answer:1. Given a parallelogram (ABCD) with obtuse angle (A), - (H) is the foot of the perpendicular dropped from point (A) to (BC). 2. Let (E) be the foot of the perpendicular dropped from point (B) to (AD). It follows that the quadrilateral (AHBE) is a rectangle because: - The angles ( angle AHB ) and ( angle BHE ) are right angles by construction. 3. Since (AHBE) is a rectangle, the opposite angles are equal. Therefore, [ angle HED = angle ABC ] 4. Notice in parallelogram (ABCD): [ angle ABC = 180^circ - angle BCD ] Thus, it ensures that the points (D, C, H, E) lie on a circle ( omega ) as: - ( angle HED = 180^circ - angle BCD ) 5. Now consider triangle (ABC) and the median (CM) from (C) to (AB). - The extension of median (CM) intersects the circumcircle of triangle (ABC) at point (K). 6. Since points (A, K, B, C) lie on the circumcircle of ( triangle ABC ), using Power of a Point theorem, we have: [ MK cdot MC = MA cdot MB ] 7. Additionally, since (M) is the midpoint of diagonal (BE) in rectangle (AHBE), [ MK cdot MC = MA cdot MB = MH cdot ME ] This equality confirms that points (C, K, H, E) lie on the same circle ( omega ). 8. Since circle ( omega ) already contains points (D, C, H, E) and point (K) as shown above: - Circle ( omega ) must also include points (K, H, C, D) because they share a common circle with at least three points being the same. Conclusion: Therefore, points (K, H, C, D) are concyclic, meaning they lie on the same circle ( omega ). (blacksquare)

answer:Let's break down the solution step by step, adhering to the rules provided: # Part 1: Maximizing the Number Displayed on the Screen Given the initial number on the screen is -1, we need to find the sequence of two keys that maximizes the number displayed on the screen. We have three keys to consider: A, B, and C. Let's analyze the effect of each key: 1. **Key A**: Increases the number by 3. 2. **Key B**: Multiplies the number by (-2). 3. **Key C**: Divides the number by 4. To maximize the number, we should first consider multiplying or adding to increase its value. Since we're starting with a negative number (-1), multiplying by a negative (Key B) will turn it positive, which is beneficial for maximizing. - **First Press (Key B)**: -1 times (-2) = 2. - **Second Press (Key A)**: After turning the number positive, adding will further increase its value. So, 2 + 3 = 5. Thus, the sequence to maximize the number displayed on the screen after pressing two keys in a row is B rightarrow A. # Part 2: Key Sequence A rightarrow B rightarrow C rightarrow A rightarrow B rightarrow C ldots after 2023 Presses Given the sequence A rightarrow B rightarrow C rightarrow A rightarrow B rightarrow C ldots and starting with -1, let's see the pattern: 1. **Pressing A**: -1 + 3 = 2. 2. **Pressing B**: 2 times (-2) = -4. 3. **Pressing C**: -4 div 4 = -1. After these three presses, the number returns to -1, indicating a cycle every 3 key presses. Given 2023 presses, we find how many complete cycles there are and what remains: - 2023 divided by 3 gives 674 complete cycles with a remainder of 1. This means after completing 674 cycles, we have pressed the keys 2022 times, and the screen shows -1. The next key press (the 2023^text{rd} press) is Key A: - **Pressing A**: -1 + 3 = 2. Therefore, after pressing the keys 2023 times following the given pattern, the number displayed on the screen is 2. Hence, the answers are: boxed{B rightarrow A} for the first part and boxed{2} for the second part.

answer:To find out the number of boys in the school, we can set up an equation. Let's call the number of boys B. We know that the number of boys is 807 more than the number of girls. We can express this relationship as: B = 34 (number of girls) + 807 Now we can solve for B: B = 34 + 807 B = 841 So, there are boxed{841} boys in the school.

answer:To find the length of the arc of a curve given by the polar equation (rho = 8(1 - cos varphi)) over the interval (-frac{2pi}{3} leq varphi leq 0), we use the formula for the arc length in polar coordinates: [ L = int_{varphi_0}^{varphi_1} sqrt{(rho(varphi))^2 + left(frac{drho}{dvarphi}right)^2} , dvarphi ] Step-by-Step Solution: 1. **Calculate (frac{drho}{dvarphi})** [ rho = 8(1 - cos varphi) ] [ frac{drho}{dvarphi} = 8 sin varphi ] 2. **Set up the integrand** [ L = int_{-frac{2pi}{3}}^0 sqrt{(8(1 - cos varphi))^2 + (8 sin varphi)^2} , dvarphi ] Simplify the expression inside the square root: [ sqrt{(8(1 - cos varphi))^2 + (8 sin varphi)^2} = sqrt{64(1 - cos varphi)^2 + 64 sin^2 varphi} ] 3. **Simplify the integrand** [ = 8 sqrt{(1 - cos varphi)^2 + sin^2 varphi} ] Note that: [ (1 - cos varphi)^2 + sin^2 varphi = 1 - 2cos varphi + cos^2 varphi + sin^2 varphi ] Recall that (sin^2 varphi + cos^2 varphi = 1): [ = 1 - 2 cos varphi + 1 = 2 - 2 cos varphi ] Hence: [ sqrt{2 - 2 cos varphi} = sqrt{2 cdot (1 - cos varphi)} = sqrt{2} cdot sqrt{1 - cos varphi} ] 4. **Simplify further using trigonometric identity** Recall: [ 1 - cos varphi = 2 sin^2 left(frac{varphi}{2}right) ] Therefore: [ sqrt{1 - cos varphi} = sqrt{2} cdot left| sin left(frac{varphi}{2}right) right| ] Thus the integrand becomes: [ L = 8 sqrt{2} int_{-frac{2pi}{3}}^0 sqrt{2} left| sin left(frac{varphi}{2}right) right| , dvarphi = 16 int_{-frac{2pi}{3}}^0 left| sin left(frac{varphi}{2}right) right| , dvarphi ] 5. **Change the variable for integration** By making the substitution ( u = frac{varphi}{2} ): [ du = frac{1}{2} dvarphi quad text{or} quad dvarphi = 2 du ] Update the limits: When ( varphi = -frac{2pi}{3} ), ( u = -frac{pi}{3} ) When ( varphi = 0 ), ( u = 0 ) [ L = 16 int_{-frac{pi}{3}}^0 left| sin u right| , 2 du = 32 int_{-frac{pi}{3}}^0 left| sin u right| , du ] 6. **Simplify and integrate** Since ( sin u ) is negative over the interval (-frac{pi}{3} le u le 0), we can remove the absolute value: [ L = 32 int_{-frac{pi}{3}}^0 -sin u , du = -32 int_{-frac{pi}{3}}^0 sin u , du ] The integration of (sin u) is (-cos u): [ L = -32 left[ -cos u right]_{-frac{pi}{3}}^0 = 32 left[ cos u right]_{-frac{pi}{3}}^0 ] Substitute the limits: [ L = 32 left( cos 0 - cos left(-frac{pi}{3}right) right) = 32 left( 1 - frac{1}{2} right) = 32 cdot frac{1}{2} = 16 ] # Conclusion: [ boxed{16} ]