answer:Given a convex quadrilateral with perpendicular diagonals, isosceles triangles are constructed on its sides such that their axes of symmetry are the perpendicular bisectors of the sides of the quadrilateral. We need to prove that the vertices of these isosceles triangles form a quadrilateral with equal diagonals. 1. **Applying the Cosine Law**: Use the cosine law twice for the quadrilateral. The cosine law for a triangle states: [ c^2 = a^2 + b^2 - 2ab cos(gamma) ] where (c) is the side opposite the angle (gamma), and (a) and (b) are the other two sides. 2. **Diagonal Relationships**: Let the quadrilateral be (ABCD), and let the diagonals (AC) and (BD) intersect at a right angle. This implies: [ AC^2 + BD^2 = AD^2 + BC^2 ] by the Pythagorean Theorem. 3. **Midlines of the Quadrilateral**: Consider the midlines (segments connecting the midpoints of opposite sides) of the quadrilateral. By definition and symmetry, the midlines are equal: [ text{Midline } EF = text{Midline } GH ] where (E, F, G, H) are midpoints of (AB, BC, CD,) and (DA) respectively. 4. **Symmetry in Isosceles Triangles**: The isosceles triangles constructed on the sides have their axes of symmetry as perpendicular bisectors of the sides. Therefore: [ angle AFB = angle EFD = angle EFG = angle GHJ = 90^circ ] This implies that (EF = GH). 5. **Equal Diagonals in Formed Quadrilateral**: Since the sums of the distances from two adjacent vertices of the quadrilateral to the midlines are equal: [ overline{AF} + overline{FC} = overline{BF} + overline{FD} ] and [ overline{BF} + overline{FD} = overline{BF} + overline{FE} ] we have: [ AF + FC = BF + FD ] 6. **Equal Summation of Squares**: Considering the squares of the distances of opposite sides: [ AB^2 + CD^2 = AD^2 + BC^2 ] # Conclusion We've shown that through the properties of the midlines and the isosceles triangles construct on the sides, the vertices of these triangles form a quadrilateral which maintains the distances, angles, and symmetry such that the diagonals of the new quadrilateral are equal. [ boxed{ The vertices of the isosceles triangles form a quadrilateral with equal diagonals. } ]

answer:To solve this problem, we'll consider a cross-section of the scenario, which simplifies to a circle (representing the ball) and a line (representing the surface of the lake where the hole was left). The circle's center is located above the line. Given that the hole is 24 cm across, the radius of the hole is half of that, which is 12 cm. The depth of the hole is 8 cm, which means the distance from the bottom of the ball to the line (the surface of the lake) is 8 cm. Let's denote the distance from the center of the circle (ball) to the line as x cm. We can form a right triangle by drawing a line from the center of the circle to the point where the circle intersects the line, and another line segment from this intersection point vertically down to the line, creating a right angle. The hypotenuse of this right triangle is the radius of the ball, which we are trying to find. One leg of the triangle is the radius of the hole, which is 12 cm, and the other leg is the distance from the center of the ball to the surface of the lake, which is x cm for the part above the surface and 8 cm for the part below the surface, totaling x + 8 cm. Using the Pythagorean theorem, we can set up the equation: [x^2 + 12^2 = (x + 8)^2] Expanding both sides of the equation gives: [x^2 + 144 = x^2 + 16x + 64] Subtracting x^2 from both sides simplifies to: [144 = 16x + 64] Subtracting 64 from both sides gives: [80 = 16x] Dividing both sides by 16 to solve for x gives: [x = 5] However, we are looking for the radius of the ball, which is x + 8 (since the radius extends x cm above the surface and 8 cm below it to the bottom of the hole). Therefore, the radius of the ball is: [5 + 8 = 13] Thus, the radius of the ball, in centimeters, is boxed{13}.

answer:Let b be the desired number. The given system of congruences is: begin{align*} b &equiv 2 pmod{3}, b &equiv 2 pmod{4}, b &equiv 3 pmod{5}. end{align*} Since gcd(3, 4) = 1, the first two congruences imply that b equiv 2 pmod{12}. Thus, there exists a non-negative integer m such that b = 2 + 12m. Substituting this into the third congruence yields: [ 2 + 12m equiv 3 pmod{5}, ] which simplifies to: [ 12m equiv 1 pmod{5}. ] Since 12 equiv 2 pmod{5}, this becomes: [ 2m equiv 1 pmod{5}. ] Multiplying both sides by the modular inverse of 2 modulo 5, which is 3 (since 2 times 3 = 6 equiv 1 pmod{5}), gives: [ m equiv 3 pmod{5}. ] Therefore, m = 3 + 5k for some integer k geq 0, and the smallest value of m is 3. Substituting m = 3 into b = 2 + 12m gives: [ b = 2 + 12 times 3 = 2 + 36 = 38. ] Checking b = 38 in all original congruences confirms: [ 38 equiv 2 pmod{3}, quad 38 equiv 2 pmod{4}, quad 38 equiv 3 pmod{5}. ] Thus, the smallest number b is boxed{38}.

answer:Since we need to create a five-digit palindrome, the form of the number will be abcba, where a, b, c can be any of the digits 5, 6, or 7. Each position a, b, c has 3 possible values, so under normal conditions without additional requirements, the number of such palindromes would be 3^3 = 27. However, there is a constraint that the number must contain the digit 5 at least once. We use complementary counting to find the number of palindromes without any 5s and subtract from the total. 1. Calculate total palindromes without any constraint: text{Total palindromes} = 3 times 3 times 3 = 27 2. Calculate palindromes without the digit 5: - If a, b, c can only be 6 or 7, then: text{Palindromes without 5} = 2 times 2 times 2 = 8 3. Subtract to find the number of valid palindromes: text{Palindromes with at least one 5} = 27 - 8 = 19 Hence, the number of five-digit palindromes using the digits 5, 6, and 7 that contain at least one digit 5 is boxed{19}.