If the circumcenter of the triangle $ABC$ is on the incircle of the triangle,then prove that $\cos A+\cos B+\cos C=\sqrt2$

How should i attempt this question?I thought over it hard but could not crack through.Some hints will help me.Please guide me.

  • $\begingroup$ consider the triangle in a coordinate system $\endgroup$ – Dr. Sonnhard Graubner Sep 2 '15 at 15:49
  • $\begingroup$ @Dr.SonnhardGraubner,what should be right way to choose coordinates,besides taking one vertex as (0,0),to simplify the calculations. $\endgroup$ – Vinod Kumar Punia Sep 2 '15 at 16:03
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    $\begingroup$ i would choose $$A(0,0),B(c,0),C(b\sin(\alpha),b\cos(\alpha))$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 2 '15 at 16:06

Using Euler's theorem in geometry,

$$r^2=R(R-2r)\iff r^2+2Rr-R^2=0$$

$$\implies r=\dfrac{-2R\pm\sqrt{4R^2+4R^2}}2=R(\sqrt2-1)$$ as $R,r>0$

Now using this, $$\dfrac rR=\cos A+\cos B+\cos C-1$$

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Let me try.

Let $d$ be a distance between the circumcentre and incentre.

From Euler's theorem, you have $$d^2= R(R-2r).$$

When the circumcentre is on the incircle, we have $d=r$. Then, $$r^2 = R^2 - 2Rr.$$

So, you get $$\left(\frac{r}{R}\right)^2 + 2\frac{r}{R} - 1 = 0.$$

Then, you get $$\frac{r}{R} = -1 + \sqrt{2}$$ (Do you see why not $-1-\sqrt{2}$?).

Recall that we have $$rR = \frac{abc}{2(a+b+c)}$$ (For example, see link)

Note that $a = 2R\sin A$, $b = 2R\sin B$ and $c = 2R\sin C$. We have

$$\frac{r}{R} = 2 \frac{\sin A\sin B\sin C}{\sin A + \sin B + \sin C}.$$

We have $$\sin A + \sin B +\sin C = 4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}.$$

Now, we have $$\frac{r}{R} = 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$$

Note that $$\cos A + \cos B +\cos C = 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} + 1.$$

So, we have $$-1+\sqrt{2} = 4 \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$$

Thus, we get $$\cos A +\cos B + \cos C = \sqrt{2}.$$

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  • $\begingroup$ i did not knew which theorems you can use, but you must also prove the given formulas $\endgroup$ – Dr. Sonnhard Graubner Sep 2 '15 at 16:17

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