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

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?

• 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. – Tavish Apr 21 at 19:44

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

• I can't undestand the first 2 lines, the formula giving you the radius – Maryam Apr 21 at 20:31
• Well, I don't know why you bother with it since you already calculate it. – User2020201 Apr 21 at 20:33
• just curious, sorry – Maryam Apr 21 at 20:33
• You don't know the power of the point? – User2020201 Apr 21 at 20:34

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,

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