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ABCD is a unit square and E is a point inside it, such that angle CED is right and $\frac {DE}{AE} = \frac {4}{5}$. Calculate the area of ECBAE (green area).

Although Geometry is not my strong point, I have tried the following: By drawing a vertical from E to AD (h = altitude of triangle AED), this splits the side AD into 2 parts, say x and y. Knowing also that $ED = 4k$ and $EA = 5k$, we apply Pythagoras twice, having also $x+y=1$. By this, we get a relation between x and y: $y-x = 9k^2$. But we have 3 unknowns, so I am not getting anywhere... Then we could also apply Pythagoras in right triangle CED and calculate EC and then get the areas of both triangles and deduct from the area of the square, which is 1.

By the way, Geogebra gives a number near 0.41 for the area in question. Thank you in advance!

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2 Answers 2

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Let $E(x,y),$ $A(0,0)$, $D(0,1),$ $B(1,0)$ and $C(1,1).$

Thus, we obtain the following system. $$\left(x-\frac{1}{2}\right)^2+(y-1)^2=\left(\frac{1}{2}\right)^2$$ and $$\frac{\sqrt{x^2+(y-1)^2}}{\sqrt{x^2+y^2}}=\frac{4}{5}.$$ Can you end it now?

I got $E\left(\frac{16}{65},\frac{37}{65}\right)$ or $E\left(\frac{16}{17},\frac{13}{17}\right)$ and from here easy to find the area.

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  • $\begingroup$ Michael Rozenberg, thank you very much - can you please explain how you derived the first equation? $\endgroup$ Oct 16, 2020 at 9:12
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    $\begingroup$ $E$ is a point on the semicircle, and its equation is given by the first equation (equation of a circle). The centre is at $(1/2, 1)$, and the radius is $1/2$. $\endgroup$
    – Toby Mak
    Oct 16, 2020 at 9:29
  • $\begingroup$ @TobyMak thank you! $\endgroup$ Oct 16, 2020 at 9:31
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    $\begingroup$ The fact that $5^2-4^2=3^2$ is key to making the coordinates rational. I can't (yet) tell if this is merely an algebraic nicety or if there's a particularly-clever way to exploit the $3$-$4$-$5$ triangle geometrically. @PradeepSuny: What's the source of this problem? Is there reason to believe that clever geometry is expected? Or is this more likely intended to be a straightforward exercise in solving a system of two circle equations? $\endgroup$
    – Blue
    Oct 16, 2020 at 9:41
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Let $DE=4k$ and $AE=5k$.

Let $F$ be a point on $AE $ such that $DE\perp AF$. $\triangle ADF\cong \triangle DCE$. Therefore, $AF=4k$ and hence $EF=3k$, $DF=k$.

$$(4k)^2+k^2=1$$

$k^2=\dfrac1{17}$ and the required area is $\displaystyle 1^2-\dfrac12(DE)(CE+AF)=1-\dfrac12(4k)(k+4k)=1-10k^2=\frac7{17}$.

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  • $\begingroup$ CY Aries, I liked your solution very much but I had already marked Michael's as most useful :) Sorry, there is only one vote!! Thanks to both of you!! $\endgroup$ Oct 18, 2020 at 10:51

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