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The area of the triangle $OCE$ is $3\sqrt{5}a^2$. Compute the area of the blue triangle $ACD$.

My thoughts: I used Heron's formula applied to triangle $OCE$, with sides $r, r+a, 4a$, ($r$ being the radius of the circle), to find the radius of the circle. I got $\frac{7}{2}a$ for the radius. After that, I could compute the length of $DH$, where $H$ is the orthogonal projection of $D$ on $OA$, just dividing the area of $ODC$ (which should be half the area of $OCE$) by the length of $OC$. Then I found a formula for $AD$ using Pythagoras, but calculations were to long to be performed. Is there any synthetic way to solve the blue triangle?

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  • $\begingroup$ Another way: Calculate $\angle AOD$ using the law of cosines and consequently the area of $\triangle AOD$, then subtract this including the area of $\triangle ODE$ (which you can get using Heron’s) from $3 \sqrt5 a^2$ to get the blue area. $\endgroup$ – Tavish Apr 21 at 19:44
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By the power of the point $C$ we have $$ a(a+2r)= 2a\cdot 4a\implies r ={7a\over 2}$$

Clearly, since $OD$ is median of triangle $OCE$ $$(OCE) =3\sqrt{5}a^2\implies (ODC)={3\sqrt{5}a^2\over 2}$$

Let $h$ altitude on OC from $D$, so $${h\cdot (r+a)\over 2}={3\sqrt{5}a^2\over 2}\implies h= {2\sqrt{5}a\over 3}$$

Now we have $$(ACD) = {h\cdot a\over 2} ={\sqrt{5}a^2\over 3}$$

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  • $\begingroup$ I can't undestand the first 2 lines, the formula giving you the radius $\endgroup$ – Maryam Apr 21 at 20:31
  • $\begingroup$ Well, I don't know why you bother with it since you already calculate it. $\endgroup$ – User2020201 Apr 21 at 20:33
  • $\begingroup$ just curious, sorry $\endgroup$ – Maryam Apr 21 at 20:33
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    $\begingroup$ You don't know the power of the point? $\endgroup$ – User2020201 Apr 21 at 20:34
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Let [] denote areas. D is the midpoint of EC, then [OCE] = 2[ODE], or

$$3\sqrt5 a^2 =2\cdot \frac 12 (2a)\sqrt{r^2-a^2}\implies \frac ra = \frac72$$

Then,

$$[Blue]= \frac{AC}{OC}[ODC] =\frac{a}{2(r+a)}[OCE] =\frac{3\sqrt5a^2}{2(\frac72+1)} =\frac{\sqrt5a^2}{3} $$

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