MHB Determine the value of A1 - A2 + A3 - A4

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The problem involves calculating the value of A1 - A2 + A3 - A4 for areas within the ellipse defined by the equation. The center of the ellipse is located at (19, 98), and it is divided into nine regions by vertical and horizontal lines. The areas A1, A2, A3, and A4 are expressed in terms of these regions, leading to the equation A1 - A2 + A3 - A4 = U. The final calculation reveals that U equals 7448, which is the solution to the problem. This solution is attributed to Professor Gregory Galperin and published in a mathematics book.
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Let $$A_1$$, $$A_2$$, $$A_3$$ and $$A_4$$ represent the areas within the ellipse $$\frac{(x-19)^2}{19}+\frac{(y-98)^2}{98}=1998$$ that are in the first, second, third and fourth quadrants respectively. Determine the value of $$A_1-A_2+A_3-A_4$$
 
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Re: Determine the value of A1-A2+A3-A4

I must mention here that the idea of this problem came from Professor Gregory Galperin of Eastern Illinois State University, U.S.A. and I feel so grateful and thankful to Professor Gregory Galperin for approving me to show his solution to this problem at this site.

Solution:

We know that the center of the ellipse is the point (19, 98). The two vertical lines $$ x=0 $$ and $$x=2\cdot19=38$$ and the two horizontal lines $$y=0$$ and $$y=2\cdot98=196$$ divide the ellipse into nine regions, as shown below.

View attachment 801

Then, in terms of the areas marked by S, T, U and V, we find that $$A_1=U+V+S+T$$, $$A_2=S+T$$, $$A_3=S$$, and $$A_4=V+S$$, and therefore

$$A_1-A_2+A_3-A_4=(U+V+T+S)-(S+T)+S-(V+S)=U.$$

Hence the answer is $$38\cdot196=7448$$.Note:
This solution was published in the book by G. Berzsenyi "International Mathematics Talent Search, Part 2 ", AMT Publishing(2011). The problem was given at the Round 29, and it's titled as Problem 5/29; the pages are: the formulation @ p.11, and the solution @ p.84.
 

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