# The side of a triangle inscribed in a given circle subtends angles $a, b,$ and $y$ at the center.

The side of a triangle inscribed in a given circle subtends angles $$a, b,$$ and $$y$$ at the center. The minimum value of the arithmetic mean mean of $$\cos (a+ \frac{\pi}{2}), \cos(b+\frac{\pi}{2})$$ and $$\cos(y+\frac{\pi}{2})$$ is . . . ?

One thing I noticed here is that either $$a + b+y = 2 \pi$$

or $$a+y = b$$ .

I used a diagram. I don't know if there is a more rigorous way to prove that. The problem can be solved if$$a + b + y = 2 \pi$$ but what if it is the other case?

My book solve this problem using AM GM inequality but all these terms are negative so how can that be valid here , also they forgot the other case???

@Drmathva helped me solved it $$with a+b+c = 2 \pi$$ case but what about the other case???

• Could you please provide a picture? I don't really understand what you mean in the first sentence... – Dr. Mathva May 5 at 19:02
• @Dr.Mathva I am on mobile,. I currently can't . This app will not allow me. I will still try. – user541396 May 5 at 19:04
• – user541396 May 5 at 19:09
• Oh, see. So you want to minimize $$\sum\cos\big(a+\frac\pi2\big)$$? – Dr. Mathva May 5 at 19:12
• @Dr.Mathva right! – user541396 May 5 at 19:14

## Case 1: $$a+b+c=2\pi$$

Observe, first of all, that we can expand $$\cos\big(a+\frac\pi2\big)=-\sin a$$

Thus, we want to maximize $$S=\sin a+\sin b+\sin c$$ under the constraint $$a+b+c=2\pi$$. Since we want to maximize $$S$$, we want all of it terms to be positive (this is not as rigorous as it might sound, so prove it!). Thus we can assume that $$a,b,c\in (0,\pi)$$. In that interval $$f(x)=\sin x\implies f''(x)=-\sin x<0$$ We now apply Jense's inequality $$\sin\bigg(\frac{a+b+c}{3}\bigg)=\frac{\sqrt 3}2\ge\frac{\sin a+\sin b+\sin c}3\iff S\le\frac{3\sqrt 3}2$$ Equality holds iff $$a=b=c=\frac{2\pi}3$$

The second case doesn't really exist

In fact, we didn't consider the angle subtended by $$\color{fuchsia}c$$ correctly! And we are done!

Consider for instance $$f(x):=x+3$$. With the same argument given by your book, $$f(x)_{min}$$ in $$(0, \infty)$$ is achieved when $$x=3$$. But this is wrong!
Of course $$x+3\ge 2\cdot \sqrt{3x}$$
and they are equal when $$x=3$$. But this doesn't imply that $$x=3$$ gives the minimum value for $$f$$...