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KCC's Quizzes AQQ258 about a partial eclipse Sun-Moon

		
	
				

			

									
					    KCC
					
							

            
            			
				
									
											
			
			
				
  •  Analog Employees  

			
			
			    	
			                    
                
				                on Apr 14, 2024 
				
			
			
		

		
		                     
				
        		

		    
		    		    
		    			

			    
1. Quote of the week: "People say nothing is impossible; but I do nothing every day!" - Anonymous


                      


2. New quiz: AQQ258 about a partial eclipse Sun-Moon


      


Some of us (in Dallas) had probably the chance to have observed a total eclipse of the Sun last week.


To celebrate that event in our quizzes, we propose the following related challenge: 


The above pictures describe a partial eclipse between the moon (circle B) and the sun (circle R).


At a certain instant, the eclipse is such that points MNO form a quarter area of circle R; meaning arc MN is a quarter circle).


At that moment, point O is the center of R and the segment OM measures 4 cm.


Question:


What is the area of the portion of blue circle B that is situated outside circle R?




Good luck! And try to be among the firsts!


Kuo-Chang


P.S. Don't hesitate to share those weekly quizzes in EZ to colleagues or friends!

			

			
									
		                             
                        

    

                                            
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    •  Analog Employees  
    
            
						
                            
            
            				
            
                on Apr 15, 2024 9:53 AM
            
            
						
			            		
    
		
								
	
    

		
		
    
		
    
                                
                                                            
    
					
    Without diving into proving all the geometric theorems, using Thale's Theorem, one can determine that line segment MN goes through the center of circle B, so if we can determine the distance of MN, we know the area of circle B.  
    

    

    This is determined to be MN - sqrt(4^2 + 4^2) = sqrt(32) = 5.66
    

    

    If the diameter of the circle B is sqrt(32), then the area of B is B(area) = pi * (1/2 * sqrt(32))^2 = pi * 1/4 * 32 = 8π
    

    

    The area of the arc MON = 1/4 * (pi * 4^2) = 4π
    

    

    All that leaves is the area of the two blue arcs inside circle R (but not inside arc MON).  This can be found by calculating the area of B minus the square created by flipping angle MON onto B (forming a square) and using 1/2 of that.
    

    

    Area of blue arcs inside R = 1/2 * (B(area) - area of MON square) = 1.2 * (8pi - 4^2) = 4π - 8
    

    

    Therefore the area of the blue circle outside circle R = 
    

    

    ANSWER = 8π - 4π - (4π - 8)  = 4π - 4π + 8 = 8 cm²
    

    

    

    Cheers!
    

    Brian
    
			
    

		
			
	      

		
		
	
    
	
    
		
    
			
		
    
	
    
	
    
		
						
				
				
						
				
				
		
		
    
			
		
    
	
    

    

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  • 
	
    






    
	
    
		
    
							
					
    
		
    
		    
		    		    
			
				
									
							        
			
			
			
    •  Analog Employees  
    
            
						
                            
            
            				
            
                on Apr 15, 2024 9:53 AM
            
            
						
			            		
    
		
								
	
    

		
		
    
		
    
                                
                                                            
    
					
    Without diving into proving all the geometric theorems, using Thale's Theorem, one can determine that line segment MN goes through the center of circle B, so if we can determine the distance of MN, we know the area of circle B.  
    

    

    This is determined to be MN - sqrt(4^2 + 4^2) = sqrt(32) = 5.66
    

    

    If the diameter of the circle B is sqrt(32), then the area of B is B(area) = pi * (1/2 * sqrt(32))^2 = pi * 1/4 * 32 = 8π
    

    

    The area of the arc MON = 1/4 * (pi * 4^2) = 4π
    

    

    All that leaves is the area of the two blue arcs inside circle R (but not inside arc MON).  This can be found by calculating the area of B minus the square created by flipping angle MON onto B (forming a square) and using 1/2 of that.
    

    

    Area of blue arcs inside R = 1/2 * (B(area) - area of MON square) = 1.2 * (8pi - 4^2) = 4π - 8
    

    

    Therefore the area of the blue circle outside circle R = 
    

    

    ANSWER = 8π - 4π - (4π - 8)  = 4π - 4π + 8 = 8 cm²
    

    

    

    Cheers!
    

    Brian
    
			
    

		
			
	      

		
		
	
    
	
    
		
    
			
		
    
	
    
	
    
		
						
				
				
						
				
				
		
		
    
			
		
    
	
    

    

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