = 6), [ a + b + c = 81 - 36 = 45, quad a^2 + b^2 + c^2 = 1989 - 36^2 = 693 ] Here ( 693 equiv 1 mod 4 ) implies: [ a = 2a_1, b = 2b_1, c = 2k-1 ] [ a_1 + b_1 + k = 23, quad a_1^2 + b_1^2 + (k-1)^2 = 173 ] Further solving, we get integers that satisfy all conditions. 6. **Verification:** - When ( n = 6 ), [ a = 12, quad b = 18, quad c = 15 quad Rightarrow a^2 + b^2 + c^2 + 36^2 = 1989 ] Conditions satisfied. # Conclusion: [ boxed{ (m, n) = (9, 6) } ]

answer:To solve the problem, we follow the temperature changes throughout the day step by step: 1. The temperature in the morning is 7^{circ}mathrm{C}. 2. It rises by 2^{circ}mathrm{C} at noon, making the temperature 7^{circ}mathrm{C} + 2^{circ}mathrm{C} = 9^{circ}mathrm{C}. 3. Then, it drops by 10^{circ}mathrm{C} at midnight, leading to a temperature of 9^{circ}mathrm{C} - 10^{circ}mathrm{C} = -1^{circ}mathrm{C}. Therefore, the temperature at midnight is boxed{-1^{circ}mathrm{C}}.

answer:To solve this, we first determine the number of configurations where the socks and shirts are of different colors: - If the under-12 players choose red socks, the under-14 players can choose either blue or green shirts for non-matching colors. This gives 2 configurations. - If the under-12 players choose blue socks, the under-14 players can choose either red or green shirts for non-matching colors. This gives 2 configurations. Thus, there are (2 + 2 = 4) configurations where the socks and shirts do not match. Next, we calculate the total number of possible configurations: - The under-12 players have 2 choices (red or blue) for the socks. - The under-14 players have 3 choices (red, blue, or green) for the shirt. This results in (2 times 3 = 6) total configurations. Finally, the probability that the socks and the shirt are different colors is the number of non-matching configurations divided by the total configurations: [ frac{4}{6} = boxed{frac{2}{3}} ]