A Problem concerning Areas

by cip1703, Feb 24, 2018, 1:56 PM

LEMA 1
Among the areas marked on the figure we have the relationship
$$S_{1}\cdot S_{3}=S_{2}\cdot S_{4}$$
http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvOS8yLzEzNDY2NDUyMTkxNDMwMTZmMThiN2JlMGJiYmQ5NDgzNTFhYjEyLmpwZw==&rn=UmVsYXRpaSBBcmlpLkpQRw==
Proof
Triangles $\Delta DAE,\Delta DCE$ have the same height, so $\frac{area(DAE)}{area(DCE)}=\frac{AE}{AC}$. Once again, triangles $\Delta BAE,\Delta BCE$ have the same height, so $\frac{area(BAE)}{area(BCE)}=\frac{AE}{AC}$. It follow $\frac {S_{1}}{S_{4}}=\frac{S_{2}}{S_{3}}$ from where the conclusion.

PROBLEM 1
Find x in the next figure.http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZi9mLzdmMDVkMGVjY2MwNjM4ZWVhOTg2ZTdlZDllZTUyNTM1MmU2NTgzLmpwZw==&rn=Q2FsY3VsdWwgdW5laSBhcmlpLkpQRw==
Answer
$$x=\frac{S_{1}\cdot S_{2}(S_{1}+S_{2}+2S_{3})}{S_{3}^{2}-S_{1}\cdot S_{2}}$$Solution
We were doing what he meant in https://artofproblemsolving.com/community/u44261h1597097p9921734.
In the following figure we have $area(FEG)=\frac{S_{1}\cdot S_{2}}{S_{3}}$.
Denoting with $y=area(AEF)$, http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvOC8yL2RkNjY2NGZhZWE4NDM0YmYwMWUwY2E2YjYxOTg0YzBkYThkMWRhLmpwZw==&rn=Q2FsY3VsdWwgYWx0ZWkgYXJpaS5KUEc=
triangles $\Delta FAE,\Delta FCE$ have the same height, so $\frac{area(FAE)}{area(FCE)}=\frac{AE}{EC}$ and once again, triangles $\Delta BAE,\Delta BCE$ have the same height, so $\frac{area(BAE)}{area(BCE)}=\frac{AE}{EC}$.
We write that the two fractions are equal, and we get the equation
$$\frac{y}{\frac{S_{1}S_{2}}{S_{3}}+S_{2}}=\frac{y+S_{1}+ \frac{S_{1}S_{2}}{S_{3}}}{S_{2}+S_{3}}$$where the value of $y$ results. Further $x=\frac{S_{1}S_{2}}{S_{3}}+y$, etc.

Here another such a problem, with the answer also a whole number
http://artofproblemsolving.com/wiki/index.php?title=File:2006amc10b23.gif#filehistory
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This post has been edited 13 times. Last edited by cip1703, Feb 24, 2018, 5:30 PM
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