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

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The value of the expression $$A_1 - A_2 + A_3 - A_4$$, derived from the areas within the ellipse defined by $$\frac{(x-19)^2}{19}+\frac{(y-98)^2}{98}=1998$$, is determined to be 7448. The areas are defined as follows: $$A_1 = U + V + S + T$$, $$A_2 = S + T$$, $$A_3 = S$$, and $$A_4 = V + S$$. The calculation simplifies to $$U$$, which equals the product of the ellipse's dimensions, specifically $$38 \cdot 196$$. This solution is attributed to Professor Gregory Galperin of Eastern Illinois State University and is published in "International Mathematics Talent Search, Part 2" by G. Berzsenyi.

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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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