The actual problem I have is:enter image description here

Calculate the area of the triangular region ABC.

I tried to relate the isosceles triangle properties with the angle-arc theorem (inscribed angle is half of the arc) but everything ends up bound to the radius. How do I solve this problem?


The distances of the circumcenter $O$ from the sides are $R-1, R-2, R-3$ and the side lengths are given by $$ a = BC = 2\sqrt{R^2-(R-2)^2} = 4\sqrt{R-1},$$ $$ b = AC = 2\sqrt{R^2-(R-3)^2} = 2\sqrt{3}\sqrt{2R-3}, $$ $$ c = AB = 2\sqrt{R^2-(R-1)^2} = 2\sqrt{2R-1} $$ and twice the area is given by $a(R-2)+b(R-3)+c(R-1)$, but also by $\frac{abc}{2R}$.
That gives a horrible equation in $R$, from which

$$R = 2\left(1+\cos\frac{\pi}{9}\right)\approx 3.87938524$$ follows.

The area is so $\Delta\approx \color{red}{17.1866}$, and none of the given options is the correct one.

An approximate construction with Geogebra also shows that the app creators are wrong: enter image description here

To worsen the situation, there is the fact that $R$ is an algebraic number of degree $3$ over $\mathbb{Q}$, hence the problem cannot even be solved by straightedge and ruler only!

  • 1
    $\begingroup$ that's too bad, it was such a pretty problem. Perhaps this was meant to be done on the surface of a sphere, whose radius is just right for one of the given answers to be correct. (LOL) $\endgroup$ – Mark Fischler Sep 22 '16 at 4:22
  • $\begingroup$ @MarkFischler: it took me some time, but there is something pretty about it, i.e. that $$ R = 2\left(1+\cos\frac{\pi}{9}\right) $$ ! $\endgroup$ – Jack D'Aurizio Sep 22 '16 at 4:25
  • $\begingroup$ I don't get why R equals that. $\endgroup$ – Lucas Henrique Sep 23 '16 at 1:11
  • $\begingroup$ @LucasHenrique: it is the circumradius of $ABC$, as usual. Such expression is derived by recognizing a peculiar polynomial factor (related with the minimal polynomial of $\cos\frac{\pi}{9}$) in the equation that gives $R$. $\endgroup$ – Jack D'Aurizio Sep 23 '16 at 1:12
  • 2
    $\begingroup$ $(R-2)^3-3(R-2)-1=0$ where $R>3$ and the minimal polynomial of $2\cos (\pi /9)$ is $x^3-3x-1$. Comparing we see that $R-2=2\cos (\pi /9)$ $\endgroup$ – Lozenges Sep 23 '16 at 3:36

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